Igenetics Study Guide
Test Bank to Accompany
iGenetics: A Molecular Approach
Third Edition by Russell / Bose
Benjamin Cummings
c.2010 3/10/09
Contents
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Genetics: An Introduction...............................................................................................1 DNA: The Genetic Material ............................................................................................9 DNA Replication............................................................................................................17 …show more content…
Gene Function ................................................................................................................26 Gene Expression: Transcription ...................................................................................35 Gene Expression: Translation.......................................................................................44 DNA Mutation, DNA Repair, and Transposable Elements .....................................53 Genomics: The Mapping and Sequencing of Genomes ............................................62 Functional and Comparative Genomics .....................................................................71 Recombinant DNA Technology ...................................................................................80 Mendelian Genetics .......................................................................................................88 Chromosomal Basis of Inheritance ..............................................................................97 Extensions of and Deviations from Mendelian Genetic Principles.......................105 Genetic Mapping in Eukaryotes ................................................................................115 Genetics of Bacteria and Bacteriophages ..................................................................126 Variations in Chromosome Structure and Number ................................................136 Regulation of Gene Expression in Bacteria and Bacteriophages ...........................145 Regulation of Gene Expression in Eukaryotes.........................................................154 Genetic Analysis of Development .............................................................................163 Genetics of Cancer .......................................................................................................172 Population Genetics.....................................................................................................181 Quantitative Genetics ..................................................................................................192 Molecular Evolution ....................................................................................................202
Chapter 1 Genetics: An Introduction
MATCHING QUESTIONS Please select the best match for each term. 1) A system for searching several linked databases
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A) PubMed B) OMIM C) GenBank D) BLAST E) Entrez
2) A tool to compare nucleotide or protein sequences
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3) A website that provides links to research articles
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4) A database of human genes and disorders
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5) A database of genetic sequences
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Answers:
1) E
2) D
3) A
4) B
5) C
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Please select the best match for each term. 6) Developed the polymerase chain reaction
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A) Boyer and Cohen B) Barbara McClintock C) Paul Berg D) Gregor Mendel E) Kary Mullis
7) Constructed the first recombinant DNA molecule
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8) Cloned the first recombinant DNA molecule
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9) Conducted breeding experiments with the pea plant
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10) Discovered transposable elements
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Answers:
6) E
7) C
8) A
9) D
10) B
MULTIPLE-CHOICE QUESTIONS 11) The field of genetics includes A) the study of heredity. B) the molecular nature of the genetic material. C) the ways in which genes control life functions. D) the distribution and behavior of genes in populations. E) All of these Answer: E
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12) Quantitative genetics is A) the quantitative study of the activity of a gene. B) the study of the genetic diversity within a large group of individuals of the same species. C) the study of the genetic diversity within members of a number of related species. D) the study of traits that are determined by a number of genes simultaneously. E) the study of the number of characteristics found in an organism. Answer: D
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Chapter 1 Genetics: An Introduction 3
13) Bacteria and Archaea are A) eukaryotic. B) prokaryotic. C) archaeotic. D) anucleotic. E) proteozoic. Answer: B
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14) Transposons are A) proteins that are stable. B) proteins that are unstable. C) DNA segments that are stable. D) DNA segments that are unstable. E) DNA segments that code for proteins. Answer: D
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15) Genes influence all aspects of life because they A) are structural elements of the cell. B) regulate movement of proteins. C) produce RNA and protein needed for different processes. D) localize to the nucleus. E) are needed for DNA synthesis. Answer: C
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16) An ideal organism for a geneticist to use as a study organism would possess which of the following characteristics? A) A relatively short life cycle B) A large degree of genetic variation among individuals C) Matings that produce large numbers of offspring D) A well-characterized genetic background, karyotype, etc. E) All of these are desirable characteristics Answer: E
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17) The organism that Mendel used for his experiments was A) E. coli. B) the fruit fly. C) maize. D) yeast. E) the pea plant. Answer: E
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18) A map unit (μ) is A) the position of a gene on the chromosome. B) the unit of genetic distance on a chromosome. C) the unit of physical distance on a chromosome. D) equivalent to a nucleotide. E) equal to the number of recombinant progeny. Answer: B
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19) Which of the following is a nematode worm that has been used as a model organism in genetics? A) Drosophila melanogaster B) Mus musculus C) Arabidopsis thaliana D) Caenorhabditis elegans E) Homo sapiens Answer: D
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20) In eukaryotic cells, ________ are energy-producing organelles that contain DNA. A) nuclei B) lysosomes C) chromosomes D) nucleoli E) mitochondria Answer: E
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21) The principles of heredity were first established through breeding experiments carried out in ________ by Gregor Mendel. A) the late 1700s B) the 1950s C) the 1860s D) the early 1900s E) the mid-1600s Answer: C
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22) Tetrahymena is a A) fungus. B) protozoa. C) green alga. D) weed. E) bacteria. Answer: B
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Chapter 1 Genetics: An Introduction 5
23) Eukaryotic chromosomes consist of A) DNA alone. B) DNA complexed with protein. C) DNA complexed with RNA. D) DNA complexed with fatty acids. E) DNA complexed with protein and RNA. Answer: B
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24) Translation of RNA into proteins occurs A) in the nucleus. B) in the endoplasmic reticulum. C) in the cytoplasm. D) in both the nucleus and the cytoplasm. E) in both the nucleus and the endoplasmic reticulum. Answer: C
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25) The branch of genetics concerned with analyzing the structure and function of genes is A) molecular genetics. B) plant genetics. C) transmission genetics. D) population genetics. E) applied genetics. Answer: A
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TRUE-FALSE QUESTIONS 26) Genetics is central to biology because genes and their functions form the basis of all life processes. Answer: TRUE
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27) There is a sharp divide between basic and applied research. Answer: FALSE Explanation: Basic and applied research employ the same information and similar techniques to answer questions.
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28) The smooth endoplasmic reticulum has ribosomes attached to it. Answer: FALSE Explanation: The rough endoplasmic reticulum has ribosomes attached to it.
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29) Centrioles are also called basal bodies. Answer: TRUE
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30) The centrosomes are a part of the chromosome that help in chromosome movement during mitosis and meiosis. Answer: FALSE Explanation: The centrosomes are a region of undifferentiated cytoplasm.
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31) Bacteria and plants both have a rigid cell wall outside the cell membrane. Answer: TRUE
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32) Membrane-bound nuclei are characteristic of both Archaea and eukaryotes. Answer: FALSE Explanation: Membrane-bound nuclei are characteristic of eukaryotes. Archaea, like Bacteria, are prokaryotes.
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33) Genetic maps show the relative location and arrangement of genes on chromosomes. Answer: TRUE
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34) Archaea are prokaryotes that are often found in inhospitable conditions such as hot springs and salt marshes. Answer: TRUE
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35) E. coli is a rod-shaped bacterium found in the human intestine. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) What is the hypothetico-deductive method of investigation? Answer: The hypothetico-deductive method of investigation consists of observing a phenomenon, forming hypotheses to try and explain the observations, making experimental predictions based on those hypotheses, and finally testing out the predictions by doing specific experiments.
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37) How do mutations help us in understanding the particular functions of a gene? Answer: Comparing mutants with normal cells gives us an idea as to which life process has been affected. This helps us locate the mutated gene in the specific biochemical pathway for that particular process.
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Chapter 1 Genetics: An Introduction 7
38) How does the knowledge of complete genomes of different organisms help us in any way? Answer: Since the function of most genes is conserved to a large degree among different organisms, the knowledge of the genomes of different species helps us compare them, leading to a better understanding of gene functions. This in turn will lead to a better understanding of human genetic diseases and will allow us to develop better cures.
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39) How can a genetic map be used?
Answer: Genetic maps can be used in the process of localizing genes and studying the distribution of genes on chromosomes and in the genome.
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40) What is recombination frequency? Answer: Recombination frequency is the percentage of recombinant progeny, that is, the percentage of the progeny in which the arrangement of alleles has switched compared to their arrangement in the parents. This is taken to be a measure of the genetic distance between genes and is used to create genetic maps.
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41) Are genetic maps based on the actual distance on chromosomes as measured by nucleotides? Answer: No. Genetic maps are based on the frequency of recombination between genes. As such, they show relative distances between genes but cannot specify actual distance in terms of nucleotides.
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42) Which organelles besides the nucleus contain their own DNA? Answer: The mitochondria in eukaryotes and the chloroplasts in plants contain DNA of their own.
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43) The nucleus separates the chromosomes from the rest of the cell. How is the information in the DNA communicated to the cell? Answer: The nuclear membrane has pores and is permeable to certain molecules. This allows selected materials, like RNA and proteins, to move in and out of the …show more content…
nucleus.
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44) What is recombinant DNA technology? Answer: Recombinant DNA technology encompasses procedures that allow scientists to join together DNA from two or more different organisms and make many identical copies of them (cloning).
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45) For geneticists, why is it important that genetic variability exist in the population under study? Answer: Genetic variation in individuals of a population is important for studying the inheritance pattern of those characteristics. If all the members of a population were identical for the trait under study, their progeny would be as well, and it would be impossible to determine how the trait was being passed on to the offspring.
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46) What characteristics does an organism have to possess to be a good genetic model? Answer: To be a good genetic model, an organism has to have a well -known genetic history, a short life cycle, produce many offspring, be easy to handle, and have genetic variability among the individuals in a population.
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47) Why are genetic databases so important to the study of modern genetics? Answer: The available information about the genetics of a large number of organisms has increased dramatically in the past few years. As a result, it has become impossible to study, learn, and remember all of it. All this information needs to be stored in such a way that it may be accessed and searched easily. This is done in computer databases that have assumed enormous importance in the study of genetics.
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48) What is the difference between PubMed and OMIM? Answer: PubMed is a website that is used to access citations and abstracts of journal articles. OMIM is a database of human genes and genetic disorders.
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49) What is the technique that is used to amplify DNA sequences? Answer: The procedure employed to amplify DNA sequences is called polymerase chain reaction (PCR).
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50) What are the differences between eukaryotes and prokaryotes? Answer: Eukaryotes possess a nucleus that is separated from the rest of the cell by a double-layered nuclear membrane. Prokaryotes lack a nucleus. Prokaryotes also lack all membrane-bound organelles, such as mitochondria, endoplasmic reticula, and Golgi bodies. Such organelles are present in eukaryotes. Prokaryotes are generally also much smaller than eukaryotes, and are usually unicellular.
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Chapter 2 DNA: The Genetic Material
MATCHING QUESTIONS Please select the best match for each term. 1) Centromere
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A) The region of a prokaryotic cell where the chromosome is located B) The basic structural unit of chromatin with ʺbead-on-a-stringʺ morphology C) A DNA molecule and associated proteins D) The region of a eukaryotic chromosome found near the attachment point of mitotic or meiotic spindle fibers E) The constituent monomer of DNA and RNA
2) Nucleoid
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3) Nucleotide
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4) Chromosome
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5) Nucleosome
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Answers:
1) D
2) A
3) E
4) C
5) B
MULTIPLE-CHOICE QUESTIONS 6) Loosely aggregated DNA bound to proteins in a eukaryotic cell is called A) chromosomes. B) chromatin. C) chromatid. D) centromere. E) nucleoid. Answer: B
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7) The C-value is the amount of DNA in a A) haploid genome. B) diploid genome. C) bacterial genome. D) eukaryotic genome. E) cellʹs nucleus. Answer: A
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8) The chromosome of most prokaryotes differs from those of eukaryotes in that A) the prokaryotic chromosome is linear, while the eukaryotic chromosome is circular. B) the prokaryotic chromosome is circular, while the eukaryotic chromosome is linear. C) the prokaryotic chromosome does not replicate before mitosis, while the eukaryotic chromosome does. D) the prokaryotic chromosome does not contain genes, while the eukaryotic chromosome does. E) the prokaryotic chromosome is not necessary for the organismʹs survival, while the eukaryotic chromosome is. Answer: B
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9) A Barr body is an example of A) constitutive euchromatin. B) facultative euchromatin. C) facultative heterochromatin. D) a nucleosome. E) constitutive heterochromatin. Answer: C
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10) The definition of transformation is A) the shift of genetic information from DNA to protein. B) the genetic alteration of an organism. C) the uptake of genetic information by a cell from the environment. D) Both B and C E) None of these Answer: D
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11) In Griffithʹs experiment involving the transformation of Streptococcus pneumoniae, A) the R strain was virulent. B) the S strain was virulent. C) both the R and S strains were virulent. D) the R strain had a protein capsule. E) the S strain had a protein capsule. Answer: B
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12) What was the transforming principle isolated in Griffithʹs experiment? A) Protein B) RNA C) DNA D) Virus E) Polysaccharide Answer: C
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Chapter 2 DNA: The Genetic Material 11
13) Who used radioactively labeled T2 bacteriophage to confirm the identity of the transforming principle? A) Griffith B) Hershey and Chase C) Avery D) Gierer and Schramm E) Beadle and Tatum Answer: B
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14) Which part of the T2 bacteriophage entered E. coli cells in the experiment which confirmed the identity of the transforming principle? A) The RNA B) The DNA C) The whole virus D) The protein coat E) No part Answer: B
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15) Certain ________ have RNA for their genetic material. A) bacteria B) viruses C) plants D) eukaryotes E) prokaryotes Answer: B
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16) What did the X-ray diffraction patterns initially reveal about the DNA molecule? A) It is of uniform diameter and has a helical structure. B) It is a helical molecule with paired bases in the center. C) It is double-stranded with antiparallel strands. D) It is acidic, phosphorus-rich, and large. E) It contains hereditary information. Answer: A
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17) What did Watson and Crick deduce about the three-dimensional structure of DNA? A) There is a repeating pattern every 3.4 nm and every 0.34 nm. B) It is a double-stranded helix. C) It contains a lot of phosphorus. D) It is a large molecule. E) It consists of supercoiled chromatin. Answer: B
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18) Which of the following is a nonhistone protein found in chromatin? A) H1 B) HMG C) H2A D) H5 E) All of these Answer: B
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19) Antiparallel means that A) the two polynucleotide chains run in opposite directions. B) each DNA molecule consists of one old and one new strand. C) opposite strands are held together by base pairing. D) the helix twists to the right. E) there is complementary base-pairing. Answer: A
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20) Complementary base-pairing allows for A) spontaneous mutations to occur. B) genes to be expressed as a phenotype. C) DNA to serve as its own template for replication. D) replication to be semiconservative. E) covalent bonds to form between the opposite bases. Answer: C
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21) Which of the following are the purine nucleotides in DNA? A) Adenine and thymine B) Cytosine and guanine C) Adenine and cytosine D) Guanine and adenine E) Thymine and uracil Answer: D
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22) Topoisomerases function to A) remove nucleotides from DNA. B) join DNA pieces together. C) twist DNA molecules. D) attach DNA loops to scaffold proteins. E) move chromosomes along spindle fibers. Answer: C
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Chapter 2 DNA: The Genetic Material 13
23) Which form of DNA is a left-handed double helix? A) A-DNA B) B-DNA C) L-DNA D) R-DNA E) Z-DNA Answer: E
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24) The displacement loop (D-loop) may be a characteristic of A) centromeres. B) telomeres. C) A-DNA. D) B-DNA. E) Z-DNA. Answer: B
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25) Which nucleotide is absent in RNA? A) Adenine B) Guanine C) Uracil D) Cytosine E) Thymine Answer: E
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TRUE-FALSE QUESTIONS 26) DNA and RNA both contain phosphate and ribose. Answer: FALSE Explanation: They both contain phosphate, but DNA contains the sugar deoxyribose rather than ribose.
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27) Hershey and Chase used radioactive sulfur to label the genetic material of bacteriophages. Answer: FALSE Explanation: The radioactive sulfur labeled the protein coat.
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28) In a strand of DNA, a hydrogen bond connects the phosphate group of one nucleotide to the sugar of the adjacent nucleotide. Answer: FALSE Explanation: A covalent phosphodiester bond connects the two adjacent nucleotides.
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29) The genome of the T-even family of bacteriophage consists of single-stranded RNA. Answer: FALSE Explanation: It consists of double-stranded DNA.
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30) Borrelia burgdorferi is a bacterium whose genome consists of one large and several small linear chromosomes. Answer: TRUE
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31) By weight, the amount of DNA in chromatin is less than that of histone. Answer: FALSE Explanation: The weights of DNA and histone in chromatin are equal.
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32) The virus first shown to have RNA as its genetic material was tobacco mosaic virus (TMV). Answer: TRUE
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33) The more condensed a part of a chromosome is, the more likely it is that the genes in that region will be active. Answer: FALSE Explanation: The genes in a region are less likely to be active the more condensed a part of a chromosome is.
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34) The genome of most prokaryotes consists of moderately repetitive DNA. Answer: FALSE Explanation: The genome of most prokaryotes consists of unique-sequence DNA.
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35) In eukaryotes, the greatest relative amount of tandemly repeated DNA is associated with centromeres and telomeres. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) In Griffithsʹ transformation experiments, under what conditions did the injected mice die? Answer: The mice died when they were injected with living, virulent bacteria, and when they were injected with living, nonvirulent bacteria mixed with heat-killed, virulent bacteria.
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37) How could you test whether the transforming ability of a cell extract was due to DNA or RNA? Answer: You could treat the extract with a DNase or RNase enzyme and test whether its transforming ability was intact.
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38) One of the strands in a DNA double helix has the nucleotide sequence 5'-ACCTGCTACGG-3' .
What is the sequence of the complementary DNA strand? Answer: 3'-TGGACGATGCC-5'
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Chapter 2 DNA: The Genetic Material 15
39) What is the function of dispersed repeated sequences such as SINEs and LINEs in eukaryotes? Answer: Little is known about the function of such sequences, but one hypothesis is that they have no function at all. Another is that they are involved in regulating gene expression.
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40) What is the C-value paradox, and what is its cause? Answer: There is also no direct relationship between the C-value (the total amount of DNA in the haploid genome) and the structural or organizational complexity of the organism. This is due in part to the amount of repetitive-sequence DNA found in the genome of some organisms.
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41) Define Chargaffʹs rules of the base composition of DNA. Answer: Chargaffʹs rules include the following: (1) the amount of adenine = the amount of thymine, (2) the amount of guanine = the amount of cytosine, and (3) the amount of purines = the amount of pyrimidines.
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recall
42) Describe the differences between heterochromatin and euchromatin in chromosomes. Are there any situations in which one can be changed into the other? Answer: Euchromatin contains actively transcribed genes and undergoes normal cycles of condensation and decondensation in the cell cycle. Heterochromatin remains condensed and contains genes that are usually transcriptionally inactive. Euchromatin can be inactivated, as in the case of Barr bodies. It is then known as facultative heterochromatin.
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43) What are the three necessary characteristics of the hereditary molecule in cells? Answer: (1) It must be able to carry information, (2) it must be able to accurately pass on the information to progeny cells (replicate), and (3) it must be capable of change (evolution).
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44) Name the constituent parts of a nucleoside and a nucleotide. Answer: A nucleoside consists of a pentose sugar covalently bonded to a nitrogenous base; a nucleotide is a nucleoside with the addition of a phosphate group.
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45) The DNA phage ΦΧ174 was found to have a ratio of bases of 25 A:33T :24G:18C. This departs from the usual A/T = 1 and G/C = 1 ratios. How can you explain this? Answer: The genome of the phage consists of single-stranded, rather than double-stranded, DNA.
Skill: Conceptual understanding
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46) If the human egg has 3 billion base pairs, how many nucleosomes will be present in the nucleus of a human somatic cell? Answer: In humans, the DNA wrapped around each nucleosome is approximately 200 bp (147 bp + 53 bp linker). As such, there will be approximately 3 × 109 /2 × 102 = 1.5 × 107 nucleosomes in a human egg nucleus. However, the egg is haploid, whereas the somatic cells are diploid. Therefore, there will be approximately 1.5 × 107 × 2 = 3 × 107 nucleosomes in the nucleus of a human somatic cell.
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47) Why are the amino acid sequences of eukaryotic histones so similar to one another, even among distantly related species? Answer: Evolutionary conservation of these sequences is a strong indicator that histones perform the same basic role in organizing the DNA in the chromosomes of all eukaryotes.
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48) Describe the packing of chromatin from the 10-nm to the 30-nm fiber stage. What is the role of histones? Answer: 10-nm chromatin fiber consists of nucleosomes–ʺbeadsʺ of DNA wound around eight core histone proteins–connected by strands of linker DNA. The 30-nm chromatin fiber is created by the binding of histone H1, which brings the linker DNA and the nucleosomes closer together. In the solenoid model of the 30 -nm fiber, the nucleosomes are brought together into a spiraling helical structure, with about six nucleosomes per complete turn.
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49) What is the role of centromeres and telomeres? Answer: Centromeres are the chromosomal regions where mitotic or meiotic spindle fibers attach. They are therefore responsible for the accurate segregation of chromosomes to daughter cells during replication. Telomeres are heterochromatic regions of chromosomes that are also required for replication and stability. They are usually found at the ends of the chromosome and are associated with the nuclear envelope.
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50) If the base pairs in a DNA helix are 0.34 nm apart, and a complete (360°) turn of the helix takes 3.4 nm, how many base pairs per turn are there in a DNA molecule? Answer: There are 10 base pairs per turn.
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Chapter 3 DNA Replication
MATCHING QUESTIONS Please select the best match for each term. 1) Primosome
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A) A complex of proteins and RNA that replicates the ends of eukaryotic chromosomes B) A complex of helicase and primase on the template DNA C) A complex of key replication proteins at the replication fork in E. coli and bacteriophage DNA D) The distance between the origin and the termination of replication where two replication forks fuse E) A DNA sequence that contains the specific region where replication begins
2) Replicon
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3) Replisome
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4) Telomerase
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5) Replicator
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Answers:
1) B
2) D
3) C
4) A
5) E
MULTIPLE-CHOICE QUESTIONS 6) During replication, the direction of synthesis of new DNA from the leading and lagging strands is A) 5ʹ to 3ʹ only. B) 3ʹ to 5ʹ only. C) from left to right only. D) both 5ʹ to 3ʹ and 3ʹ to 5ʹ. E) different, depending on whether the cell is prokaryotic or eukaryotic. Answer: A
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7) DNA replication is always A) discontinuous. B) bidirectional. C) conservative. D) semiconservative. E) dispersive. Answer: D
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8) In E. coli, replication begins at which chromosome site? A) The replication fork B) ter C) oriC D) TBP E) All of these Answer: C
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9) Which of the following did Kornberg use to detect DNA synthesis? A) Radioactively labeled E. coli cells B) Fluorescently labeled E. coli cells C) Radioactively labeled deoxynucleotide triphosphates D) Fluorescently labeled deoxynucleotide triphosphates E) Nonlabeled deoxynucleotide triphosphates Answer: C
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10) How many replication forks are produced when DNA denatures at an origin? A) 0 B) 1 C) 2 D) 3 E) The number varies Answer: C
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11) What is a replication bubble? A) A complex of replication enzymes on the DNA template strand B) A DNA sequence that initiates replication C) A tangle of denatured DNA strands near the replication fork D) A locally denatured segment of DNA where replication originates E) A localized site in the nucleus where chromosomes are replicating Answer: D
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12) Which of the following are necessary for DNA replication in vitro? A) RNA, helicase, primers, DNA polymerase B) Okazaki fragments, helicase, DNA polymerase C) Template DNA, DNA polymerase, four dNTPs, primers, magnesium ions D) Template DNA, four dNTPs, magnesium ions E) DNA canʹt replicate in vitro Answer: C
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Chapter 3 DNA Replication 19
13) In Meselson and Stahlʹs experiment, what kind of DNA molecules would be found after four replication cycles? A) Only heavy DNA (15N- 15N) B) Only intermediate DNA ( 15N- 14N) C) Only light DNA ( 14N- 14N) D) Both heavy ( 15N- 15N) and light DNA (14N- 14N) E) Both heavy ( 15N‐15N) and intermediate DNA (15N- 14N) Answer: D
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14) The two most basic steps of DNA replication are A) primase causes primer to bind the template and ligase copies the template. B) helicase unwinds the template and DNA polymerase binds the template. C) leading strand is copied first and lagging strand is copied second. D) the new strand is denatured and a template is synthesized. E) the template is denatured and a new strand is synthesized. Answer: E
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15) Where does the initiator protein bind DNA at the start of replication? A) At a replication fork B) At an origin of replication C) At any AT-rich region D) At a promoter region E) At a start codon Answer: B
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16) As helicase unwinds the DNA molecule, what keeps the strands apart? A) DNA polymerase B) Reverse transcriptase C) Replication fork D) Single-strand binding proteins E) Okazaki fragments Answer: D
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17) After removal of the RNA primers and replacement with DNA nucleotides, the single-stranded nick adjacent to Okazaki fragments is filled in through a reaction that involves which enzyme? A) DNA primase B) SSB protein C) RNA polymerase D) DNA ligase E) DNA helicase Answer: D
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18) Which enzyme elongates the new DNA strand starting at an RNA primer? A) DNA polymerase I B) DNA polymerase III C) RNA primase D) DNA ligase E) RNA polymerase Answer: B
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19) After a region of DNA has been replicated, ________ removes the RNA primers. A) DNA polymerase I B) DNA polymerase III C) DNA helicase D) RNA primase E) DNA ligase Answer: A
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20) Which enzyme replaces RNA primers with DNA after elongation? A) DNA polymerase I B) DNA polymerase III C) RNA polymerase D) RNA primase E) DNA ligase Answer: A
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21) Which kind of enzyme prevents DNA from tangling up by introducing negative supercoils as the replication fork migrates during replication? A) Helicase B) Ligase C) DNA polymerase I D) DNA polymerase III E) Topoisomerase Answer: E
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22) The base-pairing error rate remains low during replication because A) DNA repair mechanisms can fix the mispaired bases. B) bases that are mispaired can excise themselves. C) UV light radiation corrects any base mispairs. D) mispaired bases cause a cell to die before replication is complete. E) None of these; base-pairing errors are not possible Answer: A
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Chapter 3 DNA Replication 21
23) The enzymatic activity of a telomerase is best described as a A) polymerase. B) ligase. C) topoisomerase. D) reverse transcriptase. E) exonuclease. Answer: D
Skill: Factual recall
24) Rolling circle replication of DNA is characterized by the absence of A) the DNA polymerase. B) a nick in the DNA template. C) the RNA primers. D) the replication bubble. E) the Okazaki fragments. Answer: D
Skill: Conceptual understanding
25) Which enzyme activity is associated with the proofreading mechanism of DNA polymerase I? A) 5ʹ-to-3ʹ exonuclease activity B) 3ʹ-to-5ʹ exonuclease activity C) 5ʹ-to-3ʹ polymerase activity D) Both A and B E) All of these Answer: B
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TRUE-FALSE QUESTIONS 26) A new nucleotide is added to a growing strand of DNA at the 3ʹ end. Answer: TRUE
Skill: Factual recall
27) Only the leading strand of a DNA molecule serves as a template during replication. Answer: FALSE Explanation: Both the leading and lagging strands serve as templates.
Skill: Factual recall
28) At the growing end of a DNA chain, DNA polymerase catalyzes the formation of a disulfide bond between the 3ʹ-OH group of the deoxyribose on the last nucleotide and the 5ʹ-phosphate of the dNTP precursor. Answer: FALSE Explanation: DNA polymerase catalyzes the formation of a phosphodiester bond between nucleotides.
Skill: Factual recall
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22 Test Bank for iGenetics
29) DNA primase is an RNA polymerase. Answer: TRUE Explanation: DNA primase catalyzes the reaction to synthesize a short RNA primer molecule.
Skill: Factual recall
30) Okazaki fragments are made from the lagging strand of the DNA double helix. Answer: TRUE
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31) DNA polymerase III is very inaccurate at matching bases during replication, with errors in one out of every 100 base pairs. Answer: FALSE Explanation: DNA polymerase III is very accurate, causing errors in only one out of every 1,000,000 base pairs.
Skill: Factual recall
32) In eukaryotic cells, histone proteins are actively synthesized during the S phase of the cell cycle. Answer: TRUE
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33) In eukaryotes, DNA replication begins at a single origin of replication on each chromosome. Answer: FALSE Explanation: Replication begins at multiple origins of replication on each chromosome.
Skill: Factual recall
34) Topoisomerase and SSB proteins are important components of the replication process in prokaryotes, but they are not found in eukaryotes. Answer: FALSE Explanation: These proteins also play key roles in eukaryotic DNA replication.
Skill: Factual recall
35) Mg 2+ ions are required for optimal DNA polymerase activity. Answer: TRUE Explanation: DNA polymerases often require magnesium ions as cofactors.
Skill: Factual recall
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Chapter 3 DNA Replication 23
SHORT ANSWER QUESTIONS AND PROBLEMS 36) What are the key replication enzymes at the replisome, and how is DNA replication on both leading and lagging strands made efficient through the conformation of the DNA at the replisome in prokaryotes? Answer: The key replication enzymes at the replisome are helicase, primase, and DNA polymerase III. To make replication more efficient, the lagging-strand DNA is folded so that its DNA polymerase III is complexed with the DNA polymerase III on the leading strand (forming the DNA Pol III holoenzyme). The folding of the lagging-strand template also makes production of sequential Okazaki fragments more efficient by bringing the 3ʹ end of each completed Okazaki fragments near the site where the next Okazaki fragment will start.
Skill: Conceptual understanding
37) Why is DNA replication referred to as semiconservative? Answer: The progeny double helices consist of one parental DNA strand and one newly synthesized strand.
Skill: Conceptual understanding
38) Why is an AT-rich sequence characteristic of DNA replicators in all organisms? Answer: AT-rich regions of DNA are relatively easy to denature to single strands. AT base pairs are held together by only two hydrogen bonds, while GC pairs are held together by three.
Skill: Conceptual understanding
39) What are the differences in replication between leading and lagging strands in terms of continuity and directionality in relation to the replication fork? Answer: The leading strand is copied continuously from the 3ʹ end toward the replication fork, while the lagging strand is copied in fragments away from the replication fork.
Skill: Factual recall
40) What experiment demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I was essential for cell viability? Answer: In E. coli, the DNA Pol I polAex1 mutant strain survives at 37˚C but dies at 42˚C. This was shown to be because the mutant DNA Pol I enzyme had normal activity at 37˚C but a defective 5ʹ-to-3ʹ activity at 42˚C. This demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA Pol I was essential for cell survival.
Skill: Factual recall
41) A cross is made between yeast cells with different alleles for a set of linked genes: pr + q × p + rq+ . The resulting tetrads show a 3:1 ratio for r+ to r instead of the expected 2:2 ratio. Can you explain how this could have occurred? Answer: Gene conversion by a mismatch repair mechanism could have caused this deviation from expected ratios. During pairing of homologous chromosomes in meiosis, recombination between the two inner chromatids occurred, resulting in heteroduplex (mismatched) DNA strands. Both mismatches were repaired by excision and DNA synthesis to match the parent DNA with the r+ allele.
Skill: Analytical reasoning
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42) Describe the method devised by Arthur Kornberg which first successfully achieved DNA synthesis in vitro, including its components and their uses. Answer: Kornberg mixed together DNA fragments, all four dNTPs (DNA precursors), and an E. coli lysate to achieve DNA synthesis in vitro. The DNA fragments acted as a template for the synthesis of new DNA. The dNTPs were precursors of the new DNA strand, and the cell lysate contained the enzyme necessary to catalyze DNA synthesis (DNA Pol I).
Skill: Factual recall
43) The diploid set of chromosomes in Drosophila embryos replicates six times faster than the single E. coli chromosome, even though there is about 100 times more DNA in Drosophila than in E. coli and the rate of movement of the replication fork in Drosophila is much slower. How is this so? Answer: Eukaryotic chromosomes duplicate rapidly because DNA replication initiates at many origins of replication throughout the genome. In E. coli, there is only one replicon, while in eukaryotes there are multiple, smaller replicons.
Skill: Application of knowledge
44) What are some key differences in replication between E. coli DNA and λ phage DNA? Answer: Both start replication as a circular molecule, and in E. coli, the parental DNA strands remain in a circular form throughout the replication cycle. A replication bubble opens to form two replication forks, and replication proceeds bidirectionally. Lambda phage DNA is replicated by a rolling circle mechanism, in which a nick is made in one of the two strands of the circle, and the 5ʹ end of the cut DNA strand is rolled out as a free ʺtongueʺof increasing length as replication proceeds . The parental DNA circle is replicated continuously, while the linear displaced strand is replicated discontinuously. As long as the circle continues to roll, concatamers of phage DNA can be produced. This is later cut up into linear chromosomes and packaged into new phage heads.
Skill: Factual recall
45) How do the DNA polymerase repair mechanisms work? Answer: Both DNA Pol I and DNA Pol III have 3ʹ-to-5ʹ exonuclease activity and can remove nucleotides from the end of a DNA chain as part of an error correction mechanism. If an incorrect base is inserted by DNA polymerase, and the error is recognized immediately, the exonuclease activity excises the erroneous nucleotide from the new strand. After excision, the DNA polymerase resumes motion in the forward direction and inserts the correct nucleotide.
Skill: Factual recall
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Chapter 3 DNA Replication 25
46) How did Meselson and Stahl rule out the conservative model of DNA replication using equilibrium density gradient centrifugation? Answer: First, they grew E. coli cells in a medium containing only high-density nitrogen (15N). Then they allowed the cells to undergo successive rounds of replication in the presence of normal density nitrogen ( 14N). They used equilibrium density gradient centrifugation after each round of replication to separate the DNA produced by density. If the conservative model was correct, they would have found no intermediate-density DNA after a round of replication. However, after one round of replication, they found that the entire amount of DNA had a density exactly intermediate between 14N and 15N. Subsequent rounds of the experiment showed that semiconservative replication was the correct model.
Skill: Conceptual understanding
47) How were the sequences that compose the replication origins in yeast discovered? Answer: Yeast cells were grown in heavy isotope media to produce denser DNA and then shifted to media with normal, light isotopes. After a few minutes, DNA from these cells was extracted and cut into small pieces. The lighter DNA fragments (which should contain the origins of replication) were collected, fluorescently labeled, and used to hybridize a microarray of yeast sequences. Determination of the sequences to which this light, fluorescently labeled DNA hybridized led to the identification of a number of yeast replication origins.
Skill: Factual recall
48) How will DNA replication be affected if DNA polymerase I has a mutation that inactivates 5ʹ-to-3ʹ exonuclease activity? Answer: The 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I removes the RNA primers from the ends of Okazaki fragments. If this activity is missing, the primers may not be removed from the growing DNA strands.
Skill: Conceptual understanding
49) When the RNA primers are removed from the 5ʹ ends of eukaryotic chromosomes after replication, DNA polymerase is unable to fill in the gaps. What prevents the chromosomes from getting shorter and shorter with each round of replication? Answer: The enzyme telomerase maintains chromosome lengths by adding telomere repeats to the chromosome ends. Telomerase contains an RNA component that is complementary to the telomere repeat unit of the chromosome. After binding to the overhanging repeat, the telomerase RNA is used as a template to synthesize new chromosomal telomere repeats.
Skill: Factual recall
50) Why do tumor cells in mammals have telomerase activity? Answer: Tumor cells are ʺimmortalʺ cells, and the enzyme telomerase is required for long-term cell viability. It has been demonstrated that mutations in genes such as TLC1 and EST1 decrease cell longevity by continuous shortening of telomere length.
Skill: Conceptual understanding
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Chapter 4 Gene Function
MATCHING QUESTIONS Please select the best match for each term. 1) Alkaptonuria
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A) Hypoxanthine guanine phosphoribosyltransferase B) Galactose‐1‐phosphate uridyl transferase C) Hexosaminidase A D) Fructose‐1,6‐diphosphatase E) Homogentisic acid oxidase
2) Tay—Sachs disease
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3) Lesch—Nyhan syndrome
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4) Galactosemia
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5) Hypoglycemia
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Answers:
1) E
2) C
3) A
4) B
5) D
MULTIPLE-CHOICE QUESTIONS 6) Why did Garrod define alkaptonuria as a recessive genetic disease? A) It is expressed in male members of an affected family. B) It is expressed in all members of an affected family. C) It is expressed under certain environmental conditions. D) It is expressed when two alleles for the disease are present. E) It is expressed when one of the two alleles carries the disease trait. Answer: D
Skill: Factual recall
7) What was the significance of Beadle and Tatumʹs experiment? A) It resulted in the central dogma. B) It led to the discovery of bread mold. C) It resulted in the one‐gene‐one‐enzyme hypothesis. D) It led to the discovery of the genetic code. E) It proved that X‐rays were mutagens. Answer: C
Skill: Factual recall
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Chapter 4 Gene Function 27
8) What does a gene actually code for? A) A polypeptide B) An enzyme C) An amino acid D) A protein E) A nucleotide Answer: A
Skill: Factual recall
9) Which genetic disease(s) are caused by defective proteins that are not enzymes? A) Tay—Sachs disease and phenylketonuria B) Sickle‐cell anemia and cystic fibrosis C) Lesch—Nyhan syndrome. D) Albinism and alkaptonuria E) All of these Answer: B
Skill: Factual recall
10) Amniocentesis involves A) prenatal diagnosis of genetic defects in a fetus. B) tests for enzyme and protein deficiencies in a fetus. C) taking a sample of amniotic fluid with a needle. D) analysis for DNA and chromosome defects in a fetus. E) All of these Answer: E
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11) Auxotrophs are organisms that can grow A) on minimal media. B) on complete media. C) only on amino acid supplemented media. D) only on vitamin supplemented media. E) on any media. Answer: B
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12) A lysosomal storage disease is caused by A) failure of the cell to produce lysosomes. B) the inability to synthesize glycolipids. C) mutations in genes for lysosomic membrane proteins. D) an excess in the production of lysosomic digestive enzymes. E) mutations in genes that code for lysosomic digestive enzymes. Answer: E
Skill: Factual recall
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13) Cystic fibrosis is caused by a mutation in the gene that encodes which enzyme? A) Pyruvate kinase B) CFTR protein C) Hexosaminidase A D) Hemoglobin E) Tyrosine kinase Answer: B
Skill: Factual recall
14) In a molecule of hemoglobin C, an aspartic acid residue is changed into a ________ residue. A) lysine B) tyrosine C) leucine D) phenylalanine E) glutamine Answer: A
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15) A mutation that has pleiotropic consequences A) results in a single symptom in the affected person. B) may result in only slight symptoms in a person with the mutation. C) is only detected in homozygotes. D) has widespread consequences in the affected person. E) cannot be detected by enzyme assay. Answer: D
Skill: Factual recall
16) Individuals with Lesch—Nyhan syndrome exhibit A) urine that turns black when exposed to air. B) compulsive self‐mutilation behaviors. C) lighter than normal pigmentation of skin, hair, and eyes. D) severe anemia. E) a cherry‐colored spot on the retina. Answer: B
Skill: Factual recall
17) A mutagen is A) the process that generates mutants. B) the agent that induces mutants. C) the mutants that are generated by an agent. D) the experiment that generates mutants. E) the process that corrects mutants. Answer: B
Skill: Factual recall
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Chapter 4 Gene Function 29
18) In people who do not have Tay—Sachs disease, the hexA gene encodes an enzyme that A) converts purines to uric acid in the kidneys. B) removes a sugar group from phenylalanine. C) converts tyrosine to melanin in skin. D) cleaves acetylgalactosamine from gangliosides. E) synthesizes glycogen from glucose monomers. Answer: D
Skill: Factual recall
19) A person with sickle‐cell trait has ________ at the β‐globin locus. A) two wild‐type alleles B) two mutant alleles C) one wild‐type and one mutant allele D) one wild‐type and one deleted allele E) one mutant and one deleted allele Answer: C
Skill: Factual recall
20) Which of the following is a sex‐linked disorder? A) Lesch—Nyhan syndrome B) Alkaptonuria C) Albinism D) Down syndrome E) Tay—Sachs disease Answer: A
Skill: Factual recall
21) Which of the following statements about phenylketonuria (PKU) is false? A) Buildup of phenylpyruvic acid affects the nervous system. B) The amino acid tyrosine cannot be made. C) Production of the hormone thyroxin s affected. D) The skin color of those with phenylketonuria is light because of decreased levels of melanin. E) Adrenalin levels increase in the blood stream. Answer: E
Skill: Factual recall
22) The cystic fibrosis gene (CFTR) codes for a protein that is essential for A) secretion of mucus. B) ion transport across membranes. C) transport of oxygen in red blood cells. D) metabolism of certain amino acids. E) processing of gangliosides. Answer: B
Skill: Factual recall
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23) A carrier for a disease A) is homozygous for the dominant mutation. B) is homozygous for the recessive mutation. C) is heterozygous for the dominant mutation. D) is heterozygous for the recessive mutation E) is hemizygous for any mutation. Answer: D
Skill: Factual recall
24) Normal and sickle‐cell hemoglobin molecules differ A) in the number of amino acids in each molecule. B) by a single DNA point mutation that leads to the substitution of one amino acid for another. C) in the number and orientation of the amino acid chains attached to the heme portion of each molecule. D) in the number of oxygen molecules that can be carried by each molecule. E) by the type of bone marrow that produces them. Answer: B
Skill: Factual recall
25) Cystic fibrosis is an autosomal recessive disorder. What is the chance that two carrier parents will have an affected child? A) 100% B) 75% C) 50% D) 25% E) 0% Answer: D
Skill: Problem-solving
TRUE-FALSE QUESTIONS 26) A Neurospora mutant that is a tryptophan auxotroph does not need to be grown on a medium that contains tryptophan. Answer: FALSE Explanation: A tryptophan auxotroph cannot synthesize its own tryptophan. It should be grown on a medium that contains tryptophan.
Skill: Factual recall
27) Sickle‐cell disease is more common among people who are of African rather than European descent. Answer: TRUE
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28) Because Lesch—Nyhan syndrome is a recessive X‐linked disease, females who have an allele with the mutation responsible for this disease are generally symptom‐free. Answer: FALSE Explanation: If her normal allele has undergone lyonization (random X inactivation), the carrier female may display symptoms.
Skill: Factual recall
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Chapter 4 Gene Function 31
29) A person who is homozygous for the gene for hemoglobin A will have sickle‐cell disease. Answer: FALSE Explanation: This person will have normal hemoglobin and normal red blood cells.
Skill: Factual recall
30) The Guthrie test screens for excessive phenylalanine in the blood of infants. Answer: TRUE
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31) Albinism is caused by an inborn error of metabolism. Answer: TRUE
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32) Genetic counseling is the best advice that a physician can give a person about his or her risk of having a child with a genetic disorder while lacking precise statistical evidence. Answer: FALSE Explanation: If knowledge about the role of heredity in a genetic disorder is lacking, genetic counseling can only give a general estimate of risk. In many circumstances, however, the physician can give a person precise probabilities about the risk of passing on a genetic defect.
Skill: Factual recall
33) The sickle‐cell mutant allele and wild‐type β‐globin allele are codominant. Answer: TRUE
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34) Cystic fibrosis is a pleiotropic disease. Answer: TRUE
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35) Tay—Sachs disease is most prevalent among populations of the Pennsylvania Amish in the United States. Answer: FALSE Explanation: It is most prevalent among Ashkenazi Jews of central and eastern European descent.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) The following is a representation of a metabolic pathway that involves genes for enzymes A, B, C, and D: A → B → C → D → product A mutation in the B gene would result in an accumulation of which gene product? Answer: A
Skill: Analytical reasoning
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37) What is an essential amino acid? Give an example. Answer: An essential amino acid is one that humans must consume as part of their diet, because they are unable to synthesize it. An example is phenylalanine.
Skill: Conceptual understanding
38) In the laboratory, how could you distinguish between normal hemoglobin A and hemoglobin S? Answer: The two forms of hemoglobin have different electrical charges. Therefore, you could distinguish between them by their migration patterns on an electrophoresis gel.
Skill: Application of knowledge
39) A person with PKU controls the buildup of phenylpyruvic acid in his or her blood by restricting dietary intake of phenylalanine, thus avoiding the severe symptoms of mental retardation, slow growth rate, and early death. Yet he or she still exhibits the symptoms of fair skin and low adrenaline levels. Why? Answer: The PKU mutation is in the gene for phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. Tyrosine is essential for melanin and adrenaline production. A PKU individual cannot make tyrosine, and it is difficult to obtain enough tyrosine from food intake to avoid any symptoms of the disease.
Skill: Analytical reasoning
40) How do scientists, lacking the ability to make controlled genetic crosses in humans, study the inheritance of and give genetic counseling on human genetic disorders? Answer: Pedigree analysis can be done by compiling phenotypic records from both families over several generations. This can determine the likelihood that a particular allele is present in either family.
Skill: Application of knowledge
41) What is an inborn error of metabolism? Give an example. Answer: An inborn error of metabolism is a genetic disease caused by the absence of a particular enzyme necessary to fulfill a biochemical pathway. An example is alkaptonuria, in which an enzyme necessary for homogentisic acid (HA) metabolism is not made. Therefore, people with this disease cannot metabolize HA and it is excreted in their urine.
Skill: Factual recall
42) Why are genetic diseases much more common among children of marriages involving first cousins than among children of marriages between unrelated partners? Answer: This is because first cousins have many alleles in common, so the chances are greater for recessive alleles to be homozygous in children of first‐cousin marriages.
Skill: Conceptual understanding
43) What factor do the diseases PKU, albinism, and alkaptonuria all have in common? Answer: They are all marked by a block in the phenylalanine‐tyrosine metabolic pathways.
Skill: Factual recall
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Chapter 4 Gene Function 33
44) You discover a mutant strain of Neurospora crassa that will not grow on a minimal medium but that will grow on a medium supplemented by the amino acid methionine. How can you determine which step in the methionine synthesis pathway is affected by the mutation? Answer: You would grow the strain on a series of media each supplemented with a chemical intermediate in the methionine biosynthetic pathway and observe for growth. The mutant will be able to grow on any medium that contains an intermediate substance in the pathway that comes after the step controlled by its mutant gene. It will also accumulate the intermediate compound used in the step where it is blocked.
Skill: Analytical reasoning
45) Classic albinism is an autosomal recessive mutation. However, two parents with albinism may produce a child with normal pigment. How can this be so? Answer: There are a number of biochemical steps in melanin biosynthesis from tyrosine. For this reason, albinism can be caused by different enzyme deficiencies at different steps of the pathway. If each of the parentsʹ albinism stems from a different enzyme deficiency, they can therefore have a child with normal pigment.
Skill: Analytical reasoning
46) While the defective gene product in patients with PKU was identified using biochemical analysis, how was the cystic fibrosis gene identified? Answer: The cystic fibrosis gene was identified by a combination of genetic and modern molecular biology techniques. It was first localized to chromosome 7, and then molecularly cloned and sequenced. The sequence of the mutant allele was compared with the allele of a normal person to discover a three base‐pair deletion in the cystic fibrosis gene. To determine the geneʹs function, its amino acid sequence and three‐dimensional structure was deduced. This was compared with a database of protein sequences to determine the geneʹs function as an active transport protein.
Skill: Factual recall
47) How can enzyme assay be used to detect carriage of a recessive gene mutation in a person of normal phenotype? Answer: If they were a carrier, the person, as a heterozygote, would be expected to produce only half the normal enzyme or protein product for the gene in question.
Skill: Analytical reasoning
48) What advantages does chorionic villus sampling have over amniocentesis for fetal analysis of genetic defects? Answer: Chorionic villus sampling is generally performed up to four weeks earlier than amniocentesis, allowing parents to make an earlier decision regarding potential termination of pregnancies. It is also a faster test that does not require cell culture to do biochemical assays.
Skill: Factual recall
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34 Test Bank for iGenetics
49) Explain the difference between the one‐geneone‐enzyme hypothesis and the one‐gene one‐polypeptide hypothesis. Answer: A relationship is found between a mutation in a single gene and the loss of production of a single enzyme in a metabolic pathway. However, more than one gene may control each step in a pathway because some enzymes (and some non‐enzyme proteins such as hemoglobin) consist of two or more polypeptide chains, each of which is coded for by a different gene. Thus, it is more accurate to use the phrase ʺone‐geneone‐polypeptide.ʺ
Skill: Conceptual understanding
50) You are a doctor who specializes in genetic diseases. A married couple comes to see you for genetic counseling. The woman has PKU, the symptoms of which she manages by diet. The man and woman are first cousins and would like to know the probability that any of their children will have the disease. What would you tell them? Answer: Since the woman has the disease, she must be homozygous recessive for the defective autosomal gene. If her husband is not a carrier, all their children will be heterozygous carriers of the disease. However, since the man and woman are closely related, there is a high probability that the man is a carrier of the PKU gene. If he is a carrier, there is a 50% chance that any of their children will have the disease. You would perform a detailed pedigree analysis for the couple to better determine the probability that the husband carries the recessive allele.
Skill: Problem-solving
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Chapter 5 Gene Expression: Transcription
MATCHING QUESTIONS Please select the best match for each term. 1) mRNA
Skill: Factual recall
A) Brings amino acids to ribosomes during translation B) Forms complexes that are used in eukaryotic RNA processing C) Forms the initiation complex with RNA polymerase in eukaryotes D) With ribosomal proteins, makes up the ribosomes E) Encodes the amino acid sequence of a polypeptide 2) A 3) D 4) B 5) C
2) tRNA
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3) rRNA
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4) snRNA
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5) GTF
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Answers:
1) E
MULTIPLE-CHOICE QUESTIONS 6) In eukaryotes, the initiation complex binds DNA at the A) Pribnow box. B) GC box. C) CAAT box. D) TATA box. E) GTF region. Answer: D
Skill: Factual recall
7) The process by which genetic information is transferred from DNA to RNA is called A) replication. B) transcription. C) translation. D) transversion. E) transformation. Answer: B
Skill: Factual recall
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8) Which of the following proteins is found only in the RNA polymerase holoenzyme? A) Alpha B) Beta C) Epsilon D) Sigma E) Rho Answer: D
Skill: Factual recall
9) During the initiation step of transcription, which molecule binds the promoter region of a DNA molecule? A) DNA polymerase B) Reverse transcriptase C) RNA polymerase D) DNA primase E) Topoisomerase Answer: C
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10) During transcription, the synthesis of the mRNA strand proceeds in which direction? A) 5ʹ to 3ʹ only B) 3ʹ to 5ʹ only C) Both 5ʹ to 3ʹ and 3ʹ to 5ʹ D) Either 5ʹ to 3ʹ or 3ʹ to 5ʹ E) First 5ʹ to 3ʹ, and then 3ʹ to 5ʹ Answer: A
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11) Which type of RNA is found only in eukaryotes? A) mRNA B) tRNA C) rRNA D) snRNA E) All of these Answer: D
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12) Which of the following catalyzes the synthesis of ribosomal RNA in eukaryotes? A) RNA polymerase I B) RNA polymerase II C) RNA polymerase III D) Both RNA polymerase I and II E) Both RNA polymerase I and III Answer: E
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 37
13) Prokaryotic transcription begins when RNA polymerase binds to which site of the gene region? A) -35 B) -10 C) +1 D) +10 E) +35 Answer: C
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14) Sequences that contain information about the stability of a transcript are found in the A) 5ʹ UTR. B) coding sequence. C) 3ʹ UTR. D) leading strand. E) lagging strand. Answer: C
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15) Which of the following is not found in a eukaryotic promoter? A) -10 box B) TATA box C) GC box D) CAAT box E) None of these Answer: A
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16) In eukaryotes, where does transcription take place? A) In the cytoplasm B) Anywhere in the cell C) In the nucleus D) In both the cytoplasm and nucleus E) In the mitochondria Answer: C
Skill: Factual recall
17) In eukaryotes, precursor mRNA molecules are processed in the A) ribosomes. B) nucleus. C) mitochondria. D) cytoplasm. E) nucleolus. Answer: B
Skill: Factual recall
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38 Test Bank for iGenetics
18) The single strand of mRNA that is produced during transcription has the same polarity as the ________ strand of DNA. A) template B) nontemplate C) leading D) lagging E) nonsense Answer: B
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19) The poly(A) tail of an mRNA A) helps transport the pre‐mRNA to the cytoplasm. B) keeps the coding sequences from being degraded. C) is added without an RNA template. D) is important in initiation of translation. E) is removed from an mRNA after its maturation. Answer: B
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20) Self‐cleaving RNAs that function catalytically are called A) snRNPs. B) snRNAs. C) ribozymes. D) ribosomes. E) spliceosomes. Answer: C
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21) Molecules of tRNA are A) 75 to 90 nucleotides in length. B) single‐stranded with secondary structure. C) cloverleaf in shape. D) synthesized with modified bases. E) All of these Answer: E
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22) A prokaryotic mRNA transcript is A) transcribed completely before being translated. B) further processed before it is transcribed. C) transported from the nucleus before it is translated. D) modified before it is translated. E) translated as it is being transcribed. Answer: E
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 39
23) Which parts of a eukaryotic gene are transcribed? A) Only the exons B) Only the introns C) Both the exons and introns D) Exons, introns, promoter, and terminator sequence E) It depends on the gene Answer: C
Skill: Factual recall
24) Posttranscriptional insertion or deletion of nucleotides that is absent in the DNA template is called A) RNA splicing. B) RNA editing. C) RNA processing. D) RNA capping. E) RNA modification. Answer: B
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25) A rho‐dependent terminator A) forms a hairpin loop during termination. B) requires ATP to complete termination. C) is rich in AT nucleotides. D) Both A and B E) All of these Answer: B
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TRUE-FALSE QUESTIONS 26) A promoter is a coding portion of a gene. Answer: FALSE Explanation: A promoter is a noncoding, regulatory DNA sequence at the start of a gene.
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27) The number of initiations of transcription from a promoter that has the sequence 5'-TAGCAT-3' will be greater than the number of initiations from a wild‐type promoter. Answer: FALSE Explanation: It will be fewer than the number of initiations from a wild‐type promoter.
Skill: Factual recall
28) All genes encode proteins. Answer: FALSE Explanation: Some genes encode specific kinds of RNAs that are not translated into proteins. Protein coding (or structural) genes, code for mRNA transcripts.
Skill: Factual recall
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40 Test Bank for iGenetics
29) Activators are cis‐acting sequences that increase transcription from a promoter. Answer: FALSE Explanation: Activators are regulatory proteins that determine the efficiency of initiation of transcription.
Skill: Factual recall
30) Before a prokaryotic mRNA can be translated, it must be modified by the addition of a polyA tail. Answer: FALSE Explanation: Prokaryotic mRNAs are not processed before translation.
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31) Termination is carried out by RNA polymerase in a rho‐independent terminator. Answer: TRUE
Skill: Factual recall
32) The Rho protein is a topoisomerase that helps in untwisting the DNA template. Answer: FALSE Explanation: The Rho protein is a helicase.
Skill: Factual recall
33) RNA polymerase has no proofreading activity during transcription. Answer: FALSE Explanation: RNA polymerase has two different proofreading activities for correcting incorrectly inserted nucleotides.
Skill: Factual recall
34) Unlike eukaryotic mRNA, prokaryotic mRNA is polycistronic. Answer: TRUE
Skill: Factual recall
35) Group I introns, such as those found in Tetrahymena, are unique in that they are excised from mRNA by a protein‐driven catalytic reaction. Answer: FALSE Explanation: Group I introns are unique in that they are self‐splicing.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) Outline the basic differences in function among RNA polymerases in both prokaryotes and eukaryotes. Answer: Only one type of RNA polymerase is found in prokaryotes, so all classes of genes are transcribed by it. Eukaryotes have three distinct RNA polymerases, each of which transcribes different gene types. RNA polymerase I transcribes the genes for the 28S, 18S, and 5.8S ribosomal RNAs; RNA polymerase II transcribes mRNA genes and some snRNA genes; and RNA polymerase III transcribes genes for the 5S rRNAs, the tRNAs, and the other snRNAs.
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 41
37) What are the three basic steps of transcription? Answer: Initiation, elongation, and termination.
Skill: Factual recall
38) What is the function of the inverted repeat in the termination sequence of a rho‐independent terminator? What would happen if a mutation occurred in this region? Answer: RNA polymerase transcribes the inverted repeat into the RNA transcript. As a result, a hairpin loop structure forms because of the complementarity of the inverted repeat regions. This hairpin is believed to aid termination by affecting the ability of RNA polymerase to continue elongation. If a mutation occurred in this region, termination would not be successful.
Skill: Factual recall, analytical reasoning
39) What was the significance of the 1978 finding by Philip Lederʹs research group that the 0.7 kb β‐globin mRNA is not colinear with the gene that encodes it, but the nuclear 1.5 kb pre‐mRNA is? Answer: This meant that the β‐globin contains an intron, which is spliced out after pre‐mRNA processing, and the exons are spliced together to make a mature mRNA. The significance was the realization that genes could be ʺin pieces.ʺ
Skill: Conceptual understanding
40) What would be the advantage of having multiple sigma factors that each could be produced under different stress conditions in E. coli? Answer: Different sigma factors recognize different promoter region sequences. Therefore, they play a role in the regulation of gene expression. Sigma factors produced under a particular stress condition, such as excessive heat or infection by bacteriophage, bind to promoters that belong to genes that encode proteins required to cope with those stresses.
Skill: Analytical reasoning
41) What kind of experiment or analysis could you conduct to define the location of the promoter elements of protein‐coding genes? Answer: You could examine the effect of mutations that delete or alter base pairs upstream from the starting point of transcription and to see whether those mutants affect transcription. Mutations that significantly affect transcription define important promoter elements. Alternatively, you could compare the DNA sequences upstream of a number of protein‐coding genes to see whether there are any regions with similar sequences.
Skill: Analytical reasoning
42) What is the difference between an intervening sequence and an untranslated region in a eukaryotic mRNA? Answer: Untranslated regions (UTRs) are found on either side of the protein‐coding sequence of mRNAs and are part of the exons (and are thus not removed in pre‐mRNA processing). Intervening sequences, also called introns, are found within the protein‐coding sequence region and are removed in the processing of the pre‐mRNA molecule to the mature mRNA.
Skill: Factual recall
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42 Test Bank for iGenetics
43) The sequence of a template strand of DNA is 3'-CATTACGCTT-5' . What is the sequence of the corresponding mRNA? Answer: 5'-GUAAUGCGAA-3'
Skill: Problem-solving
44) What are the three regions of a prokaryotic gene? Answer: A promoter region, an RNA‐coding region, and a terminator region.
Skill: Factual recall
45) Describe the structure and organization of the rDNA repeat unit in eukaryotes. Answer: The rDNA repeat unit exists in tandem arrays of 100s to 1,000s of copies. Each unit consists of coding sequences for three rRNA genes (18S, 5.8S, and 28S) separated by two internal transcribed spacer (ITS) regions. The genes are flanked by external transcribed spacer (ETS) regions. These regions comprise the ribosomal DNA transcription unit. Upstream of the unit and separating from the previous repeat unit is a nontranscribed spacer (NTS) region that contains the promoter.
Skill: Factual recall
46) What are the functions of the 5ʹ cap and poly(A) tail of mature mRNAs in eukaryotes? Answer: Both protect the mRNA from degradation by exonucleases. The 5ʹ cap is also important for the binding of the ribosome at the beginning of translation. The poly(A) tail is required for efficient export of the mRNA from the nucleus to the cytoplasm.
Skill: Factual recall
47) Cite at least three ways in which transcription differs from DNA replication. Answer: (1) In replication, both strands of the double‐stranded DNA are copied, resulting in new double‐stranded DNA molecules. In transcription, only one strand of the double‐stranded DNA is transcribed into a single‐stranded RNA molecule. (2) In eukaryotes, while DNA replication typically occurs during only part of the cell cycle, transcription generally occurs throughout the cycle. (3) No primer is needed in transcription because the RNA polymerase can initiate new polynucleotide chains, unlike DNA polymerase, which needs a primer to begin replication.
Skill: Application of knowledge
48) In eukaryotes, what is the difference between promoter proximal elements in ʺhousekeeping genesʺ vs. cell‐specific genes, and how does this relate to gene expression? Answer: In genes that are expressed in all cell types for basic cellular functions (ʺhousekeeping genesʺ), promoter proximal elements are recognized by regulatory activator proteins that are ubiquitous to all cell types. In cell‐specific genes, the proximal elements are recognized by activators that are found only in those cells.
Skill: Conceptual understanding
49) What is the advantage of having a promoter sequence upstream of a coding sequence? Answer: The promoter ensures that the initiation of every RNA occurs at the same site. Otherwise, the ensuing polypeptide might be missing amino acids, or it might contain extra amino acids.
Skill: Analytical reasoning
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Chapter 5 Gene Expression: Transcription 43
50) How does the rate of initiation of transcription relate to the similarity of the promoter region to the consensus sequence? Answer: The ability of the sigma factor of the RNA polymerase holoenzyme to recognize and bind to a promoter sequence (i.e., the efficiency of RNA polymerase binding) is dependent on the similarity of the promoter to the consensus sequence (the sequence found most frequently at each position of the promoter). As a result, the rate at which transcription is initiated varies from gene to gene.
Skill: Analytical reasoning
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Chapter 6 Gene Expression: Translation
MATCHING QUESTIONS Please select the best match for each term. 1) Glutamic acid
Skill: Factual recall
A) Neutral, nonpolar B) Basic
2) Glycine
Skill: Factual recall
C) Acidic D) Neutral, polar E) Atypical structure
3) Arginine
Skill: Factual recall
4) Serine
Skill: Factual recall
5) Proline
Skill: Factual recall
Answers:
1) C
2) A
3) B
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) Which of the following is not a characteristic of proteins? A) High molecular weight B) Presence of nitrogen C) Presence of amino acids D) Presence of nucleic acids E) Structure built of one or more polypeptides Answer: D
Skill: Factual recall
7) Where does translation take place? A) In the cytoplasm B) In the nucleus C) In the nucleolus D) Anywhere in the cell E) In the centriole Answer: A
Skill: Factual recall
44
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Chapter 6 Gene Expression: Translation 45
8) Every amino acid contains A) an amino group and a hydroxyl group. B) an amino group and an acetyl group. C) a hydroxyl group and a carboxyl group. D) a carboxyl group and an amino group. E) a carboxyl group and a hydroxyl group. Answer: D
Skill: Factual recall
9) What is the function of a ribosome? A) To direct transcription B) To attach amino acids to the appropriate tRNAs C) To unwind the double‐stranded DNA helix to enable translation D) To hold mRNA and tRNAs in the correct positions to enable translation E) To make more mRNA Answer: D
Skill: Factual recall
10) What is the basic shape of a tRNA molecule? A) A three‐dimensional cloverleaf B) A straight strand C) A right‐handed helix D) A globular ball E) A β‐pleated sheet Answer: A
Skill: Factual recall
11) What is the role of tRNA in translation? A) It brings together two subunits of a ribosome. B) It couples an amino acid with aminoacyl‐tRNA synthetase. C) It helps fold up the finished polypeptide chain. D) It catalyzes peptidyl transferase activity. E) It binds to an mRNA codon and carries the corresponding amino acid. Answer: E
Skill: Factual recall
12) How can a carrot plant express a bacterial gene? A) Because the bacterial gene hijacks the carrotʹs cellular machinery B) Because the genetic code is the same in both organisms C) Because of the wobble phenomenon D) Because of the degeneracy of the genetic code E) A carrot cannot express a bacterial gene. Answer: B
Skill: Application of knowledge
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46 Test Bank for iGenetics
13) In the genetic code, both AAU and AAC code for asparagine. For this reason, the code is said to be A) degenerate. B) nonspecific. C) universal. D) ambiguous. E) wobbly. Answer: A
Skill: Factual recall
14) In a gene sequence, the DNA codon for tryptophan experiences a mutation at the first base position, changing it to T. What will the resulting amino acid be? A) Tryptophan (no change) B) Serine C) Arginine D) Threonine E) None (a stop codon will halt translation) Answer: C
Skill: Problem-solving
15) A mutation causes a G to be inserted after the first base of the codon for tryptophan. How will this affect the growing polypeptide chain? A) It will not be affected. B) Elongation will stop prematurely. C) There will be a single amino acid substitution. D) The reading frame will be shifted to the left, and the wrong amino acids will be added from this point on. E) None of these Answer: D
Skill: Problem-solving
16) A β‐pleated sheet is a type of A) primary structure found in a protein. B) secondary structure found in a protein. C) tertiary structure found in a protein. D) quaternary structure found in a protein. E) heme structure found in a protein. Answer: B
Skill: Factual recall
17) Which of the following events signals the termination of translation? A) The ribosome reaches the end of the mRNA. B) The ribosome reaches a start codon. C) The ribosome reaches a stop codon. D) The ribosome runs out of charged tRNAs. E) The polypeptide chain folds into a protein. Answer: C
Skill: Factual recall
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Chapter 6 Gene Expression: Translation 47
18) A peptide bond forms between the ________ of one amino acid and the ________ of another. A) amino group, carboxyl group B) amino group, R group C) carboxyl group, sulfide group D) amino group, phosphate group E) R group, R group Answer: A
Skill: Factual recall
19) The majority of naturally occurring amino acids are A) acidic and polar. B) basic and nonpolar. C) acidic and nonpolar. D) neutral and nonpolar. E) neutral and polar. Answer: D
Skill: Factual recall
20) Molecular chaperones A) help guide proteins from the nucleus to the cytoplasm. B) help guide proteins from the cytoplasm to the endoplasmic reticulum. C) help fold proteins properly. D) are proteins with structural function. E) are proteins with enzymatic function. Answer: C
Skill: Factual recall
21) The concept that a set of three nucleotides specifies a particular amino acid provides the basis for A) the one‐gene‐one‐enzyme hypothesis. B) the one‐gene‐one‐polypeptide hypothesis. C) the genetic code. D) biochemical reactions among nucleic acids. E) All of these Answer: C
Skill: Factual recall
22) In eukaryotes, the AUG initiation codon is located in the ________ sequence. A) Shine—Dalgarno B) Kozak C) Khorana D) Goldberg—Hogness E) Pribnow Answer: B
Skill: Factual recall
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48 Test Bank for iGenetics
23) How many naturally occurring amino acids are used by ribosomes to construct proteins? A) 46 B) 64 C) 3 D) 20 E) 10 Answer: D
Skill: Factual recall
24) Which of the following statements is not true of proteins destined for the ER? A) They have a signal sequence. B) The signal sequence for the ER consists of 15‐30 amino acids. C) The proteins are modified with carbohydrates in the ER. D) The signal sequence is subsequently removed from the mature protein. E) The SRP binds to the protein and prevents entry into the ER until translation is completed. Answer: E
Skill: Factual recall
25) Which of the following statements is correct? A) For every codon, there is one amino acid. B) For every amino acid, there is one codon. C) A codon may signify more than one amino acid. D) Three codons make up an amino acid. E) Three amino acids make up a codon. Answer: A
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) During translation, mRNA codons bind to complementary tRNA anticodons. Answer: TRUE
Skill: Factual recall
27) Molecules of mRNA are translated from 3ʹ to 5ʹ, which is the opposite direction from which they were made. Answer: FALSE Explanation: They are translated from 5ʹ to 3ʹ.
Skill: Factual recall
28) The Shine—Dalgarno sequence is a purine‐rich region just upstream of a start codon and is necessary for the initiation of protein synthesis in prokaryotes. Answer: TRUE
Skill: Factual recall
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Chapter 6 Gene Expression: Translation 49
29) With absolutely no exception, all organisms use the same genetic code for the production of proteins. Answer: FALSE Explanation: Some organisms have minor changes in the code, such as in the nuclear genome of Tetrahymena and the mammalian mitochondria.
Skill: Factual recall
30) There are 64 sense codons in the genetic code and 61 different types of tRNA molecules. Answer: FALSE Explanation: There are 64 codons but only 61 sense codons.
Skill: Factual recall
31) Protein synthesis begins when ribosomes bind to the UGA codon. Answer: FALSE Explanation: UGA is a stop codon. Protein synthesis begins when ribosomes bind to the AUG (start) codon.
Skill: Factual recall
32) The genetic code is comma‐free, nonoverlapping, and almost universal. Answer: TRUE
Skill: Factual recall
33) In the genetic code, when the first two nucleotides of a triplet are identical and the third letter is U or C, the codon always codes for the same amino acid. Answer: TRUE
Skill: Factual recall
34) A new initiation event cannot occur on an mRNA molecule until the first ribosome falls away. Answer: FALSE Explanation: Many ribosomes may simultaneously be translating each mRNA, forming a structure called a polysome.
Skill: Factual recall
35) In eukaryotes, the translation initiation complex forms in association with the 40S ribosomal subunit. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) A DNA codon reads GCC. Give the corresponding mRNA codon, tRNA anticodon, and amino acid. Answer: mRNA codon: CGG; tRNA anticodon: GCC; amino acid: alanine.
Skill: Problem-solving
37) If a tRNA anticodon reads GAI, to which mRNA codon(s) will it bind? What amino acid will the tRNA carry? Answer: It will bind to CUU, CUC, or CUA, which all code for leucine.
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50 Test Bank for iGenetics
38) Part of a DNA gene sequence reads CAT. If a mutation occurs that changes the T to A, will the final protein be affected? Answer: Both CAT and CAA encode the amino acid valine, so the final protein will not be affected.
Skill: Problem-solving
39) If a poly(C) is used as an mRNA and translated, a poly‐glycine polypeptide is formed. If on the other hand a poly(G) mRNA is used, a poly‐proline polypeptide is made. However, if both poly(C) and poly(G) are used in the same reaction, no polypeptide is made. Explain why. Answer: Since mRNA is single stranded, and guanines pair with cytosines, the poly(C) would pair with the poly(G) to give double stranded RNA that could not be used as a template for translation.
Skill: Conceptual understanding
40) Why is methionine the first amino acid to be added to every polypeptide chain? Answer: Because the start codon, AUG, specifies methionine. Since the start codon signals the beginning of translation, methionine is always the first amino acid added to a new polypeptide chain.
Skill: Conceptual understanding
41) Describe and differentiate among the primary, secondary, and tertiary structures of a protein. To which kinds of interactions can each of these stages be ascribed? Answer: Primary protein structure refers to the sequence of amino acids on the polypeptide chain, which is determined by the gene nucleotide sequence. Secondary structure is the regular folding and twisting of a single polypeptide chain into a variety of shapes. This can be ascribed to weak bonds (electrostatic or hydrogen bonds) between NH and CO groups that are near each other on the chain. Tertiary structure refers to the folding and twisting of the amino acid chain into a three‐dimensional conformation. This can be ascribed to interactions among R‐groups of individual amino acids.
Skill: Conceptual understanding
42) When does quaternary structure occur? Answer: Quaternary structure occurs only in proteins that consist of more than one polypeptide subunit. The folded subunits come together to form the quaternary structure of the protein.
Skill: Conceptual understanding
43) In the experiments performed on bacteriophage T4 by Francis Crick and others, in which it was shown that the genetic code was triplet, what were the researchers trying to identify? Answer: They were trying to identify revertants from mutant rII to wild‐type r+ phage that would have resulted from frameshift mutations due to the addition or deletion of a single base pair.
Skill: Factual recall
44) What are the two sulfur‐containing amino acids that contribute to a proteinʹs tertiary structure? Answer: Cysteine and tryptophan.
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
Chapter 6 Gene Expression: Translation 51
45) How did von Ehrenstein et al. demonstrate that the specificity of codon recognition lies in the tRNA molecule and not in the amino acid it carries? Answer: They chemically converted the cysteine on tRNA.Cys molecules to alanine in vitro. This Ala‐tRNA.Cys was used to synthesize hemoglobin, which normally contains a cysteine on each α and ß chain. The resulting hemoglobin molecules contained alanine at the position normally occupied by cysteine, which indicated that the tRNA had read the cysteine codon and inserted the alanine it carried.
Skill: Conceptual understanding
46) Of the codons AAU, AUG, UAA, and UAU, which is considered a nonsense codon? What does this mean? Answer: UAA is a nonsense, or stop, codon. It does not code for any amino acid and signals the end of translation.
Skill: Factual recall
47) What is the wobble hypothesis, and what implications does it have for base‐pairing rules and selective pressure on codons? Answer: The wobble hypothesis refers to the degeneracy at the third nucleotide position of a codon, in which the base at the 5ʹ of the tRNA anticodon can pair with more than one type of base at the 3ʹ end of the codon. For example, the same leucine tRNA molecule with anticodon GAG can read two different leucine codons ( CUC and CUU). This allows for neutral mutations to occur in genes at the third position, but not the first or second, of a codon. Therefore, there is said to be higher selective pressure on the first and second positions in a codon, causing mutations at those positions to be much more rare.
Skill: Analytical reasoning
48) How are proteins sorted into their appropriate cell compartments in eukaryotes? Answer: Proteins with specific ʺsignalʺ sequences (a hydrophobic amino terminal extension) are directed to the ER during translation, while proteins without signal sequences remain in the cytoplasm. Inside the space of the ER, the signal sequence is removed from the polypeptide. The protein is modified further by addition of specific carbohydrate groups that direct them to their final destination.
Skill: Factual recall
49) Would a two‐letter code with four different nucleotides be sufficient to encode the 20 amino acids found in cells? Why or why not? What does the three‐letter code imply about the nature of the genetic code? Answer: A two‐letter code would only encode 16 (4 2 ) amino acids, so it would not be sufficient. A three‐letter code generates 64 (4 3 ) possible codes, which is more than enough for the 20 amino acids found in cells. This implies that some amino acids may be specified by more than one codon.
Skill: Problem-solving
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52 Test Bank for iGenetics
50) Describe a cell‐free protein synthesizing system with which you could determine unambiguously which codons specify which amino acids. Answer: You could use a system with synthetic mRNAs that contain only one type of base and then observe which type of polypeptide is synthesized. The polypeptide should contain only one type of amino acid. For example, a poly(U) mRNA would contain only UUU triplets and would produce a polypeptide chain entirely of phenylalanine.
Skill: Analytical reasoning
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements
MATCHING QUESTIONS Please select the best match for each term. 1) Nonsense mutation
Skill: Factual recall
A) A mutation that causes an addition or deletion of one or two base pairs in a gene B) A mutation that changes a codon from one that represents an amino acid to one that signals a chain termination C) A mutation that causes a change in a single base pair D) A mutation in a gene that causes no detectable change in the protein product E) A mutation that changes a codon from one amino acid to another
2) Missense mutation
Skill: Factual recall
3) Frameshift mutation
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4) Point mutation
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5) Neutral mutation
Skill: Factual recall
Answers:
1) B
2) E
3) A
4) C
5) D
MULTIPLE-CHOICE QUESTIONS 6) Which of the following nucleotide changes leads to a transition mutation? A) Adenine to guanine B) Adenine to cytosine C) Guanine to cytosine D) Thymine to guanine E) Guanine to thymine Answer: A
Skill: Application of knowledge
7) Which of the following is not mutagenic? A) 5BU B) AZT C) Nitrous acid D) Hydroxylamine E) Acridine Answer: B
Skill: Factual recall
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54 Test Bank for iGenetics
8) Base analogs may cause mutations because A) they modify the chemical structure and properties of the normal base. B) they insert themselves between adjacent bases on the DNA strand and cause an extra base to be inserted during replication. C) they remove amino groups from bases, causing them to pair with the wrong base during replication. D) they may exist in alternate chemical states that pair with different DNA bases than the normal state during replication. E) Both A and C Answer: D
Skill: Conceptual understanding
9) The deamination of cytosine creates A) 5‐methyl cytosine. B) uracil. C) 2‐aminopurine. D) 5‐bromouracil. E) thymine. Answer: B
Skill: Factual recall
10) Thymine dimers are commonly caused by A) ultraviolet radiation. B) ionizing radiation such as X‐rays. C) tautomers. D) alkylating agents. E) intercalating agents. Answer: A
Skill: Factual recall
11) Xeroderma pigmentosum is a human genetic disease caused by A) elevated levels of cholesterol in the blood. B) failure to produce pigment that protects the skin cells from UV light exposure. C) defective DNA excision‐repair mechanisms. D) mutations that inactivate tumor suppressor genes. E) loss of genes controlling the SOS response. Answer: C
Skill: Factual recall
12) Mutation frequency is the A) number of mutations per gene per generation. B) number of mutations per nucleotide per generation. C) number of mutations per cell per generation. D) number of a specific mutation in a defined population. E) total number of mutations in a defined population. Answer: D
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 55
13) The codon 5'-AAA-3' codes for the amino acid lysine. Which of the following mutations in this codon is a neutral mutation? A) 5'-ATA-3' to isoleucine B) 5'-AGA-3' to arginine C) 5'-AAG-3' to lysine D) 5'-CAA-3' to glutamine E) 5'-AAC-3' to asparagine Answer: B
Skill: Application of knowledge
14) Spontaneous mutation rates are greatly reduced by A) exposure to ionizing radiation. B) reverse mutations. C) base‐modifying agents. D) DNA repair mechanisms. E) performing the Ames test. Answer: D
Skill: Factual recall
15) A mutation during DNA replication causes a G to be inserted after the first base of the codon for tryptophan. How will this affect the growing polypeptide chain? A) It will not be affected. B) Elongation will stop prematurely. C) There will be a single amino acid substitution. D) The reading frame will be shifted to the left, and the wrong amino acids will be added from this point on. E) An extra amino acid will be added, but the reading frame will not be affected. Answer: D
Skill: Reasoning and logic
16) In Drosophila, the wild‐type eye color is red. A mutation, vermilion, causes vermilion‐colored eyes, unless there is a mutation in another gene, suv, which, when homozygous or hemizygous, results in eyes that are the wild‐type red even in the presence of the vermilion mutation. This is an example of A) forward mutation. B) reverse mutation. C) intragenic suppression. D) intergenic suppression. E) back mutation. Answer: D
Skill: Application of knowledge
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56 Test Bank for iGenetics
17) A transposon is A) a DNA segment that can insert itself at one or more sites in a genome. B) a ʺjumping gene.ʺ C) a DNA segment that may cause mutations in genes or chromosomal rearrangements. D) a mobile genetic element that may or may not leave a copy of itself in its original site when it moves to a new site. E) All of these Answer: E
Skill: Factual recall
18) Nonsense suppressors are usually mutations in genes coding for A) proteins. B) enzymes. C) mRNA. D) tRNA. E) rRNA. Answer: D
Skill: Factual recall
19) Which of the following transposable elements are found in eukaryotes but not in prokaryotes? A) IS elements B) Families of autonomous and nonautonomous elements C) Retrotransposons D) Ty elements E) B, C, and D only Answer: E
Skill: Factual recall
20) In order for the dissociation element (Ds) mutations in corn to be stable, A) an Ac element must be present. B) an Ac element must not be present. C) the Ds must contain the gene for transposition. D) the DNA must not be replicated. E) None of these Answer: B
Skill: Factual recall
21) An IS (insertion sequence) element contains A) a transposase gene only. B) a transposase gene and additional genes. C) a transposase gene and inverted repeats at the ends. D) a transposase gene, additional genes, and inverted repeats at the ends. E) genes but not a transposase gene. Answer: C
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 57
22) What is conservative transposition? A) No net increase in the number of transposable elements in the genome. B) A net increase in the number of transposable elements in the genome. C) Movement of transposable elements with replication of the element. D) Movement of transposable elements without replication of the element. E) Transposition without disruption of normal gene product activity. Answer: D
Skill: Factual recall
23) Dissociation elements (Ds) in plants are examples of A) activator transposons that can direct their own transposition. B) nonautonomous elements that require activation by an autonomous element. C) mutator genes that increase the spontaneous mutation frequencies of other genes. D) repeated sequences that are targeted by a transposase. E) Both B and C Answer: E
Skill: Factual recall
24) Which procedure would you use to detect a nutritional mutation in microorganisms? A) Visible inspection B) Replica plating C) Controlled crosses D) Isolation at high temperature E) Plating on antibiotic‐containing medium Answer: B
Skill: Conceptual understanding
25) Genetic manipulation in Drosophila may be assisted by the use of A) P elements. B) Ty elements. C) factor. D) bacteriophage Mu. E) Tn10. Answer: A
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) LINEs and SINEs are repetitive sequences in humans that can also, as retrotransposons, insert into genes and cause disease. Answer: TRUE
Skill: Factual recall
27) Changes in heritable traits result from adaptation to environmental influences. Answer: FALSE Explanation: Changes in heritable traits result from random mutations.
Skill: Conceptual understanding
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58 Test Bank for iGenetics
28) Somatic mutations may be inherited by the next generation. Answer: FALSE Explanation: Only germ‐line mutations may be heritable; somatic mutations only affect the individual in which the mutation occurs.
Skill: Conceptual understanding
29) 5‐bromouracil (5BU) is a mutagen because it is an analog of the base thymine and may pair with guanine instead of adenine if it is incorporated into a DNA strand. Answer: TRUE
Skill: Factual recall
30) A silent mutation is a change in the DNA sequence that alters the amino acid sequence of the encoded protein but does not change its function. Answer: FALSE Explanation: A silent mutation is a change in the DNA sequence that does not change the amino acid sequence of the protein due to the redundancy of the genetic code.
Skill: Factual recall
31) A tautomer is an uncommon form of DNA base that naturally exists along with the common form. Answer: TRUE
Skill: Factual recall
32) The DNA polymerase proofreading mechanism maintains a low mutation rate in eukaryotic genes. Answer: FALSE Explanation: The DNA polymerase proofreading mechanism is found only in bacteria, where it detects the insertion of incorrect bases following DNA replication.
Skill: Conceptual understanding
33) A transposon may carry genes for proteins that enable their transposition as well as genes for other functions such as drug resistance. Answer: TRUE
Skill: Factual recall
34) Composite transposons contain a central region with genes and repeated sequences at their ends but do not terminate with IS elements. Answer: FALSE Explanation: Composite transposons contain a central region with genes flanked at both sides by IS elements as well as terminal inverted repeats.
Skill: Factual recall
35) A cointegrate is characteristic of replicative transposition. Answer: TRUE
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 59
SHORT ANSWER QUESTIONS AND PROBLEMS 36) An mRNA codon reads GUA. If a transition mutation occurs at the third base pair, will the final protein be changed? Answer: Yes. Valine will be inserted in place of an asparagine.
Skill: Application of knowledge
37) A point mutation changes a codon from UCG to UAG. What will happen to the resulting polypeptide? Answer: UAG is a stop codon, so the polypeptide will terminate prematurely.
Skill: Application of knowledge
38) What are two ways that a reverse mutation can occur? Answer: A mutation in the same gene can restore the original phenotype, or a suppressor mutation at a second site in the genome can suppress the forward mutation.
Skill: Factual recall
39) How does an intercalating agent such as ethidium bromide cause mutations? Answer: The substance fits between adjacent bases on one strand of a DNA sequence, causing the complementary strand to be miscopied during replication. This usually results in a frameshift mutation.
Skill: Factual recall
40) How can PCR be used to induce site‐specific mutations in DNA? Answer: Two nested pairs of primers are used, the internal pair having a targeted mismatch that is copied into the newly synthesized DNA fragments. The external primers are used to replicate the mutated fragments.
Skill: Application of knowledge
41) For a particular gene, if one gene in a million experiences a mutation each generation, what is its mutation frequency? Answer: 10-6
Skill: Application of knowledge
42) A tumor suppressor gene is cloned and mutagenized in vitro, then injected into mouse embryos to create knockout mice. How could you identify the heterozygous knockout mice and create mice susceptible to tumors? Answer: Reared heterozygous mice can be screened molecularly to identify the ones with the mutation. Interbreeding these mice will result in some homozygous knockout progeny, which in turn will show rapid development of tumors.
Skill: Analytical reasoning
43) Describe how the Ames test is used to determine whether a particular chemical is mutagenic. Answer: Strains of the bacteria Salmonella typhimurium that are auxotrophic for histidine are plated on a histidine‐lacking nutrient agar with rat liver cells. The unknown chemical is added to the agar plate, and the plate is observed for bacterial growth that would indicate reverse mutations caused by the chemical resulting in histidine prototrophy.
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44) What kind of mutation‐detection procedure can be used to detect the white‐eye mutation in Drosophila? Answer: Screening for this mutation can be done by inspection since it is a visible mutation.
Skill: Conceptual understanding
45) How does the nucleotide excision repair (NER) system in E. coli work, and what kinds of DNA damage does it repair? Answer: The NER system repairs thymine dimers and other serious distortions of the DNA helix. It is a system of four proteins (UvrA, UvrB, UvrC, and UvrD) that work in a complex to recognize a distortion in the DNA, make cuts on either side of the lesion, release the short single‐stranded cut segment, and allow DNA polymerase and ligase to fill in the gap with new bases.
Skill: Factual recall
46) Explain how IS elements produce target‐site duplications when they move. Answer: The enzyme transposase makes a staggered cut in the target sequence of the recipient DNA. The IS element is inserted, becoming joined to the single‐stranded ends. The gaps are then filled in by DNA replication, resulting in a copy of both the IS element and the target sequence.
Skill: Conceptual understanding
47) Barbara McClintock discovered that a controlling element was responsible for the mutant‐spotted phenotype (purple spots on white) in corn kernels. Explain how this works. Answer: A normal C gene produces purple pigment in the corn kernel cells. If this gene is disrupted by the transposon Ds (a ʺmobile controlling elementʺ), no pigment is produced, resulting in a white kernel. Reversions of this mutation can occur if Ds transposes out of the gene during development, resulting in spots of purple pigment on the kernel. The transposition of Ds is controlled by an autonomous ʺactivatorʺ element, Ac.
Skill: Conceptual understanding
48) What is a retrotransposon, and how does it differ from typical transposons? Answer: Most transposons move by a DNA‐to‐DNA mechanism, but retrotransposons (such as yeast Ty elements, Drosophila copia elements, and human LINEs and SINEs) transpose via an RNA intermediate using a transposon‐encoded reverse transcriptase.
Skill: Factual recall
49) How was the adaptive method for acquiring mutations disproved? Answer: Luria and Delbruck argued that if mutations occurred due to exposure to a particular environment, then the number of cells that had the mutation would be similar in different identical populations. However, if the mutations occurred randomly, then the number of cells with that particular mutation would differ or fluctuate in identical populations exposed to the same environmental factor. They proved the second argument to be true by infecting identical bacterial cultures with the bacteriophage T1 and counting the number of resistant colonies. The number of resistant colonies fluctuated widely in the different cultures.
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 61
50) Explain what a mutator gene is and give an example. Answer: A mutator gene is a gene in which a mutation results in a much higher‐than‐normal mutation frequency for all other genes. For example, mutD mutations in E. coli affect the function of DNA polymerase III. This causes defective 3ʹ‐to‐5ʹ proofreading repair activity and results in many incorrectly inserted nucleotides being left unrepaired.
Skill: Conceptual understanding
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes
MATCHING QUESTIONS Please select the best match for each term. 1) Restriction enzyme
Skill: Factual recall
A) Origin of replication B) Endonuclease isolated from bacteria C) Small circular fragment of DNA D) Eukaryotic cloning vector E) Molecule used in DNA sequencing
2) ARS
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3) Plasmid
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4) YAC
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5) Dideoxynucleotides
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Answers:
1) B
2) A
3) C
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) When DNA is cut with a restriction enzyme, the resulting fragments have A) 5ʹ carboxyls. B) 5ʹ hydroxyls. C) 3ʹ hydroxyls. D) 3ʹ phosphates. E) 3ʹ carboxyls. Answer: C
Skill: Factual recall
7) A plasmid used as a cloning vector in E. coli must have A) an ori sequence. B) a selectable marker. C) unique restriction sites. D) B and C E) All of these Answer: E
Skill: Factual recall
62
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 63
8) Bacteria protect their own DNA from restriction enzyme damage by adding ________ groups to certain nucleotides. A) methyl B) hydroxyl C) phosphate D) amino E) carboxyl Answer: A
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9) A polylinker region in a cloning plasmid is characterized by A) multiple methyl groups. B) multiple restriction cut sites. C) multiple reporter genes. D) multiple expression vectors. E) multiple cloning targets. Answer: B
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10) In a physical map, distances are measured in A) recombination frequencies. B) number of SNPs. C) base pairs. D) probabilities. E) microns. Answer: C
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11) In DNA electrophoresis, fragment separation is based on A) fragment size. B) fragment nucleotide content. C) fragment sequence. D) fragment charge. E) All of these Answer: A
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12) The chemical ʺlabelʺ that permits visualization of DNA fragments in an automated sequencing machine is A) 32P. B) 35S. C) fluorescent dyes. D) tritium. E) None of these Answer: C
Skill: Factual recall
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13) A DNA copy of an mRNA molecule is called A) dDNA. B) rDNA. C) mDNA. D) shDNA. E) cDNA. Answer: E
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14) The first complete nonviral genome sequenced was that of A) Escherichia coli. B) Methanococcus. C) Saccharomyces cerevisiae. D) the human mitochondrion. E) Mus musculus. Answer: D
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15) Many bacteria are naturally capable of taking up endogenous fragments of DNA in their environment, a phenomenon termed A) conversion. B) transformation. C) hybridization. D) transfection. E) conjugation. Answer: B
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16) An SNP always occurs due to A) a one base‐pair change. B) a three base‐pair change. C) a single restriction site change. D) a single amino acid change. E) None of these Answer: A
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17) The recognition sequences for many restriction enzymes are A) terminal repeats. B) tandem repeats. C) palindromic. D) inverted repeats. E) dual repeats. Answer: C
Skill: Factual recall
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 65
18) In addition to restriction enzymes, which of the following enzymes is required to insert a fragment of DNA into a cloning vector? A) DNA polymerase B) DNA ligase C) RNA polymerase D) Reverse transcriptase E) Topoisomerase Answer: B
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19) Which restriction enzyme leaves blunt ends? A) XbaI B) SmaI C) EcoR1 D) BamHI E) All of these Answer: B
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20) A restriction enzyme cuts DNA and leaves the following end: CTGCA
G
Which of the following could be the sequence of the corresponding end of the other fragment generated by the enzyme? A) G
CTGCA GTGCA G C
B) C
C) CCGAT D) GGCTA E) None of these Answer: A
Skill: Problem-solving
21) The ________ method was used by the Celera Genomics Corporation to sequence the human genome. A) FISH B) whole‐genome shotgun C) clone contig D) radiation hybrid E) pedigree analysis Answer: B
Skill: Factual recall
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22) Candidate open reading frames of a genome are identified by searching for A) a start codon ʺin frameʺ with a stop codon. B) a start codon. C) introns and exons. D) a promoter. E) All of these Answer: A
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23) Successful insertion of a DNA fragment into the polylinker region of pBluescript II is detected by A) production of the enzyme β‐galactosidase. B) absence of the enzyme β‐galactosidase. C) the increased size of the plasmid. D) autoradiography. E) None of these Answer: B
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24) In gel electrophoresis, which of the following DNA fragments would travel the farthest distance from the sample well? A) ATCCCGAT B) ATCCCG C) ATCC D) AT E) ATCCCGATTGCACGTT Answer: D
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25) Which types of cloning vector occur in nature? A) Plasmids and cosmids B) Plasmids and bacteriophages C) Plasmids and BACs D) YACs and BACs E) Bacteriophages and cosmids Answer: B
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TRUE-FALSE QUESTIONS 26) A DNA copy of a messenger RNA molecule is called complementary DNA or cDNA. Answer: TRUE
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27) The human genome consists mostly of noncoding DNA. Answer: TRUE
Skill: Factual recall
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 67
28) Cloning vectors are used to introduce novel DNA into cells. Answer: TRUE
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29) DNA molecules, regardless of size, have a net positive charge. Answer: FALSE Explanation: DNA is a negatively charged molecule.
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30) In partial restriction digestion, all available restriction sites will be cut. Answer: FALSE Explanation: Partial digestion leaves some sites untouched by chance.
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31) The quality of an assembled sequence does not depend on the sequence coverage. Answer: FALSE Explanation: The higher the coverage of the genome, the better the quality of the assembled sequence.
Skill: Conceptual understanding
32) Cloning vectors generally possess unique restriction sites and dominant selectable markers. Answer: TRUE
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33) A linear strand of DNA with two restriction enzyme cut sites will yield three fragments on digestion. Answer: TRUE
Skill: Problem-solving
34) A circular DNA molecule with two restriction enzyme cut sites will yield three fragments on digestion. Answer: FALSE Explanation: A circular molecule will yield two DNA fragments when cleaved in two places.
Skill: Problem-solving
35) Archaea are typically found in much the same habitats as bacteria. Answer: FALSE
Skill: Factual recall
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Briefly contrast the dideoxy DNA sequencing developed by Fred Sanger with pyrosequencing methods. Answer: Dideoxy DNA sequencing involves use of altered (dideoxy) nucleotides that cause termination of replication at each site where such a nucleotide is inserted. The pyrosequencing method does not require chain termination. All four deoxynucleotide precursors are present at all times in the dideoxy sequencing reaction, whereas, only one is present at a time in the pyrosequencing reaction. The DNA template is immobilized on a bead in pyrosequencing, and the method of detection is the production of light by utilizing the pyrophosphate produced after the incorporation of a nucleotide. In the dideoxy method, the DNA template is not immobilized and the sequence is read with the help of fluorescent dyes attached to the dideoxy nucleotide molecules.
Skill: Conceptual understanding
37) Describe the basic approach to shotgun genome sequencing. Answer: Shotgun sequencing involves cutting the genome into multiple, partially overlapping, fragments, each of which is cloned and sequenced. A computer algorithm is used to assemble the complete sequence by identifying contiguous fragments by their overlapping regions.
Skill: Factual recall
38) The restriction enzyme SmaI cuts DNA at the 6‐base recognition sequence 5'-CCCGGG-3'. Calculate the frequency of occurrence of this sequence. (Assume a 50% GC content.) Answer: Assuming a 1:1 AT:CG ratio in the genome and a random distribution of nucleotides, each nucleotide pair will occur with equal frequency ( ). The
4 1
6‐nucleotide sequence would then occur with a frequency of ( )6 , 0.00024, cutting once in every 4,096 base pairs.
Skill: Problem-solving
1 4
39) Describe a DNA microarray setup for assaying mutations associated with a hypothetical genetic disease. Answer: Normal and mutant gene sequences associated with the disease could be affixed to the array in known positions. DNA from a patient and a known nonmutant individual could be separately isolated and labeled (or mRNA could be used, making cDNA for labeling with reverse transcription). Both are applied to the microarray, and whichever labeled fragments that bind with the DNA fragments on the array will colorimetrically reveal which alleles the patient possesses.
Skill: Conceptual understanding
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 69
The following scenario applies to the questions below. A pBluescript II cloning vector is used to introduce a gene of interest into a bacterial strain. The vector is lacking ampR, however. 40) What problem does this result in? Answer: ampR is the gene conferring resistance to the antibiotic ampicillin. Ampicillin is used to select for bacterial colonies that have been successfully cloned. Without ampR, both successfully cloned and unsuccessfully cloned bacterial colonies will fail to grow if the selection medium contains ampicillin.
Skill: Conceptual understanding
41) Can you think of an alternative means of identifying successful bacterial transformants in this experiment? Answer: Successful insertion of the gene of interest into the polylinker region will disrupt the lacZ+ gene; such colonies cannot manufacture β‐galactosidase and will therefore remain white in the presence of X‐gal. Colonies can thus be screened using X‐gal. Of course, the bacteria would have to be grown on media without ampicillin, allowing all the colonies to grow.
Skill: Conceptual understanding
42) Exactly why, in dideoxy DNA sequencing, do dideoxynucleotides halt the replication process? Answer: Dideoxynucleotides are missing the 3ʹ hydroxyl group necessary for linkage with the 5ʹ phosphate group of new nucleotides by DNA polymerase.
Skill: Conceptual understanding
43) What is a hapmap? Answer: A hapmap is a complete description of all the haplotypes known in a population as well as the chromosomal location of each haplotype.
Skill: Factual recall
44) How can the recircularization of a restriction‐digested vector during a ligation reaction be minimized? Answer: The restriction‐digested vector can be treated with alkaline phosphatase to remove the 5ʹ phosphates without which the DNA ligase is unable to form a phosphodiester bond.
Skill: Factual recall
45) What advantage do cDNA libraries have over genomic libraries? Answer: Eukaryotic genomes tend to have more noncoding regions. Therefore, cDNA libraries offer a way to eliminate the nontranscribed regions and to look at only the transcribed regions of the genome. cDNA libraries can also be made from specific tissues.
Skill: Conceptual understanding
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46) Based on DNA sequences of single genes, genomic analysis has confirmed that prokaryotes are divided into two great lineages: the Bacteria and the Archaea. The latter may be more closely related to the Eukarya. Recount the lines of evidence that support these findings. Answer: Sequence divergence between archaeans and bacteria is considerable, and despite the general morphological similarity of these groups, the Archaea more closely resemble the Eukarya in terms of both the details of DNA replication and the expression (and, in some cases, the structure) of their genes.
Skill: Factual recall
47) Discuss the relative roles of descriptive and hypothesis‐driven science in genomics. Answer: Although sequencing genomes is a descriptive task, the data so obtained provide the raw material for framing an extensive array of hypotheses about gene and genome function, integration, and evolution. This is a good example of how descriptive knowledge is a necessary first step in understanding biological systems.
Skill: Conceptual understanding
48) Sequencing the dog genome is currently a high priority for genomics researchers. What insights do you think can be gained about the evolutionary process by analyzing the genome of dogs? Answer: Dogs are diversified into a multitude of breeds, many with striking morphological and behavioral differences. Yet, all dogs are the same species, and their DNA differs only minutely. Comparison of the genomes of different dog breeds may therefore shed light on which genetic elements are responsible for the radically different developmental pathways of different dog breeds–a process that may represent a microcosm of species‐level differentiation over evolutionary time.
Skill: Conceptual understanding
49) It is not always easy to locate or identify genes based on cDNA. How is this problem created by the differences between cDNA and DNA with respect to a given gene? Answer: cDNA is made from mRNA, which in eukaryotes is processed transcript with exons removed. The original gene may have many exons, and if so, the cDNA may little resemble it.
Skill: Conceptual understanding
50) Given the DNA oligonucleotide 5'—AGTCTAGGCT—3' , reconstruct a sequencing gel based on dideoxy sequencing, showing the relative position of each DNA fragment (band) on the gel. Answer: The gel drawn by the student should resemble that in Figure 8.11.
Skill: Conceptual understanding
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Chapter 9 Functional and Comparative Genomics
MATCHING QUESTIONS Please select the best match for each term. 1) Transgene
Skill: Factual recall
A) Transcript that has a stem and loop B) Small noncoding human RNA C) Searches for sequence matches D) Artificially introduced DNA E) Fluorescent dye
2) shRNA
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3) Cy5
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4) HAR‐1
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5) BLAST
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Answers:
1) D
2) A
3) E
4) B
5) C
MULTIPLE-CHOICE QUESTIONS 6) Which of the following statements correctly describes reverse genetics? A) A mutated phenotype in an organism leads to the associated gene. B) A mutated gene is introduced into an organism to determine the phenotype. C) A gene is studied to understand the protein it encodes. D) A transcript is used to make a protein encoded by a gene. E) A gene is identified by bioinformatics. Answer: B
Skill: Conceptual understanding
7) The subfield of genomics that deals with gene expression and interaction is A) comparative genomics. B) functional genomics. C) structural genomics. D) hypothetical genomics. E) expression genomics. Answer: B
Skill: Factual recall
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8) The polymerase chain reaction has revolutionized genetics because A) it is capable of making virtually unlimited copies of DNA for study. B) it is capable of making copies of DNA with very little starting material. C) it is capable of making large numbers of DNA copies very quickly. D) All of these E) A and B only Answer: D
Skill: Factual recall
9) RNA interference functions by A) degrading the protein encoded by a specific mRNA. B) degrading the RNA strand that is encoded by the Slicer gene. C) transcribing double‐stranded RNA into protein. D) disrupting the chromosomal copy of the gene. E) degrading the RNA complementary to the short, double‐stranded RNA. Answer: E
Skill: Factual recall
10) Which of the following triggers the RNAi pathway? A) mRNA B) rRNA C) tRNA D) shRNA E) snRNA Answer: D
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11) Which of the following is not used in PCR amplification of DNA? A) Double‐stranded DNA template B) Oligonucleotide primers C) DNA polymerase D) Buffer E) ddNTPs Answer: E
Skill: Application of knowledge
12) Which of the following does not occur during PCR? A) Denaturation at high temperatures B) Synthesis of oligonucleotide primers C) Annealing of oligonucleotide primers D) Extension of primers by DNA polymerase E) Changes in the reaction temperature by a thermocycler Answer: B
Skill: Factual recall
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Chapter 9 Functional and Comparative Genomics 73
13) If you are setting up a PCR reaction, how much extension time per cycle should you allocate for the Taq DNA polymerase to amplify a 5‐kb DNA fragment from a 10‐kb template? A) 1 minute B) 2.5 minutes C) 5 minutes D) 10 minutes E) 20 minutes Answer: C
Skill: Application of knowledge
14) The drug ________ is added to mouse cell cultures to select against transformants that do not have the desired gene knockout. A) ampicillin B) kanamycin C) neomycin D) G418 E) ganciclovir Answer: E
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15) The Alu family is an example of A) short interspersed repeat elements. B) intermediate repeat elements. C) long interspersed repeat elements. D) terminal sequences. E) gene‐flanking sequences. Answer: A
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16) A transgene is a A) hybrid gene. B) gene that contains no introns. C) bacterial gene that is found in mammalian cells. D) human gene of viral origin. E) gene introduced into an organism by artificial means. Answer: E
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17) The ________ gene is responsible for removing many of the drugs introduced into the human body. A) HAR‐1 B) CYP2D6 C) ASPM D) FOXP2 E) BRCA1 Answer: B
Skill: Factual recall
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18) Which of the following is an archaean that has been identified in the human gut? A) Escherichia coli B) Enterobacter aerogenes C) Bifidobacterium longum D) Methanobrevibacter smithii E) Caenorhabditis elegans Answer: D
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19) Pharmacogenomics is the study of how A) the genome affects responses to drugs. B) the genome leads to new drug discoveries. C) the genome affects microbial communities in the body. D) the genome is affected by drug use. E) medicines influence the genome of cancer cells. Answer: A
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20) The human‐accelerated region 1 (HAR‐1) gene A) was identified by functional genomics. B) is identical to the chimpanzee HAR‐1. C) encodes for the human accelerated region‐1 protein. D) is a small, noncoding RNA expressed in the brain. E) is a small, regulatory RNA that regulates gene expression. Answer: D
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21) Linkage disequilibrium refers to a condition in which A) a set of SNPs recombine often. B) a set of haplotypes recombine often. C) a set of specific alleles of two or more genes tend to appear together. D) a set of specific alleles of two or more genes only rarely appear together. E) a haplotype block rearranges every generation. Answer: C
Skill: Factual recall
22) A genomic region can rapidly become very common in a population due to A) positive selection. B) negative selection. C) specific selection. D) linkage equilibrium. E) homologous recombination. Answer: A
Skill: Factual recall
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Chapter 9 Functional and Comparative Genomics 75
23) The representational oligonucleotide microarray analysis (ROMA) represents A) a functional genomics approach. B) a structural genomics approach. C) a comparative genomics approach. D) a metagenomics approach. E) a pharmacogenomics approach. Answer: C
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24) Metagenomic analysis involves A) culturing microbes isolated from the environment. B) extracting DNA from cultured microbes. C) analyzing DNA from environmental samples. D) sequencing cloned DNA individually. E) determining culture conditions for different microbes. Answer: C
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25) A microbiome consists of A) all the microorganisms on Earth. B) the community of microorganisms in a particular environment. C) the transcriptome from a particular microorganism under different environments. D) the proteome from a particular microorganism under different environments. E) the proteome from different microorganisms in a particular environment. Answer: C
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TRUE-FALSE QUESTIONS 26) ʺKnockoutʺ mutants have had the activities of certain genes eliminated, which can aid the researchers in determining the effects of those genes on physiology, development, and other functions. Answer: TRUE
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27) Bioinformatics is a relatively new field that combines elements of genomics with computer science and mathematics. Answer: TRUE
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28) The transcriptome is the complete set of polypeptides produced by an organism. Answer: FALSE Explanation: The transcriptome is the complete set of mRNAs produced in a cell of an organism, while the proteome is the complete set of polypeptides.
Skill: Factual recall
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29) BLAST is a software that searches for similarities between DNA sequences only. Answer: FALSE Explanation: The BLAST software can search for similarities using either DNA or protein inputs.
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30) Both strands of the short, double‐stranded regulatory RNA molecules are required by the Slicer protein in order to target the complementary RNA for degradation. Answer: FALSE Explanation: The Slicer protein unwinds the short, double‐stranded regulatory RNA, discards one of the strands, and uses the other strand to recognize the complementary RNA to be degraded.
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31) Sequence similarity between genes plays an important role in assigning gene function. Answer: TRUE
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32) Cy3 and Cy5 are radioactive tags used in microarrays. Answer: FALSE Explanation: These are fluorescent tags, not radioactive.
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33) The 16S rRNA gene from human gut microflora was amplified and sequenced to understand how the gut microbes are related. Answer: TRUE
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34) The genome of cancer cells tends to become unstable. Answer: TRUE
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35) The FOXP2 protein appears to play a critical role in brain size regulation. Answer: FALSE Explanation: The FOXP2 protein plays a critical role in speech production.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) What are the different methods by which we can assign function to particular genes? Answer: Functions may be assigned to particular genes of an organism by searching for sequence homology with genes of other organisms whose functions are known. Functions may also be assigned by interfering with the gene (either by making knockouts or through gene silencing by RNA interference) and studying the phenotypes produced.
Skill: Analytical reasoning
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Chapter 9 Functional and Comparative Genomics 77
37) Imagine that you are using random PCR primers to amplify fragments of DNA. Would you expect to get more amplification products if you set the PCR machine at a higher or lower annealing temperature and why? Answer: Higher annealing temperatures increase ʺstringency,ʺ meaning the accuracy of base complementarity must be higher for successful annealing. Under such conditions, fewer successful amplifications are expected compared to conditions in which stringency has been lowered by using a lower annealing temperature.
Skill: Conceptual understanding
38) List the steps involved in constructing a knockout mutant in yeast. Answer: Most commonly, a ʺdeletion moduleʺ is constructed with a selectable marker (such as an antibiotic resistance gene). Transformants are selected using the antibiotic resistance gene. The module recombines with the target knockout gene, and successful replacements can be confirmed with PCR.
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39) Why are BLAST comparisons based on protein sequences often easier to interpret than those based on DNA sequences? Answer: DNA sequences often have greater mismatches because of the degeneracy of the genetic code. Protein matches are therefore easier to compare than DNA sequences.
Skill: Conceptual understanding
40) What is the difference between a ʺknockoutʺ and a ʺknockdownʺ of a gene? Answer: Both terms refer to the alteration in the transcription from a gene, the difference being the level of transcription. In a knockout, the gene is not transcribed at all, whereas in a knockdown there is some transcription, usually well below the normal level.
Skill: Conceptual understanding
41) Calculate the number of copies that would be produced from a single starting molecule after 10 rounds of PCR. Answer: Ten cycles would produce 210, or 1,024, copies of the target DNA.
Skill: Problem-solving
42) After how many PCR cycles, starting from a single starting molecule, will fragments consisting of only the target DNA (the DNA between the primers) be generated? Answer: Fragments consisting of only target DNA are not generated until the third PCR cycle. (See http://www.dnalc.org/shockwave/pcranwhole.html for animation.)
Skill: Problem-solving
43) Why are the number of proteins in the eukaryotic proteome so much greater than the number of eukaryotic genes? Answer: Eukaryotic translation is characterized by transcript processing, where, in many cases, exons are differentially spliced to yield different polypeptide products. In addition, many functional proteins have quaternary structure, the subunits of which are provided by different genes. The number of possible proteins thus greatly exceeds the base number of genes in the genome.
Skill: Conceptual understanding
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44) Why is the Taq DNA polymerase well‐suited for PCR? Answer: Taq DNA polymerase is a heat‐stable enzyme that was isolated from the bacterium Thermus aquaticus. It can maintain function and structure at very high temperatures, allowing the PCR to run without having to inject a new enzyme after each denaturation cycle.
Skill: Conceptual understanding
45) Why are two selectable markers present in the target vector used for making mouse gene knockouts? Answer: Two selectable markers are used for making mouse gene knockouts because there is a high level of random integration of the target vector into the mouse genome. One of the selectable markers selects for the gene knockout, while the other selects against transformants in which the target vector may have integrated into a random location by nonhomologous recombination.
Skill: Factual recall
46) Why does the transcriptome differ within an organismʹs cell types? Answer: All the genes of a multicellular organism are not expressed in every cell type. The various cell types express different subsets of the genome, and the makeup of the transcripts (the transcriptome) differs depending on which genes are expressed at a given time.
Skill: Factual recall
47) What is one advantage that an RNAi screen may have over a screen for gene knockouts? Answer: A screen for gene knockouts usually does not yield deletions of essential genes because they are nonviable and do not live. An RNAi screen, however, may often lead to a knockdown, rather than a knockout, of an essential gene, which may be viable. This may lead to better understanding of the function of the gene that may otherwise be difficult to determine.
Skill: Conceptual understanding
48) How can we explain the fact that the human genome contains far fewer genes than was predicted would be necessary for our survival? Answer: Although we currently do not know the complete answer to this question, one partial explanation is that the microorganisms that reside within our bodies help us by synthesizing chemicals, like vitamins, that we then use. This symbiotic relationship may be part of the reason why we have only 20,000 or so genes.
Skill: Factual recall
49) What does a large haplotype block suggest about a genomic region? Answer: A large haplotype block usually indicates a recent origin, because the older a block, the greater the chances that it would have undergone genetic recombination at some point in the past. It may also suggest that some mutation within the haplotype block imparted a fitness advantage and has therefore been selected for (positive selection).
Skill: Conceptual understanding
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Chapter 9 Functional and Comparative Genomics 79
50) How do we know that some of the microbes in our gut are beneficial to us? Answer: Individuals lacking normal intestinal microbes have defects in the function of the immune system and in wound healing, suggesting that these microbes are essential to our well‐being.
Skill: Factual recall
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Chapter 10 Recombinant DNA Technology
MATCHING QUESTIONS Please select the best match for each term. 1) EST
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A) Short oligonucleotide probe B) Repetitive sequence used in DNA fingerprinting C) A chloroplast enzyme D) Identified by mRNA E) Stems from single base substitution, for example
2) STR
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3) SNP
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4) EPSPS
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5) ASO
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Answers:
1) D
2) B
3) E
4) C
5) A
Please select the best match for each term. 6) PCR
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A) Technique used to generate copies of a DNA fragment B) Screening procedure used to assay DNA fragments C) Screening procedure used to assay mRNA fragments D) Movement of macromolecules in an electrical field E) DNA fragments generated by restriction enzymes 7) B 8) E 9) C 10) D
7) Southern blot
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8) RFLPs
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9) Northern blot
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10) Electrophoresis
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Answers:
6) A
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MULTIPLE-CHOICE QUESTIONS 11) ________ were the first class of DNA sequence polymorphisms discovered in the human genome. A) SNPs B) Microsatellites C) VNTRs D) STRs E) RFLPs Answer: C
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12) RFLPs arise due to A) changes in the number of tandem repeats. B) base changes that add or delete restriction enzyme recognition sites. C) addition or deletion of bases between restriction enzyme recognition sites. D) B and C only E) All of these Answer: D
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13) Which of the following sequences is considered an STR? A) CGTATTATGC B) TACGTACGTACG C) CAGCAGCAG D) TATTACGCCG E) A and B only Answer: C
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14) Expressed sequence tags, or ESTs, A) are genetic markers generated from cDNA. B) pertain to expressed genes only. C) are oligonucleotides derived from mRNA. D) A and C only E) All of these Answer: E
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15) Which of the following techniques is best suited to study protein interactions? A) PCR B) The yeast two‐hybrid system C) Restriction mapping D) RT‐PCR E) Southern blotting Answer: B
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16) In yeast, expression of the GAL genes is induced when there is ________ in the culture medium. A) glucose B) lactose C) galactose D) maltose E) Any of these Answer: C
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17) When yeast cells are cultured in the presence of both glucose and galactose, the GAL genes are A) transcribed and then translated. B) transcribed but not translated. C) neither transcribed nor translated. D) translated but not transcribed. E) transcribed and translated simultaneously. Answer: C
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18) For the yeast galactose metabolizing gene, GAL1, to be transcribed, the regulatory protein produced by the GAL4 gene binds to A) RNA polymerase. B) upstream activator sequence G (UAS G) region. C) a ribosome. D) the operator sequence. E) the repressor protein. Answer: B
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19) ASO probes are used for the molecular testing of A) chromosomal rearrangements. B) single nucleotide changes in a DNA region. C) RFLP fragments in a DNA region. D) defective proteins implicated in disease. E) successfully cloned Ti plasmids. Answer: B
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20) Vectors capable of entering two or more different host organisms are known as A) dual vectors. B) omnivectors. C) shuttle vectors. D) parity vectors. E) universal vectors. Answer: C
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21) ________ blots are used to study DNA fragments, while ________ blots are used to study RNA fragments. A) Western, northern B) Southern, western C) Eastern, northern D) Southern, northern E) Western, Southern Answer: D
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22) In a sequenced genome, candidate protein‐encoding genes are identified by searching for A) STRs. B) RFLPs. C) ESTs. D) SNPs. E) ORFs. Answer: E
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23) A(n) ________ is a photograph using X‐ray film that is used to identify the position of radiolabeled molecules on a gel. A) Southern blot B) northern blot C) X‐rayogram D) autoradiogram E) radiocopy Answer: D
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24) Which of the following is used to screen newborns for the genetic disease PKU? A) RFLPs B) ASOs C) Antibodies D) PCR E) All of these Answer: A
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25) Successful cotransformation in the yeast two‐hybrid system will result in the binding of proteins produced by each yeast expression vector. These fusion proteins function by A) binding to the lacZ operator, preventing transcription. B) binding to RNA polymerase, preventing transcription. C) binding to the lacZ repressor, enabling transcription. D) binding to the lacZ upstream activator sequence, enabling transcription. E) changing color in the presence of X‐gal. Answer: D
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TRUE-FALSE QUESTIONS 26) DNA markers are DNA polymorphisms that are useful for genetic mapping. Answer: TRUE
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27) RT‐PCR is a PCR technique used in amplifying DNA copies of mRNA. Answer: TRUE
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28) Somatic cell gene therapy can result in a cure for a genetic disease in an individual as well as any progeny produced by the individual. Answer: FALSE Explanation: Only changes in germline cells are heritable.
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29) Humulin is insulin created by recombinant DNA technology. Answer: TRUE
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30) DNase is used to treat hemophilia. Answer: FALSE Explanation: DNase is used to treat cystic fibrosis.
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31) The Guthrie test is used to screen all newborns in the United States for cystic fibrosis. Answer: FALSE Explanation: This test is used to screen for phenylketonuria, or PKU.
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32) The Ti plasmid is a natural plasmid in the soil bacterium Agrobacterium tumefaciens. Answer: TRUE
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33) Molecular DNA testing is used to test for aneuploids. Answer: FALSE Explanation: Molecular DNA testing examines the molecular structure of a particular disease‐associated gene to determine whether the mutations that would put an individual at high risk for the disease are present.
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34) DNA fingerprinting can be used to convict criminals, but has had little application in exonerating individuals already convicted. Answer: FALSE Explanation: An important application of DNA fingerprinting is the overturning of wrongful convictions (which this technology has revealed to be widespread) in cases where biological evidence is still available for testing.
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Chapter 10 Recombinant DNA Technology 85
35) Ti plasmids and the T‐DNA system can be used to transform any vascular plant. Answer: FALSE Explanation: Only dicots, not monocots, can be transformed with the Ti plasmid and the T‐DNA system.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) A potentially serious problem with Taq polymerase in PCR is its lack of proofreading ability. Briefly explain why this can be a problem and at which stages during the PCR process this is most and least problematic. Answer: Replication errors are introduced by Taq polymerase as a result of its lack of proofreading. The earlier they are introduced, the greater the number of descendant copies that will also have the error. Therefore, such errors are least problematic when they occur late in the PCR process.
Skill: Conceptual understanding
37) Briefly list the basic steps involved in constructing a DNA fingerprint for a paternity case. Answer: Cut the DNA of each sample (child, mother, and supposed father) with restriction enzymes; separate the fragments for each sample with electrophoresis; perform a Southern blot analysis to transfer the fragments to a filter; probe the fragments with radiolabeled DNA oligonucleotides; and develop an autoradiogram to read the fragment sizes of each individual.
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38) How does a transcribable vector differ from an expression vector? Answer: Transcribable vectors are designed for the transcription of the insert both in vivo and in vitro, whereas expression vectors are typically designed only for in vivo expression.
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39) Study of P elements in Drosophila has shown that gene regulation can be accomplished by differential mRNA transcript processing. Explain. Answer: The P element carries a single gene that encodes for the protein transposase. This protein is required for the transposition of P elements. The P element pre‐mRNA is processed (spliced) differently in normal flies than in flies undergoing hybrid dysgenesis. In the former, the mRNA transcript is larger because the third intron is not spliced. But this results in a smaller protein due to the presence of an in‐frame stop codon in the retained intron. In the germline of flies undergoing hybrid dysgenesis, all the introns are spliced out. This results in a smaller transcript that encodes for a functional transposase. This may result in transposition of the P element
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40) How does an RFLP help identify sickle‐cell anemia? Answer: The normal globin gene is associated with three DdeI sites. The mutation that causes sickle‐cell anemia changes one of those sites, leaving only two. This can be detected by PCR amplifying the globin gene and digesting with the DdeI restriction enzyme. A normal person yields two bands, while a person with sickle‐cell anemia yields only one larger band.
Skill: Conceptual understanding
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41) PCR has revolutionized the criminal justice system through DNA fingerprinting. For a given fragment of DNA, showing that a suspectʹs fingerprint differs from that found at a crime scene immediately excludes him or her from suspicion, yet finding a match does not mean that guilt can be automatically concluded. Why? Answer: Different fragment lengths clearly must have come from different sources, but having the same fragment length only means there is a match–it may be possible for multiple persons in the population to possess the same fragment length.
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42) Two STR loci were determined to have the following allele frequencies in the general population: STR A 1A: 0.1 2A: 0.55 3A: 0.25 4A: 0.1 STR B 1B: 0.4 2B: 0.4 3B: 0.2
Alleles:
A suspect in a criminal trial was genetically tested for these loci and was found to have the genotype (1A, 1A) for STR A and (2B, 2B) for STR B–homozygous for both loci. What is the probability that a randomly sampled person from this population would also have this genotype? Answer: The product rule states that the net probability is the product of the independent probabilities. Since these are homozygous genotypes, the probability of the STR A genotype is: 0.1 × 0.1 = 0.01, and the STR B genotype is 0.4 × 0.4 = 0.16. The net probability is then 0.01 × 0.16 = 0.0016. There is an approximately one in a thousand chance that a randomly selected person would share these STR genotypes.
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43) How do genetic tests differ from diagnostic tests for genetic disease? Answer: Genetic tests identify if an individual is carrying a mutation known to be associated with a genetic disease; depending on the disease, having the mutation does not mean the disease will necessarily develop. Diagnostic tests, in contrast, reveal if a disease is present and, if so, to what extent it has developed.
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44) Genetic testing is typically done with blood samples, yet red blood cells (erythrocytes) are anucleate (lack nuclei) and are thus devoid of DNA. How, then, are these tests done? Answer: Other blood cells, in particular white blood cells or phagocytes, are not anucleate, and DNA can be extracted from them.
Skill: Application of knowledge
45) A couple with STR genotypes (50, 75) and (50, 100) (where the allele numbers denote repeat length) have several children. What are the possible genotypes of their children for this STR locus? Answer: As for any single‐locus trait, simple Mendelian principles tell us what the genotypes may be: (50, 50) homozygote and (50, 100) and (75, 100) heterozygotes.
Skill: Problem-solving
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Chapter 10 Recombinant DNA Technology 87
46) Antisense mRNA can be used to regulate the expression of genes. Briefly describe how this technology was used to slow the ripening of Calgeneʹs ill‐fated Flavr‐Savr TM tomato. Answer: In engineered tomatoes, the gene for polygalacturonase, the enzyme responsible for fruit softening, was inserted backwards. This yielded an antisense mRNA copy that bonded with the sense mRNA, reducing the effective transcript level present for translation. Ripening was thus retarded relative to the normal tomato.
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47) Briefly explain how Ti plasmids induce crown gall (tumor) growth in plants parasitized by Agrobacterium tumefaciens, highlighting the specific properties that make this plasmid so useful in biotechnology. Answer: Ti plasmids are so named because they carry tumor‐inducing genes. The plasmid enters (transforms) a host cell, and a 30‐kb region called transforming‐ or T‐DNA integrates by recombination with the host DNA. The T‐DNA region carries genes for plant cell transformation, creating the crown gall where Agrobacterium flourishes under natural conditions. The T‐DNA region necessary for successful excision, transfer, and integration consists of only its 25‐bp terminal repeats, which means the rest of the DNA in this region can be replaced with DNA of interest and the whole plasmid engineered to vector this DNA.
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48) Rapid detection of SNPs, which are allelic variants, can be accomplished with DNA microarrays. If a microarray is set up for a set of genes, including all variants, outline a general procedure for quickly assaying the genotype of multiple individuals for that set of genes. Answer: DNA taken from the individuals would be cut (restricted) and labeled with fluorescent markers. Once introduced to an array, the probes hybridize where there is sequence complementarity. The fluorescent marker then reveals which allelic variants of the genes of interest are expressed by each individual.
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49) How has the Bt protein been used to increase insect‐resistance in plants? Answer: The Bt protein is a bacterial protein that has been introduced into certain crops. When insects eat these plants, the Bt protein kills or injures the insect. Purified Bt proteins and bacteria that naturally express the Bt protein have been used as insecticides in organic farming.
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50) DNA sequencing using the dideoxy sequencing method involves PCR reactions using Taq polymerase, which we learned in this chapter can introduce errors because it lacks a proofreading mechanism. What precaution(s) can be taken in genomic or other sequencing projects to minimize the possibility of producing DNA sequences with errors? Answer: The most common precaution taken is sequencing a fragment of DNA in both directions. Any replication errors will show up as mismatches, which could be resolved by resequencing that fragment of DNA.
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Chapter 11 Mendelian Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) Haploid
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A) Possessing the correct number of chromosomes B) Several complete sets of chromosomes C) One half of a complete set of chromosomes D) Possessing too few or too many copies of a single chromosome E) Two complete sets of chromosomes 2) E 3) A 4) B 5) D
2) Diploid
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3) Euploid
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4) Polyploid
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5) Aneuploid
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Answers:
1) C
Please select the best match for each term. 6) Monohybrid cross
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A) Allele that has to be homozygous to be expressed B) Genetic makeup of an organism C) Allele that is expressed in the heterozygote D) Always involves alleles of two genes E) Always involves alleles of a single gene F) One parent must be homozygous recessive G) Physical manifestation of a trait
7) Dihybrid cross
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8) Test cross
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9) Phenotype
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10) Genotype
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11) Dominant allele
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12) Recessive allele
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Answers:
6) E 11) C
7) D 12) A
8) F
9) G
10) B
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MULTIPLE-CHOICE QUESTIONS 13) In lilies, white flowers (W) are dominant to purple flowers (w). If two plants that are heterozygous for flower color are mated, the genotypic ratio of the offspring would be A) 3:1. B) 1:1. C) 1:2:1. D) 9:3:3:1. E) 1:1:1:1. Answer: C
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14) In a pea plant that is heterozygous for seed color, what proportion of gametes will carry the recessive allele? A) B) C) D) E)
1 4 1 2 3 4 1 8 1 8
Answer: B
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15) In his experiments, Mendel noted that when two traits are involved in a genetic cross, they are inherited independently of each other. The reason for this is that A) genes on the same chromosome separate during the formation of gametes. B) alleles on the same gene separate during the formation of gametes. C) genes on different chromosomes separate during the formation of gametes. D) alleles on the same chromosome separate during the formation of gametes. E) chromosomes often recombine. Answer: C
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16) Which of the following genotypes would not usually be represented in a gamete? A) AA B) AB C) Ab D) aB E) ab Answer: A
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17) The chi‐squared statistic can best be described as A) the standardized deviation of observed data from expected data. B) a statistically significant test. C) a statistical test used in genetics. D) a statistical test that compares observed and expected population means. E) None of these Answer: A
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18) In the F2 generation, how many genotypic classes could be generated from a dihybrid cross of two heterozygotes in which the genes involved show complete dominance? A) 3 B) 4 C) 8 D) 9 E) 12 Answer: D
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19) In peas, tall plants are dominant to short plants. A cross between a tall pea plant and a short pea plant results in half the progeny being tall, and the other half being short. Therefore, the tall parent plant is genotypically A) homozygous. B) heterozygous. C) hemizygous. D) monozygous. E) homogeneous. Answer: B
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20) If the results of a chi‐square test of a given set of data show a P value greater than 0.05, then the null hypothesis A) must be rejected. B) must be accepted. C) must be rephrased. D) cannot be rejected. E) cannot be accepted. Answer: D
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21) A P value in statistics is A) a measure of the accuracy of a data set. B) the probability of getting the observed data distribution by chance. C) arbitrarily set depending on a statistical test. D) a measure of the accuracy of a statistical test. E) None of these Answer: B
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Chapter 11 Mendelian Genetics 91
22) A man whose father expresses a recessive trait marries a woman whose brother expresses the same recessive trait. What is the likelihood that the newlyweds could have a child expressing the trait? A) B) C) D) E)
1 2 1 3 1 4 1 5 1 6
Answer: E
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23) In humans, brown eye color (B) is dominant to blue eyes (b). A brown‐eyed man marries a woman with brown eyes and they have three blue‐eyed daughters. What are the genotypes of the man and the woman? A) BB and Bb B) BB and BB C) Bb and bb D) Bb and Bb E) bb and bb Answer: D
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24) The probability that two parents with a family of four will have one girl and three boys is A) B) C) D) E)
1 2 1 4 1 8 1 16 1 32
Answer: B
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25) A couple with three girls is expecting a fourth child. The probability that this child is also a girl is A) B) C) D) E)
1 8 1 4 1 8 1 16 1 32
Answer: A
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26) Net or overall probabilities are obtained by multiplying separate independent probabilities. This is formally known as A) the sum rule. B) the chi‐square test. C) the product rule. D) the sign test. E) the probability rule. Answer: C
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27) Which of the following statements concerning the inheritance of a dominant trait is true? A) Every affected person must have at least one affected parent. B) The trait is observed to skip generations. C) An affected heterozygote will transmit the allele to all of his or her offspring. D) Nonaffected parents can have affected offspring. E) The trait is passed along paternal lines. Answer: A
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28) A dihybrid cross yields 320 F 2 offspring. How many are expected to resemble the homozygous recessive parental? A) 10 B) 16 C) 20 D) 60 E) 180 Answer: C
Skill: Problem-solving
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Chapter 11 Mendelian Genetics 93
29) In his monohybrid crosses for seed color in peas, Mendel reported 6,022 yellow seeds and 2,001 green seeds. How many of each color class were expected? A) 4,011 green and 4,011 yellow B) All should be green C) All should be yellow D) 6,017 yellow and 2,006 green E) 2,006 yellow and 6,017 green Answer: D
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30) In holly, serrated leaves are dominant to smooth‐edged leaves, and red berries are dominant to green berries. Two holly plants heterozygous for leaf edge shape and berry color are crossed together. Assuming that the two traits assort independently of one another, what proportion of the progeny will have smooth‐edged leaves and green berries? A) B) C) D) E)
1 4 3 4 9 16 3 16 1 16
Answer: E
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TRUE-FALSE QUESTIONS 31) A testcross with a heterozygous dominant individual will yield only heterozygous dominant offspring. Answer: FALSE Explanation: AA × aa → AA : aa in a 1:1 ratio.
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32) Two individuals can be phenotypically identical, yet have different genotypes for a given trait. Answer: TRUE
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33) The genotypic F 2 ratio expected in a dihybrid cross is 9:3:3:1. Answer: FALSE Explanation: That is the expected phenotypic ratio. The genotypic ratio should be 1:2:1:2:4:2:1:2:1.
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34) The phenotype determines the genotype. Answer: FALSE Explanation: Genotypes underlie phenotypes, not vice versa.
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35) True‐breeding individuals are produced by repeated backcrossing. Answer: TRUE
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36) Recessive alleles are usually loss‐of‐function mutations. Answer: TRUE
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37) The dominant allele of a gene is the most frequently found allele in a population. Answer: FALSE Explanation: The wild‐type allele is the most frequently found allele of a gene in a population.
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SHORT ANSWER QUESTIONS AND PROBLEMS 38) Mendel was the first scientist to describe the concept of diploidy–the idea that organisms like pea plants and humans possess two complete sets of genetic material. Summarize Mendelʹs observations and the reasoning that led him to this conclusion. Answer: The F1 and F2 phenotypic ratios that Mendel observed in different crosses led him to conjecture that each parent contributes one version of a ʺunit factorʺ for each trait during reproduction. Each individual thus possesses two such factors (diploidy), and it is combinations of unit factors that constitute the genotype.
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39) Mendel selected seven traits to analyze in his famous pea plant crosses, and all of these traits yielded expected 3:1 phenotypic F 2 ratios in monohybrid crosses. He was fortunate in his selection of these traits. How so? What problem might he have encountered that may have yielded confusing ratios? Answer: By chance, some of the traits he selected might not have been independently assorting, owing to linkage (presence on the same chromosome).
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40) Mendelʹs insights started with his approach of analyzing discrete traits, leading to the idea of ʺparticulateʺ rather than ʺblendingʺ inheritance. Yet this was an uncommon way to view the inheritance of traits because many found it counterintuitive or contrary to experience. How so? Answer: Experience with breeding pets or livestock, or even having children, suggests that parental traits seem to blend together in the offspring. This is because most traits that breeders considered are complex, polygenic traits. By focusing on simple discrete traits, Mendel was able to show that his ʺunit factorsʺ simply combined and recombined in pairs each generation and were not blended away.
Skill: Conceptual understanding
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Chapter 11 Mendelian Genetics 95
41) In snapdragons, the genes A, B, C, D, and E assort independently of one another. If an Aa Bb Cc dd EE plant is crossed with an Aa bb cc Dd Ee plant, what is the probability of obtaining an offspring that is phenotypically dominant for all five traits? Answer: The probability of obtaining an offspring that is phenotypically dominant for all five traits is 3 1 1 1 3 × × × × 1 . 4 2 2 2 32
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42) A monohybrid cross is made for flower color, where purple is dominant to white. Fifteen hundred F2 offspring are analyzed. How many white flowers are expected? Answer: One‐quarter of the offspring should be homozygous recessive, thus: (0.25) × 1,500 = 375 white flowers.
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43) Speculate on the molecular basis for dominance and recessiveness, using flower color as an example, where red is dominant over white. Answer: If the trait is controlled by a gene responsible for synthesizing red pigment, the recessive allele could be a dysfunctional mutant. Red pigment would be made as long as there was at least one functional copy of the gene present–this is the essence of dominance, where the presence of a single allele is sufficient to mask the recessive allele.
Skill: Conceptual understanding
44) You observe an individual of your favorite study organism expressing the dominant phenotype for a certain trait. How would you go about determining if the individual was homozygous dominant or heterozygous for that trait? Answer: Perform a test cross with a recessive homozygous individual: If Aa × aa offspring are Aa : aa in a 1:1 ratio, and if AA × aa → all offspring are Aa heterozygotes.
Skill: Problem-solving
45) A man with albinism and a woman with a normal phenotype have several children, one of whom displays albinism. a. What can you conclude about the genotype of the mother? b. What is the probability that the children who do not display albinism are heterozygous? Answer: (a) The mother must be heterozygous (a carrier) in order to have even one offspring with albinism. (b) With parents of genotype Aa × aa, any offspring that do not display albinism in their phenotype must be heterozygous, so the probability is 100%.
Skill: Problem-solving
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46) A man and a woman with normal phenotypes have several children, one of whom has albinism. a. What can you conclude about the genotype of the mother? b. What is the probability that the children who do not display albinism are heterozygous? Answer: (a) Both parents must be heterozygous (a carrier) in order to have an offspring with albinism. (b) With parents of genotype Aa × Aa, the genotypes possible for offspring not displaying the traits of albinism are AA, Aa, and aA. Two of these three are carriers, so there is a (66%) chance that a child who does not display albinism is a carrier.
3 2
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47) What characteristics make an organism a good candidate for Mendelian studies? Answer: Mendelian studies are based on the probability and statistical analyses of progeny phenotypes. Therefore, the candidate organism should be reach maturity and reproductive age fairly quickly, and should produce many offspring
Skill: Conceptual understanding
48) A monohybrid (one‐gene) cross yields 4 genotypic classes, and a dihybrid (two‐gene) cross yields 16. How many classes are expected from a tetrahybrid (four‐gene) cross? Answer: Following the simple relationship 4 1 = 4 and 4 2 = 16, 44 = 256 expected genotypic classes.
Skill: Conceptual understanding
49) Explain why it is expected that heterozygotes will be produced twice as frequently as either homozygote in a monohybrid F 1 cross. Answer: Combinatorials. Each individual offspring genotype has an equal likelihood of occurring (= ), but since there are two ways to make heterozygotes (Aa and aA),
4 1
together they occur with a + = frequency, twice that of either homozygote,
4 4 2
1
1
1
which occur at a frequency of each.
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1 4
50) State the key difference between Mendelʹs principle of segregation and independent assortment. Answer: Segregation refers to separation of alleles into gametes, while independent assortment refers to random combinations of such alleles from different genes occurring in gametes.
Skill: Conceptual understanding
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Chapter 12 Chromosomal Basis of Inheritance
MATCHING QUESTIONS Please select the best match for each term. 1) XXY
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A) Aneuploidy B) Klinefelter syndrome
2) XO
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C) Mammalian female D) Polyploidy E) Turner syndrome
3) XX
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4) 2N+1
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5) 3N
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Answers:
1) B
2) E
3) C
4) A
5) D
MULTIPLE-CHOICE QUESTIONS 6) A person with an XO genotype for the sex chromosomes would be phenotypically A) male. B) female. C) intersex (both male and female). D) male or female with equal probability. E) neither male nor female. Answer: B
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7) Cells obtained from a Triplo‐X woman would contain ________ Barr bodies. A) 0 B) 1 C) 2 D) 3 E) 4 Answer: C
Skill: Application of knowledge
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8) In human males, genes on the X chromosome are A) homozygous. B) heterozygous. C) homogametic. D) heterogametic. E) hemizygous. Answer: E
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9) In butterflies, males are the ________ sex. A) homozygous B) heterozygous C) homogametic D) heterogametic E) hemizygous Answer: C
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10) The key gene in mammalian sex determination is A) the testis‐determining factor gene. B) the Y‐factor gene. C) the holandric factor gene. D) the X‐inactivation center. E) the lyonization gene. Answer: A
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11) A litter produced by a calico cat is not expected to include A) solid‐colored female kittens. B) calico female kittens. C) solid‐colored male kittens. D) calico male kittens. E) Any of these Answer: D
Skill: Application of knowledge
12) A human male who undergoes a sex‐change operation will be chromosomally A) XX. B) XY. C) XXY. D) XYY. E) YYY. Answer: B
Skill: Factual recall
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Chapter 12 Chromosomal Basis of Inheritance 99
13) The chromosome theory of inheritance states that A) genes are located on chromosomes. B) chromosomes replicate before cell division. C) alleles of a gene segregate during cell division. D) chromosomes consist of DNA. E) different chromosomes assort independently of one another. Answer: A
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14) The spindle fibers attach to the centromere via a protein complex called A) tubulins. B) microtubules. C) chromatid. D) kinetochore. E) chiasma. Answer: D
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15) Plants that produce separate male and female flowers are said to be A) hermaphroditic. B) heterosexual. C) monoecious. D) dioecious. E) homosexual. Answer: D
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16) In an organism with 16 chromosomes per cell, how many DNA molecules are present in a cell undergoing anaphase of mitosis? A) 8 B) 16 C) 24 D) 32 E) 64 Answer: D
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17) Synaptonemal complex forms during A) leptonema. B) zygonema. C) pachynema. D) diplonema. E) diakinesis. Answer: B
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18) Phenotypically, a female (XX) mouse embryo that was transformed with a small piece of a Y chromosome bearing the SRY gene would develop as A) a male. B) a female. C) an intersex individual. D) a male initially, then revert to female after birth. E) a female initially, then revert to male after birth. Answer: A
Skill: Factual recall
19) Which of the following is an example of genetic mosaicism? A) Calico coat color in cats B) Albinism in mice C) Male pattern baldness in humans D) Red‐green color blindness in humans E) White eyes in Drosophila Answer: A
Skill: Factual recall
20) Individuals with Turner syndrome are A) euploid. B) aneuploid. C) polyploid. D) haploid. E) monoploid. Answer: B
Skill: Factual recall
21) Which of the following organisms has a ZZ‐ZW method of sex determination? A) House mice B) Monarch butterflies C) Fruit flies D) Humans E) Worms Answer: B
Skill: Factual recall
22) The largest human autosome is chromosome ________ and the smallest is chromosome ________. A) 1, 22 B) 1, 21 C) 1, Y D) X, 22 E) None of these Answer: B
Skill: Factual recall
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Chapter 12 Chromosomal Basis of Inheritance 101
23) Chromosomes with only one arm are known as A) metacentric. B) submetacentric. C) acrocentric. D) telocentric. E) acentric. Answer: D
Skill: Factual recall
24) Red‐green color blindness in humans is an X‐linked recessive trait. What will the progeny ratios be if a carrier female is mated with a normal male? A) All the sons will be color blind; all the daughters will be normal. B) All the sons and half the daughters will be color blind. C) Half the sons will be color blind; all the daughters will be carriers. D) Half the sons will be color blind; half the daughters will be carriers. E) All the sons and all the daughters will be normal. Answer: D
Skill: Problem-solving
25) Which of the following statements is not true? A) Many plants exhibit alternation between haploid and diploid stages. B) Chromosome number is generally constant within groups of organisms with common ancestry. C) The Y‐bearing sex is known as the heterogametic sex. D) Sex determination is chromosomal in some organisms and genic in others. E) The sex chromosome systems of birds and mammals are likely to have evolved independently. Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) A person with two X chromosomes is the homogametic sex. Answer: TRUE
Skill: Factual recall
27) A woman who is heterozygous for the hemophilia allele has a 100% chance of having an affected son. Answer: FALSE Explanation: The woman has one affected and one unaffected allele and will donate only one of these, with equal probability, to each of her sons.
Skill: Factual recall
28) Half a womanʹs somatic cells express her paternal X chromosome and half express her maternal X chromosome. Answer: FALSE Explanation: X‐inactivation is random, so the maternal and paternal copies of her X chromosomes will not be inactivated with equal frequency.
Skill: Factual recall
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29) Synapsis usually forms between nonsister chromatids. Answer: TRUE
Skill: Conceptual understanding
30) X‐linked or sex‐linked traits are expressed in males but not in females. Answer: FALSE Explanation: Such traits are expressed most commonly in males due to their hemizygosity, but can be expressed in females homozygous for the X‐linked alleles.
Skill: Factual recall
31) Barr bodies are inactivated X chromosomes. Answer: TRUE
Skill: Factual recall
32) In flowering plants, gametes are produced mitotically from spores. Answer: TRUE
Skill: Factual recall
33) The centrioles are highly conserved chromosome regions that interact with spindle fibers during cell division. Answer: FALSE Explanation: The chromosome region that interacts with spindle fibers is the centromere.
Skill: Factual recall
34) Some cells may exit the cell cycle and enter a nondividing G 0 state. Answer: TRUE
Skill: Factual recall
35) Nondisjunction is the failure of sex chromosomes to separate during gamete formation. Answer: FALSE Explanation: Nondisjunction is the failure of chromosomes to separate. This can involve either autosomes or sex chromosomes, and may occur either during mitosis or meiosis.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) Describe two dosage compensation mechanisms found in animals. Answer: X‐inactivation in females, in which one X chromosome is silenced, and X‐hyperactivation in males, in which genes on the X chromosome have increased expression.
Skill: Factual recall
37) X‐linked traits are more likely to be expressed in males. Why? Under what circumstances are they expressed in females? Answer: Males are hemizygous, with only one X chromosome; therefore, any genes on that chromosome are expressed, regardless of whether they are dominant or recessive. Females must be homozygous for a recessive X‐linked allele to be expressed.
Skill: Conceptual understanding
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Chapter 12 Chromosomal Basis of Inheritance 103
38) What is the connection between the work of Mary Lyon and Murray Barr? Answer: Mary Lyon put forth the theory of X‐inactivation, now called lyonization. Murray Barr was the first to describe Barr bodies, which we now know are inactivated X chromosomes.
Skill: Factual recall
39) Why will a human embryo develop as a male in the presence of a Y chromosome, regardless of the number of X chromosomes present? Answer: Sex is determined by genes located on the Y chromosome that lead to the development of masculine traits. Even one Y copy, regardless of the number of X chromosomes, is thus sufficient to initiate male development.
Skill: Conceptual understanding
40) Nondisjunction can lead to two forms of aneuploidy. What are they, and how do they occur? Answer: The two most common aneuploid conditions are N + 1 and N ‐ 1 (one extra and one missing chromosome, respectively). The N + 1 condition arises when two chromatids of a given chromosome fail to separate during meiosis, giving the resulting gamete an additional copy of that chromosome. Likewise, the N ‐ 1 condition is found in the gamete missing a chromatid that stayed paired with its sister.
Skill: Conceptual understanding
41) A woman who is color blind, a sex‐linked trait, has several children with a man who has normal vision. What are the possible phenotypes of their children with respect to this trait? Answer: All their daughters will have normal vision, while all their sons will be color blind.
Skill: Problem-solving
42) What are the differences between metaphase I and metaphase II of meiosis? Answer: Metaphase I of meiosis has the same number of chromosomes as the original cells, while metaphase II of meiosis has half the number of chromosomes. In metaphase I, the homologous chromosomes are arranged in pairs at the metaphase plate, whereas in metaphase II there are no homologs, and therefore, individual chromosomes are arranged separately at the metaphase plate.
Skill: Factual recall
43) The chromosome theory of inheritance sought to explain Mendelian patterns in terms of chromosome movement. Can you think of an experimental approach to test for the Mendelian principle of segregation using chromosomes? Answer: If replicating chromosomes could be chemically marked, permitting identification of each daughter chromatid, the movement of each could then be tracked. Radioisotopes are commonly used in such labeling experiments.
Skill: Analytical reasoning
44) How did chromosomes get their name? Answer: The term, literally meaning ʺcolored bodies,ʺ was coined in the late nineteenth century at a time when these nuclear structures could be only indistinctly viewed through a microscope with staining.
Skill: Conceptual understanding
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45) Describe a situation in which two aneuploid gametes would yield a euploid zygote. Answer: The aneuploidy of the gametes must be such that the same chromosomes are affected in each, one with an extra copy of the chromosome and the other only deficient in that same chromosome. Fusion of these two gametes would restore euploidy.
Skill: Analytical reasoning
46) You are a breeder of animals with an X‐linked coat‐color gene, one allele of which gives black fur and another of which gives white fur. If you cross a black‐colored female with a white‐colored male, what would their male and female offspring look like? Answer: Their male offspring would be all black, while their female offspring would be mottled, owing to X‐inactivation: males: XbY (black) XbXb × XwY → females: XbXw (some fur patches black, some white)
Skill: Problem-solving
47) Describe Calvin Bridgesʹs classic study of nondisjunction of sex chromosomes, which provided support for the chromosome theory of inheritance. Answer: Bridges performed crosses with fruit flies, tracking the sex‐linked eye‐color trait. Red‐eyed males were crossed with white‐eyed females, yielding mostly white‐eyed male offspring and red‐eyed female offspring. A handful of anomalous individuals proved informative: about 1/2,000 offspring were white‐eyed females or red‐eyed males. Karyotypic study of these anomalous individuals revealed they had suffered nondisjunction of the sex chromosomes.
Skill: Problem-solving
48) Closely related species, like chimpanzees and humans, often differ in chromosome number by only one or two chromosomes, suggesting that chromosomes can experience fission and fusion over evolutionary time. Can you think of a way to test this? Answer: Generate a karyotype for the species of interest and stain the chromosomes to obtain banding patterns. Compare the banding patterns of chromosomes that you speculate may be the product of fission or fusion–banding patterns are highly conserved–and you should be able to match up the chromosomes from one species with those of another.
Skill: Analytical reasoning
49) Drosophila fruit flies, like mammals, have an X‐Y sex‐determination system. The insect differs from the mammalian system in an important respect, however. What is it? Answer: While Drosophila has X‐Y sex determination, where XX individuals are female and XY individuals are male, studies have shown that sex determination is based on the ratio of X chromosomes to autosomes rather than on Y‐linked genes. Thus, XO Drosophila develop as males and XXY Drosophila develop as females–the same X : autosome ratio as XY and XX individuals.
Skill: Factual recall
50) What pattern of expression is expected for an X‐linked dominant trait like hereditary enamel hypoplasia? Answer: An affected father will have affected daughters and no affected sons; an affected mother will have a 50% chance of having an affected son and a 50% chance of having an affected daughter (assuming the mother is heterozygous).
Skill: Problem-solving Copyright © 2010 Pearson Education, Inc.
Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles
MATCHING QUESTIONS Please select the best match for each term. 1) Codominance
Skill: Factual recall
2) Epistasis
Skill: Factual recall
A) When heterozygotes display a phenotype intermediate between those shown by homozygotes for that allele B) When a genotype gives rise to a range of phenotypes C) When each allele causes a distinct phenotype, and heterozygotes exhibit both phenotypes of heterozygotes D) When the genotype at one locus generates phenotypic expression of a trait that overrides and conceals the phenotypic traits arising from one or more other genes E) When a genotype underlies a given phenotype, but not all individuals with that genotype display the phenotype
3) Incomplete dominance
Skill: Factual recall
4) Incomplete penetrance
Skill: Factual recall
5) Variable expressivity
Skill: Factual recall
Answers:
1) C
2) D
3) A
4) E
5) B
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Please select the best match for each term. 6) Heteroplasmon
Skill: Factual recall
A) When a cell or individual has more than one mitochondrial genome type B) A phenotype caused by a maternal nuclear genotype that affects eggs prior to fertilization, and is therefore expressed in all offspring, regardless of paternal genes C) When either the frequency or amount of a phenotype depends on the sex of the individual D) When a genotype causes a phenotype in only one sex E) When a genotype and phenotype are only inherited from the mother
7) Maternal effect
Skill: Factual recall
8) Maternal inheritance
Skill: Factual recall
9) Sex‐limited
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10) Sex‐influenced
Skill: Factual recall
Answers:
6) A
7) B
8) E
9) D
10) C
MULTIPLE-CHOICE QUESTIONS 11) The distinctive fur pattern of Siamese cats results when genes that are responsible for the coat color are influenced by A) age. B) X‐inactivation. C) gender. D) temperature. E) other genes. Answer: D
Skill: Factual recall
12) A person with type B blood could safely receive a transfusion of blood from someone with blood type A) B. B) AB. C) O. D) Both A and C E) Any of these Answer: D
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 107
13) Which of the following will result in modifications to the expected Mendelian ratios? A) Epistasis B) Incomplete dominance C) Incomplete penetrance D) Gene interaction E) All of these Answer: E
Skill: Factual recall
14) Two true‐breeding mutant strains of Drosophila have black body color instead of the wild‐type gray yellow. When the two strains are crossed, all the F 1 flies have wild‐type body color. How can these data be interpreted? A) Recombination occurred. B) Complementation occurred. C) The mutations involved are on two different genes. D) A new mutation occurred in all the offspring. E) Both B and C Answer: E
Skill: Analytical reasoning
15) Manx cats have no tails. When two Manx cats are bred together there is always a one third chance that a kitten will have a tail. When a Manx cat is bred to a cat with a normal tail there is a one‐half chance that a kitten will have a tail. Which of the following is the best explanation for this? A) The Manx genotype exhibits variable expression. B) The Manx phenotype is dominant, but the allele is a recessive lethal. C) The Manx phenotype is dominant epistatic. D) The Manx phenotype is caused by gene interactions. E) The Manx phenotype is a result of heteroplasmy. Answer: B
Skill: Analytical reasoning
16) A and B antigens in human blood are produced by the conversion of ________ by the addition of a sugar group. A) anti‐A and anti‐B antibodies B) the H antigen C) hemoglobin D) α‐N‐acetylgalactosamine E) anti‐O antibodies Answer: B
Skill: Factual recall
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17) In the multiple allelic system for eye color in Drosophila, the phenotypic expression of the alleles depends on A) how much pigment is produced by the alleles present. B) the interaction of alleles from two different genes. C) the temperature at which the flies are grown. D) the sex of the fly. E) Both A and B Answer: A
Skill: Factual recall
18) Familial hypercholesterolemia is a genetic disease that affects approximately one in 500 people worldwide. There is more than one genetic cause of the disease. Class I is often described as an autosomal dominant disease. Affected individuals have cholesterol levels >250 as children and often >300 as adults. However, homozygotes have cholesterol levels of >600 as children and can die of heart attacks in their 20s. These individuals entirely lack a functional LDL receptor. Which of the following would be the best description of the inheritance of this form of hypercholesterolemia? A) Dominant epistatic B) Codominant C) Incompletely dominant D) Complementary E) Recessive epistatic Answer: C
Skill: Analytical reasoning
19) People affected by diseases caused by mtDNA defects A) typically have cells that are heteroplasmons. B) show partial penetrance. C) are sterile. D) inherited them through maternal effect. E) have epistatic respiration. Answer: A
Skill: Factual recall
20) Two persons who have recessive genetic deafness marry and have 6 children. All the children can hear. The reason they can hear is most likely due to A) epistasis. B) pleiotropy. C) sex linkage. D) complementation. E) partial penetrance. Answer: D
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 109
21) Comb shape in chickens is controlled by A) the interaction of three alleles and codominance, resulting in four different phenotypes. B) one gene with incomplete dominance, resulting in three different phenotypes. C) the epistatic interaction of two genes, resulting in two different phenotypes. D) the interaction of two genes with complete dominance, resulting in four different phenotypes. E) None of these Answer: D
Skill: Factual recall
22) Individuals with neurofibromatosis may have a range of phenotypes, from pigment spots on the skin to tumorlike growths. This is an example of A) incomplete penetrance. B) variable expressivity. C) age of onset. D) X‐inactivation. E) gene magnification. Answer: B
Skill: Factual recall
23) Solid black, black with tan belly, and agouti coat color in mice are all caused by alleles of the agouti gene. All three colors can be true‐breeding. When using mice from true‐breeding strains: (1) Black × Agouti F1 s are all agouti, (2) Black × Black with Tan belly F 1 s are all black and tan, and (3) Black with Tan belly × Agouti F1 s are all agouti. If you were to cross the F 1 s from (1) with the F 1 s from (2), what proportion of the resulting offspring would you expect to be black with a tan belly? A) None B) C) D) E)
1 4 3 16 1 2 1 16
Answer: B
Skill: Analytical reasoning
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The following scenario applies to the questons below. Dextral (D) and sinistral (d) shell coiling in the water snail Limnaea peregra is genetically determined. From a population of mixed sinistral and dextral snails, a sinistral female and a dextral male are chosen. All the offspring are dextral. When the female offspring are mated, 50% produce only sinistral snails and the other 50% produce only dextral snails. 24) This form of inheritance is called A) maternal inheritance. B) sex‐limited inheritance. C) sex‐linked inheritance. D) maternal effect. E) partial penetrance. Answer: D
Skill: Factual recall
25) What were the genotypes of the sinistral female and dextral male that were initially chosen? A) dd female and Dd male B) Dd female and dd male C) dd female and DD male D) DD female and dd male E) Dd female and Dd male Answer: B
Skill: Problem-solving
TRUE-FALSE QUESTIONS 26) For a gene with complete dominance, the recessive allele has no effect on the phenotype of a heterozygote. Answer: TRUE
Skill: Factual recall
27) Studies performed on identical twins separated at birth have shown that many phenotypes, such as IQ or alcoholism, are influenced by the personʹs genotype as well as by their environment. Answer: TRUE
Skill: Factual recall
28) Mutants that make up a complementation group complement each other. Answer: FALSE Explanation: A complementation group is made up of individual mutants that fail to complement each other and are therefore likely to represent mutations in the same gene.
Skill: Factual recall
29) In the human ABO blood system, the allele i is dominant to I A and IB. Answer: FALSE Explanation: Both I A and IB are dominant to i.
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 111
30) Recessive lethals are often mutations in essential genes. Answer: TRUE
Skill: Factual recall
31) In a dihybrid cross with independent assortment of the two genes, any deviation from the Mendelian 9:3:3:1 ratio indicates that the phenotype is the product of the interaction of two or more genes. Answer: TRUE
Skill: Factual recall
32) Sex‐limited traits are caused by genes that are on sex chromosomes. Answer: FALSE Explanation: They are caused by autosomal genes that affect a character that appears in one sex but not the other (such as facial hair in humans).
Skill: Factual recall
33) Phenylketonuria (PKU) is an example of a phenotype that is solely influenced by genes. Answer: FALSE Explanation: PKU expression is influenced by the diet of an individual who is homozygous for the recessive allele.
Skill: Factual recall
34) In a genotype with complete penetrance, less than 100% of individuals with a particular genotype exhibit the expected phenotype. Answer: FALSE Explanation: With complete penetrance, identical known genotypes yield 100% expected phenotypes.
Skill: Factual recall
35) An allele that exhibits incomplete dominance is usually haplosufficient. Answer: FALSE Explanation: Incompletely dominant genes are haploinsufficient.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) For a gene with multiple alleles, what is the maximum number of alleles that a diploid organism may have? Answer: Two.
Skill: Conceptual understanding
37) In the ABO blood‐type system, which genotypes could confer type A blood to a person? Answer: A person with type A blood could have either an I A/I A genotype or an I A/i genotype.
Skill: Analytical reasoning
38) Human height is a trait affected by both genes and the environment. How is this so? Answer: Human height is directly influenced by a personʹs genes. However, a single genotype for height has a range of potential phenotypes that could develop depending on environmental conditions. This is known as the norm of reaction.
Skill: Conceptual understanding
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39) How could a man with blood type A and a woman with blood type B produce a child with blood type O? Answer: They must both be heterozygous, with i as one of the two alleles (IA/i and IB/i). The child would have to receive an i allele from each parent and be genotype i/i.
Skill: Analytical reasoning
40) In chickens, there is a mutant gene called ʺfrizzleʺ that results in weak, stringy, and easily broken feathers. When a frightfully frizzled fowl is bred to a normal chicken, the offspring are all mildly frizzled. If one breeds two mildly frizzled chickens to each other, the offspring have the phenotypic ratio of 1 normal: 2 mildly frizzled: 1 frightfully frizzled. What is the mode of inheritance of ʺfrizzleʺ? Answer: There are three phenotypes observed in the F 1 offspring in the ratio of 1:2:1, and one phenotype is intermediate between the other two. This is the signature of incomplete dominant inheritance. The normal chicken represents the homozygous wild‐type genotype, the mildly frizzled chickens are heterozygotes, and the frightfully frizzled chicken has homozygous mutant phenotypes.
Skill: Analytical reasoning
41) What is the number of possible genotypes in a multiple allele series with 6 alleles? How many of these are homozygous and how many are heterozygous? Answer: The total number of genotypes for n alleles is n(n + 1)/2, so for six alleles, there are 21 total genotypes. Six of these are homozygous (n) and 15 (n(n ‐ 1)/2) are heterozygous.
Skill: Problem-solving
42) Explain and differentiate among the molecular bases of codominance, incomplete dominance, and complete dominance. Answer: In codominance, gene product from both alleles of a heterozygote is produced, so both allelic phenotypes are expressed. In incomplete dominance, a homozygous dominant produces two doses of gene product, a heterozygote produces one dose, and a homozygous recessive produces none. In complete dominance, one or both of the dominant alleles is capable of producing enough gene product for the normal dominant phenotype to be expressed.
Skill: Conceptual understanding
The following scenario applies to the questions below. Feather color in chickens is the result of an incompletely dominant gene. The gene responsible for black feathers (B) when homozygous results in a muted color called ʺAndalusian blueʺ if combined with an allele for white feathers (b). When the white allele occurs in homozygous conditions, a white phenotype results. A gene affecting comb shape (R), which results in a comb called a ʺroseʺ comb, is completely dominant over (r), which in homozygous conditions results in a comb called ʺsingle.ʺ A mating between a black single rooster and white rose hen results in a number of chicks, all of which have identical phenotypes. 43) What is the phenotype and genotype of the chicks? Answer: This must represent a cross (BBrr × bbRR), which results in all the F 1 genotypes being heterozygous (BbRr). The phenotype is blue rose.
Skill: Analytical reasoning, problem-solving
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 113
44) Use a Punnett square to demonstrate a cross between two of the F 1 offspring. What are the phenotypes and ratios that occur? Answer: 6 blue rose (B/b R/‐): 3 black rose (B/B R/‐): 3 white rose (b/b R/‐): 2 blue single (B/b r/r): 1 black single (B/B r/r): 1 white single (b/b r/r)
Skill: Problem-solving
45) In cats, the dominant white gene (W) may or may not also cause deafness. Homozygotes are more likely to be deaf than heterozygotes. If you breed heterozygous white cats together and produce 100 kittens, how many would you expect to be deaf if all the proportions came out perfectly? Assume that heterozygotes have an 8% chance of being deaf, and homozygotes a 60% chance of being deaf. Answer: Homozygotes should be of 100, which is 25. 60% of 25 = 15. Heterozygotes should be of 100, which is 50. 8% of 50 = 4. Expected deaf kittens would therefore be 15 + 4, or 19 kittens.
Skill: Factual recall
1 2 1 4
46) Generally, blue eyes are recessive to dark brown eyes in humans. Individuals with albinism typically have blue eyes with a strong red reflection. A blue‐eyed woman with albinism marries a man who is blue‐eyed but does not have albinism. All their children have dark brown eyes. How can this be explained? Answer: Albino is recessive epistatic. The woman with albinism was probably also homozygous for dark brown eyes.
Skill: Conceptual understanding
47) Why is it that human males and females who are both heterozygous for the pattern‐baldness gene have different phenotypes (males are bald, females are not bald)? Answer: Pattern baldness is an example of a sex‐influenced trait. Males and females do not express the gene in the same way. Pattern baldness is controlled by an autosomal gene that acts as a dominant in males and as a recessive in females. The b/b genotype results in baldness in both sexes, the b+ /b+ phenotype results in nonbaldness for both sexes, and the b+ /b genotype results in baldness in males but nonbaldness in females.
Skill: Conceptual understanding
48) What is the cause of Tay—Sachs disease? Answer: Tay—Sachs is a lethal disease resulting from a mutation in an essential gene. The mutation results in an enzyme deficiency (hexosaminidase A) that prevents proper nerve function.
Skill: Factual recall
49) The 9:6:1 F 2 ratio found in fruit shape in summer squash is a deviation from an expected Mendelian ratio. What does this represent? Answer: Two genes are involved in the production of squash shape. A dominant allele of either gene and homozygous recessivity for the other results in a spherical fruit. Homozygous recessivity in both genes results in long‐shaped fruit. Two dominant alleles interact to give a disk‐shaped fruit.
Skill: Conceptual understanding
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50) Comb shape in chickens also involves the interaction of two genes, but the F 2 generation shows a typical 9:3:3:1 Mendelian ratio. How does this work? Answer: A dominant allele present in one gene in combination with the homozygous recessive state in the other gene produces one of two phenotypes (rose or pea). Both dominant alleles present produce a third phenotype (walnut), and both homozygous recessives produce a fourth phenotype (single). The ratio is 9 walnut : 3 rose : 3 pea : 1 single.
Skill: Analytical reasoning
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Chapter 14 Genetic Mapping in Eukaryotes
MATCHING QUESTIONS The matching questions below refer to the following experiment. A true‐breeding strain (P1) of fruit flies has mahogany colored eyes and normal (wild-type) bristles. A second true-breeding strain (P2) of fruit flies has normal (wild-type) red eyes and stubble bristles. When they are crossed, the F1 s are wild-type (red eyes and normal bristles). Those F 1 s are then crossed to flies that have both mahogany eyes and stubble bristles. The progeny of the F 1 s consist of 64 flies that have mahogany eyes and normal bristles, 76 flies that have red eyes and stubble bristles, 28 flies that have mahogany eyes and stubble bristles and 32 flies that are wild -type. Please select the best match for each term. 1) Parental type
Skill: Factual recall
A) F1 to mahogany eyes and stubble bristles B) Genes that do not segregate independently C) Mahogany eyes and normal bristles D) Wild-type red eyes and normal bristles. E) Recombinants divided by the total × 100 2) D 3) A 4) B 5) E
2) Recombinant type
Skill: Factual recall
3) Testcross
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4) Gene linkage
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5) Map unit
Skill: Factual recall
Answers:
1) C
MULTIPLE-CHOICE QUESTIONS 6) If genes are linked and an F1 is testcrossed, A) the genes cannot separate. B) they can produce only one phenotype. C) they only produce homozygotes. D) they produce more recombinant phenotypes than parental phenotypes. E) they produce more parental phenotypes than recombinant phenotypes. Answer: E
Skill: Factual recall
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7) Crossing‐over is A) the place on a homologous pair of chromosomes at which a physical exchange occurs. B) an extremely rare event. C) an event that only takes place during meiosis. D) the reciprocal exchange of homologous regions of chromatids. E) not useful in mapping genes. Answer: D
Skill: Factual recall
8) Genes that are linked A) segregate to opposite poles during meiosis. B) do not assort independently during meiosis. C) segregate independently during meiosis. D) are on nonhomologous chromosomes. E) are always on the X chromosome. Answer: B
Skill: Factual recall
9) A chi‐square analysis of linkage between genes k and f produces a P value of 0.99. What is the likelihood that these two genes are linked? A) 99% B) 10% C) 9.9% D) 1% E) None of these Answer: D
Skill: Analytical reasoning
10) If an AABB strain is mated with an aabb strain, an AB gamete produced by the resulting F 1 is A) impossible. B) parental. C) recombinant. D) the most common. E) None of these Answer: B
Skill: Factual recall
11) During a dihybrid cross involving two linked genes, 18% of the resulting gametes showed a recombinant genotype. These two linked genes are ________ map units apart. A) 0.18 B) 18 C) 32 D) 180 E) 1.8 Answer: B
Skill: Conceptual understanding
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Chapter 14 Genetic Mapping in Eukaryotes 117
12) If two genes are not linked, then the expected phenotypic ratio resulting from a testcross is A) 9:3:3:1. B) 1:2:1. C) 3:1. D) 1:1:1:1. E) 1:1. Answer: D
Skill: Factual recall
13) In a species of Drosophila, genes q and r are found on the same chromosome 20 centimorgans apart. A cross was made between the following individuals:
What proportion of the offspring would you expect to have the wild-type phenotype? A) 0.10 B) 0.20 C) 0.40 D) 0.80 E) None of these Answer: A
Skill: Analytical reasoning
14) One map unit is equal to a recombination frequency of A) 1%. B) 5%. C) 10%. D) 50%. E) 100%. Answer: A
Skill: Factual recall
15) T. H. Morgan and his colleagues found that among the offspring of genetic crosses, parental phenotypic classes were the most frequent, while recombinant classes occurred less frequently. This observation led Morgan to conclude that A) all genes are linked. B) alleles of linked genes assort independently. C) alleles of some genes assort together. D) genes on different chromosomes assort independently. E) genes on different chromosomes do not assort independently. Answer: C
Skill: Factual recall
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16) When does crossing‐over occur? A) Prophase I of meiosis B) Prophase II of meiosis C) Interphase prior to meiosis D) At any time during the second meiotic division E) At any time throughout division Answer: A
Skill: Factual recall
17) A DNA mutation that gives a distinguishable phenotype for a chromosome or gene is a A) recessive allele. B) lethal mutation. C) parental type. D) linkage group. E) genetic marker. Answer: E
Skill: Factual recall
18) The closer together two genes are on a chromosome, A) the more likely there will be a recombination event between them. B) the less likely there will be a recombination event between them. C) the greater the chance that a double crossover will occur between them. D) the more likely they are to be epistatic. E) the less likely they are to be good genetic markers. Answer: B
Skill: Conceptual understanding
19) In the offspring resulting from an F 1 testcross, what percentage of recombinant phenotypes is expected if the genes under study are independently assorting? A) 5% B) 33% C) 50% D) 90% E) The percentage cannot be predicted. Answer: C
Skill: Factual recall
20) A two‐point testcross is a cross between A) a double heterozygote for two linked genes and a double recessive genotype. B) a heterozygote for one of two linked genes and a double recessive genotype. C) a double heterozygote for two linked genes and a homozygous dominant genotype. D) a heterozygote for one gene and a homozygote for another. E) individuals with two different genetic markers. Answer: A
Skill: Factual recall
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21) How many phenotypic classes may be generated from a three‐point testcross? A) (3)2 = 9 B) (2)3 = 8 C) (3)3 = 27 D) (2)2 = 4 E) It cannot be predicted. Answer: B
Skill: Analytical reasoning
22) Interference is a phenomenon in which A) no double crossovers occur among nearby genes. B) the number of observed double crossovers is greater than the number of expected double crossovers. C) the number of observed double crossovers is less than the number of expected double crossovers. D) the occurrence of one crossover interferes with the formation of another crossover nearby. E) Both C and D Answer: E
Skill: Factual recall
The following scenario applies to the questions below. A testcross of trihybrid Drosophila produced the following phenotypes and number of offspring. A table showing phenotype and the number of offspring with each phenotype is below. A plus sign is used for wild-type phenotype; a letter indicates the mutant phenotype for that gene. +++ abc ++c ab+ +b+ a+c +bc a++ 669 653 121 139 2280 2215 3 2
23) Without performing any calculations, which of the following is the gene order? A) abc B) acb C) cab D) +++ E) None of these Answer: B
Skill: Analytical reasoning
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24) Which gene pair is closer together; i.e. there are fewer map units between them? A) a → b B) b → c C) a → c D) a → b and b → c are the same length E) a → c and b → c are the same length Answer: A
Skill: Analytical reasoning
25) The cross represented by a+ b+ c+ /abc × abc/abc is A) a two‐point testcross. B) a three‐point testcross. C) a parental cross. D) a haploid cross. E) None of these Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) Testcrosses may be used to determine if two genes are linked. Answer: TRUE
Skill: Factual recall
27) Two genes that are located far from each other on the same chromosome will show a higher frequency of recombination than two genes that are close together on the chromosome. Answer: TRUE
Skill: Factual recall
28) A linkage map illustrates the exact physical locations of genes on a chromosome. Answer: FALSE Explanation: A linkage map shows the relative positions of genes on a chromosome and an estimate of the distance between them.
Skill: Factual recall
29) Given three genes in order x y z, it is more accurate to measure the distance between x and z directly than to add up the shorter distances between x and y and y and z. Answer: FALSE Explanation: Shorter distances are more accurate than longer ones.
Skill: Factual recall
30) If in calculating the distance between two genes using data from a testcrossed dihybrid you arrive at a genetic distance of 70 map units, there is likely to be something wrong with your calculations, the experimental data, or with the experiment itself. Answer: TRUE
Skill: Factual recall
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31) Genes may show 50% recombination either when the genes are on different chromosomes or when the genes are far apart on the same chromosome. Answer: TRUE
Skill: Factual recall
32) Parental strains are always homozygous dominant for all genes or homozygous recessive for all genes. Answer: FALSE Explanation: Parental strains are simply homozygous for autosomal genes. Any given gene can be homozygous dominant or recessive, but they can be homozygous dominant for one gene and homozygous recessive for another.
Skill: Factual recall
33) A coefficient of coincidence of zero is the same as an interference value of zero. Answer: FALSE Explanation: A coefficient of coincidence of zero is the same as total interference, which is equal to an interference value of 1.
Skill: Factual recall
34) Is it possible to correct for the effects of multiple crossovers in calculating map distances by applying a linear mapping function. Answer: FALSE Explanation: A mapping function that corrects for multiple crossovers is only linear at 7 mu or less; at greater distances the function becomes curved toward the limit.
Skill: Factual recall
35) Statistical analysis of linkage between any two genes has 1 degree of freedom since there are only two classes of data, parental and recombinant. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS The following scenario applies to the questions below. X, Y, and Z are linked genes. Based on testcross data, the frequency of recombination between genes X and Y was determined to be 33.1 map units. Between genes X and Z, the distance was 11.8 mu; between genes Y and Z, the distance was 21.3 mu. 36) What is the order of these three genes on the chromosome? Answer: X‐Z‐Y
Skill: Problem-solving
37) Of the three pairs of genes, which ones are closest to each other and which ones are farthest apart? Answer: X and Z are closest, being 11.8 map units apart. The distance between Z and Y is 21.3 mu, and the distance between X and Y is 33.1 mu.
Skill: Problem-solving
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38) Is it possible for crossing‐over to occur between two linked genes, yet for all of the resulting offspring to have parental genotypes? Answer: Yes, if a double crossover occurs between the linked genes.
Skill: Analytical reasoning
39) In a particular breed of flowering plants, two genes on the same chromosome determine flower color and leaf color. For flower color, blue (B) is dominant to pink (b), and for leaf color, white (W) is dominant to green (w). A true‐breeding plant with blue flowers and white leaves is crossed with a plant that has pink flowers and green leaves. The offspring are observed and the following phenotypes are tallied. blue flowers and white leaves: 370 plants blue flowers and green leaves: 14 plants pink flowers and white leaves: 12 plants pink flowers and green leaves: 363 plants What is the frequency of recombination between these two genes? Answer: There are a total of 733 parental phenotypes and 26 recombinant phenotypes, so the percentage of recombinants is (26/733) × 100 = 3.5%.
Skill: Problem-solving
40) A female fruit fly with the recessive mutant phenotype of white eyes and miniature wings is mated with a male possessing the wild‐type phenotype of red eyes and normal wings. Among the F 1 s, all the females are wild‐type while all the males exhibit the mutant phenotype. The F 2 s resulting from a testcross exhibit predominantly (63%) parental phenotypes. How do you explain these results? Answer: Since there is a significant deviation from the expected 1:1 ratio of parental and recombinant phenotypes in the F 2 offspring of the testcross, the genes for eye color and wing type must be linked. Furthermore, the F 1 phenotypes indicate that the genes are X‐linked.
Skill: Analytical reasoning
41) Explain how Curt Stern showed, using evidence from Drosophila, that genetic recombination is associated with the physical exchange of parts between homologous chromosomes. Answer: In genetic crosses, carnation-colored eyes is a sex‐linked recessive allele, and bar‐shaped eyes is an incompletely dominant sex‐linked allele caused by a deletion. Stern used male flies that had carnation eyes and normal eye shape, and thus a normal size X chromosome. He crossed those to heterozygous female flies that had a shortened Bar X‐ chromosome with carnation eye color and a second X that was a normal sized B+ chromosome with the wild‐type (car+ ) eye color and a piece of the Y chromosome attached to the X with normal alleles of both genes. Male offspring that were recombinant with red-colored eyes in the bar shape had an X chromosome that was shorted and had a piece of the Y chromosome attached showing physical exchange on the X chromosome.
Skill: Conceptual understanding
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42) Two genes on a Drosophila chromosome are 70 mu apart. How do you think this number was calculated, and what percentage of recombination would you expect to see between them experimentally? Answer: Distances must have been calculated between these genes and other genes that are located between them. The percentage of recombination measured between genes that are over 50 mu apart will be 50%, because that is the maximum measurable.
Skill: Conceptual understanding
43) An F1 plant with the genotype DdGg is mated to a ddgg plant. The resulting seeds are planted and the following phenotypes produced in these numbers: DG Dg dG dg 47 396 412 55
a. What were the genotypes of the parental strains that produced the F 1 ? b. What is the distance between the genes? Answer: (a) DDgg and ddGG. (b) 102/910 = 0.112, or 11.2 mu.
Skill: Problem-solving
44) A red, round, true‐breeding tomato is bred with a yellow, oval, true‐breeding tomato, and the F 1 s are testcrossed to a homozygous‐recessive tester. This results in the offspring below. Calculate the chi‐squared value to test the likelihood of linkage. yellow oval red round yellow round red oval 11 14 7 8
Answer: There are 40 total offspring, so if the genes are unlinked the parentals should equal the recombinants at 20 each. The chi‐square value is (25‐20) 2 /20 + (15‐20) 2 /20 = 5. With 1 degree of freedom, the P value is less than 0.05, so it is unlikely that the results were from random assortment and likely that the genes are linked.
Skill: Problem-solving
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45) The genotypes and numbers of progeny resulting from a testcross are as follows: t+ g + l+ tgl t+ gl tg+ l+ t+ gl+ tg+ l t+ g + l tgl+ Total 72 69 21 18 9 8 2 1 200
Calculate the genetic distance between genes g and l. Answer: The distance would be the sum of the single crossovers in the region between g and l and the number of double crossovers divided by the total number of progeny. Thus, (9 + 8) + (2 + 1)/200 = 0.1 or 10%. This is the recombination frequency of genes g and l. This distance between them is 10.0 map units.
Skill: Problem-solving
46) Why does the recombination frequency often lead to an underestimation of the true map distance between linked genes? Answer: Because map units between linked genes are defined in terms of the crossover frequency, but distances are estimated from the frequency of recombinant phenotypes in genetic crosses, and these frequencies are not identical; only odd numbers of double crossover events will show up as recombinant phenotypes.
Skill: Conceptual understanding
47) What does the chi‐square test tell us about the linkage of two genes? Answer: We can use the chi‐square test to test the hypothesis that two genes are not linked. A chi‐square table gives you the statistical significance of the deviation of the observed number of recombinants in a testcross from the expected number of recombinants if the genes are not linked. If the P value obtained is less than 0.05, the observed number is significantly below the expected, and the null hypothesis is rejected.
Skill: Conceptual understanding
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48) Here is a genetic map: A 18 mu B 10 mu C . AAbbCC is crossed to aaBBcc to produce an F 1 generation. If you produce 1000 offspring from testcrossing the F 1 , what genotypes do you expect, and in what numbers? Assume that there is no interference. Answer: There will be 8 possible genotypes, listed here as 4 pairs in the following order: double crossovers, A to B single crossovers, B to C single crossovers, and finally parental types. Every individual must receive a recessive allele of each gene from the tester strain. AaBbCc for aabbcc AaBbcc represent aabbCc genotype. Aabbcc aaBbCc genotype. AabbCc 738. aaBbcc You would expect 0.18 × 0.10 × 1000 = 18 total double crossovers, 9 each of these two genotypes. A to B crossovers should = 0.18 × 1000 = 180. Eighteen of these double crossovers, so A to B only = 180 - 18 = 162, 81 for each
B to C crossovers should = 0.10 × 1000 = 100. 18 of these represent double crossovers, so B to C only = 100 - 18 = 82, 41 for each
The two parental types together should be 1000 - 18 - 162 - 82 = Therefore, each parental type genotype should be 369.
Skill: Conceptual understanding, analytical reasoning
49) It is not necessary to use a male from a tester strain to genetically map genes on the X chromosome. How can this be done and why does it work? Answer: It is necessary to start with females that are heterozygous for X‐linked genes. They can be bred with any male. Their male offspring can then be used to determine linkage distances. Male offspring will have only a single X chromosome, and that chromosome will come from the mother, so they can be used to measure recombination frequencies to construct a genetic map.
Skill: Conceptual understanding
50) What is the difference in the likelihood of linkage between two genes with a lod score of 2 and two genes with a lod score of 3? Answer: Linkage is 10 times more likely with a lod score of 3 than with a lod score of 2 because the number is based on the log 10 of the ratio.
Skill: Factual recall
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Chapter 15 Genetics of Bacteria and Bacteriophages
MATCHING QUESTIONS Please select the best match for each term. 1) Conjugation
Skill: Factual recall
A) A sex factor plasmid contained in donor bacterial cells that allows mating B) A self‐replicating, circular DNA that is distinct from the main bacterial chromosome C) The transfer of genes from one bacterium to another by phage vectors D) The unidirectional transfer of extracellular DNA into cells E) The unidirectional transfer of genetic information through direct cellular contact between a donor bacterial cell and a recipient bacterial cell
2) Transformation
Skill: Factual recall
3) Transduction
Skill: Factual recall
4) F factor
Skill: Factual recall
5) Plasmid
Skill: Factual recall
Answers:
1) E
2) D
3) C
4) A
5) B
MULTIPLE-CHOICE QUESTIONS 6) ________ bacterial strains require nutritional supplements to grow in a minimal medium. A) Autotrophic B) Prototrophic C) Phototrophic D) Heterotrophic E) Auxotrophic Answer: E
Skill: Factual recall
7) Minimal media is A) growth media used in the smallest volume in which cells can grow. B) growth media designed to minimize the growth of contaminants. C) growth media that contains the minimal nutritional requirements for normal cells. D) a way to visualize new mutations in a minimum amount of time. E) used to reveal only mutant cell colonies against a dark background. Answer: C
Skill: Factual recall
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8) An arg- strain of E. coli is transformed with a plasmid carrying the wild‐type (arg+ ) gene. The transformed cells are replica‐plated to two minimal medium plates: one supplemented with arginine and one lacking arginine. The transformed cells would grow on A) only the plate supplemented with arginine. B) only the plate lacking arginine. C) both plates. D) neither plate. E) the plate supplemented with arginine, but only if a mutation occurred. Answer: C
Skill: Factual recall
9) When Lederberg and Tatum performed their experiments on gene transfer in bacteria that lead to our understanding of conjugation, what were their controls? A) They mixed strains A and B together. B) They put strain A on one side of a U‐shaped tube and strain B on the other, with a filter between them. C) They found bio- met - cells. D) They plated strain A and strain B separately onto minimal media plates to screen for spontaneous prototrophs. E) They observed mating bridges with transmission electron microscopes. Answer: D
Skill: Conceptual understanding
10) A plasmid such as an F factor that is capable of integrating into the bacterial chromosome is a(n) A) prophage. B) episome. C) auxisome. D) perisome. E) elaiosome. Answer: B
Skill: Factual recall
11) Which of the following has episomal DNA inserted in the cellʹs chromosome? A) An F+ cell B) An Fʹ plasmid C) An Hfr cell D) An F- cell E) All except D Answer: C
Skill: Factual recall
12) Cotransductants can be detected when the transduced genes are A) closely linked. B) far apart on the same chromosome. C) on different chromosomes. D) mutant. E) Both B and D Answer: A
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
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13) mtl, polA, and xyl are bacterial genes that can be transmitted using an Hfr strain. An interrupted mating experiment resulted in xyl exconjugants appearing at 7 minutes and reaching maximum level by 30 minutes. The mtl gene function exconjugants appeared at 11 minutes and leveled off at 80% maximum by about 33 minutes, while polA did not appear until 24 minutes into the experiment and then only slowly rose to about 30%. Which of the following shows the gene order with the insertion site (arrow) location correctly? A) xyl mtl polA → B) polA mtl xyl → C) mtl xyl polA → D) mtl polA xyl → E) None of these Answer: B
Skill: Problem-solving
14) Conjugation by which of the following with an F- cell results in a cell that remains F- ? A) F+ B) Hfr C) F D) A and C E) B and C Answer: B
Skill: Conceptual understanding
15) The reason that many plasmids used in laboratories for transformation experiments contain an ampicillin resistance gene (ampR) is A) so that the plasmid can replicate in bacterial cells. B) so that the plasmid can recombine with the bacterial genome. C) to enable the recipient cell to conjugate. D) to provide a selectable marker. E) to map point mutations. Answer: D
Skill: Factual recall
16) In a genetic cross between two bacteriophage T4 rII mutants, r+ recombinants are produced. The genetic changes in these two rII mutants must be A) heterozygous. B) heteroallelic. C) homozygous. D) homoallelic. E) polymorphic. Answer: B
Skill: Factual recall
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17) In E. coli, λ DNA integrates into which site on the bacterial chromosome? A) ori B) trp C) bio D) gal E) att Answer: E
Skill: Factual recall
18) To grow on artificial medium, prototrophic bacteria require A) sunlight. B) amino acid supplements. C) vitamin supplements. D) All of these E) None of these Answer: E
Skill: Factual recall
19) E. coli strains that are Hfr A) are easily mutated. B) are susceptible to infection by bacteriophage. C) contain the F factor integrated in the bacterial chromosome. D) cannot be made competent. E) have a low frequency of recombination. Answer: C
Skill: Factual recall
20) The E. coli chromosome is A) linear. B) circular. C) single‐stranded. D) less than a megabase in length. E) compartmentalized within an intracellular membrane. Answer: B
Skill: Factual recall
21) Which of the following methods can be used to create competent bacterial cells? A) Treating the cells chemically B) Exposing cells to a strong electric field C) Making the cell membrane more permeable to DNA D) Allowing the culture to enter stationary phase growth E) A, B, and C only Answer: E
Skill: Factual recall
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22) In general, transformation of most genes occurs at a frequency of one in every ________ cells. A) 10 B) 100 C) 1,000 D) 10,000 E) 1,000,000 Answer: C
Skill: Factual recall
23) Which of the following statements is true regarding a temperate phage? A) It can reproduce at warm or cool temperatures. B) It is required for transformation. C) It can be lytic or lysogenic. D) It causes bacterial cells to conjugate. E) It reproduces by binary fission. Answer: C
Skill: Factual recall
24) In the lysogenic bacteriophage life cycle, the λ chromosome A) replicates and the phage genes take over the bacterium. B) inserts itself physically into the host cellʹs chromosome. C) expresses a repressor protein gene that inhibits the lytic pathway. D) is replaced by a piece of bacterial DNA when packaged inside phage progeny. E) Both B and C Answer: E
Skill: Factual recall
25) Cotransductant bacteria can occur if A) a bacterial cell picks up two fragments of DNA from the environment, each with a different gene. B) double crossover occurs between prophage DNA and the bacterial chromosome. C) two genes are closely linked enough so that they can be packaged into a phage head and injected into a cell by a single phage. D) two genes are introduced into the same bacterium by simultaneous infection with two different phages. E) Both C and D Answer: E
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) Genetic exchange in bacteria by transduction requires cell‐to‐cell contact. Answer: FALSE Explanation: Transduction requires a phage vector but does not require cell‐to‐cell contact. Genetic exchange by conjugation requires cell‐to‐cell contact.
Skill: Factual recall
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27) The viral chromosome of a bacteriophage can never integrate into the host bacteriumʹs chromosome. Answer: FALSE Explanation: The viral chromosomes of some bacteriophages can integrate into the host chromosome and replicate. At this stage, the virus is termed a prophage.
Skill: Factual recall
28) Genetic mapping experiments performed by Seymour Benzer on rII mutants in the T4 phage showed that the gene is indivisible by the process of mutation and recombination. Answer: FALSE Explanation: Benzerʹs experiments showed that genetic recombination can occur within genes; in other words, that the base pair, rather than the gene, is the unit of mutation and recombination in genomes.
Skill: Factual recall
29) In the case of genetic exchange in bacteria by transformation, there is an exchange of DNA back and forth between cells, and a complete diploid cell is formed. Answer: FALSE Explanation: During transformation, gene transfer is unidirectional, and the transformed cell remains haploid.
Skill: Factual recall
30) In genetic exchange by conjugation, a single Hfr strain will transfer the entire bacterial chromosome to the recipient, allowing mapping in minutes from the first gene entering to the last one before the other half of the inserted F factor. Answer: FALSE Explanation: Only fragments of a genome can be easily mapped from a single Hfr strain. Mating bridges break long before the entire genome is transferred.
Skill: Conceptual understanding
31) ʺHot spotsʺ are locations in a gene where mutations are found at a relatively high frequency. Answer: TRUE
Skill: Factual recall
32) A complete bacterial medium contains only the nutrients required for the growth of wild‐type cells. Answer: FALSE Explanation: A minimal medium contains only the nutrients required for the growth of wild‐type cells. A complete medium provides nutritional substances required by auxotrophic mutant cells to grow.
Skill: Factual recall
33) A merodiploid cell has two copies of one or a few genes and only one copy of all the others. Answer: TRUE
Skill: Factual recall
34) In F+ × F- crosses, none of the bacterial chromosome is transferred; only the F factor is. Answer: TRUE
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
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35) To enhance the efficiency of transformation, bacterial cells can be induced to take up DNA by a strong electrical field. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) An Hfr strain of E. coli with the genotype a+ b+ c+ d + e+ f+ is mated with an F- auxotrophic strain with the genotype a- b- c- d - e- f- . Conjugation is stopped at 10‐minute intervals and the genotypes of the resulting conjugants are determined. The following results are obtained: After 10 minutes After 20 minutes After 30 minutes After 40 minutes After 50 minutes After 60 minutes e+ a+ e+ a+ b+ e+ a+ b+ d + e+ a+ b+ c+ d + e+ a+ b+ c+ d + e+ f+
What is the correct order of genes on this bacterial chromosome? Answer: e- a- b- d - c- f
Skill: Problem-solving
37) What are temperate bacteriophages? Answer: Phages that can alternate between lytic and lysogenic pathways in bacterial cells.
Skill: Factual recall
38) What is the difference between generalized and specialized transduction? Give an example of specialized transduction. Answer: In generalized transduction, any gene or set of genes on the chromosome may be transferred from the donor bacterium to the recipient bacterium. In specialized transduction, only specific genes may be transferred. This is mediated by temperate phages in which prophages associate with only one site of the bacterial chromosome. For example, the λ genome integrates into the E. coli bacterial chromosome at a specific site between the gal region and the bio region, called att λ. By abnormal excision of the prophage from the host chromosome, gal or bio genes may be transferred to other bacterial cells.
Skill: Conceptual understanding
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39) For a P1 transduction experiment, an E. coli donor strain with the genotype arg+ , ser+ , ampR and a recipient strain with the genotype arg- , ser- , ampS were used. The following results were obtained for each marker: Of the transductants selected for arginine, 5% are ser+ and 30% are ampR. Of the transductants selected for serine, 3% are arg+ and 0% are ampR. What is the correct order and relative distance of these three genes on the E. coli chromosome? Answer: The order and relative distance is: ampR- arg‐‐‐‐ser. Since amp and ser are never cotransduced, they must be very far apart from each other. Arg and ser are rarely cotransduced, so they must be farther apart from each other than are arg and amp, which are more frequently cotransduced, so they must be fairly close together.
Skill: Problem-solving
40) During a cis‐trans complementation assay, two lac- strains of bacteria are crossed. The progeny cells exhibit a lac- phenotype. What does this indicate? Answer: The two mutations are in different genes that encode different products. The two mutant strains complement each other to create a functional cell that can metabolize lactose.
Skill: Conceptual understanding
41) In a bacterial species genes q, r, s, and t were mapped in minutes through interrupted mating experiments. The results placed q and s on the two outer ends, but r and t were not resolvable by minutes. You have discovered no phage that can infect this species. If you have a q r s t Hfr strain and a q- r- s- t - F- strain, how could you measure the distance between the genes in map units? Answer: There are two ways. (1) You could do a conjugation experiment in which you select for the last marker in, to be sure that all the DNA was available for recombination, and measure (recombinants in the various gene pairs/total) X 100 for map units. (2) You could do a similar experiment through transformation. To do so, you would isolate and fragment the DNA from the q r s t strain and use it to transform the q- rs- t - . In this case you would need to select for q s colonies to make sure that the entire region was likely to have entered the recipient cells.
Skill: Conceptual understanding
42) A bacterial broth culture is diluted by a factor of 100,000, and 0.1 mL of the dilution is spread over the surface of an agar plate. The next day, 123 colonies are observed on the plate. What was the concentration of bacteria in the broth? Answer: We want to determine the number of colony‐forming units (cfu) per mL. Since 123 colonies are observed on the plate, this means that there were 123 bacterial cells (cfu) in 0.1 mL of the 100,000X dilution. Thus, in the original culture, there were 123 cfu × 100,000 (dilution factor) × 10 = 1.23 × 108 cfu/mL.
Skill: Problem-solving
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43) Explain how an Fʹ × F- conjugation can produce a partially diploid bacterial cell. Answer: The Fʹ factor was derived from an integrated factor that acquired bacterial genes during excision from the chromosome by means of imprecise looping out and single crossover. During Fʹ × F- conjugation, the Fʹ factor is transferred to the F- cell. The recipient cell then contains two copies of the bacterial genes carried by the Fʹ factor.
Skill: Conceptual understanding
44) In a series of cotransformation experiments, DNA fragments from x + y + z+ donor bacteria were used to transform xyz recipient bacteria. The following transformation phenotypes were observed in the recipient cells: xy+ z xyz+ x+y+z x + yz+ What does this imply about the gene order of x, y, and z? Answer: The gene order must be y‐x‐z. Since the genes y and z are never cotransformed, they must be far apart on the DNA. However, x and y are sometimes cotransformed, and y and z are sometimes cotransformed, so each of these genes must be more closely linked to one another.
Skill: Analytical reasoning
45) How does a bacteriophage become a transducing phage? Answer: During the lytic phage life cycle, the bacterial DNA is degraded and, rarely, a piece of bacterial DNA is packaged into a progeny phage head instead of phage DNA. These phages are then transducing phages, which can carry genetic material to other bacteria.
Skill: Factual recall
46) How would you design an experiment to test whether a transduction event occurred between two different bacterial strains? Answer: You would use a selected marker to identify the transductant cells. For example, if the donor cell is thr+ and the recipient cell is thr, the prototrophic transductants can be grown on minimal medium that does not contain threonine, whereas the nontransduced cells still require threonine to grow.
Skill: Application of knowledge
47) How could you use T2 bacteriophage plaque phenotypes to determine the genetic distance between two genes, h and r, that affect plaque appearance in the phage DNA? Answer: Just as in eukaryotic organisms, a crossing experiment can be performed. You would use different strains of T2–one of which is h + r and one of which is hr + –that express different plaque phenotypes to coinfect plated bacterial cells. Inside the bacterial cells, the two different phage chromosomes can crossover to produce recombinants. The percentages of parental and recombinant plaque phenotypes can be determined. As in eukaryotic cells, the frequency of recombinants is equal to the genetic distance between the two genes.
Skill: Application of knowledge
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Chapter 15 Genetics of Bacteria and Bacteriophages 135
48) How did Seymour Benzer use deletion mapping to construct fine‐scale maps of the rII region of the T4 bacteriophage genome? Answer: He used deletion mutants, which, unlike point mutants, have lost a segment of DNA and cannot revert to a wild‐type state. He crossed unknown point mutants with standard (known) deletion mutants for different segments of the rII region to determine which crosses could result in wild‐type recombinant phage. Those that could not must have used point mutants with a mutation in the region of the deletion. This could be repeated in crosses between the point mutant and smaller secondary reference deletion mutants until the point mutation was localized to within a very small segment of the rII region.
Skill: Conceptual understanding
49) In bacteriophage T4, all rIIA mutants are found to complement all rIIB mutants. However, rIIA mutants fail to complement other rIIA mutants, and rIIB mutants fail to complement other rIIB mutants. Furthermore, mutants with deletions that span both rIIA and rIIB complementation groups do not complement either A or B mutants. What do these data mean, and what do the complementation groups represent? Answer: Complementation occurs in crosses between mutants of the different complementation groups, which represent functional units, or genes. That is, rIIA mutants produce functional B gene products, and rIIB mutants produce functional A gene products. Both A and B gene products are required to produce a wild‐type phenotype. Mutants with deletions that span both complementation regions cannot recover the wild‐type phenotype because they will always be missing part of a functional unit.
Skill: Conceptual understanding
50) Performing a complementation test through cotransduction using mutant phage genomes created by X‐ray mutagenesis (mutants a‐e), as well as ones created by chemical mutagenesis using EMS (mutants 1‐5), you get the following results: 1 2 a — + b — + c + + d + — e — — 3 + — — + + 4 — — — + + 5 — — + — —
Explain this data. Answer: X‐rays create DNA breaks that lead to deletions, and chemical mutagens such as EMS result in point mutations. Thus, mutants a‐e are deletion mutants and 1‐5 are point mutants. Analysis of the data can provide us with a comparative map of the mutants as follows (this map can be flipped horizontally).
Skill: Problem-solving, conceptual understanding Copyright © 2010 Pearson Education, Inc.
Chapter 16 Variations in Chromosome Structure and Number
MATCHING QUESTIONS Please select the best match for each term. 1) Triploid
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A) A chromosome number that is not an exact multiple of the haploid set of chromosomes B) More than two full sets of chromosomes C) A normal number of chromosomes D) Three copies of one chromosome E) Three full sets of chromosomes
2) Trisomic
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3) Euploid
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4) Polyploid
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5) Aneuploid
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Answers:
1) E
2) D
3) C
4) B
5) A
Please select the best match for each term. 6) Turner syndrome
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A) Three copies of chromosome 21 B) XO
7) Klinefelter syndrome
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C) XXY D) Three copies of chromosome 18 E) Three copies of chromosome 13
8) Down syndrome
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9) Edwards syndrome
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10) Patau syndrome
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Answers:
6) B
7) C
8) A
9) E
10) D
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Chapter 16 Variations in Chromosome Structure and Number 137
MULTIPLE-CHOICE QUESTIONS 11) Which of the following types of chromosomal mutation cannot revert to the wild‐type state? A) Duplication B) Deletion C) Translocation D) Inversion E) Any of these may revert to wild‐type Answer: B
Skill: Conceptual understanding
12) In Drosophila melanogaster, polytene chromosomes are produced when A) a chromosome breaks and rejoins with another section of the same chromosome. B) a cell divides several times before the chromosomes are duplicated. C) chromosomes are duplicated repeatedly without a corresponding nuclear division. D) a segment of a chromosome is duplicated during an unequal crossing‐over event. E) characteristic banding patterns form on the chromosomes. Answer: C
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13) Which of the following is not used to help identify individual chromosomes in a karyotype? A) Centromere position B) Point mutations C) Chromosome size D) Banding pattern on staining E) Heterochromatin regions Answer: B
Skill: Conceptual understanding
14) Many scientists believe that the evolution of multigene families, such as the genes for hemoglobin, is a result of which type of genetic rearrangement? A) Duplication B) Deletion C) Inversion D) Reciprocal translocation E) Nonreciprocal translocation Answer: A
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15) A(n) ________ inversion includes the centromere. A) concentric B) pericentric C) paracentric D) epicentric E) chromocentric Answer: B
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16) If a small region of a chromosome containing a non‐essential gene is completely deleted, and an organism has two identical chromosomes with this deletion, then the organism A) has no alleles of that gene. B) is dead. C) has more heterochromatin. D) has recessive alleles of those genes. E) will exhibit pseudodominance. Answer: D
Skill: Conceptual understanding
17) What is the genetic consequence of a homozygous translocation? A) Inviable gamete formation (semisterility) B) Gene duplications and deletions C) Formation of abnormal chromatids following crossing‐over D) Abnormal pairing during meiosis E) An alteration in the linkage relationships of genes Answer: E
Skill: Analytical reasoning
18) A human cell containing two sets of 23 chromosomes is A) aneuploid. B) polyploid. C) euploid. D) haploid. E) None of these Answer: C
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19) Nondisjunction of chromosomes may result in A) the loss of one homologous chromosome pair. B) the addition of one homologous chromosome pair. C) the addition of a single chromosome. D) the loss of a single chromosome. E) All of these Answer: E
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20) A monosomic cell would produce gametes with how many chromosomes? A) N B) N and N‐1 C) N‐1 D) N+1 E) N and N+1 Answer: B
Skill: Factual recall
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Chapter 16 Variations in Chromosome Structure and Number 139
21) An X‐linked homozygous color‐blind woman gives birth to a son who is not color blind. He could have normal vision because he A) has Down syndrome. B) has Klinefelter syndrome. C) has Turner syndrome. D) is XYY. E) has triploidy. Answer: B
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22) Which of the following traits is determined by programmed transposition? A) Pattern baldness in humans B) Mating types in yeast C) Eye color in Drosophila D) All of these E) None of these Answer: D
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23) A dicentric chromosome has two A) arms. B) centromeres. C) telomeres. D) mutations. E) centrosomes. Answer: B
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24) Which of the following types of analysis allows scientists to visualize an individualʹs chromosomal makeup? A) Karyotype analysis B) Pedigree analysis C) Genetic crosses D) Inheritance analysis E) Deletion mapping Answer: A
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25) In familial Down syndrome resulting from Robertsonian translocation, A) the affected individual has three copies of the complete chromosome 21. B) the affected individual is missing a copy of chromosome 21. C) the affected individual has three copies of the long arm of chromosome 21, one of which is attached to part of chromosome 14. D) the affected individual has two copies of the long arm of chromosome 21, one of which is attached to part of chromosome 14. E) the affected individual has three copies of the long arm of chromosome 14, one of which is attached to chromosome 21. Answer: C
Skill: Factual recall
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26) Most monoploid individuals do not survive because A) they do not have the appropriate gene dosages. B) their cells cannot divide properly. C) recessive lethal mutations cannot be masked by dominant alleles. D) they have missing genes. E) they have only one sex chromosome. Answer: C
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27) The chemical colchicine prevents the formation of microtubules. It is commonly used in certain experimental procedures to cause changes in cellular chromosomes. Which of the following changes is it likely to be used to create? A) Induction of mutant polyploid individuals B) Induction of mutant aneuploid individuals C) Prevention of nondisjunction in cell cultures D) Induction of chromosomal deletions in experimental cell lines E) Induction of chromosomal duplications in experimental cell lines Answer: A
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28) Seedless bananas are produced from A) sterile tetraploid allopolyploid plants. B) sterile triploid autopolyploid plants. C) monoploid plants grown from unfertilized seeds. D) fertile diploid plants that are unfertilized. E) Both C and D Answer: A
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29) Cultivated wheat is descended from three distinct species, each with a diploid set of 14 chromosomes. Thus, it is A) an allotriploid with 21 chromosomes. B) an allotriploid with 42 chromosomes. C) an allohexaploid with 84 chromosomes. D) an allohexaploid with 42 chromosomes. E) an autotetraploid with 28 chromosomes. Answer: D
Skill: Analytical reasoning
TRUE-FALSE QUESTIONS 30) In humans, many types of cancers are associated with chromosomal mutations. Answer: TRUE
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31) A tetraploid cell contains four chromosomes. Answer: FALSE Explanation: It contains four copies of each chromosome.
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Chapter 16 Variations in Chromosome Structure and Number 141
32) Approximately 15% of all human conceptions contain at least one chromosomal mutation. Answer: TRUE
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33) The causes of chromosomal deletion may include radiation, viral infection, chemical exposure, or transposable elements. Answer: TRUE
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34) No known living humans have one whole chromosome of a homologous pair of autosomes deleted from the genome. Answer: TRUE
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35) For eukaryotic organisms, polyploidy is always a lethal condition. Answer: FALSE Explanation: Polyploidy is lethal for most animal species, but is less consequential for plants, and in fact has played a significant role in the evolution of plant species.
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36) Both changes in chromosome number and chromosome structure are believed to have resulted in novel genomes, leading to new species. Answer: TRUE
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37) Inversion loops form to allow homologous chromosomes containing homozygous inversions to pair properly during meiosis. Answer: FALSE Explanation: Inversion loops allow inversion heterozygotes to pair properly during meiosis.
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38) Down syndrome (trisomy‐21) is the only human autosomal trisomy in which the individuals may survive to adulthood. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 39) In some fruit flies, an inversion involving the white‐eyed locus (w) moves the w+ gene from one end of the X chromosome to a location nearer to the centromere. Flies with a w+ or w+ /w genotype are normally expected to have red eyes. However, in flies with the inversion, the eyes are mottled red and white. Explain why. Answer: The mottled phenotype is caused by a position effect. During the inversion, the w+ gene is moved from a euchromatic region on the X to a region near the heterochromatin at the centromere. Sometimes genes in this region are not transcribed because the heterochromatin remains condensed throughout the cell cycle. The eye cells in which the w+ gene has become inactivated produce white spots in the eyes. The cells in which the gene is not inactivated produce red spots.
Skill: Conceptual understanding
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40) What are the gametic consequences of a lack of cytokinesis following the first or second meiotic division? Answer: The result will be a change in the number of complete set of chromosomes. If the individual is a diploid organism, and cytokinesis fails to occurs at meiosis I, half the gametes will have no sets of chromosomes and half will have two sets of chromosomes. If cytokinesis fails at meiosis II, half the gametes will have one set of chromosomes (normal condition), one‐fourth of them will have two sets, and one‐fourth of them will have no sets.
Skill: Analytical reasoning
41) How does a Philadelphia chromosome occur, and what condition does it cause? Answer: Ninety percent of patients with the fatal cancer chronic myelogenous leukemia (CML cancer) have a chromosomal mutation in their myeloblasts caused by a reciprocal translocation event between chromosome 22 and chromosome 9. The defective chromosome 22 is called the Philadelphia chromosome.
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42) How can the banding patterns on Drosophila polytene chromosomes be used to aid deletion mapping of genes on the chromosomes? Answer: The unexpected expression of a recessive trait caused by pseudodominance (the loss of a dominant allele) can be correlated with changes in band number and physical appearance of the polytene chromosomes. For example, in a polytene X bearing wild‐type alleles, deletions in certain regions can be visually observed. These can be correlated with the appearance of pseudodominant phenotypes for those alleles. The alleles can then be mapped to those missing regions on the chromosome.
Skill: Conceptual understanding
43) The diagram below shows a region near the center of a long arm of a chromosome. Assume that in the diagram below, every vertical hatch mark is 1 map unit from the next vertical hatch mark. The dashed double line represents an inversion. If you crossed this heterozygous individual with a normal homozygous recessive individual, and produced 1000 live offspring, how many would you expect to be Aa dd?
Answer: Recombination within this region would be lethal, producing no live offspring, so the distance between A and D will be effectively changed from 6 map units to 2 map units. As a result, 2% recombination is expected. Half of the recombinants will be Aadd, the other half aaDd. Therefore, we can expect the number of Aadd individuals to be 1% of 1000, or 10 individuals.
Skill: Problem-solving, conceptual understanding
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Chapter 16 Variations in Chromosome Structure and Number 143
44) What are all the possible gametic genotypes produced by a trisomic individual that is genotype A/A + /B for the trisomic chromosome? Answer: Three types of haploid gametes may be produced: A, A+ , and B; and three types of disomic gametes may be produced: A/A + , A/B, and A+ /B.
Skill: Problem-solving
45) How is hybrid sterility overcome in allopolyploid plants? Answer: Typically, allopolyploids are produced when two different species hybridize. The resulting offspring then contains one set of chromosomes from each parental species and is sterile because the sets are not homologous and cannot pair at meiosis to produce viable gametes. However, if both chromosome sets double by means of division error (nondisjunction), each diploid set can function normally during meiosis, and N1 +N2 gametes will be produced. Fusion of two such gametes will produce an allotetraploid offspring.
Skill: Conceptual understanding
46) A woman with normal vision, whose father has sex‐linked color blindness, marries a man with normal vision. They have a son who has Klinefelter syndrome and is color blind. Describe precisely what kind of event could caused have his Klinefelter syndrome. Answer: Nondisjunction must have occurred in the mother during meiosis II, when sister chromatids normally separate. With Klinefelter syndrome, the boy has an XXY karyotype. Since both of his X chromosomes contain the allele for color blindness, he likely has two copies of one homolog of his motherʹs X chromosome containing the allele for color blindness.
Skill: Problem-solving, conceptual understanding
47) Why do polyploid individuals with an even number of chromosome sets have a higher likelihood of fertility than those with an odd number of chromosome sets? Answer: Individuals with an even number of chromosome sets have the potential for homologs to be segregated equally during meiosis and are likely to produce balanced gametes. Polyploids with an odd number of chromosome sets always have an unpaired chromosome for each type and thus produce many unbalanced gametes.
Skill: Conceptual understanding
48) During meiosis in a reciprocal translocation heterozygote, what are the three ways that chromosomes may segregate at anaphase I, and what are the consequences for the resulting gametes? Answer: In alternate segregation, alternate centromeres (from nonhomologous chromosomes) migrate to the same pole. This results in normal gametes, each having a complete set of genes (one half with all normal chromosomes and one half with translocated chromosomes). In adjacent‐1 segregation, adjacent nonhomologous centromeres migrate to the same pole. This results in gametes with either duplications or deletions of genes, which are usually inviable. In adjacent‐2 segregation, adjacent homologous centromeres migrate to the same pole. This also results in inviable gametes with duplications or deletions.
Skill: Conceptual understanding
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49) A paracentric inversion heterozygote contains a normal chromosome with the long‐arm gene order GENE and an inverted chromosome with the long‐arm gene order GNEE. They form an inversion loop to pair during prophase I of meiosis. If a single crossover occurs between the G and the E segments of the loop, what will be the gene orders in the resulting recombinant dicentric chromosome and acentric fragment? Assume crossing over occurs at the four‐strand stage involving two nonsister chromatids. Answer: The dicentric chromosome gene order will become GENG, and the gene order in the acentric fragment will become EENE.
Skill: Problem-solving
50) During meiosis, what are the gametic consequences of a paracentric inversion with a single crossover event, compared to a pericentric inversion with a single crossover? Answer: The paracentric inversion results in two viable gametes: one with a normal chromosome and one with an inversion chromosome. Two gametes will contain fragments of a broken dicentric chromosome and will be inviable (since the acentric fragment is lost). The pericentric inversion also results in two viable gametes–one containing a normal chromosome and one containing an inversion chromosome–and two inviable gametes containing different deletion/duplication products.
Skill: Conceptual understanding
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages
MATCHING QUESTIONS Please select the best match for each term. 1) Operator
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A) Molecule capable of down‐regulating protein synthesis B) Adjacent genes serving one function that are transcribed and regulated together C) Regulatory region to which repressor protein binds D) Regulatory molecule that leads to gene expression E) General class of molecules that help control regulation of expression 2) A 3) D 4) E 5) B
2) Repressor protein
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3) Inducer
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4) Effector molecule
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5) Operon
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Answers:
1) C
MULTIPLE-CHOICE QUESTIONS 6) The order of the structural genes in the lac operon is A) 5ʹ‐lacY‐lacZ‐lacA‐3ʹ. B) 3ʹ‐lacZ‐lacY‐lacA‐5ʹ. C) 5ʹ‐lacZ‐lacY‐lacA‐3ʹ. D) 3ʹ‐lacY‐lacA‐lacZ‐5ʹ. E) 5ʹ‐lacA‐lacY‐lacZ‐3ʹ. Answer: C
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7) In the lac operon in E. coli, a nonsense mutation in the ________ gene will result in a loss of β‐galactosidase activity in the cell. A) lacZ B) lacI C) lacY D) Both A and B E) All of these Answer: A
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8) The inducer of the lac operon is A) galactose. B) glucose. C) allolactose. D) repressor protein. E) permease. Answer: C
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9) Attenuation of the trp operon fails to occur when A) a lack of tryptophan causes stalling of ribosomes on the leader sequence of the RNA. B) an RNA stem‐loop forms that allows transcription to continue. C) an RNA stem‐loop that causes transcription to terminate fails to form. D) the cell has enough tryptophan. E) the operator is mutated. Answer: D
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10) To inhibit the transcription of operon genes, the lacI gene product binds to A) the operator. B) the promoter. C) the repressor. D) the inducer. E) β‐galactosidase. Answer: A
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11) A lacOC mutation is A) constitutive. B) cis‐dominant. C) incompletely dominant. D) codominant E) recessive. Answer: A
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12) During transcription of the trp operon, pairing of regions 2 and 3 in the leader peptide mRNA creates a(n) ________ signal. A) continuous B) termination C) antitermination D) induction E) repression Answer: C
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 147
13) What is the underlying reason that a lambda phage would need to switch from the lysogenic state to the lytic state? A) To have its genome replicated B) To kill its host cell C) To make more of the Cro protein D) To be more efficient in its use of resources E) To escape an endangered host cell Answer: E
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14) Which of the following mutations is dominant over wild‐type lacI+ in a partial diploid experimental cell? A) lacAB) lacOC) lacID) lacIS E) C and Donly Answer: D
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15) Which positive regulatory protein is required for the production of phage coat protein in lambda? A) CI B) Cro C) N D) Q E) None of these Answer: D
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16) How many structural genes are under the control of the trp operon promoter? A) 1 B) 2 C) 3 D) 4 E) 5 Answer: E
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17) In terms of regulation of gene expression, tryptophan and allolactose are examples of ________ molecules. A) repressor B) effector C) inducer D) operator E) terminator Answer: B
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18) Which of the following can be a transdominant mutation in partial diploid cells? A) lacZB) PlacIC) lacIS D) lacYE) lacIAnswer: C
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19) The interaction between allolactose and the repressor protein results in a shape change known as a(n) A) isomer. B) mutation. C) enantiomer. D) allosteric shift. E) transformant. Answer: D
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20) Operons can best be defined as A) operational mutations affecting gene expression. B) a set of genes under common regulatory control. C) dominant alleles responsible for enzyme synthesis. D) a balance between repressor and inducer molecules. E) a common mode of gene expression in bacteria. Answer: B
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21) The ________ pathway in lambda phage leads to integration of viral DNA with that of the host, while the ________ pathway leads to induction of progeny viruses. A) lytic, lysogenic B) invasive, replicative C) repressive, inductive D) lysogenic, lytic E) replicative, lysogenic Answer: D
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22) In bacteriophage lambda, the establishment of lysogeny requires the protein product of which gene? A) N B) cII C) cIII D) B and C only E) A, B, and C Answer: D
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 149
23) To maintain lysogeny in lambda, which of the following needs to occur? A) Binding of repressor protein to OL B) Binding of repressor protein to OR C) Inactivation of PL D) Inactivation of PR E) All of these Answer: B
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24) The genetic switch for the infection pathways of lambda phage is controlled by two regulatory proteins: ________ and ________. A) lambda repressor, Cro B) lambda repressor, permease C) excisionase, integrase D) recA, Cro E) lyase, lysogenase Answer: A
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25) Attenuation in trp gene regulation is accomplished by A) reducing permeability of tryptophan, the effector molecule. B) blocking expression of trp mRNA. C) formation of secondary mRNA structures in the leader region. D) binding with RNA polymerase. E) repression of the trp operator. Answer: C
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TRUE-FALSE QUESTIONS 26) Genes required for maintaining basic cell structure, growth, and division are housekeeping genes and have regulated expression. Answer: FALSE Explanation: They are constitutive.
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27) The structural genes in the lac operon are transcribed into a single polycistronic mRNA molecule. Answer: TRUE
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28) Mutants that are lacIS do not produce lac enzymes, regardless of the presence or absence of lactose. Answer: TRUE
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29) A constitutive gene is a gene that is occasionally expressed. Answer: FALSE Explanation: By definition, a constitutive gene is continually expressed.
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30) The enzyme transacetylase is responsible for uptake of lactose. Answer: FALSE Explanation: Permease, encoded by the lacY gene, plays this role.
Skill: Factual recall
31) Mutations of lacOC lead to attenuated expression of lac structural genes. Answer: FALSE Explanation: Mutations of lacOC lead to constitutive expression.
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32) lacIS encodes superrepressors, repressors that always bind to the operator and prevent gene transcription. Answer: TRUE
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33) The purpose of regulation of the trp operon in bacteria is to ensure the cell can digest tryptophan if it is present and the cell is short on energy. Answer: FALSE Explanation: The purpose of regulation of the trp operon is only to make the gene products necessary for the synthesis of tryptophan if there is not enough tryptophan present in the cell, as it is an essential amino acid.
Skill: Conceptual understanding
34) The use of CAP‐ cAMP complex as a positive regulator is specific to the lac operon. Answer: FALSE Explanation: CAP protein and its need for cAMP are part of a general catobolite repression mechanism. Bacterial cells that have plenty of their preferred carbon source (glucose) limit the expression of genes used for alternative, non‐preferred energy sources.
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35) The tryptophan codons of the attenuator sequence help speed up transcription of downstream trp genes at high concentrations of tryptophan. Answer: FALSE Explanation: These codons are readily read by trp‐tRNA at high tryptophan concentrations, leading to the attenuator formation that terminates transcription.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) How are partial diploid E. coli constructs made? Answer: The partial diploids are a special strain of bacteria that contain an engineered plasmid (the F factor) with a set of lac operon genes. This second set makes the cell ʺdiploidʺ with respect to these genes. Partial diploids are strains of bacteria that contain an Fʹ plasmid that bears a few chromosomal genes. These chromosomal genes in the Fʹ plasmid do not necessarily have to be lac operon genes.
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 151
37) Distinguish between cis‐ and trans‐dominance. Answer: Cis‐dominant genes affect upstream or downstream genes on the same chromosome, while trans‐dominant genes can affect expression of genes located on another chromosome.
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38) For each of the following lac operon mutations, state whether transcription of β‐galactosidase will increase, decrease, or be unaffected. a. Placb. lacY+ c. lacId. lacOe. lacAAnswer: (a) decrease. (b) initially unaffected. (c) increase. (d) increase. (e) unaffected.
Skill: Problem-solving
39) If a cell has the partial diploid genotype
lacI+ lacO+ lacZ- lacY+ _______________________ lacI- l acO+ lacZ+ lacY- ,
how will its gene expression in the presence of allolactose? Answer: The lacI+ gene is dominant over lacI- , and the operator is functional. Similarly, the wild‐type (+) versions of the lacZ and lacY genes are dominant. So, this construct should function normally in the presence of allolactose.
Skill: Problem-solving
40) If a cell has the partial diploid genotype
lacI+ lacO- lacZ+ lacY+ ______________________ lacI- lacO- lacZ+ lacY- ,
how will its gene expression function in the presence of allolactose? Answer: In this construct, lacI+ gene is dominant over lacI- , which means repressor protein is produced. Both copies of the operator are defective, however, and the repressor cannot bind to it. The structural genes will therefore be constitutively produced.
Skill: Problem-solving
41) Briefly define what is meant by negative and positive control of the lac operon. Answer: In negative control, the repressor protein blocks lac gene transcription in the absence of the inducer. In positive control, the catabolite activator protein complexed with cAMP binds to the CAP site in the lac promoter, facilitating the transcription of lac genes.
Skill: Conceptual understanding
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42) Compare the action of repressor proteins in the lac operon and the trp operon. Answer: In both cases, the active repressor binds to the operator and prevents transcription of the operon. In the lac operon, the repressor is inactivated by allolactose, an isomer of lactose that functions as an inducer. Allolactose is only made when lactose is present; therefore, the operon makes the gene products for lactose catabolism only when lactose is present. On the other hand, the trp repressor is active only when bound to tryptophan. Thus, the gene product for the biosynthesis of tryptophan is not made when sufficient quantities of the amino acid are already present.
Skill: Factual recall
43) Why do lacOC mutants express in cis‐dominant and not in transdominant fashion? Answer: Because in this mutant, the repressor cannot bind to the lac operator, preventing the downstream lac structural genes from being transcribed. Thus this mutant has no bearing on the functionality of another lac operon in a partial diploid construct.
Skill: Conceptual understanding
44) What selective pressures are likely to have led to the evolution of regulatory systems like operons? Why not simply express these genes continuously? Answer: Continual production of unnecessary enzymes is energetically costly; presumably, this is the main selective pressure to evolve regulatory mechanisms that express genes only when needed.
Skill: Conceptual understanding
45) What effect would mutational damage of the following genes or gene regions have on the induction of an integrated lambda phage virus? a. damage to the cI gene b. damage to the cro gene c. damage to the xis gene Answer: (a) The cI gene encodes the lambda repressor, elimination of which permits transcription from the PL and PR promoters, leading to the lytic pathway. (b) Eliminating the Cro protein also permits transcription from the PL and PR promoters. (c) Damage to the enzyme excisionase would circumvent induction because this enzyme is necessary for excision from the bacterial chromosome.
Skill: Problem-solving
46) At the molecular level, how do proteins like repressors bind to DNA? How can binding affinities be increased or decreased? Answer: Protein‐nucleic acid interaction is mediated by hydrogen bonding, not covalent bonding. DNA is a negatively charged molecule, so binding occurs through positively charged amino acid residues in the protein. Binding affinity can be increased or decreased through amino acid substitutions that affect charge or shape.
Skill: Conceptual understanding
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 153
47) Give an overview of the mechanism by which ultraviolet radiation induces the lytic pathway in lambda phage. Answer: The DNA damage that accompanies UV irradiation may include damage to the recA gene, causing it to code for a modified protein which stimulates the lambda repressor polypeptides to cleave, inactivating them. Lack of repressor allows RNA polymerase to bind to the PR promoter, transcribing cro. Cro protein reduces synthesis of cII, which regulates lambda repressor synthesis. Eventually, enough Q proteins accumulate from the transcription from the PR promoter to initiate the lytic pathway.
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48) Should viruses like lambda phage be considered living entities? Or are they more properly viewed as parasitic fragments of DNA? Answer: The answer depends on the definition of life; viruses are not alive under the cellular definition of life, which holds that living organisms are composed of one or more cells. They do, however, evolve by natural selection, the main criterion for life as suggested by geneticist H. J. Muller.
Skill: Conceptual understanding
49) Contrast negative (repressor) and positive (CAP) control in the lac operon. Answer: Under negative control, transcription of the structural genes is shut down by a repressor protein until inactivated by the inducer (allolactose). Under positive control, transcription of the structural genes is shut down by the inducer (CAP).
Skill: Conceptual understanding
50) Microbial genomics is a rapidly growing field. Of great interest to geneticists is the search for new operons as new microbial genome sequences become available. What kind of ʺsignatureʺ or identifying feature would you look for as you searched new genomes for candidate operons? Answer: Operons are sets of genes under common regulatory control, with promoter and operator regions situated adjacent to structural gene regions. Those regions that are conserved (i.e., occur with similar structure) across multiple genomes are good candidate operons for research and can often be identified via computer‐based searches.
Skill: Problem-solving
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Chapter 18 Regulation of Gene Expression in Eukaryotes
MATCHING QUESTIONS Please select the best match for each term. 1) Alternative polyadenylation
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A) Controls transcription initiation B) Produces more than one protein from one gene C) A method of targeting proteins for degradation D) Gene regulation dependant on parent of origin E) A method for targeting RNAs for degradation
2) Genomic imprinting
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3) Enhancer
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4) Ubiquitination
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5) siRNA
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Answers:
1) B
2) D
3) A
4) C
5) E
MULTIPLE-CHOICE QUESTIONS 6) A similarity between regulation of the lac operon in bacteria and the regulation of galactose utilization in yeast is that both pathways A) transcribe all the genes necessary as one polycistronic mRNA. B) use feedback inhibition to stop making their sugars. C) use a modified version of their sugars as inducers. D) are subject to catabolite repression. E) C and D only Answer: E
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7) When a promoter element is bound by a positive regulatory protein, the result is A) activation of replication. B) activation of transcription. C) activation of translation. D) repression of replication. E) repression of transcription. Answer: B
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Chapter 18 Regulation of Gene Expression in Eukaryotes 155
8) Unlike bacterial operons, eukaryotic operons A) are all positively controlled. B) are all negatively controlled. C) are restricted to certain eukaryotic kingdoms. D) are exceedingly rare. E) None of these Answer: D
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9) Hormones can be considered A) inducers. B) repressors. C) transcription regulators. D) translation regulators. E) enzymes. Answer: C
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10) Unlike prokaryotes, most gene regulation in eukaryotes A) takes place at the transcriptional level. B) is controlled at the level of transcript processing. C) is controlled by inhibitory proteins. D) is based on posttranslational modification. E) None of these Answer: B
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11) Which of the following secondary structures may be a DNA‐binding domain? A) Zinc finger B) Leucine zipper C) Helix‐turn‐helix D) A and B only E) A, B, and C Answer: E
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12) In eukaryotic cells, steroid hormones A) bind to cytoplasmic receptors, are transported to the nucleus, bind DNA, and regulate gene expression. B) bind to cell surface receptors and send signals that result in regulation of genes in the nucleus. C) act as enhancers and bind to an assortment of activators and repressors. D) methylate DNA to regulate gene expression. E) cause maternal effect. Answer: A
Skill: Factual recall
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13) RNA silencing involves A) miRNAs. B) siRNAs. C) double‐stranded RNA. D) a protein called Dicer. E) All of these Answer: E
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14) HREs are A) hormone receptors. B) DNA sequences in promoter regions to which steroid hormone receptors bind. C) proteins that interact with histones. D) elements that lead to histone deacetylation. E) None of these Answer: B
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15) Chromosome‐level gene repression involves A) supercoiling. B) physical blockage of gene regions. C) chromatin formation. D) nucleosomes. E) All of these Answer: E
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16) Change in the DNA‐histone complex that can increase or decrease transcriptional activity is termed A) induction. B) gene regulation. C) chromatin remodeling. D) nucleosome formation. E) None of these Answer: C
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17) The protein class(es) involved in the activation of transcription is (are) A) general transcription factors (GTFs). B) transactivators. C) coactivators. D) A and B only E) A, B, and C Answer: E
Skill: Factual recall
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Chapter 18 Regulation of Gene Expression in Eukaryotes 157
18) Control of transcription initiation of eukaryotic protein‐coding genes involves A) enhancers. B) promoters. C) repressors. D) activators. E) All of these Answer: E
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19) Activators in eukaryotes A) bind to enhancers. B) may act as monomers C) may act as homodimers. D) may act as heterodimers with a DNA binding subunit and an activation subunit. E) All of these Answer: E
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20) The half‐life of a protein is directly related to A) what kind of cell it is produced in. B) its tertiary structure. C) whether it is encoded by a constitutively expressed gene. D) its N‐terminal amino acid residue. E) the number of stabilizing cross‐linkages. Answer: D
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21) Removal of the 5ʹ G‐cap of mRNA transcripts A) prevents transcript transport. B) leads to degradation by exonucleases. C) extends mRNA half‐life. D) leads to polyadelylation. E) reduces mRNA half‐life. Answer: B
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22) mRNA regulation can be achieved at the level of A) transport. B) processing. C) translation. D) life span (degradation). E) All of these Answer: E
Skill: Factual recall
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23) With respect to genetics, URS stands for A) uracil response system. B) upstream regulatory sequence. C) upstream repressor‐binding sequence. D) unregulated repression series. E) uniform response system. Answer: B
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24) Proteins that are encoded by the same gene but differ in structure and function are the product of alternative splicing in pre‐mRNA processing. Such protein variants are termed A) isoteins. B) heteromeric proteins. C) isozymes. D) protein isoforms. E) allelomorphs. Answer: D
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25) Heterochromatin is associated with A) differential gene expression. B) gene silencing. C) mRNA translation. D) heterozygotes. E) A, B, and C only Answer: B
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TRUE-FALSE QUESTIONS 26) Transcriptionally active genes show lower levels of DNA methylation compared with transcriptionally inactive genes. Answer: TRUE
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27) Gene silencing in eukaryotes is often achieved through chromatin structure. Answer: TRUE
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28) Eukaryotic repression generally occurs via physical blockage of the promoter. Answer: FALSE Explanation: In eukaryotes, repressors generally interfere with activators, preventing them from initiating transcription.
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29) Longer poly(A) tails on mRNAs are associated with less translational activity. Answer: FALSE Explanation: Shorter poly(A) tails are associated with less translational activity.
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Chapter 18 Regulation of Gene Expression in Eukaryotes 159
30) Recombinases mediate transcript processing from pre‐mRNA to mature mRNA. Answer: FALSE Explanation: The spliceosome mediates transcript processing.
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31) Steriod hormones are effector molecules. Answer: TRUE
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32) Steroid and peptide hormone receptors are typically cytoplasmic. Answer: FALSE Explanation: Steroid hormone receptors are cytoplasmic, while peptide hormone receptors are on the cell surface, in the cell membrane.
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33) Imprinting is implicated in Prader‐Willi and Angelman syndromes. Answer: TRUE
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34) Gene silencing is a form of regulation controlled by repressor proteins. Answer: FALSE Explanation: Gene silencing is a form of expression regulation either based on packaging (heterochromatin), which usually affects large numbers of genes at once, or by RNAi, which is often gene specific.
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35) Noncoding RNAs called mRNAs are present in most eukaryote genomes and are involved in negatively regulating gene expression through degradation of mRNAs with complementary sequences. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Describe how siRNA functions in regulation of expression and how this can be used as a tool in genetic research. Answer: ʺSmall interferingʺ RNA can complex with mRNA with complementary base sequences, preventing transcription. This makes siRNAs valuable tools, as they can be synthesized to experimentally block the expression of genes in specific tissue types with very precise timing, helping geneticists determine gene function and interaction.
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37) Some genes that are expressed in the fetal stage are never expressed again in that individualʹs life. What do you think is the most common mechanism for preventing the expression of such genes? Answer: Such genes are permanently silenced, typically via heterochromatic packaging. This is more efficient than relying on repression.
Skill: Conceptual understanding
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38) What essential difference between the growth and reproductive strategies of prokaryotes and multicellular eukaryotes explains why the latter have far more sophisticated regulatory mechanisms? Answer: An important point made in this chapter is that while prokaryotes grow and divide, multicellular eukaryotes develop and differentiate. Tissue differentiation and the homeostatic maintenance of the multicellular body require a diversity of flexible, precise regulatory mechanisms.
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39) Describe the use of the DNA‐degrading enzyme DNase I in experiments that explore the effect of chromatin on gene inactivation. Answer: DNA from cells with certain transcriptionally active vs. inactive genes can be isolated and treated with DNase I; then the condition of the genes are assessed with a Southern blot. Inactive, sequestered genes should remain intact, while active, exposed genes should have been degraded by the enzyme.
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40) Explain how combinatorial gene regulation permits regulation of a large diversity of genes with relatively few regulatory proteins. Answer: Although the number of different regulatory proteins is limited, in combination they can interact with a far greater number of genes with specificity.
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41) Embryonic development is orchestrated by a cascading series of genes, the expression of many of which are controlled in time and space by transcription factors and effector molecules of various kinds. But what controls the expression of the first regulatory genes to be transcribed in the newly fertilized zygote? Answer: The egg contains stored mRNAs as well as transcription factors derived from the mother, and both of these orchestrate the expression of the first regulatory genes.
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42) What would happen developmentally if a zygote with XY sex chromosome karyotype had a defective testosterone receptor? Answer: We learned in this chapter that testosterone is necessary for initiating the male developmental pathway; an XY individual lacking this receptor would develop as a female.
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43) What are enhancers? Answer: Enhancers are DNA sequences that typically have binding sites for several different activators and repressors. They are located somewhere in the vicinity of the gene, much farther away than promoter sequences, sometimes tens of kilobases away. They allow complex and nuanced control of gene expression depending on a number of different conditions such as developmental stage, tissue type, and response to extracellular signals.
Skill: Conceptual understanding
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Chapter 18 Regulation of Gene Expression in Eukaryotes 161
44) Plants produce steroids called phytosterols, which have various functions in plant physiology. Some phytosterols are thought to have a defensive function, however, specifically in repelling insect herbivores. How might plant‐derived steroids help protect plants from insect attack? Answer: In some cases, the plant‐derived steroids are sufficiently similar to insect steroid hormones that they can wreak havoc with insect physiological functioning and development when ingested. Debilitating or killing insect herbivores reduces tissue loss to the plant.
Skill: Conceptual understanding
45) Insect growth and metamorphosis is controlled by the steroid hormones ecdysone and juvenile hormone (JH). In insects like butterflies and moths that have a distinct larval (juvenile) and adult stage, the relative amount of JH determines whether the insect molts to the next juvenile stage or switches to become an adult. Can you think of a way to use this hormonal system to control caterpillar agricultural pests? Answer: The best approach would be to somehow suppress JH early in caterpillar development, forcing them to mature early. This would reduce the duration of the most destructive stage (the leaf‐eating caterpillar). The opposite treatment (extending the caterpillar stage) would have dire consequences, because the quantity of leaf biomass these insects consume increases geometrically with each larval stage.
Skill: Conceptual understanding
46) Interference RNA technology is one of the most exciting new developments in genetics in recent years. In April 2004, Cancer Research UK and the Netherlands Cancer Institute announced creation of a 24,000‐molecule iRNA library, designed to inactivate about 8,000 human genes. One hope for this library is its application in combating cancer. How might iRNA be implemented in cancer treatment? Answer: Cancers for which the genetic basis is understood might be treated by preventing gene expression through iRNA‐mediated gene silencing. Use of this technology may depend on the type of cancer and its localization.
Skill: Conceptual understanding
47) In genetic imprinting, expression depends on the parent of origin of a given allele. How might imprinting affect the expression of a hypothetical autosomal disease that expresses in dominant fashion and recessive fashion, and how might imprinting be detected? Answer: In individuals heterozygous for the dominant allele, expression of a disease phenotype depends on which allele is silenced–and thus which parent donated the disease allele. This will not matter in homozygous recessive individuals with respect to recessively expressing disease alleles. In the former case, imprinting could be detected through analysis of enough pedigrees with known male vs. female heterozygotes and known progeny phenotypes. This would reveal the influence of parent of origin on allele expression.
Skill: Problem-solving
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48) Evolutionary biologists suspect that genetic imprinting evolved through parent‐offspring conflict. What kinds of ʺconflict of interestʺ in progeny development can you imagine between mammalian parents? Answer: Fetal development requires enormous energy investment, and in many species the degree of pre‐ and postnatal investment directly influences future maternal reproduction. From the point of view of the motherʹs evolutionary fitness, there should be a balance between present and future investment. In terms of the fatherʹs evolutionary fitness, however, his unborn offspring should take as much energy from the mother as possible to safeguard his present investment, since his future investments may not necessarily be with the same mother. Imprinting may have evolved as a result of conflict between the maternal and paternal genomes. (See Reik and Walter. 2001. Nat Genet. 27:255‐256; PMID 11242103.)
Skill: Problem-solving
49) What are some advantages and disadvantages of arranging gene clusters into operons? Answer: Operons can be advantageous in that regulation of expression can be controlled by a single regulatory region, with the whole set up‐ or down‐regulated. In prokaryotes, induced operons can respond to needs quickly by transcribing a single polycistronic mRNA. Some of these features can be disadvantageous as well; control by a single regulatory region also means greater vulnerability to mutation. A mutant operator or promoter, for example, might render all genes in the operon dysfunctional.
Skill: Conceptual understanding
50) Several common protein structural motifs are now known to be involved in DNA recognition and binding. What are these, and how would researchers use genomic data to help figure out if new candidate genes encode such DNA‐binding proteins? Answer: DNA‐binding structural motifs of proteins include zinc and leucine zippers and the helix‐turn‐helix (HTH) motif. Researchers can electronically scan the sequence of new genes for the particular nucleotide sequence underlying such motifs, thereby identifying them as candidate DNA recognition or binding proteins.
Skill: Factual recall, conceptual understanding
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Chapter 19 Genetic Analysis of Development
MATCHING QUESTIONS Please select the best match for each term. 1) Differentiation
Skill: Factual recall, conceptual understanding
A) Genes that determine the number of segments in Drosophila B) Process through which vaccines increase production of specific antibodies C) The specialization of cells during development D) Formation of limbs and organs during development E) Tissue patches in larvae that will develop into adult structures
2) Morphogenesis
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3) Imaginal discs
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4) Clonal selection
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5) Pair-rule genes
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Answers:
1) C
2) D
3) E
4) B
5) A
MULTIPLE-CHOICE QUESTIONS 6) ________ cells may differentiate into any cell or tissue type. A) Pluripotent B) Omnipotent C) Totipotent D) Hyperpotent E) Potent Answer: C
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7) Which of the following cell types has the lowest developmental potential? A) Oocyte B) Red blood cell C) Embryonic stem cell D) Zygote E) All have equal developmental potential. Answer: B
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8) The developmental fate of a cell is established by the process termed A) development. B) differentiation. C) determination. D) specialization. E) A, B, and C only Answer: C
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9) ________ is the process of cell or tissue change in development. A) Transformation B) Morphogenesis C) Development D) Change E) None of these Answer: B
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10) Which of the following correctly expresses the order of development? A) Fertilization, development, maturity B) Induction, determination, differentiation C) Induction, differentiation, determination D) Fertilization, differentiation, determination E) None of these Answer: B
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11) Which of the following statements is true? A) Sex determination in fruit flies depends on the X : autosome ratio. B) Mammalian sex determination can be influenced by temperature. C) Mammals experience random Y inactivation. D) The testis‐determining factor gene is X‐linked. E) Dosage compensation occurs in the same way in most animals. Answer: A
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12) The X‐inactive specific transcripts gene is designated A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: E
Skill: Factual recall
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Chapter 19 Genetic Analysis of Development 165
13) The best model organism for studying cell fate from fertilized egg through adult is A) Arabidopsis thaliana. B) Caenorhabditis elegans. C) Drosophila melanogaster. D) Mus musculus. E) Saccharomyces cerevisiae. Answer: B
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14) Which of the following is not true of X‐inactivation? A) It is necessary for dosage compensation. B) It is the cause of calico color in cats. C) The mechanism includes transcription of the XIST gene and chromatin remodeling. D) The XIST RNA is non‐coding. E) The master regulatory gene for X‐inactivation is sxl or sex‐lethal. Answer: E
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15) In Drosophila, sex determination is carried out by A) a cascade of alternative splicing. B) the SRY gene. C) histone modification. D) the DMRT1 gene. E) All of these Answer: A
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16) Antibody genes A) contain both constant and variable regions. B) are formed by cutting and splicing DNA in B cells during development. C) are made from two heavy and two light chains, each of which has a variable region. D) each recognize one antigen and are the only antibody made by their B cell. E) All of these Answer: E
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17) The X‐inactivation center gene is A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: B
Skill: Factual recall
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18) SRY is A) a single‐region Y gene. B) a Y‐inactivation center gene. C) the sex‐determining region of the Y chromosome. D) a Y‐controlling element. E) None of these Answer: C
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19) The X‐controlling element gene is A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: C
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20) Homeobox genes A) encode proteins with a conserved sequence of amino acids. B) are important in development. C) are conserved in multicellular animals. D) are transcription factors. E) All of these Answer: E
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21) Mammalian sex determination is controlled by A) XIST. B) SRY. C) XIC. D) All of these E) None of these Answer: A
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22) The central function of X-inactivation is A) silencing imprinted genes. B) enhancing the expression of genes on the chromosome left active. C) dosage compensation. D) silencing deleterious alleles. E) None of these Answer: C
Skill: Factual recall
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The following scenario applies to the questions below. In the snail Limnaea peregra, shell coiling is determined by the nuclear alleles S (dextral) and s (sinistral). This gene is a maternal effect gene that influences egg formation, which in turn influences whether the snail that hatches from the egg has a shell that coils to the right (dextral) or left (sinistral). A female sinistral snail is mated with a sinistral male snail. The genotypes of both are unknown. All the F 1 progeny are dextral. When the members of the F 1 generation are crossed, they yield an F 2 generation that is dextral and sinistral.
4 4 3 1
23) What was the genotype of the mother of the female snail? A) SS B) Ss C) ss D) Ss or ss E) SS or Ss Answer: C
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24) What were the genotypes of the sinistral snails? A) ss female and Ss male B) Ss female and Ss male C) SS female and ss male D) Ss female and ss male E) ss female and SS male Answer: B
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25) The ________ is a conserved 180‐bp sequence found in homeotic genes. A) TATA box B) bicoid box C) Hox box D) homeobox E) Pribnow box Answer: D
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TRUE-FALSE QUESTIONS 26) The developmental trajectory of the cells of an embryo can be tracked and diagrammatically expressed as a developmental atlas. Answer: FALSE Explanation: The correct term for the diagrammatical expression of this information is fate map.
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27) Determination occurs via induction. Answer: TRUE
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28) Antibodies elicit the production of antigens. Answer: FALSE Explanation: Antigens elicit antibody production.
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29) Development occurs through the activation and inactivation of a series of specific genes in a precise sequence. Answer: TRUE
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30) Morphogens are substances that help control development. Answer: TRUE
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31) Antennapedia and bithorax belong to the class of mutants referred to as atavistic mutants. Answer: FALSE Explanation: Both antennapedia and bithorax are homeotic mutants.
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32) Retention of the full genomic structure during cell differentiation is termed DNA constancy. Answer: TRUE
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33) The major classes of antibody molecules are called antigen‐globulins. Answer: FALSE Explanation: Antibody molecules are immunoglobulins, of which there are five classes in humans.
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34) T and B cells are so named for the types of antigens they attack. Answer: FALSE Explanation: They are named for the site of production: T cells are produced in the thymus, B cells in the bone marrow.
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35) No human cells lose all of their DNA during differentiation. Answer: FALSE Explanation: Red blood cells (erythrocytes) are anucleate; that is, they have no nucleus and thus no DNA.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Why are the changes associated with development irreversible? Answer: Development is a programmed sequence of cellular‐level phenotypic events, a cascade in which activation of certain genes in turn sets into motion the activation of other genes. The changes set in motion involve morphology and physiology –phenotypic changes that are too complex to reverse.
Skill: Analytical reasoning
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Chapter 19 Genetic Analysis of Development 169
37) Vascular plant cells generally have the ability to differentiate into various types of tissue–a fact long known to horticulturists who propagate plants by cuttings. This means that plants do not display the strict separation of germline and somatic cells that animals do. Might this be expected to accelerate or decelerate evolution in plants relative to animals? Answer: Since somatic cell lines can become germline in plants (as when vegetative cells give rise to germ‐bearing flower tissue), somatic mutations can become heritable. Opportunities for the generation and propagation of heritable genetic variants would be greater in plants as a result. One consequence of this could be more rapid evolution in some cases, but a host of other factors play a role in setting evolutionary rates.
Skill: Analytical reasoning
38) If cell and tissue differentiation is controlled genetically, how can environmental factors play a role in determining the final phenotype? Answer: Environmental factors may influence the timing and degree of expression of developmental genes, thereby influencing the phenotype.
Skill: Analytical reasoning
39) The use of embryonic stem cells in federally funded research in the United States is a hotly debated topic. What are the arguments for and against use of such stem cell lines in research and treatment of disease? Answer: The arguments in favor of this research primarily focus on the research and therapeutic gains that may be made through their use–in particular, the promise of using stem cells to treat diseases that involve deterioration of certain cell types. Arguments opposed to this research largely stem from opposition to one source of stem cells: human embryos.
Skill: Conceptual understanding
40) A maternal effect mutation bicoid (bcd) is recessive. The bcd allele makes no Bicoid protein. In the absence of the Bicoid protein, product embryogenesis is not completed and eggs fail to hatch into viable larva. Consider a cross between a female who is heterozygous ( bcd+ /bcd) and a homozygous mutant male (bcd/bcd). a. How is it possible for a male homozygous for the mutation to exist? b. Predict the outcome (normal and/or failed embryogenesis) in the F 1 and F2 generations resulting from this cross. Answer: (a) The bcd/bcd male must have had a heterozygous mother. (b) The first generation offspring will all hatch normally, although half of them will be bcd/bcd. In the F2 generation, half the eggs will fail to produce larva. An F 1 female who is bcd/bcd will produce no live offspring, but the bcd+ /bcd females will produce all live offspring.
Skill: Problem-solving
41) Recount the evidence that Dolly the sheep was indeed the product of cloning, or cell fusion. Answer: Dollyʹs phenotype matched that of her nuclear mother, rather than that of the recipient mother that carried her to term. Genetic analysis (DNA fingerprinting) confirmed that Dolly matched the genetic signature of the donorʹs somatic cell type rather than the recipient motherʹs.
Skill: Factual recall
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42) Can you think of a specific example in which multiple genome datasets would be useful in the study of morphogenesis? Answer: Datasets for phenotypically variable, or plastic, organisms would be especially valuable. A good candidate group for study would be domesticated dogs, characterized by many and varied breeds.
Skill: Conceptual understanding
43) The cells of triploid animals, like certain salamanders, are individually larger, yet their overall body size is no different from that of diploids. Pose a developmental hypothesis for the control of morphogenesis in such animals. Answer: The increase in cell size with no concomitant increase in body size means that the triploids produce fewer cells for the same body size. Intercellular signals are important in morphogenesis, so perhaps size and shape of tissues and organs is sensed and communicated by chemical signals generated, say, when certain cell types come in contact. A certain size or dimension would be reached sooner with larger cells, triggering the next set of developmental genes and preserving overall body dimensions.
Skill: Problem-solving
44) Contrast the mechanisms of dosage compensation in humans with those found in insects like Drosophila. Answer: The mechanisms of dosage compensation found in humans and fruit flies represent two different solutions to the same problem. In both cases, X-linked genes are present in twice the copy number in XX females as in XY or XO males, potentially creating a dosage imbalance. In mammals like humans, this is solved through X-inactivation, while in insects like Drosophila it is solved through X-hyperactivation. Both act to restore X-expression parity between females and males.
Skill: Conceptual understanding
45) Mutations like bithorax, in which a second pair of wings develops in the place of halteres, give clues to the evolutionary origin of differentiated structures. What inferences might you draw from bithorax, or antennapedia? Answer: These otherwise different structures have a similar genetic basis, suggesting that they are likely to have a common origin (gene duplication and divergence, for example).
Skill: Analytical reasoning
46) In what ways does the cloned cat Cc demonstrate the limitations of cloning technology in obtaining desired phenotypes? Answer: Cc differs in several respects from her somatic cell‐donor mother despite being genetically identical. Her coat color differs, as does her shape and personality. These differences stem from various complicating factors, including patterns of X‐inactivation and environmental contributions to certain quantitative traits.
Skill: Factual recall
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47) Cloned animals are typically not entirely genetically identical to their cloned parent. Why? Answer: Cloned animals are usually made by transplanting a somatic cell nucleus into an enucleated oocyte from a donor. Unless the donor cell is from the same maternal lineage as the somatic cell, the mitochondria and the mitochondrial DNA will be that of the egg donor, not the clone parent.
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48) Is there a relationship between the problems encountered cloning mammals and the rarity of polyploidy in this group? Answer: Possibly so; mammals, unlike some animal groups, appear to be extremely sensitive to gene dosage and even small perturbations in developmental pathways.
Skill: Analytical reasoning
49) Your body is likely to have existing antibodies that will counter antigens you have yet to contact. How is this accomplished? Answer: Differential splicing of numerous gene elements (V, J, C, K, and D segments) underlies the tremendous antibody diversity of animals. With distinct molecules ranging to 108 , chances are good that at least one of these molecules will recognize a novel antigen.
Skill: Factual recall
50) As of 2008, human cloning was banned in the United States. What are the arguments for maintaining such a ban? Answer: Human cloning is a hugely complex ethical issue. For the time being, the extreme inefficiency and high developmental error rate associated with mammalian cloning are sufficient to justify a ban.
Skill: Conceptual understanding
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Chapter 20 Genetics of Cancer
MATCHING QUESTIONS Please select the best match for each term. 1) CDK
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A) Genes that produce abnormally active cell growth proteins B) Phosphorylate proteins important to the cell cycle C) Normal growth-promoting genes D) Proteins that can stop the cell cycle E) Are made and degraded at specific points in the cell cycle 2) E 3) A 4) C 5) D
2) Cyclins
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3) Oncogenes
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4) Proto-oncogenes
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5) Tumor suppressors
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Answers:
1) B
MULTIPLE-CHOICE QUESTIONS 6) Hereditary retinoblastoma is an autosomal dominant hereditary cancer. Cells from retinal tumors in a child who has this disease possess A) evidence of non‐disjunction in all autosomes. B) one non‐functional copy of the retinoblastoma gene. C) two non‐functional copies of the retinoblastoma gene. D) trisomy for the retinblastoma gene. E) a dominant mutation in the retinoblastoma gene. Answer: C
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7) Apoptosis is A) rapid cell division. B) caused by proto‐oncogenes. C) the change in shape of a cell when it becomes cancerous. D) programmed cell death. E) None of these Answer: D
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172
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Chapter 20 Genetics of Cancer 173
8) Protein kinases are enzymes that normally catalyze the A) degradation of cellular proteins. B) phosphorylation of cellular proteins. C) formation of peptide bonds in proteins. D) phosphorylation of ADP. E) synthesis of kinins. Answer: B
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9) A proto‐oncogene that gains function due to amplification, translocation to a more transcriptionally active area of the genome, or a mutation that causes either an increase in function or the suppression of down‐regulation, is A) a pseudo‐oncogene. B) recessive. C) dominant. D) an oncogene. E) C and D only Answer: E
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10) Metastatic cancer is A) terminally differentiated. B) malignant. C) invasive. D) B and C only E) A, B, and C Answer: D
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11) Retroviruses are responsible for which of the following conditions? A) Rous sarcoma B) AIDS C) Feline leukemia D) Mouse mammary tumors E) All of these Answer: E
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12) Mutations in a normal growth‐stimulating gene are most likely to have no effect or to cause A) the loss of responsiveness to growth stimulation in the cell with the mutant gene. B) the cell with the mutant gene to become cancerous. C) excessive growth in the cells surrounding the one with the mutant gene. D) a mutator effect in the cell with the mutant gene. E) apoptosis of the cell with the mutant gene. Answer: A
Skill: Conceptual understanding
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13) Small DNA tumor viruses such as SV40 and HPV have viral oncogenes that are not homologs of host cell genes. Some of these novel viral oncogenes make proteins that exert their tumorigenic effect by A) binding to and inactivating pRB. B) making Gag, Pol and Env proteins more active. C) causing cellular proto‐oncogenes to mutate into oncogenes. D) causing apoptosis in target tissues. E) inactivating the immune system. Answer: A
Skill: Conceptual understanding
14) ________ play a pivotal role in programmed cell death. A) Cyclins B) Telomeres C) Proto‐oncogenes D) Oncogenes E) Telomerases Answer: A
Skill: Factual recall
15) In sporadic (nonhereditary) cancers, mutations occur in ________ cells. A) germ B) somatic C) undifferentiated D) mammary E) T Answer: B
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16) Cancer is best defined as A) immune dysfunction. B) uncontrolled and abnormal cell division. C) viral malignancy. D) regulated differentiation of tissue. E) cellular deregulation. Answer: B
Skill: Factual recall
17) Agents that can induce cancer through mutagenesis are called A) destabilizers. B) cancerogens. C) mutagens. D) carcinogens. E) None of these Answer: D
Skill: Factual recall
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Chapter 20 Genetics of Cancer 175
18) In retroviruses, the pol gene product is a(n) A) DNA polymerase. B) reverse transcriptase. C) integrase. D) RNA polymerase. E) recombinase. Answer: B
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19) Cancer may arise as a result of A) mutations associated with viral infection. B) mutations associated with chemical mutagens. C) mutations associated with radiation. D) spontaneous mutations. E) All of these Answer: E
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20) v‐onc genes are found in A) vital cells. B) viral cells. C) cancer‐causing retroviruses. D) very rapidly growing cells. E) bacteriophages. Answer: C
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21) Malignant breast, colon, and lung cancers typically have A) mutated tumor suppressor genes. B) active oncogenes. C) chromosomal abnormalities. D) activated telomerase. E) All of these Answer: E
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22) Which of the following statements is not true? A) Certain types of virus can cause cancer. B) Melanomas are commonly caused by ionizing radiation. C) Neoplasia is another term for cancerous cells. D) About half the people with cancer have a mutation in a p53 gene. E) Most cancers are hereditary. Answer: E
Skill: Factual recall
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23) Which of the following is not likely to be a tumor suppressor? A) A growth factor receptor B) A checkpoint protein C) A DNA repair enzyme D) An apoptosis‐promoting factor E) The negative regulator of a CDK Answer: A
Skill: Conceptual understanding
24) Cell proliferation in culture is normally limited by A) cytophagy. B) quorum sensing. C) contact inhibition. D) competition. E) nutrient limitation. Answer: C
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25) In normal cells, p53 protein acts as A) a signal transducer. B) a tumor suppressor. C) a proto‐oncogene activator. D) a GTPase inhibitor. E) None of these Answer: E
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TRUE-FALSE QUESTIONS 26) Ionizing radiation is emitted at low levels by many natural objects, including some rocks and gases. Answer: TRUE
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27) The products of tumor suppressor genes stimulate cell proliferation, while the products of proto‐oncogenes inhibit cell proliferation. Answer: FALSE Explanation: The statement is correct the other way around: proto‐oncogene products stimulate cell proliferation, while those of tumor suppressor genes inhibit cell proliferation.
Skill: Factual recall
28) The nucleic acid of retroviral provirus is composed of RNA. Answer: FALSE Explanation: Although viral nucleic acid is composed of RNA, that of the provirus is composed of DNA.
Skill: Factual recall
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Chapter 20 Genetics of Cancer 177
29) Some carcinogens act on the genome directly, while others are converted to mutagenic substances by the cellsʹs enzymes. Answer: TRUE
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30) Terminally differentiated cells are noncancerous, normal cells. Answer: TRUE
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31) The process of relaying a growth‐stimulatory or growth‐inhibitory signal in response to an extracellular factor binding at the cell surface is called intercellular signaling. Answer: FALSE Explanation: This process is called signal transduction.
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32) Wild‐type mutator genes produce substances that induce point mutations and chromosomal rearrangements. Answer: FALSE Explanation: Wild‐type mutator gene products function in DNA replication and repair.
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33) Inherited mutant tumor suppressors produce a dominant susceptibility to cancer, but the mutant alleles are actually recessive in the cancer. Answer: TRUE
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34) The only genes necessary for the retroviral lifecycle are gag and pol. Answer: FALSE Explanation: In addition to these, the env gene is also necessary.
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35) Most cancer deaths in the United States are caused by radiation‐induced cancer. Answer: FALSE Explanation: Random mutations due to metabolic processes and chemical carcinogens, including natural and synthetic chemicals, are responsible for most cancer deaths in the United States.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) List the ways in which proto‐oncogenes may be converted to oncogenes. Answer: The mechanisms include point mutations, deletions, and gene amplification.
Skill: Analytical reasoning
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178 Test Bank for iGenetics
37) The terms proto‐oncogene and oncogene may give the misleading impression that these are ʺgenes for cancerʺ that are waiting to be turned on. What are these genes, in terms of normal function, and what name might have been given to them to reflect this function rather than their role in cancer? Answer: Proto‐oncogene products play a role in cell growth or the cell cycle. Examples discussed in this chapter include growth factors, protein kinases, and membrane‐associated G proteins. Had they not been discovered in defective form as oncogenes, they would be named more precisely for their function.
Skill: Analytical reasoning
38) Programmed cell death, or apoptosis, is a necessary and useful property of cells. Why is this so? Answer: It is useful for multicellular organisms to have the ability to have some cells selectively self‐destruct. Cells that have become irreparably genetically damaged, are infected by viruses, or have completed their developmental role may pose a danger to the organism if they live.
Skill: Conceptual understanding
39) How does telomerase, which is not a cause of cancer, nonetheless play a role in its development? Answer: The enzyme telomerase helps cancerous cells continue to divide by lengthening chromosomal telomeres.
Skill: Factual recall
40) Describe Knudsonʹs two‐mutation model: When and where are the mutations posited to occur? Answer: In Knudsonʹs model, one mutation is inherited and the second occurs in a somatic cell. The inherited mutation would be found in all cells, requiring only one more mutation in one of them to lead to cancer.
Skill: Factual recall
41) Reverse transcriptase, the enzyme responsible for copying viral RNA into cDNA prior to integration into the host genome, lacks 3ʹ‐to‐5ʹ exonuclease ʺproofreadingʺ ability. How is this advantageous to organisms that depend on such enzymes? Answer: Retroviruses depend on reverse transcriptase. Part of the successful infection and proliferation strategy of retroviruses like HIV is their high mutability. The errors made by the enzyme are useful in generating variable progeny viruses that can evade the hostʹs immune system.
Skill: Analytical reasoning
42) When oncologists (scientists who study cancer) refer to ʺtransformation,ʺ they mean something different from what the word ʺtransformationʺ signifies when used by biotechnologists. Explain the difference between the two uses of the term in genetics. Answer: In terms of cancer, cells that lose their growth control and no longer behave like normal cells are referred to as transformed. In the context of bacterial genetics and gene manipulation, however, as was discussed in Chapter 9 and Chapter 15, transformation refers to a living cell picking up naked DNA from outside the cell and incorporating it into its genome.
Skill: Conceptual understanding
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Chapter 20 Genetics of Cancer 179
43) People with the genetic disease xeroderma pigmentosum have compromised DNA repair mechanisms, making them extremely sensitive to genetic damage from sunlight. Without taking measures to minimize exposure, these individuals typically succumb to skin cancer in childhood or early adulthood–a demonstration of the mutagenicity of ultraviolet radiation. Yet, deliberate exposure in the form of sunbathing is very popular with some segments of the population. Why do you think this is so, and how might public health professionals raise awareness of the danger in which these people place themselves? Answer: The reasons for the popularity of sunbathing in some groups are probably numerous and complex, but generally relate to the wish to project a culturally conditioned image of health and vigor. This is ironic in that the practice actually causes genetic damage–decreasing health and vigor. Health professionals might try to combat this practice with advertising campaigns highlighting the effects of genetic damage, perhaps along the lines of the campaigns against cigarette smoking.
Skill: Analytical reasoning
44) The Bcl-2 gene product in its normal form can be activated to prevent cells from undergoing apoptosis (programmed cell death) in specific circumstances. It is a carefully regulated gene, being ʺturned onʺ only under specific circumstances. A null mutation in Bcl-2 leads to excessive cell loss. Some Bcl-2 mutation can cause a ʺgain of functionʺ in which the protein is active all the time. a. In the terms used in this chapter, what would the normal version be called? b. What would the gain‐of‐function mutation be called? c. How would you expect the gain‐of‐function mutation to be involved in cancer? Answer: (a) A proto‐oncogene. (b) An oncogene. (c) The gain‐of‐function mutation could cause abnormal cells to fail to commit suicide when their genome has been irreparably damaged. As a result, these cells could mutate into cancer cells.
Skill: Conceptual understanding
45) BRCA1 is a gene involved in repair of double‐stranded DNA breaks. Mutant forms of this gene are linked to a substantial proportion of familial breast cancers. A woman who inherits a certain allele of the gene has about a 60‐80% chance of getting breast cancer, as well as an elevated chance of getting ovarian cancer, in her lifetime. a. Why donʹt a higher percentage of women with the mutation get breast cancer? b. How would the breast cancer inheritance caused by BRCA1 be described in Mendelian terms? Answer: (a) BRCA1 is a tumor suppressor gene. Tumor suppressor genes generally conform to Knudsonʹs ʺtwo‐hitʺ mutational model of cancer. For a cell to become cancerous as a result of BRCA1 mutation both copies of the gene need to have become mutant and non‐functional. If a woman is born with one defective copy her odds of ending up with cells with two defective copies are greatly increased over that of a person whose cells started out with two good copies, but the chance is still less than 100%. (b) In Mendelian terms breast cancer caused by a hereditary mutation in the BRCA1 gene would appear to be dominant but with partial penetrance.
Skill: Conceptual understanding, analytical reasoning
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46) Li-Fraumeni syndrome is a rare inherited disease that greatly increases a personʹs risk of developing several types of cancer. This disease stems from a mutation in the TP53 tumor suppressor gene (Nichols et al. 2001. Cancer Epidemiol Biomarkers Prev 10:83‐87). In most cases, approximately 95% of the mutations can be detected by sequence analysis of exons 4 through 9. If you isolated and analyzed DNA from a Li -Fraumeni patient, what genotype should you find in DNA isolated from a. normal tissue? b. malignant tissue? Answer: (a) The DNA from normal tissue should reveal one loss‐of‐function TP53 mutation (TP53‐/TP53+). (b) Malignant tissue should show loss of both alleles (TP53-/TP53-).
Skill: Problem-solving
47) At the molecular level, what exactly are the mutagenic effects of UVA and UVB radiation? Answer: As discussed in Chapter 7, this radiation primarily causes point mutations by inducing tautomeric shifts and thymine dimers.
Skill: Factual recall
48) Suppose you are studying a rare spontaneous cancer in cats, which you suspect is retrovirus‐induced. What kinds of retroviral insertion could lead to cancer, and what kind of genetic signature would you look for to distinguish between them? Answer: One possible scenario is stimulation of a proto‐oncogene through insertional mutagenesis, inserting near and overexpressing a proto‐oncogene under the regulatory control of viral control elements. Another scenario is infection with a transducing retrovirus, in which expression of the v-onc gene induces tumor formation. The genetic signature of the two scenarios differs in the presence of v-onc. Isolating DNA from tumor tissue and sequencing the viral insertional site (perhaps first identified through viral genes like gag, pol, and env) would reveal whether v-onc or c-onc genes are present. Most proto‐oncogenes contain introns that are not present in the corresponding v-onc.
Skill: Conceptual understanding
49) What kinds of cancers might eventually be treated with RNAi, which you learned about in Chapter 18? Answer: siRNAs can be engineered to silence or down‐regulate genes by binding with their mRNA. A potential application may thus be treating cancers that stem from the expression or overexpression of a gene product; for example, overproduction of growth factors or protein kinases by an oncogene.
Skill: Analytical reasoning
50) What is metastasis, and why is metastatic cancer the most difficult cancer to treat? Answer: Metastasis is the spreading of cancerous cells to several locations of the body via the bloodstream. The cancerous cells are thus no longer at a point source, which might be treatable locally with surgery or radiation. Metastatic cancers must be treated systemically, as with chemotherapy or radiation.
Skill: Conceptual understanding
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Chapter 21 Population Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) Favorable alleles of genes can cause selection for an allele of an unrelated linked gene, a phenomenon known as
Skill: Factual recall
A) gene flow. B) overdominance. C) genetic hitchhiking. D) postzygotic isolation. E) genetic drift.
2) Sickle‐cell heterozygotes have a selective advantage, exhibiting
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3) Immigration of individuals with different allele frequencies into a population causes
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4) Hybrid sterility is an example of
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5) Very small, isolated populations are subject to
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Answers:
1) C
2) B
3) A
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) In the Hardy‐Weinberg model, the ideal population is A) very large. B) randomly mating. C) free of mutations. D) non‐migrating. E) All of these Answer: E
Skill: Factual recall
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7) If there are two alleles of a gene, B and b, and the frequency of the B allele (p) is 0.90, the frequency of the b allele (q) is A) 0.81. B) 0.30. C) 0.10. D) 0.09. E) 0.01. Answer: C
Skill: Problem-solving
8) For alleles of genes in Hardy‐Weinberg equilibrium, the frequency of heterozygotes is represented as A) p. B) q. C) p 2 . D) q2 . E) 2pq. Answer: E
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9) Which of the following methods is least useful for assessing levels of genetic variation in populations? A) DNA sequencing B) VNTRs C) Protein electrophoresis D) Phenotypic observation E) RFLP analysis Answer: D
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10) If the genotypic frequencies for an allele in a population at Hardy‐Weinberg equilibrium are 0.635 homozygous dominants, 0.430 heterozygotes, and 0.125 homozygous recessives, what is the frequency of the recessive allele? A) 0.35 B) 0.65 C) 0.125 D) 0.875 E) 0.01 Answer: A
Skill: Problem-solving
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Chapter 21 Population Genetics 183
11) Allele frequencies can change across the geographic range of a species according to climate, geological features, or direction in the range. For instance, a certain allele frequency might go up along with the elevation or down as you go south in the speciesʹ range. This systematic change in frequency is referred to as A) genetic drift. B) q2 . C) directional selection. D) a cline. E) heterosis. Answer: D
Skill: Factual recall
12) Which of the following is not a form of genetic drift? A) Founder effect B) Nonrandom mating C) Population bottleneck D) Heterozygote advantage E) All of these are forms of genetic drift. Answer: D
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13) Genetic drift may lead to A) loss of genetic variation in a population. B) a decrease in the mutation rate of a particular gene. C) increased heterozygosity at many loci. D) Hardy‐Weinberg equilibrium. E) All of these Answer: A
Skill: Factual recall
The following scenario applies to the questions below. For a gene A, a geneticist studying a population of butterflies found the following genotypes: 108 AA butterflies 144 Aa butterflies 48 aa butterflies 14) What is the observed genotypic frequency of Aa individuals? A) 0.144 B) 0.72 C) 0.40 D) 0.48 E) 1.44 Answer: D
Skill: Problem-solving
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184 Test Bank for iGenetics
15) What is the frequency of the a allele in the butterfly population? A) 0.16 B) 0.04 C) 0.40 D) 0.64 E) 0.40 Answer: C
Skill: Problem-solving
16) Synonymous mutations are found much more frequently than nonsynonymous mutations because A) mutations do not occur randomly B) the genetic code is redundant. C) of genetic drift. D) there is a selective advantage in looking like another species. E) changes in the amino acid sequence of proteins are likely to be harmful. Answer: E
Skill: Conceptual understanding
17) ________ is the ultimate source of genetic variation in populations. A) Natural selection B) Genetic drift C) Gene flow D) Mutation E) Recombination Answer: D
Skill: Factual recall
18) In a population of frogs, RR homozygotes produce an average of 5 offspring that survive to reproduce. Rr frogs produce 4, and rr frogs produce 1. The fitness of the three genotypes w11, w12, and w22 are respectively A) 5, 4, and 1. B) 0.5, 0.4, and 0.1. C) 1.0., 0.8, and 0.2. D) 0.5, 1.0, and 0.1. E) None of these Answer: C
Skill: Problem-solving
19) In reciprocal hybrid crosses, the heterogametic sex is typically sterile while the homogametic sex is not. This phenomenon is referred to as A) hybrid incompatibility. B) postzygotic isolation. C) Haldaneʹs rule. D) male sterility. E) None of these Answer: C
Skill: Factual recall
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Chapter 21 Population Genetics 185
20) A population genetic study of a certain insect revealed four alleles for the GPI (glucose‐phosphate isomerase) locus. How many different genotypic classes should be found? A) 4 B) 8 C) 10 D) 16 E) 20 Answer: C
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21) A sizable population was analyzed for SNPs in a certain DNA region. Four possible variants were found, with the following frequencies: SNP 1: GGTCTAGGA; frequency = 0.91 SNP 2: GGTGTAGGA; frequency = 0.03 SNP 3: GGTATAGGA; frequency = 0.03 SNP 4: GGTTTAGGA; frequency = 0.03 If the population is in Hardy‐Weinberg equilibrium, what is the total frequency of heterozygotes at this SNP locus? A) 0.828 B) 0.009 C) 0.170 D) 0.055 E) 0.0018 Answer: C
Skill: Problem-solving
22) Heterozygotes for the sickle‐cell allele show higher fitness than either homozygote genotype in malarial regions. This is an example of A) epistasis. B) pleiotropy. C) founder effect. D) overdominance. E) codominance. Answer: D
Skill: Factual recall
23) For which of the following traits is human mating expected to be nonrandom? A) Blood type B) Isocitrate dehydrogenase enzyme C) Skin color D) Presence or absence of a widowʹs peak E) Fingerprint loci Answer: C
Skill: Factual recall
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186 Test Bank for iGenetics
24) Twenty loci are screened for genetic variation in a common caterpillar species. Four loci are found to have two or more alleles. The proportion of polymorphic loci for this population is thus A) 0.10. B) 0.20. C) 0.30. D) 0.40. E) 0.50. Answer: B
Skill: Problem-solving
25) Although bees carry pollen from tulip flowers to the blossoms on cherry trees, hybrid seeds are not produced because of A) gametic isolation. B) temporal isolation. C) ecological isolation. D) mechanical isolation. E) All of these Answer: A
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) For two populations to remain genetically homogenized, a significant proportion of the population of each must be exchanged at least every other generation. Answer: FALSE Explanation: Remarkably little genetic exchange is necessary to keep two populations from diverging in allele frequencies by genetic drift. Theory suggests that as little as one migrant every other generation is sufficient to maintain similar allele frequencies.
Skill: Factual recall
27) When the population mean for a given measured trait is observed to decline over the course of several generations, it is an example of disruptive selection. Answer: FALSE Explanation: A positive or negative change in the population mean value-i.e., in one or the other direction-is defined as directional selection.
Skill: Factual recall
28) The Hardy‐Weinberg relationship cannot be used to compute allele frequencies when one or more alleles are recessive. Answer: FALSE Explanation: The mode of expression of alleles has no bearing on whether the Hardy‐Weinberg relationship can be used to compute their frequency.
Skill: Factual recall
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Chapter 21 Population Genetics 187
29) Two closely related insect species are active at different times of day, with one diurnal and the other nocturnal. This is an example of spatial isolation. Answer: FALSE Explanation: This is an example of temporal isolation.
Skill: Factual recall
30) A selection coefficient of 1 means zero fitness. Answer: TRUE
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31) Survival is the single most important process in evolution by natural selection. Answer: FALSE Explanation: Reproduction is the primary evolutionary currency. Survival clearly has a bearing on this, but individuals that survive will have zero evolutionary fitness if they do not reproduce.
Skill: Factual recall
32) According to the Hardy‐Weinberg principle, at equilibrium the allele frequencies are dependent on the genotypic frequencies. Answer: FALSE Explanation: According to the Hardy‐Weinberg principle, the genotypic frequencies of a trait follow from the frequencies of the alleles for that trait.
Skill: Factual recall
33) The frequency of a recessive X‐linked allele is equal to the phenotypic frequency of males exhibiting that allele. Answer: TRUE
Skill: Factual recall
34) The genetic variation between human ʺraces,ʺ such as Africans and Europeans, is greater than the variation within a single race. Answer: FALSE Explanation: There is less variation between the human ʺracesʺ than within a single race, showing that our concept of race has little genetic validity.
Skill: Factual recall
35) A gene is known to have three alleles: p, q, and r. The frequency of p is estimated at 0.70. Thus, the frequencies or q and r must be 0.15 each. Answer: FALSE Explanation: The three frequencies can assume any value as long as they all sum to 1.0. Moreover, for a three‐allele gene, knowing the frequency of just one allele is not sufficient to determine the frequencies of the other two alleles.
Skill: Factual recall
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188 Test Bank for iGenetics
SHORT ANSWER QUESTIONS AND PROBLEMS 36) In light of the fact that the assumptions of the Hardy‐Weinberg principle are unrealistic (no mutation, no selection, etc.), how can the principle be useful to population geneticists? Answer: In one respect, the Hardy‐Weinberg principle is a null hypothesis: It provides insight into expected allele and genotypic frequencies, and when real populations are observed to deviate substantially from Hardy‐Weinberg expectations, the reasons can be investigated in the form of alternative hypotheses. Also, it turns out that real populations are often robust in deviations from Hardy‐Weinberg assumptions, making the principle useful in estimating the frequencies of alleles and genotypes in populations.
Skill: Conceptual understanding
37) The deciduous forest of eastern North America has experienced dramatic changes over the past two to three centuries, transitioning from a nearly continuous forested area to increasingly patchy areas of forest broken up by farming and development. Some species are more sensitive to severe habitat fragmentation than others. What are some characteristics that might make a species more susceptible to the effects of habitat and population fragmentation? Answer: Habitat fragmentation leads to ʺislandsʺ of habitat, introducing the potential for inbreeding and genetic drift. More sensitive species might be those that, for whatever reason, are not highly mobile, and thus tend to become marooned in subpopulations that may experience drift effects that carry them below the minimum viable population size. Highly mobile species–certain birds, for example–would be better able to maintain gene flow between the habitat islands. A related characteristic of sensitive species might be dependence on a specialized resource that is itself increasingly scarce as a result of habitat fragmentation. Generalists may have an easier time finding the resources they need in whatever habitat island they find themselves.
Skill: Conceptual understanding
38) Observed genotypic frequencies in populations rarely match Hardy‐Weinberg expectations exactly. Why is this so, and how do scientists determine if the frequencies they observe depart from the Hardy‐Weinberg equilibrium to a biologically meaningful degree? Answer: There will always be chance deviations in observed data owing to sampling error or the random element of the mating process. Statistical testing is used to compute the probability of observing a given level of discrepancy between observed and expected values. The chi‐square test is most often applied to these kinds of genetic data.
Skill: Conceptual understanding
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Chapter 21 Population Genetics 189
39) With respect to a gene with two alleles, for each generation of complete inbreeding the proportion of heterozygotes is expected to be reduced by while the proportion of each homozygous class is expected to increase in frequency by . Explain why this is so. Answer: The three genotypes for our two‐allele gene are AA, aa, and Aa. Complete inbreeding for AA and aa will yield the same genotypes. Each cross of Aa × Aa, however, will yield AA, Aa, and aa in a 1:2:1 ratio [ : : ]. Thus, the heterozygotes are reduced by
1 1 , while each homozygous class increases by . 2 4 1 1 1 4 2 4 1 4 1 2
Skill: Problem-solving
40) Populations that suffer significant reductions in number may experience two population-genetic consequences that together can hasten their decline. Which two processes are likely to act in concert in small populations, and what is their effect? Answer: The two processes are inbreeding and genetic drift. As populations decrease in size, inbreeding is expected to increase. This in turn may further accelerate population decline by increasing the expression of deleterious recessive alleles, decreasing survivorship or fecundity. Genetic drift leads to further loss of genetic variation, accelerating inbreeding.
Skill: Conceptual understanding
41) Show algebraically that under Hardy‐Weinberg equilibrium, allele frequencies will not change from one generation to another. Answer: The demonstration is simple. Following allele p, start with the Hardy‐Weinberg genotypic frequencies p 2 + 2pq + q2 . The proportion of genotypes contributing p to the next generation is: (p 2 + pq) / (p 2 + 2pq + q2 ). p is transmitted via genotype p 2 and one‐half the frequency of the 2pq individuals, or p 2 + pq. Rewriting p 2 as p(1 - q) enables us reduce this to simply p as follows: p 2 + pq = p(1 - q) + pq = p - pq + pq = p. This means that the allele is transmitted to the next generation at frequency p, unchanged.
Skill: Problem-solving
42) Two neighboring Caribbean islands harbor iguana populations. One island (island A) has a total iguana population of 800, while the other (island B) has a population less than half that size, numbering 250. On island A, the frequency of an allele p is estimated at 0.75, while this same allele is present at a frequency of 0.90 on island B. During a hurricane tracking across the islands, a group of about 100 iguanas get rafted from island A to island B. What effect does this have on the frequency of allele p on island B? Answer: The migrant subpopulation is 100/800, or 0.125. This can be taken as m, the migration coefficient. The frequency of p in island A is 0.75; call this p A . p B is given as 0.9. Using the relationship Δp = m(p A - p B), Δp = 0.125(0.75 - 0.9) = 0.125(-0.15) = -0.019. The frequency of p on island B thus declines by 0.019, to 0.9 - 0.019 = 0.881.
Skill: Problem-solving
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190 Test Bank for iGenetics
43) Red‐green color blindness is an X‐linked trait. In a population genetics study, 1,000 people (500 men and 500 women) were tested for this trait, and 35 men were found to be color blind. Use this information to compute the frequency of the allele for color blindness and the wild‐type allele in this population, and estimate the expected number of carrier females. Answer: 35/1,000 = 0.035 = q; this is the frequency of the trait in this population. Since p = (1 q) = (1 - 0.035) = 0.965, you can use p and q to compute 2pq at 2(0.035)(0.965) = 0.068; this is the frequency of heterozygotes, which can only be female. This gives an estimated 33 female carriers in this population.
Skill: Problem-solving
44) The northern elephant seal, Mirounga angustirostris, suffered a significant population bottleneck in the late nineteenth century, when hunting reduced their population size to as few as 20 individuals. It has since rebounded to several tens of thousands. The related southern elephant seal, M. leonine, was not hunted as intensively. What prediction can you make about the relative levels of heterozygosity in populations of these two species? Answer: Much genetic variation would have been lost in the northern elephant seal population as a result of such a severe bottleneck, so heterozygosity will be very low. The southern elephant seal, in contrast, should still have significant genetic variation and correspondingly higher heterozygosity.
Skill: Conceptual understanding
45) Two populations experience equally severe bottlenecks, reducing each to one‐tenth of its original size. One population is in the bottleneck for one generation, and the other is in the bottleneck for five generations. You assess the heterozygosity of both populations after they have been restored to their prebottleneck size. What do you expect to find, and why? Answer: The population that was bottlenecked for five generations is likely to have lost much more genetic variation than the one bottlenecked for only one generation. This is because each generation in a bottleneck state is equivalent to a separate bout of genetic drift, losing variants by chance each time.
Skill: Conceptual understanding
46) In the early twentieth century, evolutionary biologists argued whether there was enough genetic variation in populations to explain the diversity of life by natural selection. With the advent of protein electrophoresis and DNA‐level analysis, the problem turned on its head: the neutralist school arose from the idea that there was too much genetic variation to be maintained by natural selection. What are some of the arguments and observations neutralists cited to support the idea that much variation is neither favored nor disfavored by natural selection? Answer: More genetic variation is expected in noncoding regions, such as intergenic areas and introns, and has indeed been found to be the case. Also, owing to redundancy in the genetic code, the third codon position can often be variable and still specify the same amino acid. More variation might therefore be expected at third base‐pair positions in codons, which has also been found to be the case.
Skill: Conceptual understanding
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Chapter 21 Population Genetics 191
47) Why do we expect that postzygotic isolation will evolutionarily precede prezygotic isolation in the speciation process? Answer: Postzygotic isolation is fitness‐reducing in that mating effort is wasted on offspring that have low survivorship or fecundity. Natural selection should act to reduce this waste by favoring, for example, mate recognition signals or spatial or temporal partitioning that prevent such matings in the first place.
Skill: Conceptual understanding
48) Some recessive mutations can be exceedingly debilitating or lethal when expressed in homozygotes. If their effects are so severe, why doesnʹt natural selection simply purge such alleles from the population completely? Answer: Natural selection does act on these alleles when they are expressed, but for the most part, when alleles are rare they occur in heterozygotes and are thus masked. Another reason selection cannot eliminate such alleles completely is that they arise continually by mutation. Finally, any recessive alleles that are expressed late in life will be shielded from natural selection, since they will not affect an individualʹs chances of producing viable offspring.
Skill: Conceptual understanding
49) This chapter discussed speciation by genetic divergence following geographic isolation, which is expected to lead to reduced gene flow, a process known as allopatric speciation. A more controversial form of speciation is the genetic divergence of populations without physical isolation, a process known as sympatric speciation. Can you envision a mechanism or process that would permit two coexisting populations of the same species to begin to diverge without being isolated from one another? Answer: Sympatric speciation is thought to be uncommon but almost certainly has occurred. In one general scenario, a form of isolation ʺin place,ʺ or in sympatry, may occur if certain males or females begin (by drift, perhaps) to prefer different resources–say, a special site for mating or laying eggs. If their mates follow in this ʺimprintingʺ process, the two preference populations may begin to experience reduced gene flow between them, which if maintained long enough may bring about a degree of genetic divergence that would be selectively reinforced should some of the individuals from the two populations ever hybridize.
Skill: Analytical reasoning
50) Remote oceanic islands are characterized as having disharmonic biota, meaning the number and relative proportions of taxa living on the island differ significantly from the number and relative proportions of taxa on the nearest continental mainland area. Each island system is disharmonic in its own unique way. Why? Answer: This pattern stems from the random nature of the ancestral colonization of those islands. Successfully colonizing a remote island is rare, but in the vastness of time such events occur regularly. Precisely what groups make it is a matter of chance, but some have a far greater likelihood than others. Flying animals, for example, are more likely colonists than large or sensitive terrestrial animals (and in fact no native mammals or amphibians are found on the remotest islands). Since the colonization process is random, each island system experiences a unique colonization history. Thus, each will have its own complement of taxa from continental areas, which, depending on the timing of ancestral colonization and the available niches, will spread and diversify to different degrees on each.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) VG/V P
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A) Narrow‐sense heritability B) Genetic variance C) Determines whether differences in means are significant D) Variance E) Broad‐sense heritability
2) V A /V P
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3) V A + V D + V I
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4) s2
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5) ANOVA
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Answers:
1) E
2) A
3) B
4) D
5) C
MULTIPLE-CHOICE QUESTIONS 6) Which of the following is a source of genetic variation in a population caused by epistatic interactions between genes? A) Additive genetic variance B) Dominance variance C) General environmental effects D) Interaction variance E) Maternal effect Answer: D
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7) Quantitative traits typically have multiple genetic influences in which alleles of a number of genes contribute individual amounts to the final phenotype without classical dominance or epistasis. The term used for the observed variation that results from this is A) V A . B) V D . C) V E. D) V G. E) V I. Answer: A
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192
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Chapter 22 Quantitative Genetics 193
8) A female cat that is small due to nutritional deficiencies experienced in early life will give birth to smaller kittens than a genetically identical cat that is larger in size due to better nutrition in early life, even if both female cats were bred to the same male cat and both had good nutrition during pregnancy. The term that is used in measuring this type of variation is A) V P . B) V E. C) V EG. D) V ES. E) V EM. Answer: E
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9) The distribution of continuous traits can best be described as A) linear. B) bell‐shaped. C) hyperbolic. D) logarithmic. E) bimodal. Answer: B
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10) Ornamental hydrangeas are well known to produce either blue or magenta flowers, depending on soil pH. This may be a good example of a discrete trait influenced by A) genetics. B) environment. C) multiple genes. D) A and B only E) A, B, and C Answer: B
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11) Which of the following statements about heritability is true? A) Broad‐sense heritability defines the complete genetic basis of a trait. B) Heritability is fixed for a given trait. C) Traits shared by members of the same family do not necessarily have high heritability. D) Heritability is the measure of the proportion of an individualʹs phenotype that is due to genetics. E) If a population displays little variation in a given trait, then that trait has low heritability Answer: C
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12) When graphing a distribution of phenotypes for a continuous trait, a broad curve implies A) little variation in the phenotypes. B) a large standard deviation. C) a small variance. D) a large correlation coefficient. E) a zero regression line slope. Answer: B
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13) The proportion of phenotypic variance that results from additive genetic variance is known as ________ heritability. A) broad‐sense B) narrow‐sense C) multifactorial D) generic E) genetic Answer: B
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14) A broad‐sense heritability of 0.8 for a particular trait indicates that A) phenotypic variation stems from environmental influences. B) the trait is 80% heritable. C) genes and environment play equal roles in shaping the trait. D) genetic differences among individuals account for much of the phenotypic variation for that trait. E) None of these Answer: D
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15) A Mendelian cross for a three‐gene trait is expected to yield ________ F 2 genotypes. A) 4 B) 16 C) 27 D) 64 E) 265 Answer: C
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16) In parent‐offspring regression, the average value of the parents is termed the A) mean parental value. B) parental average. C) midparent value. D) median parental value. E) modal parental value. Answer: C
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Chapter 22 Quantitative Genetics 195
17) A ________ trait is typically influenced by one or two genes. A) quantitative B) continuous C) polygenic D) multifactorial E) discrete Answer: E
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18) In maize, some alleles at the seed coat locus influence our ability to visually determine the endosperm color phenotype. Specifically, an opaque seed coat prevents us from seeing the endosperm within, while we can readily see the endosperm in seeds with transparent coats. The opaque coat color allele thus appears to mask the expression of endosperm color alleles, providing an example of A) codominance. B) pleiotropy. C) polygenic inheritance. D) epistasis. E) maternal effect. Answer: D
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19) In the general regression equation y = mx + b, variables m and b represent ________ and ________, respectively. A) the x‐intercept, the y‐intercept B) the regression line slope, the y‐intercept C) the y‐intercept, the regression line slope D) the y‐intercept, the x‐intercept E) narrow‐sense heritability, broad‐sense heritability Answer: B
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20) Assume that the length of a type of cucumber at maturity is controlled by three genes ( A, B, and C), each of which has two alleles. The A, B, and C alleles each add 2 inches of cucumber growth, while the a, b, and c alleles add only 1 inch. A true‐breeding 6‐inch strain is bred to a true‐breeding 12‐inch strain. All the F 1 s produce 9‐inch cucumbers. What sizes would you expect to see in the mature F 2 s? A) Discrete lengths of 6, 9, and 12 inches B) Discrete lengths of 6 and 12 inches C) Discrete lengths of 9 inches D) Discrete lengths of 6, 7, 8, 9, 10, 11, and 12 inches E) Lengths ranging continuously from 6 to 12 inches Answer: D
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21) Analysis of specific genetic contributions to quantitative traits has been done by A) altering the environment of individuals and observing phenotype alterations. B) completing sequencing of specific genes for various species. C) measuring V G by adding V A , V D , and V I. D) conducting ANOVA on genetic markers and phenotypes to identify linkage to QTLs. E) measuring specific environmental effects, such as sunlight, on phenotype. Answer: D
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22) Which of the following problems might arise in a statistical study as a result of a population sample of inadequate size? A) A non‐representative sample B) Sampling error C) A non‐random sample D) A sample drawn from a biased subset of data points E) All of these Answer: E
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23) Heritability can only be measured for A) traits with phenotypic variance. B) traits with genotypic variance. C) discrete traits. D) polygenic traits. E) None of these Answer: A
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24) Populations of the snail Cepaea nemoralis exhibit three color morphs, a good example of a ________ trait. A) continuous B) discontinuous C) quantitative D) polygenic E) B, C, and D only Answer: E
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25) A quantitative trait influenced by four loci, each with two alleles, would be expected to show ________ possible genotypes. A) 81 B) 16 C) 8 D) 256 E) None of these Answer: A
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Chapter 22 Quantitative Genetics 197
TRUE-FALSE QUESTIONS 26) If a populations shows very little phenotypic variance for a trait, then that trait is probably controlled principally by the environment rather than genetics. Answer: FALSE Explanation: A population could be homozygous for alleles that control a trait, or have genetic variations that do not appreciably affect phenotype for that trait, yet the trait itself might still be influenced entirely by genes. As an example, the human trait of possessing two arms and two legs is determined genetically and is quite constant.
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27) Narrow‐sense heritability is a more useful measure than broad‐sense heritability in agricultural selective breeding. Answer: TRUE
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28) The mean value of a quantitative trait in F 1 offspring is generally smaller than either of the values of the parentals. Answer: FALSE Explanation: The F1 value will be intermediate between the two parentals.
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29) Narrow‐sense heritability for a trait can be obtained from a regression of midparent value on offspring value. Answer: TRUE
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30) If members of the same family display the same phenotypic traits, these traits have high heritability. Answer: FALSE Explanation: Even if the phenotypic traits of members of the same family are correlated, they are not necessarily influenced by genetics. The common familial environment may lead to any such correlation.
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31) Continuous traits are often influenced by multiple simple (Mendelian) genes. Answer: TRUE
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32) Evolution is defined as genetic change in populations over time. Answer: TRUE
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33) A gene that influences several traits is termed pleiotropic. Answer: TRUE
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34) As the number of genes influencing a quantitative trait increases, the range of phenotypic variation is expected to increase. Answer: FALSE Explanation: The range of phenotypic variation may have little to do with the number of genes, although the number of observable phenotypes is expected to increase.
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35) In descriptive statistics, the median is defined as the value exactly in the center of a distribution, with half the values lying above and half lying below. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Continuously varying traits typically show a bell‐shaped frequency distribution. Explain why this is the case, and why intermediate phenotypes are more frequent than phenotypes at either end of the distribution. Answer: The greater frequency of intermediate phenotypes is a consequence of a greater frequency of intermediate genotypes stemming from allelic combinatorials. Even a single two-allele gene possesses twice as many ways to make a heterozygote ( Aa and aA), the number of which increase exponentially as the number of genes increase.
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37) Distinguish between polygenic and multifactorial traits. Answer: Polygenic traits are quantitative traits, shaped by multiple genes. Multifactorial traits are shaped both by multiple genes and environmental factors.
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38) Explain why narrow‐sense heritability is a more accurate indicator of responsiveness to selection than is broad‐sense heritability. Answer: Narrow‐sense heritability is defined in terms of additive genetic variance, meaning that each allele of each contributing gene has a phenotypic effect. Broad-sense heritability is based on genetic variance in general. Since selective breeding is based on phenotypes, insofar as additive genetic variance is a more precise reflection of genotype underlying phenotype, this measure will help in making better selective breeding decisions.
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39) The correlation coefficient is a measure of the strength of association between two traits. Correlation, however, does not indicate causation. Think of one or two variables (not necessarily phenotypes) that are correlated and are causally linked, and one or two variables that may be correlated but are not likely to be causally linked. Answer: Traits like possession of a Y chromosome and possession of testes or a heavy beard are causally linked, owing to the testis‐determining factor gene on the Y chromosome. Blonde hair and blue eyes are often associated but are not necessarily causally linked.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics 199
40) Define, in your own words, the biological meaning of heritability. Answer: Heritability is often imprecisely defined as a measure of the genetic component of a trait. More precisely, it is a statistical measure that quantifies the influence of genes on phenotypic variance with respect to a given trait.
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41) A botanist studying a plant with geographically separated and morphologically distinct populations wants to know the degree to which the differences between the populations are attributable to genetic factors. Outline a simple experiment she can perform to address this question. Answer: Two classic experimental approaches would be useful here. The first is reciprocal transplant, which involves switching the environments of the plants, for example, by creating test plots in each area and planting seeds or seedlings collected from the other site. The second would be to create a common garden, in which seeds or seedlings of plants from each population are reared under identical conditions, for example, side by side in a greenhouse. Comparison of key morphological traits across the populations will provide a way to estimate heritability of the traits.
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42) If you measured the narrow‐sense heritability of a crop plant trait at 0.25, what would that value mean in terms of the ease of selecting the trait in a selective breeding program? Answer: This heritability value is rather low but does indicate that the observed phenotypic variation is influenced by genetics. The trait is selectable, albeit perhaps slowly.
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43) A regression analysis of two traits yields a slope of 0. Explain the biological meaning of this value. Answer: Phenotypic variation in the traits is not correlated.
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44) The broad‐sense heritability of a trait is estimated at 0.8. If the total phenotypic variance is estimated at 32.20, what is the genetic variance? Answer: V G = 25.76; broad‐sense heritability = V G / V P
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45) Human fingerprints are a good example of a multifactorial trait. What evidence can you cite to suggest that fingerprints are partially influenced by genes? How about evidence suggesting that they are partially influenced by environmental factors? Answer: Observations suggesting there is a genetic component to fingerprint patterns include consistent population differences in the frequency of basic patterns (whorls, arches, loops). However, even identical twins have unique fingerprints, indicating that they are also determined by environmental factors.
Skill: Analytical reasoning
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200 Test Bank for iGenetics
46) In a controlled cross for a quantitative trait, 600 F 2 offspring are obtained, 10 of which exhibit the phenotype of one of the parentals. How many genes are estimated to control this trait based on these results? Answer: 1/(4 n) gives the fraction of F2 offspring resembling either parental for a polygenic trait in which n is the number of genes involved. 10/600 is approximately corresponding with a value of n = 3 genes.
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1 64
,
47) In Mendelʹs day, ʺparticulateʺ inheritance would have made less sense to people than blending inheritance. Why? Answer: At the time Mendel performed his experiments, it was unusual to focus on specific traits. If one looks at an entire organism, it is easy to see how offspring could appear to have been produced by blending the traits of each parent, which was the standard assumption among Mendelʹs contemporaries. Mendelʹs insight came from focusing on the inheritance of specific traits, leading him to uncover patterns of mathematical regularity in their occurrence in his controlled breeding experiments.
Skill: Analytical reasoning
48) The average adult height in the U.S. population has increased by several centimeters over the past two generations. Is this likely to reflect a change in the genetic makeup of the population, a change in environmental factors, or both? Explain. Answer: There has been no directional change in the genetic makeup of the U.S. population in this time interval. Rather, environmental factors–nutritional improvement, in particular-underlie this trend.
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49) Identical (monozygotic) and fraternal (dizygotic) twins provide a natural experiment for investigating the influence of genetics on traits. Outline a study approach using twins, assuming you could identify enough pairs, in which you are examining the relative contribution of genetics on blood cholesterol level. Answer: If there is a genetic predisposition to achieving a certain blood cholesterol level, monozygotic twins might be expected to show a stronger statistical association in cholesterol level than dizygotic twins. You would need to control for environment, however: ideally, your twins should be given a controlled diet for a specified period.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics 201
50) An isolated population of wild turkeys has adult males that weigh an average of 18 pounds and females that weigh an average of 10 pounds. Another isolated population living 100 miles north has a population where adult males weigh 13 pounds and females 8 pounds on average. Both populations have a very low phenotypic variance. How could you determine genetic contributions to the weight differences between the populations? Answer: The phenotypic variation (V P) is dependent on genotypic variation (V G) plus environmental variation (V E) and gene‐environment interactions (2COV G,E + V G ×
E). To isolate V G members of both populations could be captured and bred in
captivity under identical circumstances. After a few generations, environmental influences, including V M, should have equalized. If the separate populations have achieved identical V P then V G contributions are limited. If the separate populations maintain a significant size difference, the populations could be crossed. If the F 1 s are intermediate in weight and the F 2 s are average but have a much wider variance, then quantitative genetic factors are more important and narrow‐sense heritability can be measured as H N = V A /V P. If, instead, the F 1 s have one parental type and the F2 s are of that type and of the other parental type, a single gene with a dominant and recessive phenotype could cause the size difference. A number of other possibilities exist depending on one or two other genes that might have incompletely dominant or additive alleles.
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3 4 1 4 2
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Chapter 23 Molecular Evolution
MATCHING QUESTIONS Please select the best match for each term. 1) Transition
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A) A mutation that alters the encoded amino acid B) A purine-to-pyrimidine mutation C) Nonfunctional evolutionary remnant of a gene D) An A‐to‐G or C-to‐T mutation E) A mutation that does not alter the encoded amino acid.
2) Transversion
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3) Synonymous
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4) Nonsynonymous
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5) Pseudogene
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Answers:
1) D
2) B
3) E
4) A
5) C
MULTIPLE-CHOICE QUESTIONS 6) Homologous proteins A) share a common ancestor. B) share a common amino acid sequence. C) are determined by alleles on homologous chromosomes. D) A and B only E) A, B, and C Answer: A
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7) Synonymous changes that arise within coding sequences of genes A) generally have no effect on the phenotype. B) are usually detrimental to fitness. C) may be eliminated by natural selection. D) occur more frequently than nonsynonymous changes. E) B and C only Answer: A
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202
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Chapter 23 Molecular Evolution 203
8) Indels are A) index deletions used to identify a species. B) parsimonious trees. C) transitions and transversions combined. D) small deletions or insertions that produce gaps in sequence alignments. E) the result of codon usage bias. Answer: D
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9) Higher substitution rates are seen in ________ than in ________. A) coding regions, noncoding regions B) 5ʹ flanking regions, 3ʹ flanking regions C) first codon positions, second codon positions D) introns, exons E) charged amino acid codons, noncharged amino acid codons Answer: D
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10) Nonsynonymous substitutions A) do not lead to amino acid change. B) lead to amino acid change. C) refer to substitutions in unrelated lineages. D) affect only certain amino acids. E) lead to transversions. Answer: B
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11) What is the number of possible unrooted trees for 5 taxa? A) 5 B) 10 C) 15 D) 20 E) 25 Answer: C
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12) Which of the following statements about molecular evolution is true? A) Homologous genes often evolve in the same way. B) Different regions of the same gene evolve at different rates. C) Genes in the same organism evolve at about the same rate. D) Genes in different organisms evolve at about the same rate. E) Nucleotide substitutions are nonrandom. Answer: B
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204 Test Bank for iGenetics
13) Which of the following does not provide molecular evidence for evolution? A) The 1,000‐fold differences in nonsynonymous substitutions between some genes B) The existence of pseudogenes C) The fact that high‐expression genes use amino acids that are less energetically costly than those used by low‐expression genes D) The fact that there is a lower rate of nucleotide substitutions in coding regions than in noncoding regions of genes E) All these are molecular evidence for evolution Answer: E
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14) With regards to the sequence alignment
G T C C A T C G G G T C G A T C — G,
which of the following statements is not true? A) There has been a base deletion. B) There has been a base substitution. C) There has been a gene conversion. D) There has been a transversion. E) There is evidence of an indel. Answer: C
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15) Which of the following mutations is considered a synonymous change? A) ACC to AUG B) CUU to AUU C) CUU to CUA D) CUU to CUA E) UUU to UUA Answer: D
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16) Codon bias suggests that A) some codons evolved earlier than others. B) not all nonsynonymous substitutions are neutral. C) some codons are selectively favored over others. D) small fitness effects can make a large difference over time. E) some tRNAs are more abundant than others. Answer: C
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Chapter 23 Molecular Evolution 205
17) From the standpoint of evolutionary fitness, it is expected that genes with a high rate of expression will make use of amino acids that are ________ than those amino acids used by genes with a low rate of expression. A) more common B) more energetically economical C) more energetically costly D) rarer E) None of these Answer: B
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18) Which of the following types of proteins evolves at the slowest rate? A) Peptide hormones B) Collagen C) Insulin D) Histones E) Lipoprotein Answer: D
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19) Which of the following factors does not interfere with the interpretation of evolutionary time using the molecular clock? A) Horizontal gene transfer B) Differences in generation time C) Mutations due to background cosmic radiation D) Different selection pressures on different taxonomic groups E) Differences in DNA repair efficiency in different taxonomic groups Answer: C
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20) The greatest genetic differences found in humans are A) found between individuals from different continents. B) found between individuals of different racial groups. C) found between island populations and mainland populations. D) found between individuals of African descent. E) found between individuals of European descent. Answer: D
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21) The average synonymous substitution rate in mammalian mitochondrial genes is ________ the average rate for nuclear genes. A) 2 times B) 5 times C) 10 times D) 50 times E) 100 times Answer: C
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206 Test Bank for iGenetics
22) In the unweighted pair group method (UPGMA) of determining evolutionary relationships, A) each set of taxa is connected by the shortest genetic distance. B) a constant rate of evolution across all lineages is assumed. C) a distance matrix is first constructed. D) A and B only E) A, B, and C Answer: E
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23) Reconstructions of the evolutionary relationships of a set of taxa are called A) classifications. B) phylogenetic trees. C) UPGMA networks. D) networks. E) evolutionary scenarios. Answer: B
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24) The differences in evolutionary rates of change revealed by relative rate tests may be attributable to A) differential strength of selection. B) generation time. C) average DNA repair efficiency. D) differential exposure to mutagens. E) All of these Answer: E
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25) _______ tend(s) to occur more frequently than ________. A) Synonymous substitutions, nonsynonymous substitutions B) Genetic divergence, genetic convergence C) Transitions, transversions D) A and B only E) A, B, and C Answer: E
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TRUE-FALSE QUESTIONS 26) The noncoding 5ʹ flanking region of genes is more constrained by selection than the noncoding 3ʹ flanking region. Answer: TRUE
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27) Many viruses are helped along in generating antibody‐evading capsid protein variations by replication error. Answer: TRUE
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Chapter 23 Molecular Evolution 207
28) Pseudogenes tend to have lower average substitution rates than functional genes. Answer: FALSE Explanation: Pseudogenes are nonfunctional and are therefore not subject to natural selection. Substitutions thus accumulate more rapidly than in functional genes.
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29) The counterpart to the use of mtDNA to reconstruct maternal lineages is the use of chloroplast DNA to reconstruct paternal lineages. Answer: FALSE Explanation: The patrilineally inherited Y chromosome is the male counterpart to mtDNA.
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30) High molecular similarity always reflects evolutionary relatedness. Answer: FALSE Explanation: Molecular data are less susceptible to convergence than certain morphological traits, but it is possible that a given pair of similar sequences (particularly coding sequences) from unrelated groups may have converged owing to natural selection. Horizontal (or lateral) transfer also occurs.
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31) Maximum confidence is a procedure for reconstructing phylogenetic trees invoking the fewest number of steps or mutations. Answer: FALSE Explanation: The procedure invoking the fewest steps or mutations is maximum parsimony because the shortest tree is also the most parsimonious in requiring the fewest evolutionary events to obtain it.
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32) Nuclear autosomal DNA sequence is more abundant than mtDNA and is therefore preferable for use in reconstructing events like domestication or human migration. Answer: FALSE Explanation: Recombination and slower overall base substitution rates make nuclear DNA less useful for such purposes than mtDNA.
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33) The fact that approximately 80% of the six codons that encode the amino acid leucine in yeast are UUG is an example of codon bias. Answer: TRUE
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34) Genetic distance refers to the proportion of dissimilar nucleotides between two homologous sequences. Answer: TRUE
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208 Test Bank for iGenetics
35) Mitochondria and chloroplasts have genomes that have evolved from the nuclear genomes in eukaryotes. Answer: FALSE Explanation: Organellar genomes have many similarities to bacterial genomes and are thought to have arisen from the endosymbiosis of bacteria by the ancestors of modern eukaryotes.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) What is the rationale behind the bootstrap procedure, in which trees are repeatedly reconstructed following resampling with replacement of the original data? Answer: Phylogenetic bootstrapping gives an estimate of the robustness of each node in an inferred tree by seeing to what extent the node is affected by random resampling of the data. A very weakly supported node would be more likely to ʺcollapseʺ with resampling (showing that not much of the ʺsignalʺ in the original data supported it), while a strongly supported node will be reconstructed time and again with resampling because it is based on support from a wider array of informative sites.
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37) Given the following aligned sequences Sequence 1 Sequence 2 Sequence 3 Sequence 4
ATA ATA ATC ATC
TGT TGT TGT TGG
ATA TTA ATA ATA
AAA AAG AAA AAG,
identify any transitions and transversions as well as any synonymous and nonsynonymous substitutions. Answer: First codon: A to C: Second codon: T to G: Third codon: A to T: Fourth codon: A to G: transversion, synonymous substitution (Ile - Ile) transversion, nonsynonymous substitution (Cys - Trp) transversion, nonsynonymous substitution (Ile - Leu) transition, synonymous substitution (Lys - Lys)
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38) What are the major weaknesses of the parsimony and distance matrix methods of phylogeny reconstruction? Answer: The main problem with parsimony is that, by definition, it assumes the most parsimonious (shortest or simplest) evolutionary pathway, while it is almost certainly true that this is not invariably the pathway taken. Still, as a guiding principle, parsimony has much to recommend it. Distance methods make no such assumptions about how evolution should work, but on the other hand assume that simple unweighted similarity and dissimilarity reflect evolutionary relatedness and that evolutionary rate is constant across lineages. We know that this is not true, owing to selective constraints, convergence, etc.
Skill: Conceptual understanding
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Chapter 23 Molecular Evolution 209
39) The McDonald‐Kreitman test is an important tool in molecular evolutionary analysis. What is this test, and what kinds of insights can it provide? Answer: The test compares synonymous and nonsynonymous sites in gene-coding regions. If the nonsynonymous : synonymous substitution ratio within species is the same as that between species, then the substitutions are likely to be neutral. If the ratios differ, natural selection is likely to have been responsible by favoring certain substitutions over others. Insights the test provides include identifying genes on which natural selection may be acting for further study, and testing for neutrality for population genetic analyses that assume or require neutrality.
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40) What is codon bias? If synonymous substitutions give the same amino acid, why should there be a bias in which codons encode that amino acid? Answer: Codon bias is the disproportionate use of certain synonymous codons over others. Some synonymous codons pair with different tRNAs that carry the same amino acid. One explanation for this concerns efficiency in translation: While a mutation to a synonymous codon does not change the amino acid, it may change the tRNA used by a ribosome during translation. Over evolutionary time, natural selection may favor use of codons corresponding with the most readily available tRNA.
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41) In most genes, the rate of synonymous mutations is significantly higher than that of nonsynonymous mutation. The MHC genes exhibit the reverse, having over twice as many nonsynonymous as synonymous mutations. Explain this phenomenon. Answer: Synonymous mutations do not change the amino acid composition of a gene product. Nonsynonymous mutations do. Most changes in amino acids in a protein could have a deleterious effect and would be selected against. The MHC relies on genetic diversity to optimize immune responses both for the individual and for the species. Therefore, nonsynonymous mutations could provide a selective advantage.
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42) In the 1960s and 1970s, organisms were classified into five kingdoms, one of which was prokaryotes. In the 1980s, organisms were reorganized into three domains, with the prokaryotes split into Archaea and Bacteria, and all of the other four kingdoms lumped together into Eukarya. Why did this happen? Answer: In the original classification, prokaryotes were separated from all other organisms because of cell structure. With the advent of DNA sequencing, gene sequence comparisons became possible. Carl Woese, Norm Pace, and others compared sequences of conserved genes such as ribosomal RNAs and discovered that prokaryotes fall into two distinct molecular classes that are as different from each other as either is from the eukaryotes.
Skill: Factual recall
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210 Test Bank for iGenetics
43) In the following alignment of 12 base pairs, which sites are phylogenetically informative? Sequence 1 Sequence 2 Sequence 3 Sequence 4
ATTGGTATAAAA ATTAGTATAAAA ATTGGTATCAAA ATTAGTATCAAA
Answer: Only the fourth and ninth sites show variation, and are thus potentially informative sites.
Skill: Conceptual understanding
44) It is not unusual for a phylogeny reconstruction to result in several equally robust phylogenetic trees that differ in their branching patterns. How can this be the case if evolution only took one true pathway? Answer: Phylogeny reconstruction is in part a statistical procedure, and the reconstructions depend on the amount and nature of the input data and the assumptions inherent in the particular reconstruction method (maximum parsimony, maximum likelihood, etc.). Given these constraints, it is expected that phylogenetic analysis of a given data set will, at times, produce multiple, equally well‐supported trees.
Skill: Analytical reasoning
45) One problem commonly experienced in phylogeny reconstruction arises when two or more of the groups being compared have accumulated far more genetic changes (mutations) than the others in the comparison. This situation often results in a phenomenon known as ʺlong branch attraction,ʺ in which the highly mutated lineages end up grouping together in a phylogenetic analysis even if they are distantly related. What do you think causes long branch attraction? How might it be compensated for? Answer: Since nucleotides can change only among four states ( A,C,T, and G), in lineages with disproportionately many base substitutions the odds of becoming convergently similar (or at least obscuring the underlying pattern of homology with their true nearest relative) increases greatly. This problem may be compensated for by using data from several different genes or regions of the genome.
Skill: Conceptual understanding
46) Distinguish between a gene tree and a species tree. Answer: Trees based on a single homologous gene reflect the evolutionary history of that gene, but its pattern of divergence may not exactly match that of the species carrying that gene. Single‐gene phylogenies are gene trees, while several such genes can be collectively analyzed to infer a more accurate picture of species‐level phylogeny, or species tree.
Skill: Conceptual understanding
47) In a molecular‐evolutionary analysis, you estimate the number of base substitutions per site (K) at 0.82. Given an estimated divergence time ( T) of 10 million years based on the fossil record, what is the substitution rate (r)? Answer: Using the relationship r = K/(2T) to compute r, the rate is: 0.82/2(1 × 10 7 ) = 0.82 × 10 -8 .
Skill: Problem-solving
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Chapter 23 Molecular Evolution 211
48) One of the lines of evidence supporting the theory that modern humans originated in Africa is that the modern‐day human population of Africa contains the greatest diversity of genetic variants of any human population. Why is this consistent with the idea that humans first evolved in Africa? Answer: The reasoning is that populations arising in and anciently occupying an area will have had more time to accumulate genetic variants. Derivative populations migrating elsewhere would experience a founder effect, which would lead to loss of some variation.
Skill: Analytical reasoning
49) Humans have an extremely high level of genetic similarity, with greater than 99% sequence identity across populations. Yet, many people appear very different phenotypically. How do you reconcile these observations? Answer: These observations are very easily reconciled. The phenotypic differences are actually genetically trivial. Differences in body shape, skin color, eye color, body shape, and so on, undoubtedly result from variation in a scant few coding loci. In addition, much phenotypic variation is caused by environmental factors, such as nutrition. The human genome consists of some three billion nucleotides, so the few loci that underlie phenotypic differences constitute so trivial a subset of these that geneticists generally do not accept the idea of a genetic basis for ʺrace.ʺ
Skill: Analytical reasoning
50) Describe briefly the mechanisms that are thought to give rise to gene duplication and multigene families. Answer: Gene duplication, arising from unequal crossing‐over, can give rise to tandem gene copies. Some of these may then evolve different but related functions. Transposition can also create duplicate copies of genes, but these are less likely to occur in tandem, typically inserting elsewhere in the genome.
Skill: Factual recall
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iGenetics: A Molecular Approach
Third Edition by Russell / Bose
Benjamin Cummings
c.2010 3/10/09
Contents
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Genetics: An Introduction...............................................................................................1 DNA: The Genetic Material ............................................................................................9 DNA Replication............................................................................................................17 …show more content…
Gene Function ................................................................................................................26 Gene Expression: Transcription ...................................................................................35 Gene Expression: Translation.......................................................................................44 DNA Mutation, DNA Repair, and Transposable Elements .....................................53 Genomics: The Mapping and Sequencing of Genomes ............................................62 Functional and Comparative Genomics .....................................................................71 Recombinant DNA Technology ...................................................................................80 Mendelian Genetics .......................................................................................................88 Chromosomal Basis of Inheritance ..............................................................................97 Extensions of and Deviations from Mendelian Genetic Principles.......................105 Genetic Mapping in Eukaryotes ................................................................................115 Genetics of Bacteria and Bacteriophages ..................................................................126 Variations in Chromosome Structure and Number ................................................136 Regulation of Gene Expression in Bacteria and Bacteriophages ...........................145 Regulation of Gene Expression in Eukaryotes.........................................................154 Genetic Analysis of Development .............................................................................163 Genetics of Cancer .......................................................................................................172 Population Genetics.....................................................................................................181 Quantitative Genetics ..................................................................................................192 Molecular Evolution ....................................................................................................202
Chapter 1 Genetics: An Introduction
MATCHING QUESTIONS Please select the best match for each term. 1) A system for searching several linked databases
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A) PubMed B) OMIM C) GenBank D) BLAST E) Entrez
2) A tool to compare nucleotide or protein sequences
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3) A website that provides links to research articles
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4) A database of human genes and disorders
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5) A database of genetic sequences
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Answers:
1) E
2) D
3) A
4) B
5) C
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1
2 Test Bank for iGenetics
Please select the best match for each term. 6) Developed the polymerase chain reaction
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A) Boyer and Cohen B) Barbara McClintock C) Paul Berg D) Gregor Mendel E) Kary Mullis
7) Constructed the first recombinant DNA molecule
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8) Cloned the first recombinant DNA molecule
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9) Conducted breeding experiments with the pea plant
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10) Discovered transposable elements
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Answers:
6) E
7) C
8) A
9) D
10) B
MULTIPLE-CHOICE QUESTIONS 11) The field of genetics includes A) the study of heredity. B) the molecular nature of the genetic material. C) the ways in which genes control life functions. D) the distribution and behavior of genes in populations. E) All of these Answer: E
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12) Quantitative genetics is A) the quantitative study of the activity of a gene. B) the study of the genetic diversity within a large group of individuals of the same species. C) the study of the genetic diversity within members of a number of related species. D) the study of traits that are determined by a number of genes simultaneously. E) the study of the number of characteristics found in an organism. Answer: D
Skill: Factual recall
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Chapter 1 Genetics: An Introduction 3
13) Bacteria and Archaea are A) eukaryotic. B) prokaryotic. C) archaeotic. D) anucleotic. E) proteozoic. Answer: B
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14) Transposons are A) proteins that are stable. B) proteins that are unstable. C) DNA segments that are stable. D) DNA segments that are unstable. E) DNA segments that code for proteins. Answer: D
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15) Genes influence all aspects of life because they A) are structural elements of the cell. B) regulate movement of proteins. C) produce RNA and protein needed for different processes. D) localize to the nucleus. E) are needed for DNA synthesis. Answer: C
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16) An ideal organism for a geneticist to use as a study organism would possess which of the following characteristics? A) A relatively short life cycle B) A large degree of genetic variation among individuals C) Matings that produce large numbers of offspring D) A well-characterized genetic background, karyotype, etc. E) All of these are desirable characteristics Answer: E
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17) The organism that Mendel used for his experiments was A) E. coli. B) the fruit fly. C) maize. D) yeast. E) the pea plant. Answer: E
Skill: Factual recall
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4 Test Bank for iGenetics
18) A map unit (μ) is A) the position of a gene on the chromosome. B) the unit of genetic distance on a chromosome. C) the unit of physical distance on a chromosome. D) equivalent to a nucleotide. E) equal to the number of recombinant progeny. Answer: B
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19) Which of the following is a nematode worm that has been used as a model organism in genetics? A) Drosophila melanogaster B) Mus musculus C) Arabidopsis thaliana D) Caenorhabditis elegans E) Homo sapiens Answer: D
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20) In eukaryotic cells, ________ are energy-producing organelles that contain DNA. A) nuclei B) lysosomes C) chromosomes D) nucleoli E) mitochondria Answer: E
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21) The principles of heredity were first established through breeding experiments carried out in ________ by Gregor Mendel. A) the late 1700s B) the 1950s C) the 1860s D) the early 1900s E) the mid-1600s Answer: C
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22) Tetrahymena is a A) fungus. B) protozoa. C) green alga. D) weed. E) bacteria. Answer: B
Skill: Factual recall
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Chapter 1 Genetics: An Introduction 5
23) Eukaryotic chromosomes consist of A) DNA alone. B) DNA complexed with protein. C) DNA complexed with RNA. D) DNA complexed with fatty acids. E) DNA complexed with protein and RNA. Answer: B
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24) Translation of RNA into proteins occurs A) in the nucleus. B) in the endoplasmic reticulum. C) in the cytoplasm. D) in both the nucleus and the cytoplasm. E) in both the nucleus and the endoplasmic reticulum. Answer: C
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25) The branch of genetics concerned with analyzing the structure and function of genes is A) molecular genetics. B) plant genetics. C) transmission genetics. D) population genetics. E) applied genetics. Answer: A
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TRUE-FALSE QUESTIONS 26) Genetics is central to biology because genes and their functions form the basis of all life processes. Answer: TRUE
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27) There is a sharp divide between basic and applied research. Answer: FALSE Explanation: Basic and applied research employ the same information and similar techniques to answer questions.
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28) The smooth endoplasmic reticulum has ribosomes attached to it. Answer: FALSE Explanation: The rough endoplasmic reticulum has ribosomes attached to it.
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29) Centrioles are also called basal bodies. Answer: TRUE
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6 Test Bank for iGenetics
30) The centrosomes are a part of the chromosome that help in chromosome movement during mitosis and meiosis. Answer: FALSE Explanation: The centrosomes are a region of undifferentiated cytoplasm.
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31) Bacteria and plants both have a rigid cell wall outside the cell membrane. Answer: TRUE
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32) Membrane-bound nuclei are characteristic of both Archaea and eukaryotes. Answer: FALSE Explanation: Membrane-bound nuclei are characteristic of eukaryotes. Archaea, like Bacteria, are prokaryotes.
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33) Genetic maps show the relative location and arrangement of genes on chromosomes. Answer: TRUE
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34) Archaea are prokaryotes that are often found in inhospitable conditions such as hot springs and salt marshes. Answer: TRUE
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35) E. coli is a rod-shaped bacterium found in the human intestine. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) What is the hypothetico-deductive method of investigation? Answer: The hypothetico-deductive method of investigation consists of observing a phenomenon, forming hypotheses to try and explain the observations, making experimental predictions based on those hypotheses, and finally testing out the predictions by doing specific experiments.
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37) How do mutations help us in understanding the particular functions of a gene? Answer: Comparing mutants with normal cells gives us an idea as to which life process has been affected. This helps us locate the mutated gene in the specific biochemical pathway for that particular process.
Skill: Conceptual understanding
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Chapter 1 Genetics: An Introduction 7
38) How does the knowledge of complete genomes of different organisms help us in any way? Answer: Since the function of most genes is conserved to a large degree among different organisms, the knowledge of the genomes of different species helps us compare them, leading to a better understanding of gene functions. This in turn will lead to a better understanding of human genetic diseases and will allow us to develop better cures.
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39) How can a genetic map be used?
Answer: Genetic maps can be used in the process of localizing genes and studying the distribution of genes on chromosomes and in the genome.
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40) What is recombination frequency? Answer: Recombination frequency is the percentage of recombinant progeny, that is, the percentage of the progeny in which the arrangement of alleles has switched compared to their arrangement in the parents. This is taken to be a measure of the genetic distance between genes and is used to create genetic maps.
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41) Are genetic maps based on the actual distance on chromosomes as measured by nucleotides? Answer: No. Genetic maps are based on the frequency of recombination between genes. As such, they show relative distances between genes but cannot specify actual distance in terms of nucleotides.
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42) Which organelles besides the nucleus contain their own DNA? Answer: The mitochondria in eukaryotes and the chloroplasts in plants contain DNA of their own.
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43) The nucleus separates the chromosomes from the rest of the cell. How is the information in the DNA communicated to the cell? Answer: The nuclear membrane has pores and is permeable to certain molecules. This allows selected materials, like RNA and proteins, to move in and out of the …show more content…
nucleus.
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44) What is recombinant DNA technology? Answer: Recombinant DNA technology encompasses procedures that allow scientists to join together DNA from two or more different organisms and make many identical copies of them (cloning).
Skill: Factual recall
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8 Test Bank for iGenetics
45) For geneticists, why is it important that genetic variability exist in the population under study? Answer: Genetic variation in individuals of a population is important for studying the inheritance pattern of those characteristics. If all the members of a population were identical for the trait under study, their progeny would be as well, and it would be impossible to determine how the trait was being passed on to the offspring.
Skill: Conceptual understanding
46) What characteristics does an organism have to possess to be a good genetic model? Answer: To be a good genetic model, an organism has to have a well -known genetic history, a short life cycle, produce many offspring, be easy to handle, and have genetic variability among the individuals in a population.
Skill: Factual recall
47) Why are genetic databases so important to the study of modern genetics? Answer: The available information about the genetics of a large number of organisms has increased dramatically in the past few years. As a result, it has become impossible to study, learn, and remember all of it. All this information needs to be stored in such a way that it may be accessed and searched easily. This is done in computer databases that have assumed enormous importance in the study of genetics.
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48) What is the difference between PubMed and OMIM? Answer: PubMed is a website that is used to access citations and abstracts of journal articles. OMIM is a database of human genes and genetic disorders.
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49) What is the technique that is used to amplify DNA sequences? Answer: The procedure employed to amplify DNA sequences is called polymerase chain reaction (PCR).
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50) What are the differences between eukaryotes and prokaryotes? Answer: Eukaryotes possess a nucleus that is separated from the rest of the cell by a double-layered nuclear membrane. Prokaryotes lack a nucleus. Prokaryotes also lack all membrane-bound organelles, such as mitochondria, endoplasmic reticula, and Golgi bodies. Such organelles are present in eukaryotes. Prokaryotes are generally also much smaller than eukaryotes, and are usually unicellular.
Skill: Factual recall
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Chapter 2 DNA: The Genetic Material
MATCHING QUESTIONS Please select the best match for each term. 1) Centromere
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A) The region of a prokaryotic cell where the chromosome is located B) The basic structural unit of chromatin with ʺbead-on-a-stringʺ morphology C) A DNA molecule and associated proteins D) The region of a eukaryotic chromosome found near the attachment point of mitotic or meiotic spindle fibers E) The constituent monomer of DNA and RNA
2) Nucleoid
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3) Nucleotide
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4) Chromosome
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5) Nucleosome
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Answers:
1) D
2) A
3) E
4) C
5) B
MULTIPLE-CHOICE QUESTIONS 6) Loosely aggregated DNA bound to proteins in a eukaryotic cell is called A) chromosomes. B) chromatin. C) chromatid. D) centromere. E) nucleoid. Answer: B
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7) The C-value is the amount of DNA in a A) haploid genome. B) diploid genome. C) bacterial genome. D) eukaryotic genome. E) cellʹs nucleus. Answer: A
Skill: Factual recall
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10 Test Bank for iGenetics
8) The chromosome of most prokaryotes differs from those of eukaryotes in that A) the prokaryotic chromosome is linear, while the eukaryotic chromosome is circular. B) the prokaryotic chromosome is circular, while the eukaryotic chromosome is linear. C) the prokaryotic chromosome does not replicate before mitosis, while the eukaryotic chromosome does. D) the prokaryotic chromosome does not contain genes, while the eukaryotic chromosome does. E) the prokaryotic chromosome is not necessary for the organismʹs survival, while the eukaryotic chromosome is. Answer: B
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9) A Barr body is an example of A) constitutive euchromatin. B) facultative euchromatin. C) facultative heterochromatin. D) a nucleosome. E) constitutive heterochromatin. Answer: C
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10) The definition of transformation is A) the shift of genetic information from DNA to protein. B) the genetic alteration of an organism. C) the uptake of genetic information by a cell from the environment. D) Both B and C E) None of these Answer: D
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11) In Griffithʹs experiment involving the transformation of Streptococcus pneumoniae, A) the R strain was virulent. B) the S strain was virulent. C) both the R and S strains were virulent. D) the R strain had a protein capsule. E) the S strain had a protein capsule. Answer: B
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12) What was the transforming principle isolated in Griffithʹs experiment? A) Protein B) RNA C) DNA D) Virus E) Polysaccharide Answer: C
Skill: Factual recall
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Chapter 2 DNA: The Genetic Material 11
13) Who used radioactively labeled T2 bacteriophage to confirm the identity of the transforming principle? A) Griffith B) Hershey and Chase C) Avery D) Gierer and Schramm E) Beadle and Tatum Answer: B
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14) Which part of the T2 bacteriophage entered E. coli cells in the experiment which confirmed the identity of the transforming principle? A) The RNA B) The DNA C) The whole virus D) The protein coat E) No part Answer: B
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15) Certain ________ have RNA for their genetic material. A) bacteria B) viruses C) plants D) eukaryotes E) prokaryotes Answer: B
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16) What did the X-ray diffraction patterns initially reveal about the DNA molecule? A) It is of uniform diameter and has a helical structure. B) It is a helical molecule with paired bases in the center. C) It is double-stranded with antiparallel strands. D) It is acidic, phosphorus-rich, and large. E) It contains hereditary information. Answer: A
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17) What did Watson and Crick deduce about the three-dimensional structure of DNA? A) There is a repeating pattern every 3.4 nm and every 0.34 nm. B) It is a double-stranded helix. C) It contains a lot of phosphorus. D) It is a large molecule. E) It consists of supercoiled chromatin. Answer: B
Skill: Factual recall
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12 Test Bank for iGenetics
18) Which of the following is a nonhistone protein found in chromatin? A) H1 B) HMG C) H2A D) H5 E) All of these Answer: B
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19) Antiparallel means that A) the two polynucleotide chains run in opposite directions. B) each DNA molecule consists of one old and one new strand. C) opposite strands are held together by base pairing. D) the helix twists to the right. E) there is complementary base-pairing. Answer: A
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20) Complementary base-pairing allows for A) spontaneous mutations to occur. B) genes to be expressed as a phenotype. C) DNA to serve as its own template for replication. D) replication to be semiconservative. E) covalent bonds to form between the opposite bases. Answer: C
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21) Which of the following are the purine nucleotides in DNA? A) Adenine and thymine B) Cytosine and guanine C) Adenine and cytosine D) Guanine and adenine E) Thymine and uracil Answer: D
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22) Topoisomerases function to A) remove nucleotides from DNA. B) join DNA pieces together. C) twist DNA molecules. D) attach DNA loops to scaffold proteins. E) move chromosomes along spindle fibers. Answer: C
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Chapter 2 DNA: The Genetic Material 13
23) Which form of DNA is a left-handed double helix? A) A-DNA B) B-DNA C) L-DNA D) R-DNA E) Z-DNA Answer: E
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24) The displacement loop (D-loop) may be a characteristic of A) centromeres. B) telomeres. C) A-DNA. D) B-DNA. E) Z-DNA. Answer: B
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25) Which nucleotide is absent in RNA? A) Adenine B) Guanine C) Uracil D) Cytosine E) Thymine Answer: E
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TRUE-FALSE QUESTIONS 26) DNA and RNA both contain phosphate and ribose. Answer: FALSE Explanation: They both contain phosphate, but DNA contains the sugar deoxyribose rather than ribose.
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27) Hershey and Chase used radioactive sulfur to label the genetic material of bacteriophages. Answer: FALSE Explanation: The radioactive sulfur labeled the protein coat.
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28) In a strand of DNA, a hydrogen bond connects the phosphate group of one nucleotide to the sugar of the adjacent nucleotide. Answer: FALSE Explanation: A covalent phosphodiester bond connects the two adjacent nucleotides.
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29) The genome of the T-even family of bacteriophage consists of single-stranded RNA. Answer: FALSE Explanation: It consists of double-stranded DNA.
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14 Test Bank for iGenetics
30) Borrelia burgdorferi is a bacterium whose genome consists of one large and several small linear chromosomes. Answer: TRUE
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31) By weight, the amount of DNA in chromatin is less than that of histone. Answer: FALSE Explanation: The weights of DNA and histone in chromatin are equal.
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32) The virus first shown to have RNA as its genetic material was tobacco mosaic virus (TMV). Answer: TRUE
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33) The more condensed a part of a chromosome is, the more likely it is that the genes in that region will be active. Answer: FALSE Explanation: The genes in a region are less likely to be active the more condensed a part of a chromosome is.
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34) The genome of most prokaryotes consists of moderately repetitive DNA. Answer: FALSE Explanation: The genome of most prokaryotes consists of unique-sequence DNA.
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35) In eukaryotes, the greatest relative amount of tandemly repeated DNA is associated with centromeres and telomeres. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) In Griffithsʹ transformation experiments, under what conditions did the injected mice die? Answer: The mice died when they were injected with living, virulent bacteria, and when they were injected with living, nonvirulent bacteria mixed with heat-killed, virulent bacteria.
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37) How could you test whether the transforming ability of a cell extract was due to DNA or RNA? Answer: You could treat the extract with a DNase or RNase enzyme and test whether its transforming ability was intact.
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38) One of the strands in a DNA double helix has the nucleotide sequence 5'-ACCTGCTACGG-3' .
What is the sequence of the complementary DNA strand? Answer: 3'-TGGACGATGCC-5'
Skill: Problem-solving
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Chapter 2 DNA: The Genetic Material 15
39) What is the function of dispersed repeated sequences such as SINEs and LINEs in eukaryotes? Answer: Little is known about the function of such sequences, but one hypothesis is that they have no function at all. Another is that they are involved in regulating gene expression.
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40) What is the C-value paradox, and what is its cause? Answer: There is also no direct relationship between the C-value (the total amount of DNA in the haploid genome) and the structural or organizational complexity of the organism. This is due in part to the amount of repetitive-sequence DNA found in the genome of some organisms.
Skill: Conceptual understanding
41) Define Chargaffʹs rules of the base composition of DNA. Answer: Chargaffʹs rules include the following: (1) the amount of adenine = the amount of thymine, (2) the amount of guanine = the amount of cytosine, and (3) the amount of purines = the amount of pyrimidines.
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recall
42) Describe the differences between heterochromatin and euchromatin in chromosomes. Are there any situations in which one can be changed into the other? Answer: Euchromatin contains actively transcribed genes and undergoes normal cycles of condensation and decondensation in the cell cycle. Heterochromatin remains condensed and contains genes that are usually transcriptionally inactive. Euchromatin can be inactivated, as in the case of Barr bodies. It is then known as facultative heterochromatin.
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43) What are the three necessary characteristics of the hereditary molecule in cells? Answer: (1) It must be able to carry information, (2) it must be able to accurately pass on the information to progeny cells (replicate), and (3) it must be capable of change (evolution).
Skill: Conceptual understanding
44) Name the constituent parts of a nucleoside and a nucleotide. Answer: A nucleoside consists of a pentose sugar covalently bonded to a nitrogenous base; a nucleotide is a nucleoside with the addition of a phosphate group.
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45) The DNA phage ΦΧ174 was found to have a ratio of bases of 25 A:33T :24G:18C. This departs from the usual A/T = 1 and G/C = 1 ratios. How can you explain this? Answer: The genome of the phage consists of single-stranded, rather than double-stranded, DNA.
Skill: Conceptual understanding
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16 Test Bank for iGenetics
46) If the human egg has 3 billion base pairs, how many nucleosomes will be present in the nucleus of a human somatic cell? Answer: In humans, the DNA wrapped around each nucleosome is approximately 200 bp (147 bp + 53 bp linker). As such, there will be approximately 3 × 109 /2 × 102 = 1.5 × 107 nucleosomes in a human egg nucleus. However, the egg is haploid, whereas the somatic cells are diploid. Therefore, there will be approximately 1.5 × 107 × 2 = 3 × 107 nucleosomes in the nucleus of a human somatic cell.
Skill: Problem-solving
47) Why are the amino acid sequences of eukaryotic histones so similar to one another, even among distantly related species? Answer: Evolutionary conservation of these sequences is a strong indicator that histones perform the same basic role in organizing the DNA in the chromosomes of all eukaryotes.
Skill: Conceptual understanding
48) Describe the packing of chromatin from the 10-nm to the 30-nm fiber stage. What is the role of histones? Answer: 10-nm chromatin fiber consists of nucleosomes–ʺbeadsʺ of DNA wound around eight core histone proteins–connected by strands of linker DNA. The 30-nm chromatin fiber is created by the binding of histone H1, which brings the linker DNA and the nucleosomes closer together. In the solenoid model of the 30 -nm fiber, the nucleosomes are brought together into a spiraling helical structure, with about six nucleosomes per complete turn.
Skill: Conceptual understanding
49) What is the role of centromeres and telomeres? Answer: Centromeres are the chromosomal regions where mitotic or meiotic spindle fibers attach. They are therefore responsible for the accurate segregation of chromosomes to daughter cells during replication. Telomeres are heterochromatic regions of chromosomes that are also required for replication and stability. They are usually found at the ends of the chromosome and are associated with the nuclear envelope.
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50) If the base pairs in a DNA helix are 0.34 nm apart, and a complete (360°) turn of the helix takes 3.4 nm, how many base pairs per turn are there in a DNA molecule? Answer: There are 10 base pairs per turn.
Skill: Problem-solving
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Chapter 3 DNA Replication
MATCHING QUESTIONS Please select the best match for each term. 1) Primosome
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A) A complex of proteins and RNA that replicates the ends of eukaryotic chromosomes B) A complex of helicase and primase on the template DNA C) A complex of key replication proteins at the replication fork in E. coli and bacteriophage DNA D) The distance between the origin and the termination of replication where two replication forks fuse E) A DNA sequence that contains the specific region where replication begins
2) Replicon
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3) Replisome
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4) Telomerase
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5) Replicator
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Answers:
1) B
2) D
3) C
4) A
5) E
MULTIPLE-CHOICE QUESTIONS 6) During replication, the direction of synthesis of new DNA from the leading and lagging strands is A) 5ʹ to 3ʹ only. B) 3ʹ to 5ʹ only. C) from left to right only. D) both 5ʹ to 3ʹ and 3ʹ to 5ʹ. E) different, depending on whether the cell is prokaryotic or eukaryotic. Answer: A
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7) DNA replication is always A) discontinuous. B) bidirectional. C) conservative. D) semiconservative. E) dispersive. Answer: D
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8) In E. coli, replication begins at which chromosome site? A) The replication fork B) ter C) oriC D) TBP E) All of these Answer: C
Skill: Factual recall
9) Which of the following did Kornberg use to detect DNA synthesis? A) Radioactively labeled E. coli cells B) Fluorescently labeled E. coli cells C) Radioactively labeled deoxynucleotide triphosphates D) Fluorescently labeled deoxynucleotide triphosphates E) Nonlabeled deoxynucleotide triphosphates Answer: C
Skill: Factual recall
10) How many replication forks are produced when DNA denatures at an origin? A) 0 B) 1 C) 2 D) 3 E) The number varies Answer: C
Skill: Factual recall
11) What is a replication bubble? A) A complex of replication enzymes on the DNA template strand B) A DNA sequence that initiates replication C) A tangle of denatured DNA strands near the replication fork D) A locally denatured segment of DNA where replication originates E) A localized site in the nucleus where chromosomes are replicating Answer: D
Skill: Factual recall
12) Which of the following are necessary for DNA replication in vitro? A) RNA, helicase, primers, DNA polymerase B) Okazaki fragments, helicase, DNA polymerase C) Template DNA, DNA polymerase, four dNTPs, primers, magnesium ions D) Template DNA, four dNTPs, magnesium ions E) DNA canʹt replicate in vitro Answer: C
Skill: Factual recall
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Chapter 3 DNA Replication 19
13) In Meselson and Stahlʹs experiment, what kind of DNA molecules would be found after four replication cycles? A) Only heavy DNA (15N- 15N) B) Only intermediate DNA ( 15N- 14N) C) Only light DNA ( 14N- 14N) D) Both heavy ( 15N- 15N) and light DNA (14N- 14N) E) Both heavy ( 15N‐15N) and intermediate DNA (15N- 14N) Answer: D
Skill: Conceptual understanding
14) The two most basic steps of DNA replication are A) primase causes primer to bind the template and ligase copies the template. B) helicase unwinds the template and DNA polymerase binds the template. C) leading strand is copied first and lagging strand is copied second. D) the new strand is denatured and a template is synthesized. E) the template is denatured and a new strand is synthesized. Answer: E
Skill: Conceptual understanding
15) Where does the initiator protein bind DNA at the start of replication? A) At a replication fork B) At an origin of replication C) At any AT-rich region D) At a promoter region E) At a start codon Answer: B
Skill: Factual recall
16) As helicase unwinds the DNA molecule, what keeps the strands apart? A) DNA polymerase B) Reverse transcriptase C) Replication fork D) Single-strand binding proteins E) Okazaki fragments Answer: D
Skill: Factual recall
17) After removal of the RNA primers and replacement with DNA nucleotides, the single-stranded nick adjacent to Okazaki fragments is filled in through a reaction that involves which enzyme? A) DNA primase B) SSB protein C) RNA polymerase D) DNA ligase E) DNA helicase Answer: D
Skill: Factual recall
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18) Which enzyme elongates the new DNA strand starting at an RNA primer? A) DNA polymerase I B) DNA polymerase III C) RNA primase D) DNA ligase E) RNA polymerase Answer: B
Skill: Factual recall
19) After a region of DNA has been replicated, ________ removes the RNA primers. A) DNA polymerase I B) DNA polymerase III C) DNA helicase D) RNA primase E) DNA ligase Answer: A
Skill: Factual recall
20) Which enzyme replaces RNA primers with DNA after elongation? A) DNA polymerase I B) DNA polymerase III C) RNA polymerase D) RNA primase E) DNA ligase Answer: A
Skill: Factual recall
21) Which kind of enzyme prevents DNA from tangling up by introducing negative supercoils as the replication fork migrates during replication? A) Helicase B) Ligase C) DNA polymerase I D) DNA polymerase III E) Topoisomerase Answer: E
Skill: Factual recall
22) The base-pairing error rate remains low during replication because A) DNA repair mechanisms can fix the mispaired bases. B) bases that are mispaired can excise themselves. C) UV light radiation corrects any base mispairs. D) mispaired bases cause a cell to die before replication is complete. E) None of these; base-pairing errors are not possible Answer: A
Skill: Factual recall
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Chapter 3 DNA Replication 21
23) The enzymatic activity of a telomerase is best described as a A) polymerase. B) ligase. C) topoisomerase. D) reverse transcriptase. E) exonuclease. Answer: D
Skill: Factual recall
24) Rolling circle replication of DNA is characterized by the absence of A) the DNA polymerase. B) a nick in the DNA template. C) the RNA primers. D) the replication bubble. E) the Okazaki fragments. Answer: D
Skill: Conceptual understanding
25) Which enzyme activity is associated with the proofreading mechanism of DNA polymerase I? A) 5ʹ-to-3ʹ exonuclease activity B) 3ʹ-to-5ʹ exonuclease activity C) 5ʹ-to-3ʹ polymerase activity D) Both A and B E) All of these Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) A new nucleotide is added to a growing strand of DNA at the 3ʹ end. Answer: TRUE
Skill: Factual recall
27) Only the leading strand of a DNA molecule serves as a template during replication. Answer: FALSE Explanation: Both the leading and lagging strands serve as templates.
Skill: Factual recall
28) At the growing end of a DNA chain, DNA polymerase catalyzes the formation of a disulfide bond between the 3ʹ-OH group of the deoxyribose on the last nucleotide and the 5ʹ-phosphate of the dNTP precursor. Answer: FALSE Explanation: DNA polymerase catalyzes the formation of a phosphodiester bond between nucleotides.
Skill: Factual recall
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29) DNA primase is an RNA polymerase. Answer: TRUE Explanation: DNA primase catalyzes the reaction to synthesize a short RNA primer molecule.
Skill: Factual recall
30) Okazaki fragments are made from the lagging strand of the DNA double helix. Answer: TRUE
Skill: Factual recall
31) DNA polymerase III is very inaccurate at matching bases during replication, with errors in one out of every 100 base pairs. Answer: FALSE Explanation: DNA polymerase III is very accurate, causing errors in only one out of every 1,000,000 base pairs.
Skill: Factual recall
32) In eukaryotic cells, histone proteins are actively synthesized during the S phase of the cell cycle. Answer: TRUE
Skill: Factual recall
33) In eukaryotes, DNA replication begins at a single origin of replication on each chromosome. Answer: FALSE Explanation: Replication begins at multiple origins of replication on each chromosome.
Skill: Factual recall
34) Topoisomerase and SSB proteins are important components of the replication process in prokaryotes, but they are not found in eukaryotes. Answer: FALSE Explanation: These proteins also play key roles in eukaryotic DNA replication.
Skill: Factual recall
35) Mg 2+ ions are required for optimal DNA polymerase activity. Answer: TRUE Explanation: DNA polymerases often require magnesium ions as cofactors.
Skill: Factual recall
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Chapter 3 DNA Replication 23
SHORT ANSWER QUESTIONS AND PROBLEMS 36) What are the key replication enzymes at the replisome, and how is DNA replication on both leading and lagging strands made efficient through the conformation of the DNA at the replisome in prokaryotes? Answer: The key replication enzymes at the replisome are helicase, primase, and DNA polymerase III. To make replication more efficient, the lagging-strand DNA is folded so that its DNA polymerase III is complexed with the DNA polymerase III on the leading strand (forming the DNA Pol III holoenzyme). The folding of the lagging-strand template also makes production of sequential Okazaki fragments more efficient by bringing the 3ʹ end of each completed Okazaki fragments near the site where the next Okazaki fragment will start.
Skill: Conceptual understanding
37) Why is DNA replication referred to as semiconservative? Answer: The progeny double helices consist of one parental DNA strand and one newly synthesized strand.
Skill: Conceptual understanding
38) Why is an AT-rich sequence characteristic of DNA replicators in all organisms? Answer: AT-rich regions of DNA are relatively easy to denature to single strands. AT base pairs are held together by only two hydrogen bonds, while GC pairs are held together by three.
Skill: Conceptual understanding
39) What are the differences in replication between leading and lagging strands in terms of continuity and directionality in relation to the replication fork? Answer: The leading strand is copied continuously from the 3ʹ end toward the replication fork, while the lagging strand is copied in fragments away from the replication fork.
Skill: Factual recall
40) What experiment demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I was essential for cell viability? Answer: In E. coli, the DNA Pol I polAex1 mutant strain survives at 37˚C but dies at 42˚C. This was shown to be because the mutant DNA Pol I enzyme had normal activity at 37˚C but a defective 5ʹ-to-3ʹ activity at 42˚C. This demonstrated that the 5ʹ-to-3ʹ exonuclease activity of DNA Pol I was essential for cell survival.
Skill: Factual recall
41) A cross is made between yeast cells with different alleles for a set of linked genes: pr + q × p + rq+ . The resulting tetrads show a 3:1 ratio for r+ to r instead of the expected 2:2 ratio. Can you explain how this could have occurred? Answer: Gene conversion by a mismatch repair mechanism could have caused this deviation from expected ratios. During pairing of homologous chromosomes in meiosis, recombination between the two inner chromatids occurred, resulting in heteroduplex (mismatched) DNA strands. Both mismatches were repaired by excision and DNA synthesis to match the parent DNA with the r+ allele.
Skill: Analytical reasoning
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42) Describe the method devised by Arthur Kornberg which first successfully achieved DNA synthesis in vitro, including its components and their uses. Answer: Kornberg mixed together DNA fragments, all four dNTPs (DNA precursors), and an E. coli lysate to achieve DNA synthesis in vitro. The DNA fragments acted as a template for the synthesis of new DNA. The dNTPs were precursors of the new DNA strand, and the cell lysate contained the enzyme necessary to catalyze DNA synthesis (DNA Pol I).
Skill: Factual recall
43) The diploid set of chromosomes in Drosophila embryos replicates six times faster than the single E. coli chromosome, even though there is about 100 times more DNA in Drosophila than in E. coli and the rate of movement of the replication fork in Drosophila is much slower. How is this so? Answer: Eukaryotic chromosomes duplicate rapidly because DNA replication initiates at many origins of replication throughout the genome. In E. coli, there is only one replicon, while in eukaryotes there are multiple, smaller replicons.
Skill: Application of knowledge
44) What are some key differences in replication between E. coli DNA and λ phage DNA? Answer: Both start replication as a circular molecule, and in E. coli, the parental DNA strands remain in a circular form throughout the replication cycle. A replication bubble opens to form two replication forks, and replication proceeds bidirectionally. Lambda phage DNA is replicated by a rolling circle mechanism, in which a nick is made in one of the two strands of the circle, and the 5ʹ end of the cut DNA strand is rolled out as a free ʺtongueʺof increasing length as replication proceeds . The parental DNA circle is replicated continuously, while the linear displaced strand is replicated discontinuously. As long as the circle continues to roll, concatamers of phage DNA can be produced. This is later cut up into linear chromosomes and packaged into new phage heads.
Skill: Factual recall
45) How do the DNA polymerase repair mechanisms work? Answer: Both DNA Pol I and DNA Pol III have 3ʹ-to-5ʹ exonuclease activity and can remove nucleotides from the end of a DNA chain as part of an error correction mechanism. If an incorrect base is inserted by DNA polymerase, and the error is recognized immediately, the exonuclease activity excises the erroneous nucleotide from the new strand. After excision, the DNA polymerase resumes motion in the forward direction and inserts the correct nucleotide.
Skill: Factual recall
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Chapter 3 DNA Replication 25
46) How did Meselson and Stahl rule out the conservative model of DNA replication using equilibrium density gradient centrifugation? Answer: First, they grew E. coli cells in a medium containing only high-density nitrogen (15N). Then they allowed the cells to undergo successive rounds of replication in the presence of normal density nitrogen ( 14N). They used equilibrium density gradient centrifugation after each round of replication to separate the DNA produced by density. If the conservative model was correct, they would have found no intermediate-density DNA after a round of replication. However, after one round of replication, they found that the entire amount of DNA had a density exactly intermediate between 14N and 15N. Subsequent rounds of the experiment showed that semiconservative replication was the correct model.
Skill: Conceptual understanding
47) How were the sequences that compose the replication origins in yeast discovered? Answer: Yeast cells were grown in heavy isotope media to produce denser DNA and then shifted to media with normal, light isotopes. After a few minutes, DNA from these cells was extracted and cut into small pieces. The lighter DNA fragments (which should contain the origins of replication) were collected, fluorescently labeled, and used to hybridize a microarray of yeast sequences. Determination of the sequences to which this light, fluorescently labeled DNA hybridized led to the identification of a number of yeast replication origins.
Skill: Factual recall
48) How will DNA replication be affected if DNA polymerase I has a mutation that inactivates 5ʹ-to-3ʹ exonuclease activity? Answer: The 5ʹ-to-3ʹ exonuclease activity of DNA polymerase I removes the RNA primers from the ends of Okazaki fragments. If this activity is missing, the primers may not be removed from the growing DNA strands.
Skill: Conceptual understanding
49) When the RNA primers are removed from the 5ʹ ends of eukaryotic chromosomes after replication, DNA polymerase is unable to fill in the gaps. What prevents the chromosomes from getting shorter and shorter with each round of replication? Answer: The enzyme telomerase maintains chromosome lengths by adding telomere repeats to the chromosome ends. Telomerase contains an RNA component that is complementary to the telomere repeat unit of the chromosome. After binding to the overhanging repeat, the telomerase RNA is used as a template to synthesize new chromosomal telomere repeats.
Skill: Factual recall
50) Why do tumor cells in mammals have telomerase activity? Answer: Tumor cells are ʺimmortalʺ cells, and the enzyme telomerase is required for long-term cell viability. It has been demonstrated that mutations in genes such as TLC1 and EST1 decrease cell longevity by continuous shortening of telomere length.
Skill: Conceptual understanding
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Chapter 4 Gene Function
MATCHING QUESTIONS Please select the best match for each term. 1) Alkaptonuria
Skill: Factual recall
A) Hypoxanthine guanine phosphoribosyltransferase B) Galactose‐1‐phosphate uridyl transferase C) Hexosaminidase A D) Fructose‐1,6‐diphosphatase E) Homogentisic acid oxidase
2) Tay—Sachs disease
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3) Lesch—Nyhan syndrome
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4) Galactosemia
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5) Hypoglycemia
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Answers:
1) E
2) C
3) A
4) B
5) D
MULTIPLE-CHOICE QUESTIONS 6) Why did Garrod define alkaptonuria as a recessive genetic disease? A) It is expressed in male members of an affected family. B) It is expressed in all members of an affected family. C) It is expressed under certain environmental conditions. D) It is expressed when two alleles for the disease are present. E) It is expressed when one of the two alleles carries the disease trait. Answer: D
Skill: Factual recall
7) What was the significance of Beadle and Tatumʹs experiment? A) It resulted in the central dogma. B) It led to the discovery of bread mold. C) It resulted in the one‐gene‐one‐enzyme hypothesis. D) It led to the discovery of the genetic code. E) It proved that X‐rays were mutagens. Answer: C
Skill: Factual recall
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Chapter 4 Gene Function 27
8) What does a gene actually code for? A) A polypeptide B) An enzyme C) An amino acid D) A protein E) A nucleotide Answer: A
Skill: Factual recall
9) Which genetic disease(s) are caused by defective proteins that are not enzymes? A) Tay—Sachs disease and phenylketonuria B) Sickle‐cell anemia and cystic fibrosis C) Lesch—Nyhan syndrome. D) Albinism and alkaptonuria E) All of these Answer: B
Skill: Factual recall
10) Amniocentesis involves A) prenatal diagnosis of genetic defects in a fetus. B) tests for enzyme and protein deficiencies in a fetus. C) taking a sample of amniotic fluid with a needle. D) analysis for DNA and chromosome defects in a fetus. E) All of these Answer: E
Skill: Factual recall
11) Auxotrophs are organisms that can grow A) on minimal media. B) on complete media. C) only on amino acid supplemented media. D) only on vitamin supplemented media. E) on any media. Answer: B
Skill: Factual recall
12) A lysosomal storage disease is caused by A) failure of the cell to produce lysosomes. B) the inability to synthesize glycolipids. C) mutations in genes for lysosomic membrane proteins. D) an excess in the production of lysosomic digestive enzymes. E) mutations in genes that code for lysosomic digestive enzymes. Answer: E
Skill: Factual recall
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13) Cystic fibrosis is caused by a mutation in the gene that encodes which enzyme? A) Pyruvate kinase B) CFTR protein C) Hexosaminidase A D) Hemoglobin E) Tyrosine kinase Answer: B
Skill: Factual recall
14) In a molecule of hemoglobin C, an aspartic acid residue is changed into a ________ residue. A) lysine B) tyrosine C) leucine D) phenylalanine E) glutamine Answer: A
Skill: Factual recall
15) A mutation that has pleiotropic consequences A) results in a single symptom in the affected person. B) may result in only slight symptoms in a person with the mutation. C) is only detected in homozygotes. D) has widespread consequences in the affected person. E) cannot be detected by enzyme assay. Answer: D
Skill: Factual recall
16) Individuals with Lesch—Nyhan syndrome exhibit A) urine that turns black when exposed to air. B) compulsive self‐mutilation behaviors. C) lighter than normal pigmentation of skin, hair, and eyes. D) severe anemia. E) a cherry‐colored spot on the retina. Answer: B
Skill: Factual recall
17) A mutagen is A) the process that generates mutants. B) the agent that induces mutants. C) the mutants that are generated by an agent. D) the experiment that generates mutants. E) the process that corrects mutants. Answer: B
Skill: Factual recall
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Chapter 4 Gene Function 29
18) In people who do not have Tay—Sachs disease, the hexA gene encodes an enzyme that A) converts purines to uric acid in the kidneys. B) removes a sugar group from phenylalanine. C) converts tyrosine to melanin in skin. D) cleaves acetylgalactosamine from gangliosides. E) synthesizes glycogen from glucose monomers. Answer: D
Skill: Factual recall
19) A person with sickle‐cell trait has ________ at the β‐globin locus. A) two wild‐type alleles B) two mutant alleles C) one wild‐type and one mutant allele D) one wild‐type and one deleted allele E) one mutant and one deleted allele Answer: C
Skill: Factual recall
20) Which of the following is a sex‐linked disorder? A) Lesch—Nyhan syndrome B) Alkaptonuria C) Albinism D) Down syndrome E) Tay—Sachs disease Answer: A
Skill: Factual recall
21) Which of the following statements about phenylketonuria (PKU) is false? A) Buildup of phenylpyruvic acid affects the nervous system. B) The amino acid tyrosine cannot be made. C) Production of the hormone thyroxin s affected. D) The skin color of those with phenylketonuria is light because of decreased levels of melanin. E) Adrenalin levels increase in the blood stream. Answer: E
Skill: Factual recall
22) The cystic fibrosis gene (CFTR) codes for a protein that is essential for A) secretion of mucus. B) ion transport across membranes. C) transport of oxygen in red blood cells. D) metabolism of certain amino acids. E) processing of gangliosides. Answer: B
Skill: Factual recall
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23) A carrier for a disease A) is homozygous for the dominant mutation. B) is homozygous for the recessive mutation. C) is heterozygous for the dominant mutation. D) is heterozygous for the recessive mutation E) is hemizygous for any mutation. Answer: D
Skill: Factual recall
24) Normal and sickle‐cell hemoglobin molecules differ A) in the number of amino acids in each molecule. B) by a single DNA point mutation that leads to the substitution of one amino acid for another. C) in the number and orientation of the amino acid chains attached to the heme portion of each molecule. D) in the number of oxygen molecules that can be carried by each molecule. E) by the type of bone marrow that produces them. Answer: B
Skill: Factual recall
25) Cystic fibrosis is an autosomal recessive disorder. What is the chance that two carrier parents will have an affected child? A) 100% B) 75% C) 50% D) 25% E) 0% Answer: D
Skill: Problem-solving
TRUE-FALSE QUESTIONS 26) A Neurospora mutant that is a tryptophan auxotroph does not need to be grown on a medium that contains tryptophan. Answer: FALSE Explanation: A tryptophan auxotroph cannot synthesize its own tryptophan. It should be grown on a medium that contains tryptophan.
Skill: Factual recall
27) Sickle‐cell disease is more common among people who are of African rather than European descent. Answer: TRUE
Skill: Factual recall
28) Because Lesch—Nyhan syndrome is a recessive X‐linked disease, females who have an allele with the mutation responsible for this disease are generally symptom‐free. Answer: FALSE Explanation: If her normal allele has undergone lyonization (random X inactivation), the carrier female may display symptoms.
Skill: Factual recall
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Chapter 4 Gene Function 31
29) A person who is homozygous for the gene for hemoglobin A will have sickle‐cell disease. Answer: FALSE Explanation: This person will have normal hemoglobin and normal red blood cells.
Skill: Factual recall
30) The Guthrie test screens for excessive phenylalanine in the blood of infants. Answer: TRUE
Skill: Factual recall
31) Albinism is caused by an inborn error of metabolism. Answer: TRUE
Skill: Factual recall
32) Genetic counseling is the best advice that a physician can give a person about his or her risk of having a child with a genetic disorder while lacking precise statistical evidence. Answer: FALSE Explanation: If knowledge about the role of heredity in a genetic disorder is lacking, genetic counseling can only give a general estimate of risk. In many circumstances, however, the physician can give a person precise probabilities about the risk of passing on a genetic defect.
Skill: Factual recall
33) The sickle‐cell mutant allele and wild‐type β‐globin allele are codominant. Answer: TRUE
Skill: Factual recall
34) Cystic fibrosis is a pleiotropic disease. Answer: TRUE
Skill: Factual recall
35) Tay—Sachs disease is most prevalent among populations of the Pennsylvania Amish in the United States. Answer: FALSE Explanation: It is most prevalent among Ashkenazi Jews of central and eastern European descent.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) The following is a representation of a metabolic pathway that involves genes for enzymes A, B, C, and D: A → B → C → D → product A mutation in the B gene would result in an accumulation of which gene product? Answer: A
Skill: Analytical reasoning
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37) What is an essential amino acid? Give an example. Answer: An essential amino acid is one that humans must consume as part of their diet, because they are unable to synthesize it. An example is phenylalanine.
Skill: Conceptual understanding
38) In the laboratory, how could you distinguish between normal hemoglobin A and hemoglobin S? Answer: The two forms of hemoglobin have different electrical charges. Therefore, you could distinguish between them by their migration patterns on an electrophoresis gel.
Skill: Application of knowledge
39) A person with PKU controls the buildup of phenylpyruvic acid in his or her blood by restricting dietary intake of phenylalanine, thus avoiding the severe symptoms of mental retardation, slow growth rate, and early death. Yet he or she still exhibits the symptoms of fair skin and low adrenaline levels. Why? Answer: The PKU mutation is in the gene for phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. Tyrosine is essential for melanin and adrenaline production. A PKU individual cannot make tyrosine, and it is difficult to obtain enough tyrosine from food intake to avoid any symptoms of the disease.
Skill: Analytical reasoning
40) How do scientists, lacking the ability to make controlled genetic crosses in humans, study the inheritance of and give genetic counseling on human genetic disorders? Answer: Pedigree analysis can be done by compiling phenotypic records from both families over several generations. This can determine the likelihood that a particular allele is present in either family.
Skill: Application of knowledge
41) What is an inborn error of metabolism? Give an example. Answer: An inborn error of metabolism is a genetic disease caused by the absence of a particular enzyme necessary to fulfill a biochemical pathway. An example is alkaptonuria, in which an enzyme necessary for homogentisic acid (HA) metabolism is not made. Therefore, people with this disease cannot metabolize HA and it is excreted in their urine.
Skill: Factual recall
42) Why are genetic diseases much more common among children of marriages involving first cousins than among children of marriages between unrelated partners? Answer: This is because first cousins have many alleles in common, so the chances are greater for recessive alleles to be homozygous in children of first‐cousin marriages.
Skill: Conceptual understanding
43) What factor do the diseases PKU, albinism, and alkaptonuria all have in common? Answer: They are all marked by a block in the phenylalanine‐tyrosine metabolic pathways.
Skill: Factual recall
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Chapter 4 Gene Function 33
44) You discover a mutant strain of Neurospora crassa that will not grow on a minimal medium but that will grow on a medium supplemented by the amino acid methionine. How can you determine which step in the methionine synthesis pathway is affected by the mutation? Answer: You would grow the strain on a series of media each supplemented with a chemical intermediate in the methionine biosynthetic pathway and observe for growth. The mutant will be able to grow on any medium that contains an intermediate substance in the pathway that comes after the step controlled by its mutant gene. It will also accumulate the intermediate compound used in the step where it is blocked.
Skill: Analytical reasoning
45) Classic albinism is an autosomal recessive mutation. However, two parents with albinism may produce a child with normal pigment. How can this be so? Answer: There are a number of biochemical steps in melanin biosynthesis from tyrosine. For this reason, albinism can be caused by different enzyme deficiencies at different steps of the pathway. If each of the parentsʹ albinism stems from a different enzyme deficiency, they can therefore have a child with normal pigment.
Skill: Analytical reasoning
46) While the defective gene product in patients with PKU was identified using biochemical analysis, how was the cystic fibrosis gene identified? Answer: The cystic fibrosis gene was identified by a combination of genetic and modern molecular biology techniques. It was first localized to chromosome 7, and then molecularly cloned and sequenced. The sequence of the mutant allele was compared with the allele of a normal person to discover a three base‐pair deletion in the cystic fibrosis gene. To determine the geneʹs function, its amino acid sequence and three‐dimensional structure was deduced. This was compared with a database of protein sequences to determine the geneʹs function as an active transport protein.
Skill: Factual recall
47) How can enzyme assay be used to detect carriage of a recessive gene mutation in a person of normal phenotype? Answer: If they were a carrier, the person, as a heterozygote, would be expected to produce only half the normal enzyme or protein product for the gene in question.
Skill: Analytical reasoning
48) What advantages does chorionic villus sampling have over amniocentesis for fetal analysis of genetic defects? Answer: Chorionic villus sampling is generally performed up to four weeks earlier than amniocentesis, allowing parents to make an earlier decision regarding potential termination of pregnancies. It is also a faster test that does not require cell culture to do biochemical assays.
Skill: Factual recall
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34 Test Bank for iGenetics
49) Explain the difference between the one‐geneone‐enzyme hypothesis and the one‐gene one‐polypeptide hypothesis. Answer: A relationship is found between a mutation in a single gene and the loss of production of a single enzyme in a metabolic pathway. However, more than one gene may control each step in a pathway because some enzymes (and some non‐enzyme proteins such as hemoglobin) consist of two or more polypeptide chains, each of which is coded for by a different gene. Thus, it is more accurate to use the phrase ʺone‐geneone‐polypeptide.ʺ
Skill: Conceptual understanding
50) You are a doctor who specializes in genetic diseases. A married couple comes to see you for genetic counseling. The woman has PKU, the symptoms of which she manages by diet. The man and woman are first cousins and would like to know the probability that any of their children will have the disease. What would you tell them? Answer: Since the woman has the disease, she must be homozygous recessive for the defective autosomal gene. If her husband is not a carrier, all their children will be heterozygous carriers of the disease. However, since the man and woman are closely related, there is a high probability that the man is a carrier of the PKU gene. If he is a carrier, there is a 50% chance that any of their children will have the disease. You would perform a detailed pedigree analysis for the couple to better determine the probability that the husband carries the recessive allele.
Skill: Problem-solving
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Chapter 5 Gene Expression: Transcription
MATCHING QUESTIONS Please select the best match for each term. 1) mRNA
Skill: Factual recall
A) Brings amino acids to ribosomes during translation B) Forms complexes that are used in eukaryotic RNA processing C) Forms the initiation complex with RNA polymerase in eukaryotes D) With ribosomal proteins, makes up the ribosomes E) Encodes the amino acid sequence of a polypeptide 2) A 3) D 4) B 5) C
2) tRNA
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3) rRNA
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4) snRNA
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5) GTF
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Answers:
1) E
MULTIPLE-CHOICE QUESTIONS 6) In eukaryotes, the initiation complex binds DNA at the A) Pribnow box. B) GC box. C) CAAT box. D) TATA box. E) GTF region. Answer: D
Skill: Factual recall
7) The process by which genetic information is transferred from DNA to RNA is called A) replication. B) transcription. C) translation. D) transversion. E) transformation. Answer: B
Skill: Factual recall
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8) Which of the following proteins is found only in the RNA polymerase holoenzyme? A) Alpha B) Beta C) Epsilon D) Sigma E) Rho Answer: D
Skill: Factual recall
9) During the initiation step of transcription, which molecule binds the promoter region of a DNA molecule? A) DNA polymerase B) Reverse transcriptase C) RNA polymerase D) DNA primase E) Topoisomerase Answer: C
Skill: Factual recall
10) During transcription, the synthesis of the mRNA strand proceeds in which direction? A) 5ʹ to 3ʹ only B) 3ʹ to 5ʹ only C) Both 5ʹ to 3ʹ and 3ʹ to 5ʹ D) Either 5ʹ to 3ʹ or 3ʹ to 5ʹ E) First 5ʹ to 3ʹ, and then 3ʹ to 5ʹ Answer: A
Skill: Factual recall
11) Which type of RNA is found only in eukaryotes? A) mRNA B) tRNA C) rRNA D) snRNA E) All of these Answer: D
Skill: Factual recall
12) Which of the following catalyzes the synthesis of ribosomal RNA in eukaryotes? A) RNA polymerase I B) RNA polymerase II C) RNA polymerase III D) Both RNA polymerase I and II E) Both RNA polymerase I and III Answer: E
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 37
13) Prokaryotic transcription begins when RNA polymerase binds to which site of the gene region? A) -35 B) -10 C) +1 D) +10 E) +35 Answer: C
Skill: Factual recall
14) Sequences that contain information about the stability of a transcript are found in the A) 5ʹ UTR. B) coding sequence. C) 3ʹ UTR. D) leading strand. E) lagging strand. Answer: C
Skill: Factual recall
15) Which of the following is not found in a eukaryotic promoter? A) -10 box B) TATA box C) GC box D) CAAT box E) None of these Answer: A
Skill: Factual recall
16) In eukaryotes, where does transcription take place? A) In the cytoplasm B) Anywhere in the cell C) In the nucleus D) In both the cytoplasm and nucleus E) In the mitochondria Answer: C
Skill: Factual recall
17) In eukaryotes, precursor mRNA molecules are processed in the A) ribosomes. B) nucleus. C) mitochondria. D) cytoplasm. E) nucleolus. Answer: B
Skill: Factual recall
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38 Test Bank for iGenetics
18) The single strand of mRNA that is produced during transcription has the same polarity as the ________ strand of DNA. A) template B) nontemplate C) leading D) lagging E) nonsense Answer: B
Skill: Factual recall
19) The poly(A) tail of an mRNA A) helps transport the pre‐mRNA to the cytoplasm. B) keeps the coding sequences from being degraded. C) is added without an RNA template. D) is important in initiation of translation. E) is removed from an mRNA after its maturation. Answer: B
Skill: Factual recall
20) Self‐cleaving RNAs that function catalytically are called A) snRNPs. B) snRNAs. C) ribozymes. D) ribosomes. E) spliceosomes. Answer: C
Skill: Factual recall
21) Molecules of tRNA are A) 75 to 90 nucleotides in length. B) single‐stranded with secondary structure. C) cloverleaf in shape. D) synthesized with modified bases. E) All of these Answer: E
Skill: Factual recall
22) A prokaryotic mRNA transcript is A) transcribed completely before being translated. B) further processed before it is transcribed. C) transported from the nucleus before it is translated. D) modified before it is translated. E) translated as it is being transcribed. Answer: E
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 39
23) Which parts of a eukaryotic gene are transcribed? A) Only the exons B) Only the introns C) Both the exons and introns D) Exons, introns, promoter, and terminator sequence E) It depends on the gene Answer: C
Skill: Factual recall
24) Posttranscriptional insertion or deletion of nucleotides that is absent in the DNA template is called A) RNA splicing. B) RNA editing. C) RNA processing. D) RNA capping. E) RNA modification. Answer: B
Skill: Factual recall
25) A rho‐dependent terminator A) forms a hairpin loop during termination. B) requires ATP to complete termination. C) is rich in AT nucleotides. D) Both A and B E) All of these Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) A promoter is a coding portion of a gene. Answer: FALSE Explanation: A promoter is a noncoding, regulatory DNA sequence at the start of a gene.
Skill: Factual recall
27) The number of initiations of transcription from a promoter that has the sequence 5'-TAGCAT-3' will be greater than the number of initiations from a wild‐type promoter. Answer: FALSE Explanation: It will be fewer than the number of initiations from a wild‐type promoter.
Skill: Factual recall
28) All genes encode proteins. Answer: FALSE Explanation: Some genes encode specific kinds of RNAs that are not translated into proteins. Protein coding (or structural) genes, code for mRNA transcripts.
Skill: Factual recall
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40 Test Bank for iGenetics
29) Activators are cis‐acting sequences that increase transcription from a promoter. Answer: FALSE Explanation: Activators are regulatory proteins that determine the efficiency of initiation of transcription.
Skill: Factual recall
30) Before a prokaryotic mRNA can be translated, it must be modified by the addition of a polyA tail. Answer: FALSE Explanation: Prokaryotic mRNAs are not processed before translation.
Skill: Factual recall
31) Termination is carried out by RNA polymerase in a rho‐independent terminator. Answer: TRUE
Skill: Factual recall
32) The Rho protein is a topoisomerase that helps in untwisting the DNA template. Answer: FALSE Explanation: The Rho protein is a helicase.
Skill: Factual recall
33) RNA polymerase has no proofreading activity during transcription. Answer: FALSE Explanation: RNA polymerase has two different proofreading activities for correcting incorrectly inserted nucleotides.
Skill: Factual recall
34) Unlike eukaryotic mRNA, prokaryotic mRNA is polycistronic. Answer: TRUE
Skill: Factual recall
35) Group I introns, such as those found in Tetrahymena, are unique in that they are excised from mRNA by a protein‐driven catalytic reaction. Answer: FALSE Explanation: Group I introns are unique in that they are self‐splicing.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) Outline the basic differences in function among RNA polymerases in both prokaryotes and eukaryotes. Answer: Only one type of RNA polymerase is found in prokaryotes, so all classes of genes are transcribed by it. Eukaryotes have three distinct RNA polymerases, each of which transcribes different gene types. RNA polymerase I transcribes the genes for the 28S, 18S, and 5.8S ribosomal RNAs; RNA polymerase II transcribes mRNA genes and some snRNA genes; and RNA polymerase III transcribes genes for the 5S rRNAs, the tRNAs, and the other snRNAs.
Skill: Factual recall
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Chapter 5 Gene Expression: Transcription 41
37) What are the three basic steps of transcription? Answer: Initiation, elongation, and termination.
Skill: Factual recall
38) What is the function of the inverted repeat in the termination sequence of a rho‐independent terminator? What would happen if a mutation occurred in this region? Answer: RNA polymerase transcribes the inverted repeat into the RNA transcript. As a result, a hairpin loop structure forms because of the complementarity of the inverted repeat regions. This hairpin is believed to aid termination by affecting the ability of RNA polymerase to continue elongation. If a mutation occurred in this region, termination would not be successful.
Skill: Factual recall, analytical reasoning
39) What was the significance of the 1978 finding by Philip Lederʹs research group that the 0.7 kb β‐globin mRNA is not colinear with the gene that encodes it, but the nuclear 1.5 kb pre‐mRNA is? Answer: This meant that the β‐globin contains an intron, which is spliced out after pre‐mRNA processing, and the exons are spliced together to make a mature mRNA. The significance was the realization that genes could be ʺin pieces.ʺ
Skill: Conceptual understanding
40) What would be the advantage of having multiple sigma factors that each could be produced under different stress conditions in E. coli? Answer: Different sigma factors recognize different promoter region sequences. Therefore, they play a role in the regulation of gene expression. Sigma factors produced under a particular stress condition, such as excessive heat or infection by bacteriophage, bind to promoters that belong to genes that encode proteins required to cope with those stresses.
Skill: Analytical reasoning
41) What kind of experiment or analysis could you conduct to define the location of the promoter elements of protein‐coding genes? Answer: You could examine the effect of mutations that delete or alter base pairs upstream from the starting point of transcription and to see whether those mutants affect transcription. Mutations that significantly affect transcription define important promoter elements. Alternatively, you could compare the DNA sequences upstream of a number of protein‐coding genes to see whether there are any regions with similar sequences.
Skill: Analytical reasoning
42) What is the difference between an intervening sequence and an untranslated region in a eukaryotic mRNA? Answer: Untranslated regions (UTRs) are found on either side of the protein‐coding sequence of mRNAs and are part of the exons (and are thus not removed in pre‐mRNA processing). Intervening sequences, also called introns, are found within the protein‐coding sequence region and are removed in the processing of the pre‐mRNA molecule to the mature mRNA.
Skill: Factual recall
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42 Test Bank for iGenetics
43) The sequence of a template strand of DNA is 3'-CATTACGCTT-5' . What is the sequence of the corresponding mRNA? Answer: 5'-GUAAUGCGAA-3'
Skill: Problem-solving
44) What are the three regions of a prokaryotic gene? Answer: A promoter region, an RNA‐coding region, and a terminator region.
Skill: Factual recall
45) Describe the structure and organization of the rDNA repeat unit in eukaryotes. Answer: The rDNA repeat unit exists in tandem arrays of 100s to 1,000s of copies. Each unit consists of coding sequences for three rRNA genes (18S, 5.8S, and 28S) separated by two internal transcribed spacer (ITS) regions. The genes are flanked by external transcribed spacer (ETS) regions. These regions comprise the ribosomal DNA transcription unit. Upstream of the unit and separating from the previous repeat unit is a nontranscribed spacer (NTS) region that contains the promoter.
Skill: Factual recall
46) What are the functions of the 5ʹ cap and poly(A) tail of mature mRNAs in eukaryotes? Answer: Both protect the mRNA from degradation by exonucleases. The 5ʹ cap is also important for the binding of the ribosome at the beginning of translation. The poly(A) tail is required for efficient export of the mRNA from the nucleus to the cytoplasm.
Skill: Factual recall
47) Cite at least three ways in which transcription differs from DNA replication. Answer: (1) In replication, both strands of the double‐stranded DNA are copied, resulting in new double‐stranded DNA molecules. In transcription, only one strand of the double‐stranded DNA is transcribed into a single‐stranded RNA molecule. (2) In eukaryotes, while DNA replication typically occurs during only part of the cell cycle, transcription generally occurs throughout the cycle. (3) No primer is needed in transcription because the RNA polymerase can initiate new polynucleotide chains, unlike DNA polymerase, which needs a primer to begin replication.
Skill: Application of knowledge
48) In eukaryotes, what is the difference between promoter proximal elements in ʺhousekeeping genesʺ vs. cell‐specific genes, and how does this relate to gene expression? Answer: In genes that are expressed in all cell types for basic cellular functions (ʺhousekeeping genesʺ), promoter proximal elements are recognized by regulatory activator proteins that are ubiquitous to all cell types. In cell‐specific genes, the proximal elements are recognized by activators that are found only in those cells.
Skill: Conceptual understanding
49) What is the advantage of having a promoter sequence upstream of a coding sequence? Answer: The promoter ensures that the initiation of every RNA occurs at the same site. Otherwise, the ensuing polypeptide might be missing amino acids, or it might contain extra amino acids.
Skill: Analytical reasoning
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Chapter 5 Gene Expression: Transcription 43
50) How does the rate of initiation of transcription relate to the similarity of the promoter region to the consensus sequence? Answer: The ability of the sigma factor of the RNA polymerase holoenzyme to recognize and bind to a promoter sequence (i.e., the efficiency of RNA polymerase binding) is dependent on the similarity of the promoter to the consensus sequence (the sequence found most frequently at each position of the promoter). As a result, the rate at which transcription is initiated varies from gene to gene.
Skill: Analytical reasoning
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Chapter 6 Gene Expression: Translation
MATCHING QUESTIONS Please select the best match for each term. 1) Glutamic acid
Skill: Factual recall
A) Neutral, nonpolar B) Basic
2) Glycine
Skill: Factual recall
C) Acidic D) Neutral, polar E) Atypical structure
3) Arginine
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4) Serine
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5) Proline
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Answers:
1) C
2) A
3) B
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) Which of the following is not a characteristic of proteins? A) High molecular weight B) Presence of nitrogen C) Presence of amino acids D) Presence of nucleic acids E) Structure built of one or more polypeptides Answer: D
Skill: Factual recall
7) Where does translation take place? A) In the cytoplasm B) In the nucleus C) In the nucleolus D) Anywhere in the cell E) In the centriole Answer: A
Skill: Factual recall
44
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Chapter 6 Gene Expression: Translation 45
8) Every amino acid contains A) an amino group and a hydroxyl group. B) an amino group and an acetyl group. C) a hydroxyl group and a carboxyl group. D) a carboxyl group and an amino group. E) a carboxyl group and a hydroxyl group. Answer: D
Skill: Factual recall
9) What is the function of a ribosome? A) To direct transcription B) To attach amino acids to the appropriate tRNAs C) To unwind the double‐stranded DNA helix to enable translation D) To hold mRNA and tRNAs in the correct positions to enable translation E) To make more mRNA Answer: D
Skill: Factual recall
10) What is the basic shape of a tRNA molecule? A) A three‐dimensional cloverleaf B) A straight strand C) A right‐handed helix D) A globular ball E) A β‐pleated sheet Answer: A
Skill: Factual recall
11) What is the role of tRNA in translation? A) It brings together two subunits of a ribosome. B) It couples an amino acid with aminoacyl‐tRNA synthetase. C) It helps fold up the finished polypeptide chain. D) It catalyzes peptidyl transferase activity. E) It binds to an mRNA codon and carries the corresponding amino acid. Answer: E
Skill: Factual recall
12) How can a carrot plant express a bacterial gene? A) Because the bacterial gene hijacks the carrotʹs cellular machinery B) Because the genetic code is the same in both organisms C) Because of the wobble phenomenon D) Because of the degeneracy of the genetic code E) A carrot cannot express a bacterial gene. Answer: B
Skill: Application of knowledge
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46 Test Bank for iGenetics
13) In the genetic code, both AAU and AAC code for asparagine. For this reason, the code is said to be A) degenerate. B) nonspecific. C) universal. D) ambiguous. E) wobbly. Answer: A
Skill: Factual recall
14) In a gene sequence, the DNA codon for tryptophan experiences a mutation at the first base position, changing it to T. What will the resulting amino acid be? A) Tryptophan (no change) B) Serine C) Arginine D) Threonine E) None (a stop codon will halt translation) Answer: C
Skill: Problem-solving
15) A mutation causes a G to be inserted after the first base of the codon for tryptophan. How will this affect the growing polypeptide chain? A) It will not be affected. B) Elongation will stop prematurely. C) There will be a single amino acid substitution. D) The reading frame will be shifted to the left, and the wrong amino acids will be added from this point on. E) None of these Answer: D
Skill: Problem-solving
16) A β‐pleated sheet is a type of A) primary structure found in a protein. B) secondary structure found in a protein. C) tertiary structure found in a protein. D) quaternary structure found in a protein. E) heme structure found in a protein. Answer: B
Skill: Factual recall
17) Which of the following events signals the termination of translation? A) The ribosome reaches the end of the mRNA. B) The ribosome reaches a start codon. C) The ribosome reaches a stop codon. D) The ribosome runs out of charged tRNAs. E) The polypeptide chain folds into a protein. Answer: C
Skill: Factual recall
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Chapter 6 Gene Expression: Translation 47
18) A peptide bond forms between the ________ of one amino acid and the ________ of another. A) amino group, carboxyl group B) amino group, R group C) carboxyl group, sulfide group D) amino group, phosphate group E) R group, R group Answer: A
Skill: Factual recall
19) The majority of naturally occurring amino acids are A) acidic and polar. B) basic and nonpolar. C) acidic and nonpolar. D) neutral and nonpolar. E) neutral and polar. Answer: D
Skill: Factual recall
20) Molecular chaperones A) help guide proteins from the nucleus to the cytoplasm. B) help guide proteins from the cytoplasm to the endoplasmic reticulum. C) help fold proteins properly. D) are proteins with structural function. E) are proteins with enzymatic function. Answer: C
Skill: Factual recall
21) The concept that a set of three nucleotides specifies a particular amino acid provides the basis for A) the one‐gene‐one‐enzyme hypothesis. B) the one‐gene‐one‐polypeptide hypothesis. C) the genetic code. D) biochemical reactions among nucleic acids. E) All of these Answer: C
Skill: Factual recall
22) In eukaryotes, the AUG initiation codon is located in the ________ sequence. A) Shine—Dalgarno B) Kozak C) Khorana D) Goldberg—Hogness E) Pribnow Answer: B
Skill: Factual recall
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48 Test Bank for iGenetics
23) How many naturally occurring amino acids are used by ribosomes to construct proteins? A) 46 B) 64 C) 3 D) 20 E) 10 Answer: D
Skill: Factual recall
24) Which of the following statements is not true of proteins destined for the ER? A) They have a signal sequence. B) The signal sequence for the ER consists of 15‐30 amino acids. C) The proteins are modified with carbohydrates in the ER. D) The signal sequence is subsequently removed from the mature protein. E) The SRP binds to the protein and prevents entry into the ER until translation is completed. Answer: E
Skill: Factual recall
25) Which of the following statements is correct? A) For every codon, there is one amino acid. B) For every amino acid, there is one codon. C) A codon may signify more than one amino acid. D) Three codons make up an amino acid. E) Three amino acids make up a codon. Answer: A
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) During translation, mRNA codons bind to complementary tRNA anticodons. Answer: TRUE
Skill: Factual recall
27) Molecules of mRNA are translated from 3ʹ to 5ʹ, which is the opposite direction from which they were made. Answer: FALSE Explanation: They are translated from 5ʹ to 3ʹ.
Skill: Factual recall
28) The Shine—Dalgarno sequence is a purine‐rich region just upstream of a start codon and is necessary for the initiation of protein synthesis in prokaryotes. Answer: TRUE
Skill: Factual recall
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Chapter 6 Gene Expression: Translation 49
29) With absolutely no exception, all organisms use the same genetic code for the production of proteins. Answer: FALSE Explanation: Some organisms have minor changes in the code, such as in the nuclear genome of Tetrahymena and the mammalian mitochondria.
Skill: Factual recall
30) There are 64 sense codons in the genetic code and 61 different types of tRNA molecules. Answer: FALSE Explanation: There are 64 codons but only 61 sense codons.
Skill: Factual recall
31) Protein synthesis begins when ribosomes bind to the UGA codon. Answer: FALSE Explanation: UGA is a stop codon. Protein synthesis begins when ribosomes bind to the AUG (start) codon.
Skill: Factual recall
32) The genetic code is comma‐free, nonoverlapping, and almost universal. Answer: TRUE
Skill: Factual recall
33) In the genetic code, when the first two nucleotides of a triplet are identical and the third letter is U or C, the codon always codes for the same amino acid. Answer: TRUE
Skill: Factual recall
34) A new initiation event cannot occur on an mRNA molecule until the first ribosome falls away. Answer: FALSE Explanation: Many ribosomes may simultaneously be translating each mRNA, forming a structure called a polysome.
Skill: Factual recall
35) In eukaryotes, the translation initiation complex forms in association with the 40S ribosomal subunit. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) A DNA codon reads GCC. Give the corresponding mRNA codon, tRNA anticodon, and amino acid. Answer: mRNA codon: CGG; tRNA anticodon: GCC; amino acid: alanine.
Skill: Problem-solving
37) If a tRNA anticodon reads GAI, to which mRNA codon(s) will it bind? What amino acid will the tRNA carry? Answer: It will bind to CUU, CUC, or CUA, which all code for leucine.
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50 Test Bank for iGenetics
38) Part of a DNA gene sequence reads CAT. If a mutation occurs that changes the T to A, will the final protein be affected? Answer: Both CAT and CAA encode the amino acid valine, so the final protein will not be affected.
Skill: Problem-solving
39) If a poly(C) is used as an mRNA and translated, a poly‐glycine polypeptide is formed. If on the other hand a poly(G) mRNA is used, a poly‐proline polypeptide is made. However, if both poly(C) and poly(G) are used in the same reaction, no polypeptide is made. Explain why. Answer: Since mRNA is single stranded, and guanines pair with cytosines, the poly(C) would pair with the poly(G) to give double stranded RNA that could not be used as a template for translation.
Skill: Conceptual understanding
40) Why is methionine the first amino acid to be added to every polypeptide chain? Answer: Because the start codon, AUG, specifies methionine. Since the start codon signals the beginning of translation, methionine is always the first amino acid added to a new polypeptide chain.
Skill: Conceptual understanding
41) Describe and differentiate among the primary, secondary, and tertiary structures of a protein. To which kinds of interactions can each of these stages be ascribed? Answer: Primary protein structure refers to the sequence of amino acids on the polypeptide chain, which is determined by the gene nucleotide sequence. Secondary structure is the regular folding and twisting of a single polypeptide chain into a variety of shapes. This can be ascribed to weak bonds (electrostatic or hydrogen bonds) between NH and CO groups that are near each other on the chain. Tertiary structure refers to the folding and twisting of the amino acid chain into a three‐dimensional conformation. This can be ascribed to interactions among R‐groups of individual amino acids.
Skill: Conceptual understanding
42) When does quaternary structure occur? Answer: Quaternary structure occurs only in proteins that consist of more than one polypeptide subunit. The folded subunits come together to form the quaternary structure of the protein.
Skill: Conceptual understanding
43) In the experiments performed on bacteriophage T4 by Francis Crick and others, in which it was shown that the genetic code was triplet, what were the researchers trying to identify? Answer: They were trying to identify revertants from mutant rII to wild‐type r+ phage that would have resulted from frameshift mutations due to the addition or deletion of a single base pair.
Skill: Factual recall
44) What are the two sulfur‐containing amino acids that contribute to a proteinʹs tertiary structure? Answer: Cysteine and tryptophan.
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
Chapter 6 Gene Expression: Translation 51
45) How did von Ehrenstein et al. demonstrate that the specificity of codon recognition lies in the tRNA molecule and not in the amino acid it carries? Answer: They chemically converted the cysteine on tRNA.Cys molecules to alanine in vitro. This Ala‐tRNA.Cys was used to synthesize hemoglobin, which normally contains a cysteine on each α and ß chain. The resulting hemoglobin molecules contained alanine at the position normally occupied by cysteine, which indicated that the tRNA had read the cysteine codon and inserted the alanine it carried.
Skill: Conceptual understanding
46) Of the codons AAU, AUG, UAA, and UAU, which is considered a nonsense codon? What does this mean? Answer: UAA is a nonsense, or stop, codon. It does not code for any amino acid and signals the end of translation.
Skill: Factual recall
47) What is the wobble hypothesis, and what implications does it have for base‐pairing rules and selective pressure on codons? Answer: The wobble hypothesis refers to the degeneracy at the third nucleotide position of a codon, in which the base at the 5ʹ of the tRNA anticodon can pair with more than one type of base at the 3ʹ end of the codon. For example, the same leucine tRNA molecule with anticodon GAG can read two different leucine codons ( CUC and CUU). This allows for neutral mutations to occur in genes at the third position, but not the first or second, of a codon. Therefore, there is said to be higher selective pressure on the first and second positions in a codon, causing mutations at those positions to be much more rare.
Skill: Analytical reasoning
48) How are proteins sorted into their appropriate cell compartments in eukaryotes? Answer: Proteins with specific ʺsignalʺ sequences (a hydrophobic amino terminal extension) are directed to the ER during translation, while proteins without signal sequences remain in the cytoplasm. Inside the space of the ER, the signal sequence is removed from the polypeptide. The protein is modified further by addition of specific carbohydrate groups that direct them to their final destination.
Skill: Factual recall
49) Would a two‐letter code with four different nucleotides be sufficient to encode the 20 amino acids found in cells? Why or why not? What does the three‐letter code imply about the nature of the genetic code? Answer: A two‐letter code would only encode 16 (4 2 ) amino acids, so it would not be sufficient. A three‐letter code generates 64 (4 3 ) possible codes, which is more than enough for the 20 amino acids found in cells. This implies that some amino acids may be specified by more than one codon.
Skill: Problem-solving
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52 Test Bank for iGenetics
50) Describe a cell‐free protein synthesizing system with which you could determine unambiguously which codons specify which amino acids. Answer: You could use a system with synthetic mRNAs that contain only one type of base and then observe which type of polypeptide is synthesized. The polypeptide should contain only one type of amino acid. For example, a poly(U) mRNA would contain only UUU triplets and would produce a polypeptide chain entirely of phenylalanine.
Skill: Analytical reasoning
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements
MATCHING QUESTIONS Please select the best match for each term. 1) Nonsense mutation
Skill: Factual recall
A) A mutation that causes an addition or deletion of one or two base pairs in a gene B) A mutation that changes a codon from one that represents an amino acid to one that signals a chain termination C) A mutation that causes a change in a single base pair D) A mutation in a gene that causes no detectable change in the protein product E) A mutation that changes a codon from one amino acid to another
2) Missense mutation
Skill: Factual recall
3) Frameshift mutation
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4) Point mutation
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5) Neutral mutation
Skill: Factual recall
Answers:
1) B
2) E
3) A
4) C
5) D
MULTIPLE-CHOICE QUESTIONS 6) Which of the following nucleotide changes leads to a transition mutation? A) Adenine to guanine B) Adenine to cytosine C) Guanine to cytosine D) Thymine to guanine E) Guanine to thymine Answer: A
Skill: Application of knowledge
7) Which of the following is not mutagenic? A) 5BU B) AZT C) Nitrous acid D) Hydroxylamine E) Acridine Answer: B
Skill: Factual recall
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54 Test Bank for iGenetics
8) Base analogs may cause mutations because A) they modify the chemical structure and properties of the normal base. B) they insert themselves between adjacent bases on the DNA strand and cause an extra base to be inserted during replication. C) they remove amino groups from bases, causing them to pair with the wrong base during replication. D) they may exist in alternate chemical states that pair with different DNA bases than the normal state during replication. E) Both A and C Answer: D
Skill: Conceptual understanding
9) The deamination of cytosine creates A) 5‐methyl cytosine. B) uracil. C) 2‐aminopurine. D) 5‐bromouracil. E) thymine. Answer: B
Skill: Factual recall
10) Thymine dimers are commonly caused by A) ultraviolet radiation. B) ionizing radiation such as X‐rays. C) tautomers. D) alkylating agents. E) intercalating agents. Answer: A
Skill: Factual recall
11) Xeroderma pigmentosum is a human genetic disease caused by A) elevated levels of cholesterol in the blood. B) failure to produce pigment that protects the skin cells from UV light exposure. C) defective DNA excision‐repair mechanisms. D) mutations that inactivate tumor suppressor genes. E) loss of genes controlling the SOS response. Answer: C
Skill: Factual recall
12) Mutation frequency is the A) number of mutations per gene per generation. B) number of mutations per nucleotide per generation. C) number of mutations per cell per generation. D) number of a specific mutation in a defined population. E) total number of mutations in a defined population. Answer: D
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 55
13) The codon 5'-AAA-3' codes for the amino acid lysine. Which of the following mutations in this codon is a neutral mutation? A) 5'-ATA-3' to isoleucine B) 5'-AGA-3' to arginine C) 5'-AAG-3' to lysine D) 5'-CAA-3' to glutamine E) 5'-AAC-3' to asparagine Answer: B
Skill: Application of knowledge
14) Spontaneous mutation rates are greatly reduced by A) exposure to ionizing radiation. B) reverse mutations. C) base‐modifying agents. D) DNA repair mechanisms. E) performing the Ames test. Answer: D
Skill: Factual recall
15) A mutation during DNA replication causes a G to be inserted after the first base of the codon for tryptophan. How will this affect the growing polypeptide chain? A) It will not be affected. B) Elongation will stop prematurely. C) There will be a single amino acid substitution. D) The reading frame will be shifted to the left, and the wrong amino acids will be added from this point on. E) An extra amino acid will be added, but the reading frame will not be affected. Answer: D
Skill: Reasoning and logic
16) In Drosophila, the wild‐type eye color is red. A mutation, vermilion, causes vermilion‐colored eyes, unless there is a mutation in another gene, suv, which, when homozygous or hemizygous, results in eyes that are the wild‐type red even in the presence of the vermilion mutation. This is an example of A) forward mutation. B) reverse mutation. C) intragenic suppression. D) intergenic suppression. E) back mutation. Answer: D
Skill: Application of knowledge
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56 Test Bank for iGenetics
17) A transposon is A) a DNA segment that can insert itself at one or more sites in a genome. B) a ʺjumping gene.ʺ C) a DNA segment that may cause mutations in genes or chromosomal rearrangements. D) a mobile genetic element that may or may not leave a copy of itself in its original site when it moves to a new site. E) All of these Answer: E
Skill: Factual recall
18) Nonsense suppressors are usually mutations in genes coding for A) proteins. B) enzymes. C) mRNA. D) tRNA. E) rRNA. Answer: D
Skill: Factual recall
19) Which of the following transposable elements are found in eukaryotes but not in prokaryotes? A) IS elements B) Families of autonomous and nonautonomous elements C) Retrotransposons D) Ty elements E) B, C, and D only Answer: E
Skill: Factual recall
20) In order for the dissociation element (Ds) mutations in corn to be stable, A) an Ac element must be present. B) an Ac element must not be present. C) the Ds must contain the gene for transposition. D) the DNA must not be replicated. E) None of these Answer: B
Skill: Factual recall
21) An IS (insertion sequence) element contains A) a transposase gene only. B) a transposase gene and additional genes. C) a transposase gene and inverted repeats at the ends. D) a transposase gene, additional genes, and inverted repeats at the ends. E) genes but not a transposase gene. Answer: C
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 57
22) What is conservative transposition? A) No net increase in the number of transposable elements in the genome. B) A net increase in the number of transposable elements in the genome. C) Movement of transposable elements with replication of the element. D) Movement of transposable elements without replication of the element. E) Transposition without disruption of normal gene product activity. Answer: D
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23) Dissociation elements (Ds) in plants are examples of A) activator transposons that can direct their own transposition. B) nonautonomous elements that require activation by an autonomous element. C) mutator genes that increase the spontaneous mutation frequencies of other genes. D) repeated sequences that are targeted by a transposase. E) Both B and C Answer: E
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24) Which procedure would you use to detect a nutritional mutation in microorganisms? A) Visible inspection B) Replica plating C) Controlled crosses D) Isolation at high temperature E) Plating on antibiotic‐containing medium Answer: B
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25) Genetic manipulation in Drosophila may be assisted by the use of A) P elements. B) Ty elements. C) factor. D) bacteriophage Mu. E) Tn10. Answer: A
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TRUE-FALSE QUESTIONS 26) LINEs and SINEs are repetitive sequences in humans that can also, as retrotransposons, insert into genes and cause disease. Answer: TRUE
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27) Changes in heritable traits result from adaptation to environmental influences. Answer: FALSE Explanation: Changes in heritable traits result from random mutations.
Skill: Conceptual understanding
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28) Somatic mutations may be inherited by the next generation. Answer: FALSE Explanation: Only germ‐line mutations may be heritable; somatic mutations only affect the individual in which the mutation occurs.
Skill: Conceptual understanding
29) 5‐bromouracil (5BU) is a mutagen because it is an analog of the base thymine and may pair with guanine instead of adenine if it is incorporated into a DNA strand. Answer: TRUE
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30) A silent mutation is a change in the DNA sequence that alters the amino acid sequence of the encoded protein but does not change its function. Answer: FALSE Explanation: A silent mutation is a change in the DNA sequence that does not change the amino acid sequence of the protein due to the redundancy of the genetic code.
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31) A tautomer is an uncommon form of DNA base that naturally exists along with the common form. Answer: TRUE
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32) The DNA polymerase proofreading mechanism maintains a low mutation rate in eukaryotic genes. Answer: FALSE Explanation: The DNA polymerase proofreading mechanism is found only in bacteria, where it detects the insertion of incorrect bases following DNA replication.
Skill: Conceptual understanding
33) A transposon may carry genes for proteins that enable their transposition as well as genes for other functions such as drug resistance. Answer: TRUE
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34) Composite transposons contain a central region with genes and repeated sequences at their ends but do not terminate with IS elements. Answer: FALSE Explanation: Composite transposons contain a central region with genes flanked at both sides by IS elements as well as terminal inverted repeats.
Skill: Factual recall
35) A cointegrate is characteristic of replicative transposition. Answer: TRUE
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 59
SHORT ANSWER QUESTIONS AND PROBLEMS 36) An mRNA codon reads GUA. If a transition mutation occurs at the third base pair, will the final protein be changed? Answer: Yes. Valine will be inserted in place of an asparagine.
Skill: Application of knowledge
37) A point mutation changes a codon from UCG to UAG. What will happen to the resulting polypeptide? Answer: UAG is a stop codon, so the polypeptide will terminate prematurely.
Skill: Application of knowledge
38) What are two ways that a reverse mutation can occur? Answer: A mutation in the same gene can restore the original phenotype, or a suppressor mutation at a second site in the genome can suppress the forward mutation.
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39) How does an intercalating agent such as ethidium bromide cause mutations? Answer: The substance fits between adjacent bases on one strand of a DNA sequence, causing the complementary strand to be miscopied during replication. This usually results in a frameshift mutation.
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40) How can PCR be used to induce site‐specific mutations in DNA? Answer: Two nested pairs of primers are used, the internal pair having a targeted mismatch that is copied into the newly synthesized DNA fragments. The external primers are used to replicate the mutated fragments.
Skill: Application of knowledge
41) For a particular gene, if one gene in a million experiences a mutation each generation, what is its mutation frequency? Answer: 10-6
Skill: Application of knowledge
42) A tumor suppressor gene is cloned and mutagenized in vitro, then injected into mouse embryos to create knockout mice. How could you identify the heterozygous knockout mice and create mice susceptible to tumors? Answer: Reared heterozygous mice can be screened molecularly to identify the ones with the mutation. Interbreeding these mice will result in some homozygous knockout progeny, which in turn will show rapid development of tumors.
Skill: Analytical reasoning
43) Describe how the Ames test is used to determine whether a particular chemical is mutagenic. Answer: Strains of the bacteria Salmonella typhimurium that are auxotrophic for histidine are plated on a histidine‐lacking nutrient agar with rat liver cells. The unknown chemical is added to the agar plate, and the plate is observed for bacterial growth that would indicate reverse mutations caused by the chemical resulting in histidine prototrophy.
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44) What kind of mutation‐detection procedure can be used to detect the white‐eye mutation in Drosophila? Answer: Screening for this mutation can be done by inspection since it is a visible mutation.
Skill: Conceptual understanding
45) How does the nucleotide excision repair (NER) system in E. coli work, and what kinds of DNA damage does it repair? Answer: The NER system repairs thymine dimers and other serious distortions of the DNA helix. It is a system of four proteins (UvrA, UvrB, UvrC, and UvrD) that work in a complex to recognize a distortion in the DNA, make cuts on either side of the lesion, release the short single‐stranded cut segment, and allow DNA polymerase and ligase to fill in the gap with new bases.
Skill: Factual recall
46) Explain how IS elements produce target‐site duplications when they move. Answer: The enzyme transposase makes a staggered cut in the target sequence of the recipient DNA. The IS element is inserted, becoming joined to the single‐stranded ends. The gaps are then filled in by DNA replication, resulting in a copy of both the IS element and the target sequence.
Skill: Conceptual understanding
47) Barbara McClintock discovered that a controlling element was responsible for the mutant‐spotted phenotype (purple spots on white) in corn kernels. Explain how this works. Answer: A normal C gene produces purple pigment in the corn kernel cells. If this gene is disrupted by the transposon Ds (a ʺmobile controlling elementʺ), no pigment is produced, resulting in a white kernel. Reversions of this mutation can occur if Ds transposes out of the gene during development, resulting in spots of purple pigment on the kernel. The transposition of Ds is controlled by an autonomous ʺactivatorʺ element, Ac.
Skill: Conceptual understanding
48) What is a retrotransposon, and how does it differ from typical transposons? Answer: Most transposons move by a DNA‐to‐DNA mechanism, but retrotransposons (such as yeast Ty elements, Drosophila copia elements, and human LINEs and SINEs) transpose via an RNA intermediate using a transposon‐encoded reverse transcriptase.
Skill: Factual recall
49) How was the adaptive method for acquiring mutations disproved? Answer: Luria and Delbruck argued that if mutations occurred due to exposure to a particular environment, then the number of cells that had the mutation would be similar in different identical populations. However, if the mutations occurred randomly, then the number of cells with that particular mutation would differ or fluctuate in identical populations exposed to the same environmental factor. They proved the second argument to be true by infecting identical bacterial cultures with the bacteriophage T1 and counting the number of resistant colonies. The number of resistant colonies fluctuated widely in the different cultures.
Skill: Factual recall
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Chapter 7 DNA Mutation, DNA Repair, and Transposable Elements 61
50) Explain what a mutator gene is and give an example. Answer: A mutator gene is a gene in which a mutation results in a much higher‐than‐normal mutation frequency for all other genes. For example, mutD mutations in E. coli affect the function of DNA polymerase III. This causes defective 3ʹ‐to‐5ʹ proofreading repair activity and results in many incorrectly inserted nucleotides being left unrepaired.
Skill: Conceptual understanding
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes
MATCHING QUESTIONS Please select the best match for each term. 1) Restriction enzyme
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A) Origin of replication B) Endonuclease isolated from bacteria C) Small circular fragment of DNA D) Eukaryotic cloning vector E) Molecule used in DNA sequencing
2) ARS
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3) Plasmid
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4) YAC
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5) Dideoxynucleotides
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Answers:
1) B
2) A
3) C
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) When DNA is cut with a restriction enzyme, the resulting fragments have A) 5ʹ carboxyls. B) 5ʹ hydroxyls. C) 3ʹ hydroxyls. D) 3ʹ phosphates. E) 3ʹ carboxyls. Answer: C
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7) A plasmid used as a cloning vector in E. coli must have A) an ori sequence. B) a selectable marker. C) unique restriction sites. D) B and C E) All of these Answer: E
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 63
8) Bacteria protect their own DNA from restriction enzyme damage by adding ________ groups to certain nucleotides. A) methyl B) hydroxyl C) phosphate D) amino E) carboxyl Answer: A
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9) A polylinker region in a cloning plasmid is characterized by A) multiple methyl groups. B) multiple restriction cut sites. C) multiple reporter genes. D) multiple expression vectors. E) multiple cloning targets. Answer: B
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10) In a physical map, distances are measured in A) recombination frequencies. B) number of SNPs. C) base pairs. D) probabilities. E) microns. Answer: C
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11) In DNA electrophoresis, fragment separation is based on A) fragment size. B) fragment nucleotide content. C) fragment sequence. D) fragment charge. E) All of these Answer: A
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12) The chemical ʺlabelʺ that permits visualization of DNA fragments in an automated sequencing machine is A) 32P. B) 35S. C) fluorescent dyes. D) tritium. E) None of these Answer: C
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13) A DNA copy of an mRNA molecule is called A) dDNA. B) rDNA. C) mDNA. D) shDNA. E) cDNA. Answer: E
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14) The first complete nonviral genome sequenced was that of A) Escherichia coli. B) Methanococcus. C) Saccharomyces cerevisiae. D) the human mitochondrion. E) Mus musculus. Answer: D
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15) Many bacteria are naturally capable of taking up endogenous fragments of DNA in their environment, a phenomenon termed A) conversion. B) transformation. C) hybridization. D) transfection. E) conjugation. Answer: B
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16) An SNP always occurs due to A) a one base‐pair change. B) a three base‐pair change. C) a single restriction site change. D) a single amino acid change. E) None of these Answer: A
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17) The recognition sequences for many restriction enzymes are A) terminal repeats. B) tandem repeats. C) palindromic. D) inverted repeats. E) dual repeats. Answer: C
Skill: Factual recall
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 65
18) In addition to restriction enzymes, which of the following enzymes is required to insert a fragment of DNA into a cloning vector? A) DNA polymerase B) DNA ligase C) RNA polymerase D) Reverse transcriptase E) Topoisomerase Answer: B
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19) Which restriction enzyme leaves blunt ends? A) XbaI B) SmaI C) EcoR1 D) BamHI E) All of these Answer: B
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20) A restriction enzyme cuts DNA and leaves the following end: CTGCA
G
Which of the following could be the sequence of the corresponding end of the other fragment generated by the enzyme? A) G
CTGCA GTGCA G C
B) C
C) CCGAT D) GGCTA E) None of these Answer: A
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21) The ________ method was used by the Celera Genomics Corporation to sequence the human genome. A) FISH B) whole‐genome shotgun C) clone contig D) radiation hybrid E) pedigree analysis Answer: B
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22) Candidate open reading frames of a genome are identified by searching for A) a start codon ʺin frameʺ with a stop codon. B) a start codon. C) introns and exons. D) a promoter. E) All of these Answer: A
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23) Successful insertion of a DNA fragment into the polylinker region of pBluescript II is detected by A) production of the enzyme β‐galactosidase. B) absence of the enzyme β‐galactosidase. C) the increased size of the plasmid. D) autoradiography. E) None of these Answer: B
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24) In gel electrophoresis, which of the following DNA fragments would travel the farthest distance from the sample well? A) ATCCCGAT B) ATCCCG C) ATCC D) AT E) ATCCCGATTGCACGTT Answer: D
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25) Which types of cloning vector occur in nature? A) Plasmids and cosmids B) Plasmids and bacteriophages C) Plasmids and BACs D) YACs and BACs E) Bacteriophages and cosmids Answer: B
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TRUE-FALSE QUESTIONS 26) A DNA copy of a messenger RNA molecule is called complementary DNA or cDNA. Answer: TRUE
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27) The human genome consists mostly of noncoding DNA. Answer: TRUE
Skill: Factual recall
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 67
28) Cloning vectors are used to introduce novel DNA into cells. Answer: TRUE
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29) DNA molecules, regardless of size, have a net positive charge. Answer: FALSE Explanation: DNA is a negatively charged molecule.
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30) In partial restriction digestion, all available restriction sites will be cut. Answer: FALSE Explanation: Partial digestion leaves some sites untouched by chance.
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31) The quality of an assembled sequence does not depend on the sequence coverage. Answer: FALSE Explanation: The higher the coverage of the genome, the better the quality of the assembled sequence.
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32) Cloning vectors generally possess unique restriction sites and dominant selectable markers. Answer: TRUE
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33) A linear strand of DNA with two restriction enzyme cut sites will yield three fragments on digestion. Answer: TRUE
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34) A circular DNA molecule with two restriction enzyme cut sites will yield three fragments on digestion. Answer: FALSE Explanation: A circular molecule will yield two DNA fragments when cleaved in two places.
Skill: Problem-solving
35) Archaea are typically found in much the same habitats as bacteria. Answer: FALSE
Skill: Factual recall
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Briefly contrast the dideoxy DNA sequencing developed by Fred Sanger with pyrosequencing methods. Answer: Dideoxy DNA sequencing involves use of altered (dideoxy) nucleotides that cause termination of replication at each site where such a nucleotide is inserted. The pyrosequencing method does not require chain termination. All four deoxynucleotide precursors are present at all times in the dideoxy sequencing reaction, whereas, only one is present at a time in the pyrosequencing reaction. The DNA template is immobilized on a bead in pyrosequencing, and the method of detection is the production of light by utilizing the pyrophosphate produced after the incorporation of a nucleotide. In the dideoxy method, the DNA template is not immobilized and the sequence is read with the help of fluorescent dyes attached to the dideoxy nucleotide molecules.
Skill: Conceptual understanding
37) Describe the basic approach to shotgun genome sequencing. Answer: Shotgun sequencing involves cutting the genome into multiple, partially overlapping, fragments, each of which is cloned and sequenced. A computer algorithm is used to assemble the complete sequence by identifying contiguous fragments by their overlapping regions.
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38) The restriction enzyme SmaI cuts DNA at the 6‐base recognition sequence 5'-CCCGGG-3'. Calculate the frequency of occurrence of this sequence. (Assume a 50% GC content.) Answer: Assuming a 1:1 AT:CG ratio in the genome and a random distribution of nucleotides, each nucleotide pair will occur with equal frequency ( ). The
4 1
6‐nucleotide sequence would then occur with a frequency of ( )6 , 0.00024, cutting once in every 4,096 base pairs.
Skill: Problem-solving
1 4
39) Describe a DNA microarray setup for assaying mutations associated with a hypothetical genetic disease. Answer: Normal and mutant gene sequences associated with the disease could be affixed to the array in known positions. DNA from a patient and a known nonmutant individual could be separately isolated and labeled (or mRNA could be used, making cDNA for labeling with reverse transcription). Both are applied to the microarray, and whichever labeled fragments that bind with the DNA fragments on the array will colorimetrically reveal which alleles the patient possesses.
Skill: Conceptual understanding
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Chapter 8 Genomics: The Mapping and Sequencing of Genomes 69
The following scenario applies to the questions below. A pBluescript II cloning vector is used to introduce a gene of interest into a bacterial strain. The vector is lacking ampR, however. 40) What problem does this result in? Answer: ampR is the gene conferring resistance to the antibiotic ampicillin. Ampicillin is used to select for bacterial colonies that have been successfully cloned. Without ampR, both successfully cloned and unsuccessfully cloned bacterial colonies will fail to grow if the selection medium contains ampicillin.
Skill: Conceptual understanding
41) Can you think of an alternative means of identifying successful bacterial transformants in this experiment? Answer: Successful insertion of the gene of interest into the polylinker region will disrupt the lacZ+ gene; such colonies cannot manufacture β‐galactosidase and will therefore remain white in the presence of X‐gal. Colonies can thus be screened using X‐gal. Of course, the bacteria would have to be grown on media without ampicillin, allowing all the colonies to grow.
Skill: Conceptual understanding
42) Exactly why, in dideoxy DNA sequencing, do dideoxynucleotides halt the replication process? Answer: Dideoxynucleotides are missing the 3ʹ hydroxyl group necessary for linkage with the 5ʹ phosphate group of new nucleotides by DNA polymerase.
Skill: Conceptual understanding
43) What is a hapmap? Answer: A hapmap is a complete description of all the haplotypes known in a population as well as the chromosomal location of each haplotype.
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44) How can the recircularization of a restriction‐digested vector during a ligation reaction be minimized? Answer: The restriction‐digested vector can be treated with alkaline phosphatase to remove the 5ʹ phosphates without which the DNA ligase is unable to form a phosphodiester bond.
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45) What advantage do cDNA libraries have over genomic libraries? Answer: Eukaryotic genomes tend to have more noncoding regions. Therefore, cDNA libraries offer a way to eliminate the nontranscribed regions and to look at only the transcribed regions of the genome. cDNA libraries can also be made from specific tissues.
Skill: Conceptual understanding
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46) Based on DNA sequences of single genes, genomic analysis has confirmed that prokaryotes are divided into two great lineages: the Bacteria and the Archaea. The latter may be more closely related to the Eukarya. Recount the lines of evidence that support these findings. Answer: Sequence divergence between archaeans and bacteria is considerable, and despite the general morphological similarity of these groups, the Archaea more closely resemble the Eukarya in terms of both the details of DNA replication and the expression (and, in some cases, the structure) of their genes.
Skill: Factual recall
47) Discuss the relative roles of descriptive and hypothesis‐driven science in genomics. Answer: Although sequencing genomes is a descriptive task, the data so obtained provide the raw material for framing an extensive array of hypotheses about gene and genome function, integration, and evolution. This is a good example of how descriptive knowledge is a necessary first step in understanding biological systems.
Skill: Conceptual understanding
48) Sequencing the dog genome is currently a high priority for genomics researchers. What insights do you think can be gained about the evolutionary process by analyzing the genome of dogs? Answer: Dogs are diversified into a multitude of breeds, many with striking morphological and behavioral differences. Yet, all dogs are the same species, and their DNA differs only minutely. Comparison of the genomes of different dog breeds may therefore shed light on which genetic elements are responsible for the radically different developmental pathways of different dog breeds–a process that may represent a microcosm of species‐level differentiation over evolutionary time.
Skill: Conceptual understanding
49) It is not always easy to locate or identify genes based on cDNA. How is this problem created by the differences between cDNA and DNA with respect to a given gene? Answer: cDNA is made from mRNA, which in eukaryotes is processed transcript with exons removed. The original gene may have many exons, and if so, the cDNA may little resemble it.
Skill: Conceptual understanding
50) Given the DNA oligonucleotide 5'—AGTCTAGGCT—3' , reconstruct a sequencing gel based on dideoxy sequencing, showing the relative position of each DNA fragment (band) on the gel. Answer: The gel drawn by the student should resemble that in Figure 8.11.
Skill: Conceptual understanding
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Chapter 9 Functional and Comparative Genomics
MATCHING QUESTIONS Please select the best match for each term. 1) Transgene
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A) Transcript that has a stem and loop B) Small noncoding human RNA C) Searches for sequence matches D) Artificially introduced DNA E) Fluorescent dye
2) shRNA
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3) Cy5
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4) HAR‐1
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5) BLAST
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Answers:
1) D
2) A
3) E
4) B
5) C
MULTIPLE-CHOICE QUESTIONS 6) Which of the following statements correctly describes reverse genetics? A) A mutated phenotype in an organism leads to the associated gene. B) A mutated gene is introduced into an organism to determine the phenotype. C) A gene is studied to understand the protein it encodes. D) A transcript is used to make a protein encoded by a gene. E) A gene is identified by bioinformatics. Answer: B
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7) The subfield of genomics that deals with gene expression and interaction is A) comparative genomics. B) functional genomics. C) structural genomics. D) hypothetical genomics. E) expression genomics. Answer: B
Skill: Factual recall
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8) The polymerase chain reaction has revolutionized genetics because A) it is capable of making virtually unlimited copies of DNA for study. B) it is capable of making copies of DNA with very little starting material. C) it is capable of making large numbers of DNA copies very quickly. D) All of these E) A and B only Answer: D
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9) RNA interference functions by A) degrading the protein encoded by a specific mRNA. B) degrading the RNA strand that is encoded by the Slicer gene. C) transcribing double‐stranded RNA into protein. D) disrupting the chromosomal copy of the gene. E) degrading the RNA complementary to the short, double‐stranded RNA. Answer: E
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10) Which of the following triggers the RNAi pathway? A) mRNA B) rRNA C) tRNA D) shRNA E) snRNA Answer: D
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11) Which of the following is not used in PCR amplification of DNA? A) Double‐stranded DNA template B) Oligonucleotide primers C) DNA polymerase D) Buffer E) ddNTPs Answer: E
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12) Which of the following does not occur during PCR? A) Denaturation at high temperatures B) Synthesis of oligonucleotide primers C) Annealing of oligonucleotide primers D) Extension of primers by DNA polymerase E) Changes in the reaction temperature by a thermocycler Answer: B
Skill: Factual recall
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Chapter 9 Functional and Comparative Genomics 73
13) If you are setting up a PCR reaction, how much extension time per cycle should you allocate for the Taq DNA polymerase to amplify a 5‐kb DNA fragment from a 10‐kb template? A) 1 minute B) 2.5 minutes C) 5 minutes D) 10 minutes E) 20 minutes Answer: C
Skill: Application of knowledge
14) The drug ________ is added to mouse cell cultures to select against transformants that do not have the desired gene knockout. A) ampicillin B) kanamycin C) neomycin D) G418 E) ganciclovir Answer: E
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15) The Alu family is an example of A) short interspersed repeat elements. B) intermediate repeat elements. C) long interspersed repeat elements. D) terminal sequences. E) gene‐flanking sequences. Answer: A
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16) A transgene is a A) hybrid gene. B) gene that contains no introns. C) bacterial gene that is found in mammalian cells. D) human gene of viral origin. E) gene introduced into an organism by artificial means. Answer: E
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17) The ________ gene is responsible for removing many of the drugs introduced into the human body. A) HAR‐1 B) CYP2D6 C) ASPM D) FOXP2 E) BRCA1 Answer: B
Skill: Factual recall
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18) Which of the following is an archaean that has been identified in the human gut? A) Escherichia coli B) Enterobacter aerogenes C) Bifidobacterium longum D) Methanobrevibacter smithii E) Caenorhabditis elegans Answer: D
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19) Pharmacogenomics is the study of how A) the genome affects responses to drugs. B) the genome leads to new drug discoveries. C) the genome affects microbial communities in the body. D) the genome is affected by drug use. E) medicines influence the genome of cancer cells. Answer: A
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20) The human‐accelerated region 1 (HAR‐1) gene A) was identified by functional genomics. B) is identical to the chimpanzee HAR‐1. C) encodes for the human accelerated region‐1 protein. D) is a small, noncoding RNA expressed in the brain. E) is a small, regulatory RNA that regulates gene expression. Answer: D
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21) Linkage disequilibrium refers to a condition in which A) a set of SNPs recombine often. B) a set of haplotypes recombine often. C) a set of specific alleles of two or more genes tend to appear together. D) a set of specific alleles of two or more genes only rarely appear together. E) a haplotype block rearranges every generation. Answer: C
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22) A genomic region can rapidly become very common in a population due to A) positive selection. B) negative selection. C) specific selection. D) linkage equilibrium. E) homologous recombination. Answer: A
Skill: Factual recall
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Chapter 9 Functional and Comparative Genomics 75
23) The representational oligonucleotide microarray analysis (ROMA) represents A) a functional genomics approach. B) a structural genomics approach. C) a comparative genomics approach. D) a metagenomics approach. E) a pharmacogenomics approach. Answer: C
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24) Metagenomic analysis involves A) culturing microbes isolated from the environment. B) extracting DNA from cultured microbes. C) analyzing DNA from environmental samples. D) sequencing cloned DNA individually. E) determining culture conditions for different microbes. Answer: C
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25) A microbiome consists of A) all the microorganisms on Earth. B) the community of microorganisms in a particular environment. C) the transcriptome from a particular microorganism under different environments. D) the proteome from a particular microorganism under different environments. E) the proteome from different microorganisms in a particular environment. Answer: C
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TRUE-FALSE QUESTIONS 26) ʺKnockoutʺ mutants have had the activities of certain genes eliminated, which can aid the researchers in determining the effects of those genes on physiology, development, and other functions. Answer: TRUE
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27) Bioinformatics is a relatively new field that combines elements of genomics with computer science and mathematics. Answer: TRUE
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28) The transcriptome is the complete set of polypeptides produced by an organism. Answer: FALSE Explanation: The transcriptome is the complete set of mRNAs produced in a cell of an organism, while the proteome is the complete set of polypeptides.
Skill: Factual recall
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29) BLAST is a software that searches for similarities between DNA sequences only. Answer: FALSE Explanation: The BLAST software can search for similarities using either DNA or protein inputs.
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30) Both strands of the short, double‐stranded regulatory RNA molecules are required by the Slicer protein in order to target the complementary RNA for degradation. Answer: FALSE Explanation: The Slicer protein unwinds the short, double‐stranded regulatory RNA, discards one of the strands, and uses the other strand to recognize the complementary RNA to be degraded.
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31) Sequence similarity between genes plays an important role in assigning gene function. Answer: TRUE
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32) Cy3 and Cy5 are radioactive tags used in microarrays. Answer: FALSE Explanation: These are fluorescent tags, not radioactive.
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33) The 16S rRNA gene from human gut microflora was amplified and sequenced to understand how the gut microbes are related. Answer: TRUE
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34) The genome of cancer cells tends to become unstable. Answer: TRUE
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35) The FOXP2 protein appears to play a critical role in brain size regulation. Answer: FALSE Explanation: The FOXP2 protein plays a critical role in speech production.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) What are the different methods by which we can assign function to particular genes? Answer: Functions may be assigned to particular genes of an organism by searching for sequence homology with genes of other organisms whose functions are known. Functions may also be assigned by interfering with the gene (either by making knockouts or through gene silencing by RNA interference) and studying the phenotypes produced.
Skill: Analytical reasoning
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Chapter 9 Functional and Comparative Genomics 77
37) Imagine that you are using random PCR primers to amplify fragments of DNA. Would you expect to get more amplification products if you set the PCR machine at a higher or lower annealing temperature and why? Answer: Higher annealing temperatures increase ʺstringency,ʺ meaning the accuracy of base complementarity must be higher for successful annealing. Under such conditions, fewer successful amplifications are expected compared to conditions in which stringency has been lowered by using a lower annealing temperature.
Skill: Conceptual understanding
38) List the steps involved in constructing a knockout mutant in yeast. Answer: Most commonly, a ʺdeletion moduleʺ is constructed with a selectable marker (such as an antibiotic resistance gene). Transformants are selected using the antibiotic resistance gene. The module recombines with the target knockout gene, and successful replacements can be confirmed with PCR.
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39) Why are BLAST comparisons based on protein sequences often easier to interpret than those based on DNA sequences? Answer: DNA sequences often have greater mismatches because of the degeneracy of the genetic code. Protein matches are therefore easier to compare than DNA sequences.
Skill: Conceptual understanding
40) What is the difference between a ʺknockoutʺ and a ʺknockdownʺ of a gene? Answer: Both terms refer to the alteration in the transcription from a gene, the difference being the level of transcription. In a knockout, the gene is not transcribed at all, whereas in a knockdown there is some transcription, usually well below the normal level.
Skill: Conceptual understanding
41) Calculate the number of copies that would be produced from a single starting molecule after 10 rounds of PCR. Answer: Ten cycles would produce 210, or 1,024, copies of the target DNA.
Skill: Problem-solving
42) After how many PCR cycles, starting from a single starting molecule, will fragments consisting of only the target DNA (the DNA between the primers) be generated? Answer: Fragments consisting of only target DNA are not generated until the third PCR cycle. (See http://www.dnalc.org/shockwave/pcranwhole.html for animation.)
Skill: Problem-solving
43) Why are the number of proteins in the eukaryotic proteome so much greater than the number of eukaryotic genes? Answer: Eukaryotic translation is characterized by transcript processing, where, in many cases, exons are differentially spliced to yield different polypeptide products. In addition, many functional proteins have quaternary structure, the subunits of which are provided by different genes. The number of possible proteins thus greatly exceeds the base number of genes in the genome.
Skill: Conceptual understanding
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44) Why is the Taq DNA polymerase well‐suited for PCR? Answer: Taq DNA polymerase is a heat‐stable enzyme that was isolated from the bacterium Thermus aquaticus. It can maintain function and structure at very high temperatures, allowing the PCR to run without having to inject a new enzyme after each denaturation cycle.
Skill: Conceptual understanding
45) Why are two selectable markers present in the target vector used for making mouse gene knockouts? Answer: Two selectable markers are used for making mouse gene knockouts because there is a high level of random integration of the target vector into the mouse genome. One of the selectable markers selects for the gene knockout, while the other selects against transformants in which the target vector may have integrated into a random location by nonhomologous recombination.
Skill: Factual recall
46) Why does the transcriptome differ within an organismʹs cell types? Answer: All the genes of a multicellular organism are not expressed in every cell type. The various cell types express different subsets of the genome, and the makeup of the transcripts (the transcriptome) differs depending on which genes are expressed at a given time.
Skill: Factual recall
47) What is one advantage that an RNAi screen may have over a screen for gene knockouts? Answer: A screen for gene knockouts usually does not yield deletions of essential genes because they are nonviable and do not live. An RNAi screen, however, may often lead to a knockdown, rather than a knockout, of an essential gene, which may be viable. This may lead to better understanding of the function of the gene that may otherwise be difficult to determine.
Skill: Conceptual understanding
48) How can we explain the fact that the human genome contains far fewer genes than was predicted would be necessary for our survival? Answer: Although we currently do not know the complete answer to this question, one partial explanation is that the microorganisms that reside within our bodies help us by synthesizing chemicals, like vitamins, that we then use. This symbiotic relationship may be part of the reason why we have only 20,000 or so genes.
Skill: Factual recall
49) What does a large haplotype block suggest about a genomic region? Answer: A large haplotype block usually indicates a recent origin, because the older a block, the greater the chances that it would have undergone genetic recombination at some point in the past. It may also suggest that some mutation within the haplotype block imparted a fitness advantage and has therefore been selected for (positive selection).
Skill: Conceptual understanding
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Chapter 9 Functional and Comparative Genomics 79
50) How do we know that some of the microbes in our gut are beneficial to us? Answer: Individuals lacking normal intestinal microbes have defects in the function of the immune system and in wound healing, suggesting that these microbes are essential to our well‐being.
Skill: Factual recall
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Chapter 10 Recombinant DNA Technology
MATCHING QUESTIONS Please select the best match for each term. 1) EST
Skill: Factual recall
A) Short oligonucleotide probe B) Repetitive sequence used in DNA fingerprinting C) A chloroplast enzyme D) Identified by mRNA E) Stems from single base substitution, for example
2) STR
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3) SNP
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4) EPSPS
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5) ASO
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Answers:
1) D
2) B
3) E
4) C
5) A
Please select the best match for each term. 6) PCR
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A) Technique used to generate copies of a DNA fragment B) Screening procedure used to assay DNA fragments C) Screening procedure used to assay mRNA fragments D) Movement of macromolecules in an electrical field E) DNA fragments generated by restriction enzymes 7) B 8) E 9) C 10) D
7) Southern blot
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8) RFLPs
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9) Northern blot
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10) Electrophoresis
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Answers:
6) A
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MULTIPLE-CHOICE QUESTIONS 11) ________ were the first class of DNA sequence polymorphisms discovered in the human genome. A) SNPs B) Microsatellites C) VNTRs D) STRs E) RFLPs Answer: C
Skill: Factual recall
12) RFLPs arise due to A) changes in the number of tandem repeats. B) base changes that add or delete restriction enzyme recognition sites. C) addition or deletion of bases between restriction enzyme recognition sites. D) B and C only E) All of these Answer: D
Skill: Factual recall
13) Which of the following sequences is considered an STR? A) CGTATTATGC B) TACGTACGTACG C) CAGCAGCAG D) TATTACGCCG E) A and B only Answer: C
Skill: Factual recall
14) Expressed sequence tags, or ESTs, A) are genetic markers generated from cDNA. B) pertain to expressed genes only. C) are oligonucleotides derived from mRNA. D) A and C only E) All of these Answer: E
Skill: Factual recall
15) Which of the following techniques is best suited to study protein interactions? A) PCR B) The yeast two‐hybrid system C) Restriction mapping D) RT‐PCR E) Southern blotting Answer: B
Skill: Factual recall
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16) In yeast, expression of the GAL genes is induced when there is ________ in the culture medium. A) glucose B) lactose C) galactose D) maltose E) Any of these Answer: C
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17) When yeast cells are cultured in the presence of both glucose and galactose, the GAL genes are A) transcribed and then translated. B) transcribed but not translated. C) neither transcribed nor translated. D) translated but not transcribed. E) transcribed and translated simultaneously. Answer: C
Skill: Factual recall
18) For the yeast galactose metabolizing gene, GAL1, to be transcribed, the regulatory protein produced by the GAL4 gene binds to A) RNA polymerase. B) upstream activator sequence G (UAS G) region. C) a ribosome. D) the operator sequence. E) the repressor protein. Answer: B
Skill: Factual recall
19) ASO probes are used for the molecular testing of A) chromosomal rearrangements. B) single nucleotide changes in a DNA region. C) RFLP fragments in a DNA region. D) defective proteins implicated in disease. E) successfully cloned Ti plasmids. Answer: B
Skill: Factual recall
20) Vectors capable of entering two or more different host organisms are known as A) dual vectors. B) omnivectors. C) shuttle vectors. D) parity vectors. E) universal vectors. Answer: C
Skill: Factual recall
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Chapter 10 Recombinant DNA Technology 83
21) ________ blots are used to study DNA fragments, while ________ blots are used to study RNA fragments. A) Western, northern B) Southern, western C) Eastern, northern D) Southern, northern E) Western, Southern Answer: D
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22) In a sequenced genome, candidate protein‐encoding genes are identified by searching for A) STRs. B) RFLPs. C) ESTs. D) SNPs. E) ORFs. Answer: E
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23) A(n) ________ is a photograph using X‐ray film that is used to identify the position of radiolabeled molecules on a gel. A) Southern blot B) northern blot C) X‐rayogram D) autoradiogram E) radiocopy Answer: D
Skill: Factual recall
24) Which of the following is used to screen newborns for the genetic disease PKU? A) RFLPs B) ASOs C) Antibodies D) PCR E) All of these Answer: A
Skill: Factual recall
25) Successful cotransformation in the yeast two‐hybrid system will result in the binding of proteins produced by each yeast expression vector. These fusion proteins function by A) binding to the lacZ operator, preventing transcription. B) binding to RNA polymerase, preventing transcription. C) binding to the lacZ repressor, enabling transcription. D) binding to the lacZ upstream activator sequence, enabling transcription. E) changing color in the presence of X‐gal. Answer: D
Skill: Factual recall
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TRUE-FALSE QUESTIONS 26) DNA markers are DNA polymorphisms that are useful for genetic mapping. Answer: TRUE
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27) RT‐PCR is a PCR technique used in amplifying DNA copies of mRNA. Answer: TRUE
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28) Somatic cell gene therapy can result in a cure for a genetic disease in an individual as well as any progeny produced by the individual. Answer: FALSE Explanation: Only changes in germline cells are heritable.
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29) Humulin is insulin created by recombinant DNA technology. Answer: TRUE
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30) DNase is used to treat hemophilia. Answer: FALSE Explanation: DNase is used to treat cystic fibrosis.
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31) The Guthrie test is used to screen all newborns in the United States for cystic fibrosis. Answer: FALSE Explanation: This test is used to screen for phenylketonuria, or PKU.
Skill: Factual recall
32) The Ti plasmid is a natural plasmid in the soil bacterium Agrobacterium tumefaciens. Answer: TRUE
Skill: Factual recall
33) Molecular DNA testing is used to test for aneuploids. Answer: FALSE Explanation: Molecular DNA testing examines the molecular structure of a particular disease‐associated gene to determine whether the mutations that would put an individual at high risk for the disease are present.
Skill: Factual recall
34) DNA fingerprinting can be used to convict criminals, but has had little application in exonerating individuals already convicted. Answer: FALSE Explanation: An important application of DNA fingerprinting is the overturning of wrongful convictions (which this technology has revealed to be widespread) in cases where biological evidence is still available for testing.
Skill: Factual recall
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Chapter 10 Recombinant DNA Technology 85
35) Ti plasmids and the T‐DNA system can be used to transform any vascular plant. Answer: FALSE Explanation: Only dicots, not monocots, can be transformed with the Ti plasmid and the T‐DNA system.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) A potentially serious problem with Taq polymerase in PCR is its lack of proofreading ability. Briefly explain why this can be a problem and at which stages during the PCR process this is most and least problematic. Answer: Replication errors are introduced by Taq polymerase as a result of its lack of proofreading. The earlier they are introduced, the greater the number of descendant copies that will also have the error. Therefore, such errors are least problematic when they occur late in the PCR process.
Skill: Conceptual understanding
37) Briefly list the basic steps involved in constructing a DNA fingerprint for a paternity case. Answer: Cut the DNA of each sample (child, mother, and supposed father) with restriction enzymes; separate the fragments for each sample with electrophoresis; perform a Southern blot analysis to transfer the fragments to a filter; probe the fragments with radiolabeled DNA oligonucleotides; and develop an autoradiogram to read the fragment sizes of each individual.
Skill: Factual recall
38) How does a transcribable vector differ from an expression vector? Answer: Transcribable vectors are designed for the transcription of the insert both in vivo and in vitro, whereas expression vectors are typically designed only for in vivo expression.
Skill: Factual recall
39) Study of P elements in Drosophila has shown that gene regulation can be accomplished by differential mRNA transcript processing. Explain. Answer: The P element carries a single gene that encodes for the protein transposase. This protein is required for the transposition of P elements. The P element pre‐mRNA is processed (spliced) differently in normal flies than in flies undergoing hybrid dysgenesis. In the former, the mRNA transcript is larger because the third intron is not spliced. But this results in a smaller protein due to the presence of an in‐frame stop codon in the retained intron. In the germline of flies undergoing hybrid dysgenesis, all the introns are spliced out. This results in a smaller transcript that encodes for a functional transposase. This may result in transposition of the P element
Skill: Factual recall
40) How does an RFLP help identify sickle‐cell anemia? Answer: The normal globin gene is associated with three DdeI sites. The mutation that causes sickle‐cell anemia changes one of those sites, leaving only two. This can be detected by PCR amplifying the globin gene and digesting with the DdeI restriction enzyme. A normal person yields two bands, while a person with sickle‐cell anemia yields only one larger band.
Skill: Conceptual understanding
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41) PCR has revolutionized the criminal justice system through DNA fingerprinting. For a given fragment of DNA, showing that a suspectʹs fingerprint differs from that found at a crime scene immediately excludes him or her from suspicion, yet finding a match does not mean that guilt can be automatically concluded. Why? Answer: Different fragment lengths clearly must have come from different sources, but having the same fragment length only means there is a match–it may be possible for multiple persons in the population to possess the same fragment length.
Skill: Conceptual understanding
42) Two STR loci were determined to have the following allele frequencies in the general population: STR A 1A: 0.1 2A: 0.55 3A: 0.25 4A: 0.1 STR B 1B: 0.4 2B: 0.4 3B: 0.2
Alleles:
A suspect in a criminal trial was genetically tested for these loci and was found to have the genotype (1A, 1A) for STR A and (2B, 2B) for STR B–homozygous for both loci. What is the probability that a randomly sampled person from this population would also have this genotype? Answer: The product rule states that the net probability is the product of the independent probabilities. Since these are homozygous genotypes, the probability of the STR A genotype is: 0.1 × 0.1 = 0.01, and the STR B genotype is 0.4 × 0.4 = 0.16. The net probability is then 0.01 × 0.16 = 0.0016. There is an approximately one in a thousand chance that a randomly selected person would share these STR genotypes.
Skill: Problem-solving
43) How do genetic tests differ from diagnostic tests for genetic disease? Answer: Genetic tests identify if an individual is carrying a mutation known to be associated with a genetic disease; depending on the disease, having the mutation does not mean the disease will necessarily develop. Diagnostic tests, in contrast, reveal if a disease is present and, if so, to what extent it has developed.
Skill: Conceptual understanding
44) Genetic testing is typically done with blood samples, yet red blood cells (erythrocytes) are anucleate (lack nuclei) and are thus devoid of DNA. How, then, are these tests done? Answer: Other blood cells, in particular white blood cells or phagocytes, are not anucleate, and DNA can be extracted from them.
Skill: Application of knowledge
45) A couple with STR genotypes (50, 75) and (50, 100) (where the allele numbers denote repeat length) have several children. What are the possible genotypes of their children for this STR locus? Answer: As for any single‐locus trait, simple Mendelian principles tell us what the genotypes may be: (50, 50) homozygote and (50, 100) and (75, 100) heterozygotes.
Skill: Problem-solving
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Chapter 10 Recombinant DNA Technology 87
46) Antisense mRNA can be used to regulate the expression of genes. Briefly describe how this technology was used to slow the ripening of Calgeneʹs ill‐fated Flavr‐Savr TM tomato. Answer: In engineered tomatoes, the gene for polygalacturonase, the enzyme responsible for fruit softening, was inserted backwards. This yielded an antisense mRNA copy that bonded with the sense mRNA, reducing the effective transcript level present for translation. Ripening was thus retarded relative to the normal tomato.
Skill: Conceptual understanding
47) Briefly explain how Ti plasmids induce crown gall (tumor) growth in plants parasitized by Agrobacterium tumefaciens, highlighting the specific properties that make this plasmid so useful in biotechnology. Answer: Ti plasmids are so named because they carry tumor‐inducing genes. The plasmid enters (transforms) a host cell, and a 30‐kb region called transforming‐ or T‐DNA integrates by recombination with the host DNA. The T‐DNA region carries genes for plant cell transformation, creating the crown gall where Agrobacterium flourishes under natural conditions. The T‐DNA region necessary for successful excision, transfer, and integration consists of only its 25‐bp terminal repeats, which means the rest of the DNA in this region can be replaced with DNA of interest and the whole plasmid engineered to vector this DNA.
Skill: Conceptual understanding
48) Rapid detection of SNPs, which are allelic variants, can be accomplished with DNA microarrays. If a microarray is set up for a set of genes, including all variants, outline a general procedure for quickly assaying the genotype of multiple individuals for that set of genes. Answer: DNA taken from the individuals would be cut (restricted) and labeled with fluorescent markers. Once introduced to an array, the probes hybridize where there is sequence complementarity. The fluorescent marker then reveals which allelic variants of the genes of interest are expressed by each individual.
Skill: Factual recall
49) How has the Bt protein been used to increase insect‐resistance in plants? Answer: The Bt protein is a bacterial protein that has been introduced into certain crops. When insects eat these plants, the Bt protein kills or injures the insect. Purified Bt proteins and bacteria that naturally express the Bt protein have been used as insecticides in organic farming.
Skill: Factual recall
50) DNA sequencing using the dideoxy sequencing method involves PCR reactions using Taq polymerase, which we learned in this chapter can introduce errors because it lacks a proofreading mechanism. What precaution(s) can be taken in genomic or other sequencing projects to minimize the possibility of producing DNA sequences with errors? Answer: The most common precaution taken is sequencing a fragment of DNA in both directions. Any replication errors will show up as mismatches, which could be resolved by resequencing that fragment of DNA.
Skill: Conceptual understanding
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Chapter 11 Mendelian Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) Haploid
Skill: Factual recall
A) Possessing the correct number of chromosomes B) Several complete sets of chromosomes C) One half of a complete set of chromosomes D) Possessing too few or too many copies of a single chromosome E) Two complete sets of chromosomes 2) E 3) A 4) B 5) D
2) Diploid
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3) Euploid
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4) Polyploid
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5) Aneuploid
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Answers:
1) C
Please select the best match for each term. 6) Monohybrid cross
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A) Allele that has to be homozygous to be expressed B) Genetic makeup of an organism C) Allele that is expressed in the heterozygote D) Always involves alleles of two genes E) Always involves alleles of a single gene F) One parent must be homozygous recessive G) Physical manifestation of a trait
7) Dihybrid cross
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8) Test cross
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9) Phenotype
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10) Genotype
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11) Dominant allele
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12) Recessive allele
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Answers:
6) E 11) C
7) D 12) A
8) F
9) G
10) B
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Chapter 11 Mendelian Genetics 89
MULTIPLE-CHOICE QUESTIONS 13) In lilies, white flowers (W) are dominant to purple flowers (w). If two plants that are heterozygous for flower color are mated, the genotypic ratio of the offspring would be A) 3:1. B) 1:1. C) 1:2:1. D) 9:3:3:1. E) 1:1:1:1. Answer: C
Skill: Problem-solving
14) In a pea plant that is heterozygous for seed color, what proportion of gametes will carry the recessive allele? A) B) C) D) E)
1 4 1 2 3 4 1 8 1 8
Answer: B
Skill: Problem-solving
15) In his experiments, Mendel noted that when two traits are involved in a genetic cross, they are inherited independently of each other. The reason for this is that A) genes on the same chromosome separate during the formation of gametes. B) alleles on the same gene separate during the formation of gametes. C) genes on different chromosomes separate during the formation of gametes. D) alleles on the same chromosome separate during the formation of gametes. E) chromosomes often recombine. Answer: C
Skill: Factual recall
16) Which of the following genotypes would not usually be represented in a gamete? A) AA B) AB C) Ab D) aB E) ab Answer: A
Skill: Conceptual understanding
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17) The chi‐squared statistic can best be described as A) the standardized deviation of observed data from expected data. B) a statistically significant test. C) a statistical test used in genetics. D) a statistical test that compares observed and expected population means. E) None of these Answer: A
Skill: Conceptual understanding
18) In the F2 generation, how many genotypic classes could be generated from a dihybrid cross of two heterozygotes in which the genes involved show complete dominance? A) 3 B) 4 C) 8 D) 9 E) 12 Answer: D
Skill: Problem-solving
19) In peas, tall plants are dominant to short plants. A cross between a tall pea plant and a short pea plant results in half the progeny being tall, and the other half being short. Therefore, the tall parent plant is genotypically A) homozygous. B) heterozygous. C) hemizygous. D) monozygous. E) homogeneous. Answer: B
Skill: Problem-solving
20) If the results of a chi‐square test of a given set of data show a P value greater than 0.05, then the null hypothesis A) must be rejected. B) must be accepted. C) must be rephrased. D) cannot be rejected. E) cannot be accepted. Answer: D
Skill: Factual recall
21) A P value in statistics is A) a measure of the accuracy of a data set. B) the probability of getting the observed data distribution by chance. C) arbitrarily set depending on a statistical test. D) a measure of the accuracy of a statistical test. E) None of these Answer: B
Skill: Factual recall
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Chapter 11 Mendelian Genetics 91
22) A man whose father expresses a recessive trait marries a woman whose brother expresses the same recessive trait. What is the likelihood that the newlyweds could have a child expressing the trait? A) B) C) D) E)
1 2 1 3 1 4 1 5 1 6
Answer: E
Skill: Problem-solving
23) In humans, brown eye color (B) is dominant to blue eyes (b). A brown‐eyed man marries a woman with brown eyes and they have three blue‐eyed daughters. What are the genotypes of the man and the woman? A) BB and Bb B) BB and BB C) Bb and bb D) Bb and Bb E) bb and bb Answer: D
Skill: Problem-solving
24) The probability that two parents with a family of four will have one girl and three boys is A) B) C) D) E)
1 2 1 4 1 8 1 16 1 32
Answer: B
Skill: Problem-solving
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25) A couple with three girls is expecting a fourth child. The probability that this child is also a girl is A) B) C) D) E)
1 8 1 4 1 8 1 16 1 32
Answer: A
Skill: Problem-solving
26) Net or overall probabilities are obtained by multiplying separate independent probabilities. This is formally known as A) the sum rule. B) the chi‐square test. C) the product rule. D) the sign test. E) the probability rule. Answer: C
Skill: Factual recall
27) Which of the following statements concerning the inheritance of a dominant trait is true? A) Every affected person must have at least one affected parent. B) The trait is observed to skip generations. C) An affected heterozygote will transmit the allele to all of his or her offspring. D) Nonaffected parents can have affected offspring. E) The trait is passed along paternal lines. Answer: A
Skill: Conceptual understanding
28) A dihybrid cross yields 320 F 2 offspring. How many are expected to resemble the homozygous recessive parental? A) 10 B) 16 C) 20 D) 60 E) 180 Answer: C
Skill: Problem-solving
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Chapter 11 Mendelian Genetics 93
29) In his monohybrid crosses for seed color in peas, Mendel reported 6,022 yellow seeds and 2,001 green seeds. How many of each color class were expected? A) 4,011 green and 4,011 yellow B) All should be green C) All should be yellow D) 6,017 yellow and 2,006 green E) 2,006 yellow and 6,017 green Answer: D
Skill: Problem-solving
30) In holly, serrated leaves are dominant to smooth‐edged leaves, and red berries are dominant to green berries. Two holly plants heterozygous for leaf edge shape and berry color are crossed together. Assuming that the two traits assort independently of one another, what proportion of the progeny will have smooth‐edged leaves and green berries? A) B) C) D) E)
1 4 3 4 9 16 3 16 1 16
Answer: E
Skill: Problem-solving
TRUE-FALSE QUESTIONS 31) A testcross with a heterozygous dominant individual will yield only heterozygous dominant offspring. Answer: FALSE Explanation: AA × aa → AA : aa in a 1:1 ratio.
Skill: Factual recall
32) Two individuals can be phenotypically identical, yet have different genotypes for a given trait. Answer: TRUE
Skill: Factual recall
33) The genotypic F 2 ratio expected in a dihybrid cross is 9:3:3:1. Answer: FALSE Explanation: That is the expected phenotypic ratio. The genotypic ratio should be 1:2:1:2:4:2:1:2:1.
Skill: Factual recall
34) The phenotype determines the genotype. Answer: FALSE Explanation: Genotypes underlie phenotypes, not vice versa.
Skill: Factual recall
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35) True‐breeding individuals are produced by repeated backcrossing. Answer: TRUE
Skill: Factual recall
36) Recessive alleles are usually loss‐of‐function mutations. Answer: TRUE
Skill: Factual recall
37) The dominant allele of a gene is the most frequently found allele in a population. Answer: FALSE Explanation: The wild‐type allele is the most frequently found allele of a gene in a population.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 38) Mendel was the first scientist to describe the concept of diploidy–the idea that organisms like pea plants and humans possess two complete sets of genetic material. Summarize Mendelʹs observations and the reasoning that led him to this conclusion. Answer: The F1 and F2 phenotypic ratios that Mendel observed in different crosses led him to conjecture that each parent contributes one version of a ʺunit factorʺ for each trait during reproduction. Each individual thus possesses two such factors (diploidy), and it is combinations of unit factors that constitute the genotype.
Skill: Conceptual understanding
39) Mendel selected seven traits to analyze in his famous pea plant crosses, and all of these traits yielded expected 3:1 phenotypic F 2 ratios in monohybrid crosses. He was fortunate in his selection of these traits. How so? What problem might he have encountered that may have yielded confusing ratios? Answer: By chance, some of the traits he selected might not have been independently assorting, owing to linkage (presence on the same chromosome).
Skill: Conceptual understanding
40) Mendelʹs insights started with his approach of analyzing discrete traits, leading to the idea of ʺparticulateʺ rather than ʺblendingʺ inheritance. Yet this was an uncommon way to view the inheritance of traits because many found it counterintuitive or contrary to experience. How so? Answer: Experience with breeding pets or livestock, or even having children, suggests that parental traits seem to blend together in the offspring. This is because most traits that breeders considered are complex, polygenic traits. By focusing on simple discrete traits, Mendel was able to show that his ʺunit factorsʺ simply combined and recombined in pairs each generation and were not blended away.
Skill: Conceptual understanding
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Chapter 11 Mendelian Genetics 95
41) In snapdragons, the genes A, B, C, D, and E assort independently of one another. If an Aa Bb Cc dd EE plant is crossed with an Aa bb cc Dd Ee plant, what is the probability of obtaining an offspring that is phenotypically dominant for all five traits? Answer: The probability of obtaining an offspring that is phenotypically dominant for all five traits is 3 1 1 1 3 × × × × 1 . 4 2 2 2 32
Skill: Problem-solving
42) A monohybrid cross is made for flower color, where purple is dominant to white. Fifteen hundred F2 offspring are analyzed. How many white flowers are expected? Answer: One‐quarter of the offspring should be homozygous recessive, thus: (0.25) × 1,500 = 375 white flowers.
Skill: Problem-solving
43) Speculate on the molecular basis for dominance and recessiveness, using flower color as an example, where red is dominant over white. Answer: If the trait is controlled by a gene responsible for synthesizing red pigment, the recessive allele could be a dysfunctional mutant. Red pigment would be made as long as there was at least one functional copy of the gene present–this is the essence of dominance, where the presence of a single allele is sufficient to mask the recessive allele.
Skill: Conceptual understanding
44) You observe an individual of your favorite study organism expressing the dominant phenotype for a certain trait. How would you go about determining if the individual was homozygous dominant or heterozygous for that trait? Answer: Perform a test cross with a recessive homozygous individual: If Aa × aa offspring are Aa : aa in a 1:1 ratio, and if AA × aa → all offspring are Aa heterozygotes.
Skill: Problem-solving
45) A man with albinism and a woman with a normal phenotype have several children, one of whom displays albinism. a. What can you conclude about the genotype of the mother? b. What is the probability that the children who do not display albinism are heterozygous? Answer: (a) The mother must be heterozygous (a carrier) in order to have even one offspring with albinism. (b) With parents of genotype Aa × aa, any offspring that do not display albinism in their phenotype must be heterozygous, so the probability is 100%.
Skill: Problem-solving
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46) A man and a woman with normal phenotypes have several children, one of whom has albinism. a. What can you conclude about the genotype of the mother? b. What is the probability that the children who do not display albinism are heterozygous? Answer: (a) Both parents must be heterozygous (a carrier) in order to have an offspring with albinism. (b) With parents of genotype Aa × Aa, the genotypes possible for offspring not displaying the traits of albinism are AA, Aa, and aA. Two of these three are carriers, so there is a (66%) chance that a child who does not display albinism is a carrier.
3 2
Skill: Problem-solving
47) What characteristics make an organism a good candidate for Mendelian studies? Answer: Mendelian studies are based on the probability and statistical analyses of progeny phenotypes. Therefore, the candidate organism should be reach maturity and reproductive age fairly quickly, and should produce many offspring
Skill: Conceptual understanding
48) A monohybrid (one‐gene) cross yields 4 genotypic classes, and a dihybrid (two‐gene) cross yields 16. How many classes are expected from a tetrahybrid (four‐gene) cross? Answer: Following the simple relationship 4 1 = 4 and 4 2 = 16, 44 = 256 expected genotypic classes.
Skill: Conceptual understanding
49) Explain why it is expected that heterozygotes will be produced twice as frequently as either homozygote in a monohybrid F 1 cross. Answer: Combinatorials. Each individual offspring genotype has an equal likelihood of occurring (= ), but since there are two ways to make heterozygotes (Aa and aA),
4 1
together they occur with a + = frequency, twice that of either homozygote,
4 4 2
1
1
1
which occur at a frequency of each.
Skill: Conceptual understanding
1 4
50) State the key difference between Mendelʹs principle of segregation and independent assortment. Answer: Segregation refers to separation of alleles into gametes, while independent assortment refers to random combinations of such alleles from different genes occurring in gametes.
Skill: Conceptual understanding
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Chapter 12 Chromosomal Basis of Inheritance
MATCHING QUESTIONS Please select the best match for each term. 1) XXY
Skill: Factual recall
A) Aneuploidy B) Klinefelter syndrome
2) XO
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C) Mammalian female D) Polyploidy E) Turner syndrome
3) XX
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4) 2N+1
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5) 3N
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Answers:
1) B
2) E
3) C
4) A
5) D
MULTIPLE-CHOICE QUESTIONS 6) A person with an XO genotype for the sex chromosomes would be phenotypically A) male. B) female. C) intersex (both male and female). D) male or female with equal probability. E) neither male nor female. Answer: B
Skill: Factual recall
7) Cells obtained from a Triplo‐X woman would contain ________ Barr bodies. A) 0 B) 1 C) 2 D) 3 E) 4 Answer: C
Skill: Application of knowledge
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8) In human males, genes on the X chromosome are A) homozygous. B) heterozygous. C) homogametic. D) heterogametic. E) hemizygous. Answer: E
Skill: Factual recall
9) In butterflies, males are the ________ sex. A) homozygous B) heterozygous C) homogametic D) heterogametic E) hemizygous Answer: C
Skill: Factual recall
10) The key gene in mammalian sex determination is A) the testis‐determining factor gene. B) the Y‐factor gene. C) the holandric factor gene. D) the X‐inactivation center. E) the lyonization gene. Answer: A
Skill: Factual recall
11) A litter produced by a calico cat is not expected to include A) solid‐colored female kittens. B) calico female kittens. C) solid‐colored male kittens. D) calico male kittens. E) Any of these Answer: D
Skill: Application of knowledge
12) A human male who undergoes a sex‐change operation will be chromosomally A) XX. B) XY. C) XXY. D) XYY. E) YYY. Answer: B
Skill: Factual recall
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Chapter 12 Chromosomal Basis of Inheritance 99
13) The chromosome theory of inheritance states that A) genes are located on chromosomes. B) chromosomes replicate before cell division. C) alleles of a gene segregate during cell division. D) chromosomes consist of DNA. E) different chromosomes assort independently of one another. Answer: A
Skill: Factual recall
14) The spindle fibers attach to the centromere via a protein complex called A) tubulins. B) microtubules. C) chromatid. D) kinetochore. E) chiasma. Answer: D
Skill: Factual recall
15) Plants that produce separate male and female flowers are said to be A) hermaphroditic. B) heterosexual. C) monoecious. D) dioecious. E) homosexual. Answer: D
Skill: Factual recall
16) In an organism with 16 chromosomes per cell, how many DNA molecules are present in a cell undergoing anaphase of mitosis? A) 8 B) 16 C) 24 D) 32 E) 64 Answer: D
Skill: Application of knowledge
17) Synaptonemal complex forms during A) leptonema. B) zygonema. C) pachynema. D) diplonema. E) diakinesis. Answer: B
Skill: Factual recall
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18) Phenotypically, a female (XX) mouse embryo that was transformed with a small piece of a Y chromosome bearing the SRY gene would develop as A) a male. B) a female. C) an intersex individual. D) a male initially, then revert to female after birth. E) a female initially, then revert to male after birth. Answer: A
Skill: Factual recall
19) Which of the following is an example of genetic mosaicism? A) Calico coat color in cats B) Albinism in mice C) Male pattern baldness in humans D) Red‐green color blindness in humans E) White eyes in Drosophila Answer: A
Skill: Factual recall
20) Individuals with Turner syndrome are A) euploid. B) aneuploid. C) polyploid. D) haploid. E) monoploid. Answer: B
Skill: Factual recall
21) Which of the following organisms has a ZZ‐ZW method of sex determination? A) House mice B) Monarch butterflies C) Fruit flies D) Humans E) Worms Answer: B
Skill: Factual recall
22) The largest human autosome is chromosome ________ and the smallest is chromosome ________. A) 1, 22 B) 1, 21 C) 1, Y D) X, 22 E) None of these Answer: B
Skill: Factual recall
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Chapter 12 Chromosomal Basis of Inheritance 101
23) Chromosomes with only one arm are known as A) metacentric. B) submetacentric. C) acrocentric. D) telocentric. E) acentric. Answer: D
Skill: Factual recall
24) Red‐green color blindness in humans is an X‐linked recessive trait. What will the progeny ratios be if a carrier female is mated with a normal male? A) All the sons will be color blind; all the daughters will be normal. B) All the sons and half the daughters will be color blind. C) Half the sons will be color blind; all the daughters will be carriers. D) Half the sons will be color blind; half the daughters will be carriers. E) All the sons and all the daughters will be normal. Answer: D
Skill: Problem-solving
25) Which of the following statements is not true? A) Many plants exhibit alternation between haploid and diploid stages. B) Chromosome number is generally constant within groups of organisms with common ancestry. C) The Y‐bearing sex is known as the heterogametic sex. D) Sex determination is chromosomal in some organisms and genic in others. E) The sex chromosome systems of birds and mammals are likely to have evolved independently. Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) A person with two X chromosomes is the homogametic sex. Answer: TRUE
Skill: Factual recall
27) A woman who is heterozygous for the hemophilia allele has a 100% chance of having an affected son. Answer: FALSE Explanation: The woman has one affected and one unaffected allele and will donate only one of these, with equal probability, to each of her sons.
Skill: Factual recall
28) Half a womanʹs somatic cells express her paternal X chromosome and half express her maternal X chromosome. Answer: FALSE Explanation: X‐inactivation is random, so the maternal and paternal copies of her X chromosomes will not be inactivated with equal frequency.
Skill: Factual recall
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29) Synapsis usually forms between nonsister chromatids. Answer: TRUE
Skill: Conceptual understanding
30) X‐linked or sex‐linked traits are expressed in males but not in females. Answer: FALSE Explanation: Such traits are expressed most commonly in males due to their hemizygosity, but can be expressed in females homozygous for the X‐linked alleles.
Skill: Factual recall
31) Barr bodies are inactivated X chromosomes. Answer: TRUE
Skill: Factual recall
32) In flowering plants, gametes are produced mitotically from spores. Answer: TRUE
Skill: Factual recall
33) The centrioles are highly conserved chromosome regions that interact with spindle fibers during cell division. Answer: FALSE Explanation: The chromosome region that interacts with spindle fibers is the centromere.
Skill: Factual recall
34) Some cells may exit the cell cycle and enter a nondividing G 0 state. Answer: TRUE
Skill: Factual recall
35) Nondisjunction is the failure of sex chromosomes to separate during gamete formation. Answer: FALSE Explanation: Nondisjunction is the failure of chromosomes to separate. This can involve either autosomes or sex chromosomes, and may occur either during mitosis or meiosis.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) Describe two dosage compensation mechanisms found in animals. Answer: X‐inactivation in females, in which one X chromosome is silenced, and X‐hyperactivation in males, in which genes on the X chromosome have increased expression.
Skill: Factual recall
37) X‐linked traits are more likely to be expressed in males. Why? Under what circumstances are they expressed in females? Answer: Males are hemizygous, with only one X chromosome; therefore, any genes on that chromosome are expressed, regardless of whether they are dominant or recessive. Females must be homozygous for a recessive X‐linked allele to be expressed.
Skill: Conceptual understanding
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Chapter 12 Chromosomal Basis of Inheritance 103
38) What is the connection between the work of Mary Lyon and Murray Barr? Answer: Mary Lyon put forth the theory of X‐inactivation, now called lyonization. Murray Barr was the first to describe Barr bodies, which we now know are inactivated X chromosomes.
Skill: Factual recall
39) Why will a human embryo develop as a male in the presence of a Y chromosome, regardless of the number of X chromosomes present? Answer: Sex is determined by genes located on the Y chromosome that lead to the development of masculine traits. Even one Y copy, regardless of the number of X chromosomes, is thus sufficient to initiate male development.
Skill: Conceptual understanding
40) Nondisjunction can lead to two forms of aneuploidy. What are they, and how do they occur? Answer: The two most common aneuploid conditions are N + 1 and N ‐ 1 (one extra and one missing chromosome, respectively). The N + 1 condition arises when two chromatids of a given chromosome fail to separate during meiosis, giving the resulting gamete an additional copy of that chromosome. Likewise, the N ‐ 1 condition is found in the gamete missing a chromatid that stayed paired with its sister.
Skill: Conceptual understanding
41) A woman who is color blind, a sex‐linked trait, has several children with a man who has normal vision. What are the possible phenotypes of their children with respect to this trait? Answer: All their daughters will have normal vision, while all their sons will be color blind.
Skill: Problem-solving
42) What are the differences between metaphase I and metaphase II of meiosis? Answer: Metaphase I of meiosis has the same number of chromosomes as the original cells, while metaphase II of meiosis has half the number of chromosomes. In metaphase I, the homologous chromosomes are arranged in pairs at the metaphase plate, whereas in metaphase II there are no homologs, and therefore, individual chromosomes are arranged separately at the metaphase plate.
Skill: Factual recall
43) The chromosome theory of inheritance sought to explain Mendelian patterns in terms of chromosome movement. Can you think of an experimental approach to test for the Mendelian principle of segregation using chromosomes? Answer: If replicating chromosomes could be chemically marked, permitting identification of each daughter chromatid, the movement of each could then be tracked. Radioisotopes are commonly used in such labeling experiments.
Skill: Analytical reasoning
44) How did chromosomes get their name? Answer: The term, literally meaning ʺcolored bodies,ʺ was coined in the late nineteenth century at a time when these nuclear structures could be only indistinctly viewed through a microscope with staining.
Skill: Conceptual understanding
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45) Describe a situation in which two aneuploid gametes would yield a euploid zygote. Answer: The aneuploidy of the gametes must be such that the same chromosomes are affected in each, one with an extra copy of the chromosome and the other only deficient in that same chromosome. Fusion of these two gametes would restore euploidy.
Skill: Analytical reasoning
46) You are a breeder of animals with an X‐linked coat‐color gene, one allele of which gives black fur and another of which gives white fur. If you cross a black‐colored female with a white‐colored male, what would their male and female offspring look like? Answer: Their male offspring would be all black, while their female offspring would be mottled, owing to X‐inactivation: males: XbY (black) XbXb × XwY → females: XbXw (some fur patches black, some white)
Skill: Problem-solving
47) Describe Calvin Bridgesʹs classic study of nondisjunction of sex chromosomes, which provided support for the chromosome theory of inheritance. Answer: Bridges performed crosses with fruit flies, tracking the sex‐linked eye‐color trait. Red‐eyed males were crossed with white‐eyed females, yielding mostly white‐eyed male offspring and red‐eyed female offspring. A handful of anomalous individuals proved informative: about 1/2,000 offspring were white‐eyed females or red‐eyed males. Karyotypic study of these anomalous individuals revealed they had suffered nondisjunction of the sex chromosomes.
Skill: Problem-solving
48) Closely related species, like chimpanzees and humans, often differ in chromosome number by only one or two chromosomes, suggesting that chromosomes can experience fission and fusion over evolutionary time. Can you think of a way to test this? Answer: Generate a karyotype for the species of interest and stain the chromosomes to obtain banding patterns. Compare the banding patterns of chromosomes that you speculate may be the product of fission or fusion–banding patterns are highly conserved–and you should be able to match up the chromosomes from one species with those of another.
Skill: Analytical reasoning
49) Drosophila fruit flies, like mammals, have an X‐Y sex‐determination system. The insect differs from the mammalian system in an important respect, however. What is it? Answer: While Drosophila has X‐Y sex determination, where XX individuals are female and XY individuals are male, studies have shown that sex determination is based on the ratio of X chromosomes to autosomes rather than on Y‐linked genes. Thus, XO Drosophila develop as males and XXY Drosophila develop as females–the same X : autosome ratio as XY and XX individuals.
Skill: Factual recall
50) What pattern of expression is expected for an X‐linked dominant trait like hereditary enamel hypoplasia? Answer: An affected father will have affected daughters and no affected sons; an affected mother will have a 50% chance of having an affected son and a 50% chance of having an affected daughter (assuming the mother is heterozygous).
Skill: Problem-solving Copyright © 2010 Pearson Education, Inc.
Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles
MATCHING QUESTIONS Please select the best match for each term. 1) Codominance
Skill: Factual recall
2) Epistasis
Skill: Factual recall
A) When heterozygotes display a phenotype intermediate between those shown by homozygotes for that allele B) When a genotype gives rise to a range of phenotypes C) When each allele causes a distinct phenotype, and heterozygotes exhibit both phenotypes of heterozygotes D) When the genotype at one locus generates phenotypic expression of a trait that overrides and conceals the phenotypic traits arising from one or more other genes E) When a genotype underlies a given phenotype, but not all individuals with that genotype display the phenotype
3) Incomplete dominance
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4) Incomplete penetrance
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5) Variable expressivity
Skill: Factual recall
Answers:
1) C
2) D
3) A
4) E
5) B
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Please select the best match for each term. 6) Heteroplasmon
Skill: Factual recall
A) When a cell or individual has more than one mitochondrial genome type B) A phenotype caused by a maternal nuclear genotype that affects eggs prior to fertilization, and is therefore expressed in all offspring, regardless of paternal genes C) When either the frequency or amount of a phenotype depends on the sex of the individual D) When a genotype causes a phenotype in only one sex E) When a genotype and phenotype are only inherited from the mother
7) Maternal effect
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8) Maternal inheritance
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9) Sex‐limited
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10) Sex‐influenced
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Answers:
6) A
7) B
8) E
9) D
10) C
MULTIPLE-CHOICE QUESTIONS 11) The distinctive fur pattern of Siamese cats results when genes that are responsible for the coat color are influenced by A) age. B) X‐inactivation. C) gender. D) temperature. E) other genes. Answer: D
Skill: Factual recall
12) A person with type B blood could safely receive a transfusion of blood from someone with blood type A) B. B) AB. C) O. D) Both A and C E) Any of these Answer: D
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 107
13) Which of the following will result in modifications to the expected Mendelian ratios? A) Epistasis B) Incomplete dominance C) Incomplete penetrance D) Gene interaction E) All of these Answer: E
Skill: Factual recall
14) Two true‐breeding mutant strains of Drosophila have black body color instead of the wild‐type gray yellow. When the two strains are crossed, all the F 1 flies have wild‐type body color. How can these data be interpreted? A) Recombination occurred. B) Complementation occurred. C) The mutations involved are on two different genes. D) A new mutation occurred in all the offspring. E) Both B and C Answer: E
Skill: Analytical reasoning
15) Manx cats have no tails. When two Manx cats are bred together there is always a one third chance that a kitten will have a tail. When a Manx cat is bred to a cat with a normal tail there is a one‐half chance that a kitten will have a tail. Which of the following is the best explanation for this? A) The Manx genotype exhibits variable expression. B) The Manx phenotype is dominant, but the allele is a recessive lethal. C) The Manx phenotype is dominant epistatic. D) The Manx phenotype is caused by gene interactions. E) The Manx phenotype is a result of heteroplasmy. Answer: B
Skill: Analytical reasoning
16) A and B antigens in human blood are produced by the conversion of ________ by the addition of a sugar group. A) anti‐A and anti‐B antibodies B) the H antigen C) hemoglobin D) α‐N‐acetylgalactosamine E) anti‐O antibodies Answer: B
Skill: Factual recall
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17) In the multiple allelic system for eye color in Drosophila, the phenotypic expression of the alleles depends on A) how much pigment is produced by the alleles present. B) the interaction of alleles from two different genes. C) the temperature at which the flies are grown. D) the sex of the fly. E) Both A and B Answer: A
Skill: Factual recall
18) Familial hypercholesterolemia is a genetic disease that affects approximately one in 500 people worldwide. There is more than one genetic cause of the disease. Class I is often described as an autosomal dominant disease. Affected individuals have cholesterol levels >250 as children and often >300 as adults. However, homozygotes have cholesterol levels of >600 as children and can die of heart attacks in their 20s. These individuals entirely lack a functional LDL receptor. Which of the following would be the best description of the inheritance of this form of hypercholesterolemia? A) Dominant epistatic B) Codominant C) Incompletely dominant D) Complementary E) Recessive epistatic Answer: C
Skill: Analytical reasoning
19) People affected by diseases caused by mtDNA defects A) typically have cells that are heteroplasmons. B) show partial penetrance. C) are sterile. D) inherited them through maternal effect. E) have epistatic respiration. Answer: A
Skill: Factual recall
20) Two persons who have recessive genetic deafness marry and have 6 children. All the children can hear. The reason they can hear is most likely due to A) epistasis. B) pleiotropy. C) sex linkage. D) complementation. E) partial penetrance. Answer: D
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 109
21) Comb shape in chickens is controlled by A) the interaction of three alleles and codominance, resulting in four different phenotypes. B) one gene with incomplete dominance, resulting in three different phenotypes. C) the epistatic interaction of two genes, resulting in two different phenotypes. D) the interaction of two genes with complete dominance, resulting in four different phenotypes. E) None of these Answer: D
Skill: Factual recall
22) Individuals with neurofibromatosis may have a range of phenotypes, from pigment spots on the skin to tumorlike growths. This is an example of A) incomplete penetrance. B) variable expressivity. C) age of onset. D) X‐inactivation. E) gene magnification. Answer: B
Skill: Factual recall
23) Solid black, black with tan belly, and agouti coat color in mice are all caused by alleles of the agouti gene. All three colors can be true‐breeding. When using mice from true‐breeding strains: (1) Black × Agouti F1 s are all agouti, (2) Black × Black with Tan belly F 1 s are all black and tan, and (3) Black with Tan belly × Agouti F1 s are all agouti. If you were to cross the F 1 s from (1) with the F 1 s from (2), what proportion of the resulting offspring would you expect to be black with a tan belly? A) None B) C) D) E)
1 4 3 16 1 2 1 16
Answer: B
Skill: Analytical reasoning
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110 Test Bank for iGenetics
The following scenario applies to the questons below. Dextral (D) and sinistral (d) shell coiling in the water snail Limnaea peregra is genetically determined. From a population of mixed sinistral and dextral snails, a sinistral female and a dextral male are chosen. All the offspring are dextral. When the female offspring are mated, 50% produce only sinistral snails and the other 50% produce only dextral snails. 24) This form of inheritance is called A) maternal inheritance. B) sex‐limited inheritance. C) sex‐linked inheritance. D) maternal effect. E) partial penetrance. Answer: D
Skill: Factual recall
25) What were the genotypes of the sinistral female and dextral male that were initially chosen? A) dd female and Dd male B) Dd female and dd male C) dd female and DD male D) DD female and dd male E) Dd female and Dd male Answer: B
Skill: Problem-solving
TRUE-FALSE QUESTIONS 26) For a gene with complete dominance, the recessive allele has no effect on the phenotype of a heterozygote. Answer: TRUE
Skill: Factual recall
27) Studies performed on identical twins separated at birth have shown that many phenotypes, such as IQ or alcoholism, are influenced by the personʹs genotype as well as by their environment. Answer: TRUE
Skill: Factual recall
28) Mutants that make up a complementation group complement each other. Answer: FALSE Explanation: A complementation group is made up of individual mutants that fail to complement each other and are therefore likely to represent mutations in the same gene.
Skill: Factual recall
29) In the human ABO blood system, the allele i is dominant to I A and IB. Answer: FALSE Explanation: Both I A and IB are dominant to i.
Skill: Factual recall
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 111
30) Recessive lethals are often mutations in essential genes. Answer: TRUE
Skill: Factual recall
31) In a dihybrid cross with independent assortment of the two genes, any deviation from the Mendelian 9:3:3:1 ratio indicates that the phenotype is the product of the interaction of two or more genes. Answer: TRUE
Skill: Factual recall
32) Sex‐limited traits are caused by genes that are on sex chromosomes. Answer: FALSE Explanation: They are caused by autosomal genes that affect a character that appears in one sex but not the other (such as facial hair in humans).
Skill: Factual recall
33) Phenylketonuria (PKU) is an example of a phenotype that is solely influenced by genes. Answer: FALSE Explanation: PKU expression is influenced by the diet of an individual who is homozygous for the recessive allele.
Skill: Factual recall
34) In a genotype with complete penetrance, less than 100% of individuals with a particular genotype exhibit the expected phenotype. Answer: FALSE Explanation: With complete penetrance, identical known genotypes yield 100% expected phenotypes.
Skill: Factual recall
35) An allele that exhibits incomplete dominance is usually haplosufficient. Answer: FALSE Explanation: Incompletely dominant genes are haploinsufficient.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) For a gene with multiple alleles, what is the maximum number of alleles that a diploid organism may have? Answer: Two.
Skill: Conceptual understanding
37) In the ABO blood‐type system, which genotypes could confer type A blood to a person? Answer: A person with type A blood could have either an I A/I A genotype or an I A/i genotype.
Skill: Analytical reasoning
38) Human height is a trait affected by both genes and the environment. How is this so? Answer: Human height is directly influenced by a personʹs genes. However, a single genotype for height has a range of potential phenotypes that could develop depending on environmental conditions. This is known as the norm of reaction.
Skill: Conceptual understanding
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39) How could a man with blood type A and a woman with blood type B produce a child with blood type O? Answer: They must both be heterozygous, with i as one of the two alleles (IA/i and IB/i). The child would have to receive an i allele from each parent and be genotype i/i.
Skill: Analytical reasoning
40) In chickens, there is a mutant gene called ʺfrizzleʺ that results in weak, stringy, and easily broken feathers. When a frightfully frizzled fowl is bred to a normal chicken, the offspring are all mildly frizzled. If one breeds two mildly frizzled chickens to each other, the offspring have the phenotypic ratio of 1 normal: 2 mildly frizzled: 1 frightfully frizzled. What is the mode of inheritance of ʺfrizzleʺ? Answer: There are three phenotypes observed in the F 1 offspring in the ratio of 1:2:1, and one phenotype is intermediate between the other two. This is the signature of incomplete dominant inheritance. The normal chicken represents the homozygous wild‐type genotype, the mildly frizzled chickens are heterozygotes, and the frightfully frizzled chicken has homozygous mutant phenotypes.
Skill: Analytical reasoning
41) What is the number of possible genotypes in a multiple allele series with 6 alleles? How many of these are homozygous and how many are heterozygous? Answer: The total number of genotypes for n alleles is n(n + 1)/2, so for six alleles, there are 21 total genotypes. Six of these are homozygous (n) and 15 (n(n ‐ 1)/2) are heterozygous.
Skill: Problem-solving
42) Explain and differentiate among the molecular bases of codominance, incomplete dominance, and complete dominance. Answer: In codominance, gene product from both alleles of a heterozygote is produced, so both allelic phenotypes are expressed. In incomplete dominance, a homozygous dominant produces two doses of gene product, a heterozygote produces one dose, and a homozygous recessive produces none. In complete dominance, one or both of the dominant alleles is capable of producing enough gene product for the normal dominant phenotype to be expressed.
Skill: Conceptual understanding
The following scenario applies to the questions below. Feather color in chickens is the result of an incompletely dominant gene. The gene responsible for black feathers (B) when homozygous results in a muted color called ʺAndalusian blueʺ if combined with an allele for white feathers (b). When the white allele occurs in homozygous conditions, a white phenotype results. A gene affecting comb shape (R), which results in a comb called a ʺroseʺ comb, is completely dominant over (r), which in homozygous conditions results in a comb called ʺsingle.ʺ A mating between a black single rooster and white rose hen results in a number of chicks, all of which have identical phenotypes. 43) What is the phenotype and genotype of the chicks? Answer: This must represent a cross (BBrr × bbRR), which results in all the F 1 genotypes being heterozygous (BbRr). The phenotype is blue rose.
Skill: Analytical reasoning, problem-solving
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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles 113
44) Use a Punnett square to demonstrate a cross between two of the F 1 offspring. What are the phenotypes and ratios that occur? Answer: 6 blue rose (B/b R/‐): 3 black rose (B/B R/‐): 3 white rose (b/b R/‐): 2 blue single (B/b r/r): 1 black single (B/B r/r): 1 white single (b/b r/r)
Skill: Problem-solving
45) In cats, the dominant white gene (W) may or may not also cause deafness. Homozygotes are more likely to be deaf than heterozygotes. If you breed heterozygous white cats together and produce 100 kittens, how many would you expect to be deaf if all the proportions came out perfectly? Assume that heterozygotes have an 8% chance of being deaf, and homozygotes a 60% chance of being deaf. Answer: Homozygotes should be of 100, which is 25. 60% of 25 = 15. Heterozygotes should be of 100, which is 50. 8% of 50 = 4. Expected deaf kittens would therefore be 15 + 4, or 19 kittens.
Skill: Factual recall
1 2 1 4
46) Generally, blue eyes are recessive to dark brown eyes in humans. Individuals with albinism typically have blue eyes with a strong red reflection. A blue‐eyed woman with albinism marries a man who is blue‐eyed but does not have albinism. All their children have dark brown eyes. How can this be explained? Answer: Albino is recessive epistatic. The woman with albinism was probably also homozygous for dark brown eyes.
Skill: Conceptual understanding
47) Why is it that human males and females who are both heterozygous for the pattern‐baldness gene have different phenotypes (males are bald, females are not bald)? Answer: Pattern baldness is an example of a sex‐influenced trait. Males and females do not express the gene in the same way. Pattern baldness is controlled by an autosomal gene that acts as a dominant in males and as a recessive in females. The b/b genotype results in baldness in both sexes, the b+ /b+ phenotype results in nonbaldness for both sexes, and the b+ /b genotype results in baldness in males but nonbaldness in females.
Skill: Conceptual understanding
48) What is the cause of Tay—Sachs disease? Answer: Tay—Sachs is a lethal disease resulting from a mutation in an essential gene. The mutation results in an enzyme deficiency (hexosaminidase A) that prevents proper nerve function.
Skill: Factual recall
49) The 9:6:1 F 2 ratio found in fruit shape in summer squash is a deviation from an expected Mendelian ratio. What does this represent? Answer: Two genes are involved in the production of squash shape. A dominant allele of either gene and homozygous recessivity for the other results in a spherical fruit. Homozygous recessivity in both genes results in long‐shaped fruit. Two dominant alleles interact to give a disk‐shaped fruit.
Skill: Conceptual understanding
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50) Comb shape in chickens also involves the interaction of two genes, but the F 2 generation shows a typical 9:3:3:1 Mendelian ratio. How does this work? Answer: A dominant allele present in one gene in combination with the homozygous recessive state in the other gene produces one of two phenotypes (rose or pea). Both dominant alleles present produce a third phenotype (walnut), and both homozygous recessives produce a fourth phenotype (single). The ratio is 9 walnut : 3 rose : 3 pea : 1 single.
Skill: Analytical reasoning
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Chapter 14 Genetic Mapping in Eukaryotes
MATCHING QUESTIONS The matching questions below refer to the following experiment. A true‐breeding strain (P1) of fruit flies has mahogany colored eyes and normal (wild-type) bristles. A second true-breeding strain (P2) of fruit flies has normal (wild-type) red eyes and stubble bristles. When they are crossed, the F1 s are wild-type (red eyes and normal bristles). Those F 1 s are then crossed to flies that have both mahogany eyes and stubble bristles. The progeny of the F 1 s consist of 64 flies that have mahogany eyes and normal bristles, 76 flies that have red eyes and stubble bristles, 28 flies that have mahogany eyes and stubble bristles and 32 flies that are wild -type. Please select the best match for each term. 1) Parental type
Skill: Factual recall
A) F1 to mahogany eyes and stubble bristles B) Genes that do not segregate independently C) Mahogany eyes and normal bristles D) Wild-type red eyes and normal bristles. E) Recombinants divided by the total × 100 2) D 3) A 4) B 5) E
2) Recombinant type
Skill: Factual recall
3) Testcross
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4) Gene linkage
Skill: Factual recall
5) Map unit
Skill: Factual recall
Answers:
1) C
MULTIPLE-CHOICE QUESTIONS 6) If genes are linked and an F1 is testcrossed, A) the genes cannot separate. B) they can produce only one phenotype. C) they only produce homozygotes. D) they produce more recombinant phenotypes than parental phenotypes. E) they produce more parental phenotypes than recombinant phenotypes. Answer: E
Skill: Factual recall
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7) Crossing‐over is A) the place on a homologous pair of chromosomes at which a physical exchange occurs. B) an extremely rare event. C) an event that only takes place during meiosis. D) the reciprocal exchange of homologous regions of chromatids. E) not useful in mapping genes. Answer: D
Skill: Factual recall
8) Genes that are linked A) segregate to opposite poles during meiosis. B) do not assort independently during meiosis. C) segregate independently during meiosis. D) are on nonhomologous chromosomes. E) are always on the X chromosome. Answer: B
Skill: Factual recall
9) A chi‐square analysis of linkage between genes k and f produces a P value of 0.99. What is the likelihood that these two genes are linked? A) 99% B) 10% C) 9.9% D) 1% E) None of these Answer: D
Skill: Analytical reasoning
10) If an AABB strain is mated with an aabb strain, an AB gamete produced by the resulting F 1 is A) impossible. B) parental. C) recombinant. D) the most common. E) None of these Answer: B
Skill: Factual recall
11) During a dihybrid cross involving two linked genes, 18% of the resulting gametes showed a recombinant genotype. These two linked genes are ________ map units apart. A) 0.18 B) 18 C) 32 D) 180 E) 1.8 Answer: B
Skill: Conceptual understanding
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12) If two genes are not linked, then the expected phenotypic ratio resulting from a testcross is A) 9:3:3:1. B) 1:2:1. C) 3:1. D) 1:1:1:1. E) 1:1. Answer: D
Skill: Factual recall
13) In a species of Drosophila, genes q and r are found on the same chromosome 20 centimorgans apart. A cross was made between the following individuals:
What proportion of the offspring would you expect to have the wild-type phenotype? A) 0.10 B) 0.20 C) 0.40 D) 0.80 E) None of these Answer: A
Skill: Analytical reasoning
14) One map unit is equal to a recombination frequency of A) 1%. B) 5%. C) 10%. D) 50%. E) 100%. Answer: A
Skill: Factual recall
15) T. H. Morgan and his colleagues found that among the offspring of genetic crosses, parental phenotypic classes were the most frequent, while recombinant classes occurred less frequently. This observation led Morgan to conclude that A) all genes are linked. B) alleles of linked genes assort independently. C) alleles of some genes assort together. D) genes on different chromosomes assort independently. E) genes on different chromosomes do not assort independently. Answer: C
Skill: Factual recall
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16) When does crossing‐over occur? A) Prophase I of meiosis B) Prophase II of meiosis C) Interphase prior to meiosis D) At any time during the second meiotic division E) At any time throughout division Answer: A
Skill: Factual recall
17) A DNA mutation that gives a distinguishable phenotype for a chromosome or gene is a A) recessive allele. B) lethal mutation. C) parental type. D) linkage group. E) genetic marker. Answer: E
Skill: Factual recall
18) The closer together two genes are on a chromosome, A) the more likely there will be a recombination event between them. B) the less likely there will be a recombination event between them. C) the greater the chance that a double crossover will occur between them. D) the more likely they are to be epistatic. E) the less likely they are to be good genetic markers. Answer: B
Skill: Conceptual understanding
19) In the offspring resulting from an F 1 testcross, what percentage of recombinant phenotypes is expected if the genes under study are independently assorting? A) 5% B) 33% C) 50% D) 90% E) The percentage cannot be predicted. Answer: C
Skill: Factual recall
20) A two‐point testcross is a cross between A) a double heterozygote for two linked genes and a double recessive genotype. B) a heterozygote for one of two linked genes and a double recessive genotype. C) a double heterozygote for two linked genes and a homozygous dominant genotype. D) a heterozygote for one gene and a homozygote for another. E) individuals with two different genetic markers. Answer: A
Skill: Factual recall
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21) How many phenotypic classes may be generated from a three‐point testcross? A) (3)2 = 9 B) (2)3 = 8 C) (3)3 = 27 D) (2)2 = 4 E) It cannot be predicted. Answer: B
Skill: Analytical reasoning
22) Interference is a phenomenon in which A) no double crossovers occur among nearby genes. B) the number of observed double crossovers is greater than the number of expected double crossovers. C) the number of observed double crossovers is less than the number of expected double crossovers. D) the occurrence of one crossover interferes with the formation of another crossover nearby. E) Both C and D Answer: E
Skill: Factual recall
The following scenario applies to the questions below. A testcross of trihybrid Drosophila produced the following phenotypes and number of offspring. A table showing phenotype and the number of offspring with each phenotype is below. A plus sign is used for wild-type phenotype; a letter indicates the mutant phenotype for that gene. +++ abc ++c ab+ +b+ a+c +bc a++ 669 653 121 139 2280 2215 3 2
23) Without performing any calculations, which of the following is the gene order? A) abc B) acb C) cab D) +++ E) None of these Answer: B
Skill: Analytical reasoning
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24) Which gene pair is closer together; i.e. there are fewer map units between them? A) a → b B) b → c C) a → c D) a → b and b → c are the same length E) a → c and b → c are the same length Answer: A
Skill: Analytical reasoning
25) The cross represented by a+ b+ c+ /abc × abc/abc is A) a two‐point testcross. B) a three‐point testcross. C) a parental cross. D) a haploid cross. E) None of these Answer: B
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) Testcrosses may be used to determine if two genes are linked. Answer: TRUE
Skill: Factual recall
27) Two genes that are located far from each other on the same chromosome will show a higher frequency of recombination than two genes that are close together on the chromosome. Answer: TRUE
Skill: Factual recall
28) A linkage map illustrates the exact physical locations of genes on a chromosome. Answer: FALSE Explanation: A linkage map shows the relative positions of genes on a chromosome and an estimate of the distance between them.
Skill: Factual recall
29) Given three genes in order x y z, it is more accurate to measure the distance between x and z directly than to add up the shorter distances between x and y and y and z. Answer: FALSE Explanation: Shorter distances are more accurate than longer ones.
Skill: Factual recall
30) If in calculating the distance between two genes using data from a testcrossed dihybrid you arrive at a genetic distance of 70 map units, there is likely to be something wrong with your calculations, the experimental data, or with the experiment itself. Answer: TRUE
Skill: Factual recall
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31) Genes may show 50% recombination either when the genes are on different chromosomes or when the genes are far apart on the same chromosome. Answer: TRUE
Skill: Factual recall
32) Parental strains are always homozygous dominant for all genes or homozygous recessive for all genes. Answer: FALSE Explanation: Parental strains are simply homozygous for autosomal genes. Any given gene can be homozygous dominant or recessive, but they can be homozygous dominant for one gene and homozygous recessive for another.
Skill: Factual recall
33) A coefficient of coincidence of zero is the same as an interference value of zero. Answer: FALSE Explanation: A coefficient of coincidence of zero is the same as total interference, which is equal to an interference value of 1.
Skill: Factual recall
34) Is it possible to correct for the effects of multiple crossovers in calculating map distances by applying a linear mapping function. Answer: FALSE Explanation: A mapping function that corrects for multiple crossovers is only linear at 7 mu or less; at greater distances the function becomes curved toward the limit.
Skill: Factual recall
35) Statistical analysis of linkage between any two genes has 1 degree of freedom since there are only two classes of data, parental and recombinant. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS The following scenario applies to the questions below. X, Y, and Z are linked genes. Based on testcross data, the frequency of recombination between genes X and Y was determined to be 33.1 map units. Between genes X and Z, the distance was 11.8 mu; between genes Y and Z, the distance was 21.3 mu. 36) What is the order of these three genes on the chromosome? Answer: X‐Z‐Y
Skill: Problem-solving
37) Of the three pairs of genes, which ones are closest to each other and which ones are farthest apart? Answer: X and Z are closest, being 11.8 map units apart. The distance between Z and Y is 21.3 mu, and the distance between X and Y is 33.1 mu.
Skill: Problem-solving
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38) Is it possible for crossing‐over to occur between two linked genes, yet for all of the resulting offspring to have parental genotypes? Answer: Yes, if a double crossover occurs between the linked genes.
Skill: Analytical reasoning
39) In a particular breed of flowering plants, two genes on the same chromosome determine flower color and leaf color. For flower color, blue (B) is dominant to pink (b), and for leaf color, white (W) is dominant to green (w). A true‐breeding plant with blue flowers and white leaves is crossed with a plant that has pink flowers and green leaves. The offspring are observed and the following phenotypes are tallied. blue flowers and white leaves: 370 plants blue flowers and green leaves: 14 plants pink flowers and white leaves: 12 plants pink flowers and green leaves: 363 plants What is the frequency of recombination between these two genes? Answer: There are a total of 733 parental phenotypes and 26 recombinant phenotypes, so the percentage of recombinants is (26/733) × 100 = 3.5%.
Skill: Problem-solving
40) A female fruit fly with the recessive mutant phenotype of white eyes and miniature wings is mated with a male possessing the wild‐type phenotype of red eyes and normal wings. Among the F 1 s, all the females are wild‐type while all the males exhibit the mutant phenotype. The F 2 s resulting from a testcross exhibit predominantly (63%) parental phenotypes. How do you explain these results? Answer: Since there is a significant deviation from the expected 1:1 ratio of parental and recombinant phenotypes in the F 2 offspring of the testcross, the genes for eye color and wing type must be linked. Furthermore, the F 1 phenotypes indicate that the genes are X‐linked.
Skill: Analytical reasoning
41) Explain how Curt Stern showed, using evidence from Drosophila, that genetic recombination is associated with the physical exchange of parts between homologous chromosomes. Answer: In genetic crosses, carnation-colored eyes is a sex‐linked recessive allele, and bar‐shaped eyes is an incompletely dominant sex‐linked allele caused by a deletion. Stern used male flies that had carnation eyes and normal eye shape, and thus a normal size X chromosome. He crossed those to heterozygous female flies that had a shortened Bar X‐ chromosome with carnation eye color and a second X that was a normal sized B+ chromosome with the wild‐type (car+ ) eye color and a piece of the Y chromosome attached to the X with normal alleles of both genes. Male offspring that were recombinant with red-colored eyes in the bar shape had an X chromosome that was shorted and had a piece of the Y chromosome attached showing physical exchange on the X chromosome.
Skill: Conceptual understanding
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42) Two genes on a Drosophila chromosome are 70 mu apart. How do you think this number was calculated, and what percentage of recombination would you expect to see between them experimentally? Answer: Distances must have been calculated between these genes and other genes that are located between them. The percentage of recombination measured between genes that are over 50 mu apart will be 50%, because that is the maximum measurable.
Skill: Conceptual understanding
43) An F1 plant with the genotype DdGg is mated to a ddgg plant. The resulting seeds are planted and the following phenotypes produced in these numbers: DG Dg dG dg 47 396 412 55
a. What were the genotypes of the parental strains that produced the F 1 ? b. What is the distance between the genes? Answer: (a) DDgg and ddGG. (b) 102/910 = 0.112, or 11.2 mu.
Skill: Problem-solving
44) A red, round, true‐breeding tomato is bred with a yellow, oval, true‐breeding tomato, and the F 1 s are testcrossed to a homozygous‐recessive tester. This results in the offspring below. Calculate the chi‐squared value to test the likelihood of linkage. yellow oval red round yellow round red oval 11 14 7 8
Answer: There are 40 total offspring, so if the genes are unlinked the parentals should equal the recombinants at 20 each. The chi‐square value is (25‐20) 2 /20 + (15‐20) 2 /20 = 5. With 1 degree of freedom, the P value is less than 0.05, so it is unlikely that the results were from random assortment and likely that the genes are linked.
Skill: Problem-solving
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45) The genotypes and numbers of progeny resulting from a testcross are as follows: t+ g + l+ tgl t+ gl tg+ l+ t+ gl+ tg+ l t+ g + l tgl+ Total 72 69 21 18 9 8 2 1 200
Calculate the genetic distance between genes g and l. Answer: The distance would be the sum of the single crossovers in the region between g and l and the number of double crossovers divided by the total number of progeny. Thus, (9 + 8) + (2 + 1)/200 = 0.1 or 10%. This is the recombination frequency of genes g and l. This distance between them is 10.0 map units.
Skill: Problem-solving
46) Why does the recombination frequency often lead to an underestimation of the true map distance between linked genes? Answer: Because map units between linked genes are defined in terms of the crossover frequency, but distances are estimated from the frequency of recombinant phenotypes in genetic crosses, and these frequencies are not identical; only odd numbers of double crossover events will show up as recombinant phenotypes.
Skill: Conceptual understanding
47) What does the chi‐square test tell us about the linkage of two genes? Answer: We can use the chi‐square test to test the hypothesis that two genes are not linked. A chi‐square table gives you the statistical significance of the deviation of the observed number of recombinants in a testcross from the expected number of recombinants if the genes are not linked. If the P value obtained is less than 0.05, the observed number is significantly below the expected, and the null hypothesis is rejected.
Skill: Conceptual understanding
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48) Here is a genetic map: A 18 mu B 10 mu C . AAbbCC is crossed to aaBBcc to produce an F 1 generation. If you produce 1000 offspring from testcrossing the F 1 , what genotypes do you expect, and in what numbers? Assume that there is no interference. Answer: There will be 8 possible genotypes, listed here as 4 pairs in the following order: double crossovers, A to B single crossovers, B to C single crossovers, and finally parental types. Every individual must receive a recessive allele of each gene from the tester strain. AaBbCc for aabbcc AaBbcc represent aabbCc genotype. Aabbcc aaBbCc genotype. AabbCc 738. aaBbcc You would expect 0.18 × 0.10 × 1000 = 18 total double crossovers, 9 each of these two genotypes. A to B crossovers should = 0.18 × 1000 = 180. Eighteen of these double crossovers, so A to B only = 180 - 18 = 162, 81 for each
B to C crossovers should = 0.10 × 1000 = 100. 18 of these represent double crossovers, so B to C only = 100 - 18 = 82, 41 for each
The two parental types together should be 1000 - 18 - 162 - 82 = Therefore, each parental type genotype should be 369.
Skill: Conceptual understanding, analytical reasoning
49) It is not necessary to use a male from a tester strain to genetically map genes on the X chromosome. How can this be done and why does it work? Answer: It is necessary to start with females that are heterozygous for X‐linked genes. They can be bred with any male. Their male offspring can then be used to determine linkage distances. Male offspring will have only a single X chromosome, and that chromosome will come from the mother, so they can be used to measure recombination frequencies to construct a genetic map.
Skill: Conceptual understanding
50) What is the difference in the likelihood of linkage between two genes with a lod score of 2 and two genes with a lod score of 3? Answer: Linkage is 10 times more likely with a lod score of 3 than with a lod score of 2 because the number is based on the log 10 of the ratio.
Skill: Factual recall
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Chapter 15 Genetics of Bacteria and Bacteriophages
MATCHING QUESTIONS Please select the best match for each term. 1) Conjugation
Skill: Factual recall
A) A sex factor plasmid contained in donor bacterial cells that allows mating B) A self‐replicating, circular DNA that is distinct from the main bacterial chromosome C) The transfer of genes from one bacterium to another by phage vectors D) The unidirectional transfer of extracellular DNA into cells E) The unidirectional transfer of genetic information through direct cellular contact between a donor bacterial cell and a recipient bacterial cell
2) Transformation
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3) Transduction
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4) F factor
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5) Plasmid
Skill: Factual recall
Answers:
1) E
2) D
3) C
4) A
5) B
MULTIPLE-CHOICE QUESTIONS 6) ________ bacterial strains require nutritional supplements to grow in a minimal medium. A) Autotrophic B) Prototrophic C) Phototrophic D) Heterotrophic E) Auxotrophic Answer: E
Skill: Factual recall
7) Minimal media is A) growth media used in the smallest volume in which cells can grow. B) growth media designed to minimize the growth of contaminants. C) growth media that contains the minimal nutritional requirements for normal cells. D) a way to visualize new mutations in a minimum amount of time. E) used to reveal only mutant cell colonies against a dark background. Answer: C
Skill: Factual recall
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8) An arg- strain of E. coli is transformed with a plasmid carrying the wild‐type (arg+ ) gene. The transformed cells are replica‐plated to two minimal medium plates: one supplemented with arginine and one lacking arginine. The transformed cells would grow on A) only the plate supplemented with arginine. B) only the plate lacking arginine. C) both plates. D) neither plate. E) the plate supplemented with arginine, but only if a mutation occurred. Answer: C
Skill: Factual recall
9) When Lederberg and Tatum performed their experiments on gene transfer in bacteria that lead to our understanding of conjugation, what were their controls? A) They mixed strains A and B together. B) They put strain A on one side of a U‐shaped tube and strain B on the other, with a filter between them. C) They found bio- met - cells. D) They plated strain A and strain B separately onto minimal media plates to screen for spontaneous prototrophs. E) They observed mating bridges with transmission electron microscopes. Answer: D
Skill: Conceptual understanding
10) A plasmid such as an F factor that is capable of integrating into the bacterial chromosome is a(n) A) prophage. B) episome. C) auxisome. D) perisome. E) elaiosome. Answer: B
Skill: Factual recall
11) Which of the following has episomal DNA inserted in the cellʹs chromosome? A) An F+ cell B) An Fʹ plasmid C) An Hfr cell D) An F- cell E) All except D Answer: C
Skill: Factual recall
12) Cotransductants can be detected when the transduced genes are A) closely linked. B) far apart on the same chromosome. C) on different chromosomes. D) mutant. E) Both B and D Answer: A
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
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13) mtl, polA, and xyl are bacterial genes that can be transmitted using an Hfr strain. An interrupted mating experiment resulted in xyl exconjugants appearing at 7 minutes and reaching maximum level by 30 minutes. The mtl gene function exconjugants appeared at 11 minutes and leveled off at 80% maximum by about 33 minutes, while polA did not appear until 24 minutes into the experiment and then only slowly rose to about 30%. Which of the following shows the gene order with the insertion site (arrow) location correctly? A) xyl mtl polA → B) polA mtl xyl → C) mtl xyl polA → D) mtl polA xyl → E) None of these Answer: B
Skill: Problem-solving
14) Conjugation by which of the following with an F- cell results in a cell that remains F- ? A) F+ B) Hfr C) F D) A and C E) B and C Answer: B
Skill: Conceptual understanding
15) The reason that many plasmids used in laboratories for transformation experiments contain an ampicillin resistance gene (ampR) is A) so that the plasmid can replicate in bacterial cells. B) so that the plasmid can recombine with the bacterial genome. C) to enable the recipient cell to conjugate. D) to provide a selectable marker. E) to map point mutations. Answer: D
Skill: Factual recall
16) In a genetic cross between two bacteriophage T4 rII mutants, r+ recombinants are produced. The genetic changes in these two rII mutants must be A) heterozygous. B) heteroallelic. C) homozygous. D) homoallelic. E) polymorphic. Answer: B
Skill: Factual recall
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17) In E. coli, λ DNA integrates into which site on the bacterial chromosome? A) ori B) trp C) bio D) gal E) att Answer: E
Skill: Factual recall
18) To grow on artificial medium, prototrophic bacteria require A) sunlight. B) amino acid supplements. C) vitamin supplements. D) All of these E) None of these Answer: E
Skill: Factual recall
19) E. coli strains that are Hfr A) are easily mutated. B) are susceptible to infection by bacteriophage. C) contain the F factor integrated in the bacterial chromosome. D) cannot be made competent. E) have a low frequency of recombination. Answer: C
Skill: Factual recall
20) The E. coli chromosome is A) linear. B) circular. C) single‐stranded. D) less than a megabase in length. E) compartmentalized within an intracellular membrane. Answer: B
Skill: Factual recall
21) Which of the following methods can be used to create competent bacterial cells? A) Treating the cells chemically B) Exposing cells to a strong electric field C) Making the cell membrane more permeable to DNA D) Allowing the culture to enter stationary phase growth E) A, B, and C only Answer: E
Skill: Factual recall
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22) In general, transformation of most genes occurs at a frequency of one in every ________ cells. A) 10 B) 100 C) 1,000 D) 10,000 E) 1,000,000 Answer: C
Skill: Factual recall
23) Which of the following statements is true regarding a temperate phage? A) It can reproduce at warm or cool temperatures. B) It is required for transformation. C) It can be lytic or lysogenic. D) It causes bacterial cells to conjugate. E) It reproduces by binary fission. Answer: C
Skill: Factual recall
24) In the lysogenic bacteriophage life cycle, the λ chromosome A) replicates and the phage genes take over the bacterium. B) inserts itself physically into the host cellʹs chromosome. C) expresses a repressor protein gene that inhibits the lytic pathway. D) is replaced by a piece of bacterial DNA when packaged inside phage progeny. E) Both B and C Answer: E
Skill: Factual recall
25) Cotransductant bacteria can occur if A) a bacterial cell picks up two fragments of DNA from the environment, each with a different gene. B) double crossover occurs between prophage DNA and the bacterial chromosome. C) two genes are closely linked enough so that they can be packaged into a phage head and injected into a cell by a single phage. D) two genes are introduced into the same bacterium by simultaneous infection with two different phages. E) Both C and D Answer: E
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) Genetic exchange in bacteria by transduction requires cell‐to‐cell contact. Answer: FALSE Explanation: Transduction requires a phage vector but does not require cell‐to‐cell contact. Genetic exchange by conjugation requires cell‐to‐cell contact.
Skill: Factual recall
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27) The viral chromosome of a bacteriophage can never integrate into the host bacteriumʹs chromosome. Answer: FALSE Explanation: The viral chromosomes of some bacteriophages can integrate into the host chromosome and replicate. At this stage, the virus is termed a prophage.
Skill: Factual recall
28) Genetic mapping experiments performed by Seymour Benzer on rII mutants in the T4 phage showed that the gene is indivisible by the process of mutation and recombination. Answer: FALSE Explanation: Benzerʹs experiments showed that genetic recombination can occur within genes; in other words, that the base pair, rather than the gene, is the unit of mutation and recombination in genomes.
Skill: Factual recall
29) In the case of genetic exchange in bacteria by transformation, there is an exchange of DNA back and forth between cells, and a complete diploid cell is formed. Answer: FALSE Explanation: During transformation, gene transfer is unidirectional, and the transformed cell remains haploid.
Skill: Factual recall
30) In genetic exchange by conjugation, a single Hfr strain will transfer the entire bacterial chromosome to the recipient, allowing mapping in minutes from the first gene entering to the last one before the other half of the inserted F factor. Answer: FALSE Explanation: Only fragments of a genome can be easily mapped from a single Hfr strain. Mating bridges break long before the entire genome is transferred.
Skill: Conceptual understanding
31) ʺHot spotsʺ are locations in a gene where mutations are found at a relatively high frequency. Answer: TRUE
Skill: Factual recall
32) A complete bacterial medium contains only the nutrients required for the growth of wild‐type cells. Answer: FALSE Explanation: A minimal medium contains only the nutrients required for the growth of wild‐type cells. A complete medium provides nutritional substances required by auxotrophic mutant cells to grow.
Skill: Factual recall
33) A merodiploid cell has two copies of one or a few genes and only one copy of all the others. Answer: TRUE
Skill: Factual recall
34) In F+ × F- crosses, none of the bacterial chromosome is transferred; only the F factor is. Answer: TRUE
Skill: Factual recall Copyright © 2010 Pearson Education, Inc.
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35) To enhance the efficiency of transformation, bacterial cells can be induced to take up DNA by a strong electrical field. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) An Hfr strain of E. coli with the genotype a+ b+ c+ d + e+ f+ is mated with an F- auxotrophic strain with the genotype a- b- c- d - e- f- . Conjugation is stopped at 10‐minute intervals and the genotypes of the resulting conjugants are determined. The following results are obtained: After 10 minutes After 20 minutes After 30 minutes After 40 minutes After 50 minutes After 60 minutes e+ a+ e+ a+ b+ e+ a+ b+ d + e+ a+ b+ c+ d + e+ a+ b+ c+ d + e+ f+
What is the correct order of genes on this bacterial chromosome? Answer: e- a- b- d - c- f
Skill: Problem-solving
37) What are temperate bacteriophages? Answer: Phages that can alternate between lytic and lysogenic pathways in bacterial cells.
Skill: Factual recall
38) What is the difference between generalized and specialized transduction? Give an example of specialized transduction. Answer: In generalized transduction, any gene or set of genes on the chromosome may be transferred from the donor bacterium to the recipient bacterium. In specialized transduction, only specific genes may be transferred. This is mediated by temperate phages in which prophages associate with only one site of the bacterial chromosome. For example, the λ genome integrates into the E. coli bacterial chromosome at a specific site between the gal region and the bio region, called att λ. By abnormal excision of the prophage from the host chromosome, gal or bio genes may be transferred to other bacterial cells.
Skill: Conceptual understanding
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39) For a P1 transduction experiment, an E. coli donor strain with the genotype arg+ , ser+ , ampR and a recipient strain with the genotype arg- , ser- , ampS were used. The following results were obtained for each marker: Of the transductants selected for arginine, 5% are ser+ and 30% are ampR. Of the transductants selected for serine, 3% are arg+ and 0% are ampR. What is the correct order and relative distance of these three genes on the E. coli chromosome? Answer: The order and relative distance is: ampR- arg‐‐‐‐ser. Since amp and ser are never cotransduced, they must be very far apart from each other. Arg and ser are rarely cotransduced, so they must be farther apart from each other than are arg and amp, which are more frequently cotransduced, so they must be fairly close together.
Skill: Problem-solving
40) During a cis‐trans complementation assay, two lac- strains of bacteria are crossed. The progeny cells exhibit a lac- phenotype. What does this indicate? Answer: The two mutations are in different genes that encode different products. The two mutant strains complement each other to create a functional cell that can metabolize lactose.
Skill: Conceptual understanding
41) In a bacterial species genes q, r, s, and t were mapped in minutes through interrupted mating experiments. The results placed q and s on the two outer ends, but r and t were not resolvable by minutes. You have discovered no phage that can infect this species. If you have a q r s t Hfr strain and a q- r- s- t - F- strain, how could you measure the distance between the genes in map units? Answer: There are two ways. (1) You could do a conjugation experiment in which you select for the last marker in, to be sure that all the DNA was available for recombination, and measure (recombinants in the various gene pairs/total) X 100 for map units. (2) You could do a similar experiment through transformation. To do so, you would isolate and fragment the DNA from the q r s t strain and use it to transform the q- rs- t - . In this case you would need to select for q s colonies to make sure that the entire region was likely to have entered the recipient cells.
Skill: Conceptual understanding
42) A bacterial broth culture is diluted by a factor of 100,000, and 0.1 mL of the dilution is spread over the surface of an agar plate. The next day, 123 colonies are observed on the plate. What was the concentration of bacteria in the broth? Answer: We want to determine the number of colony‐forming units (cfu) per mL. Since 123 colonies are observed on the plate, this means that there were 123 bacterial cells (cfu) in 0.1 mL of the 100,000X dilution. Thus, in the original culture, there were 123 cfu × 100,000 (dilution factor) × 10 = 1.23 × 108 cfu/mL.
Skill: Problem-solving
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43) Explain how an Fʹ × F- conjugation can produce a partially diploid bacterial cell. Answer: The Fʹ factor was derived from an integrated factor that acquired bacterial genes during excision from the chromosome by means of imprecise looping out and single crossover. During Fʹ × F- conjugation, the Fʹ factor is transferred to the F- cell. The recipient cell then contains two copies of the bacterial genes carried by the Fʹ factor.
Skill: Conceptual understanding
44) In a series of cotransformation experiments, DNA fragments from x + y + z+ donor bacteria were used to transform xyz recipient bacteria. The following transformation phenotypes were observed in the recipient cells: xy+ z xyz+ x+y+z x + yz+ What does this imply about the gene order of x, y, and z? Answer: The gene order must be y‐x‐z. Since the genes y and z are never cotransformed, they must be far apart on the DNA. However, x and y are sometimes cotransformed, and y and z are sometimes cotransformed, so each of these genes must be more closely linked to one another.
Skill: Analytical reasoning
45) How does a bacteriophage become a transducing phage? Answer: During the lytic phage life cycle, the bacterial DNA is degraded and, rarely, a piece of bacterial DNA is packaged into a progeny phage head instead of phage DNA. These phages are then transducing phages, which can carry genetic material to other bacteria.
Skill: Factual recall
46) How would you design an experiment to test whether a transduction event occurred between two different bacterial strains? Answer: You would use a selected marker to identify the transductant cells. For example, if the donor cell is thr+ and the recipient cell is thr, the prototrophic transductants can be grown on minimal medium that does not contain threonine, whereas the nontransduced cells still require threonine to grow.
Skill: Application of knowledge
47) How could you use T2 bacteriophage plaque phenotypes to determine the genetic distance between two genes, h and r, that affect plaque appearance in the phage DNA? Answer: Just as in eukaryotic organisms, a crossing experiment can be performed. You would use different strains of T2–one of which is h + r and one of which is hr + –that express different plaque phenotypes to coinfect plated bacterial cells. Inside the bacterial cells, the two different phage chromosomes can crossover to produce recombinants. The percentages of parental and recombinant plaque phenotypes can be determined. As in eukaryotic cells, the frequency of recombinants is equal to the genetic distance between the two genes.
Skill: Application of knowledge
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48) How did Seymour Benzer use deletion mapping to construct fine‐scale maps of the rII region of the T4 bacteriophage genome? Answer: He used deletion mutants, which, unlike point mutants, have lost a segment of DNA and cannot revert to a wild‐type state. He crossed unknown point mutants with standard (known) deletion mutants for different segments of the rII region to determine which crosses could result in wild‐type recombinant phage. Those that could not must have used point mutants with a mutation in the region of the deletion. This could be repeated in crosses between the point mutant and smaller secondary reference deletion mutants until the point mutation was localized to within a very small segment of the rII region.
Skill: Conceptual understanding
49) In bacteriophage T4, all rIIA mutants are found to complement all rIIB mutants. However, rIIA mutants fail to complement other rIIA mutants, and rIIB mutants fail to complement other rIIB mutants. Furthermore, mutants with deletions that span both rIIA and rIIB complementation groups do not complement either A or B mutants. What do these data mean, and what do the complementation groups represent? Answer: Complementation occurs in crosses between mutants of the different complementation groups, which represent functional units, or genes. That is, rIIA mutants produce functional B gene products, and rIIB mutants produce functional A gene products. Both A and B gene products are required to produce a wild‐type phenotype. Mutants with deletions that span both complementation regions cannot recover the wild‐type phenotype because they will always be missing part of a functional unit.
Skill: Conceptual understanding
50) Performing a complementation test through cotransduction using mutant phage genomes created by X‐ray mutagenesis (mutants a‐e), as well as ones created by chemical mutagenesis using EMS (mutants 1‐5), you get the following results: 1 2 a — + b — + c + + d + — e — — 3 + — — + + 4 — — — + + 5 — — + — —
Explain this data. Answer: X‐rays create DNA breaks that lead to deletions, and chemical mutagens such as EMS result in point mutations. Thus, mutants a‐e are deletion mutants and 1‐5 are point mutants. Analysis of the data can provide us with a comparative map of the mutants as follows (this map can be flipped horizontally).
Skill: Problem-solving, conceptual understanding Copyright © 2010 Pearson Education, Inc.
Chapter 16 Variations in Chromosome Structure and Number
MATCHING QUESTIONS Please select the best match for each term. 1) Triploid
Skill: Factual recall
A) A chromosome number that is not an exact multiple of the haploid set of chromosomes B) More than two full sets of chromosomes C) A normal number of chromosomes D) Three copies of one chromosome E) Three full sets of chromosomes
2) Trisomic
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3) Euploid
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4) Polyploid
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5) Aneuploid
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Answers:
1) E
2) D
3) C
4) B
5) A
Please select the best match for each term. 6) Turner syndrome
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A) Three copies of chromosome 21 B) XO
7) Klinefelter syndrome
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C) XXY D) Three copies of chromosome 18 E) Three copies of chromosome 13
8) Down syndrome
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9) Edwards syndrome
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10) Patau syndrome
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Answers:
6) B
7) C
8) A
9) E
10) D
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Chapter 16 Variations in Chromosome Structure and Number 137
MULTIPLE-CHOICE QUESTIONS 11) Which of the following types of chromosomal mutation cannot revert to the wild‐type state? A) Duplication B) Deletion C) Translocation D) Inversion E) Any of these may revert to wild‐type Answer: B
Skill: Conceptual understanding
12) In Drosophila melanogaster, polytene chromosomes are produced when A) a chromosome breaks and rejoins with another section of the same chromosome. B) a cell divides several times before the chromosomes are duplicated. C) chromosomes are duplicated repeatedly without a corresponding nuclear division. D) a segment of a chromosome is duplicated during an unequal crossing‐over event. E) characteristic banding patterns form on the chromosomes. Answer: C
Skill: Factual recall
13) Which of the following is not used to help identify individual chromosomes in a karyotype? A) Centromere position B) Point mutations C) Chromosome size D) Banding pattern on staining E) Heterochromatin regions Answer: B
Skill: Conceptual understanding
14) Many scientists believe that the evolution of multigene families, such as the genes for hemoglobin, is a result of which type of genetic rearrangement? A) Duplication B) Deletion C) Inversion D) Reciprocal translocation E) Nonreciprocal translocation Answer: A
Skill: Factual recall
15) A(n) ________ inversion includes the centromere. A) concentric B) pericentric C) paracentric D) epicentric E) chromocentric Answer: B
Skill: Factual recall
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16) If a small region of a chromosome containing a non‐essential gene is completely deleted, and an organism has two identical chromosomes with this deletion, then the organism A) has no alleles of that gene. B) is dead. C) has more heterochromatin. D) has recessive alleles of those genes. E) will exhibit pseudodominance. Answer: D
Skill: Conceptual understanding
17) What is the genetic consequence of a homozygous translocation? A) Inviable gamete formation (semisterility) B) Gene duplications and deletions C) Formation of abnormal chromatids following crossing‐over D) Abnormal pairing during meiosis E) An alteration in the linkage relationships of genes Answer: E
Skill: Analytical reasoning
18) A human cell containing two sets of 23 chromosomes is A) aneuploid. B) polyploid. C) euploid. D) haploid. E) None of these Answer: C
Skill: Factual recall
19) Nondisjunction of chromosomes may result in A) the loss of one homologous chromosome pair. B) the addition of one homologous chromosome pair. C) the addition of a single chromosome. D) the loss of a single chromosome. E) All of these Answer: E
Skill: Factual recall
20) A monosomic cell would produce gametes with how many chromosomes? A) N B) N and N‐1 C) N‐1 D) N+1 E) N and N+1 Answer: B
Skill: Factual recall
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Chapter 16 Variations in Chromosome Structure and Number 139
21) An X‐linked homozygous color‐blind woman gives birth to a son who is not color blind. He could have normal vision because he A) has Down syndrome. B) has Klinefelter syndrome. C) has Turner syndrome. D) is XYY. E) has triploidy. Answer: B
Skill: Conceptual understanding
22) Which of the following traits is determined by programmed transposition? A) Pattern baldness in humans B) Mating types in yeast C) Eye color in Drosophila D) All of these E) None of these Answer: D
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23) A dicentric chromosome has two A) arms. B) centromeres. C) telomeres. D) mutations. E) centrosomes. Answer: B
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24) Which of the following types of analysis allows scientists to visualize an individualʹs chromosomal makeup? A) Karyotype analysis B) Pedigree analysis C) Genetic crosses D) Inheritance analysis E) Deletion mapping Answer: A
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25) In familial Down syndrome resulting from Robertsonian translocation, A) the affected individual has three copies of the complete chromosome 21. B) the affected individual is missing a copy of chromosome 21. C) the affected individual has three copies of the long arm of chromosome 21, one of which is attached to part of chromosome 14. D) the affected individual has two copies of the long arm of chromosome 21, one of which is attached to part of chromosome 14. E) the affected individual has three copies of the long arm of chromosome 14, one of which is attached to chromosome 21. Answer: C
Skill: Factual recall
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26) Most monoploid individuals do not survive because A) they do not have the appropriate gene dosages. B) their cells cannot divide properly. C) recessive lethal mutations cannot be masked by dominant alleles. D) they have missing genes. E) they have only one sex chromosome. Answer: C
Skill: Conceptual understanding
27) The chemical colchicine prevents the formation of microtubules. It is commonly used in certain experimental procedures to cause changes in cellular chromosomes. Which of the following changes is it likely to be used to create? A) Induction of mutant polyploid individuals B) Induction of mutant aneuploid individuals C) Prevention of nondisjunction in cell cultures D) Induction of chromosomal deletions in experimental cell lines E) Induction of chromosomal duplications in experimental cell lines Answer: A
Skill: Factual recall
28) Seedless bananas are produced from A) sterile tetraploid allopolyploid plants. B) sterile triploid autopolyploid plants. C) monoploid plants grown from unfertilized seeds. D) fertile diploid plants that are unfertilized. E) Both C and D Answer: A
Skill: Factual recall
29) Cultivated wheat is descended from three distinct species, each with a diploid set of 14 chromosomes. Thus, it is A) an allotriploid with 21 chromosomes. B) an allotriploid with 42 chromosomes. C) an allohexaploid with 84 chromosomes. D) an allohexaploid with 42 chromosomes. E) an autotetraploid with 28 chromosomes. Answer: D
Skill: Analytical reasoning
TRUE-FALSE QUESTIONS 30) In humans, many types of cancers are associated with chromosomal mutations. Answer: TRUE
Skill: Factual recall
31) A tetraploid cell contains four chromosomes. Answer: FALSE Explanation: It contains four copies of each chromosome.
Skill: Factual recall
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Chapter 16 Variations in Chromosome Structure and Number 141
32) Approximately 15% of all human conceptions contain at least one chromosomal mutation. Answer: TRUE
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33) The causes of chromosomal deletion may include radiation, viral infection, chemical exposure, or transposable elements. Answer: TRUE
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34) No known living humans have one whole chromosome of a homologous pair of autosomes deleted from the genome. Answer: TRUE
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35) For eukaryotic organisms, polyploidy is always a lethal condition. Answer: FALSE Explanation: Polyploidy is lethal for most animal species, but is less consequential for plants, and in fact has played a significant role in the evolution of plant species.
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36) Both changes in chromosome number and chromosome structure are believed to have resulted in novel genomes, leading to new species. Answer: TRUE
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37) Inversion loops form to allow homologous chromosomes containing homozygous inversions to pair properly during meiosis. Answer: FALSE Explanation: Inversion loops allow inversion heterozygotes to pair properly during meiosis.
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38) Down syndrome (trisomy‐21) is the only human autosomal trisomy in which the individuals may survive to adulthood. Answer: TRUE
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 39) In some fruit flies, an inversion involving the white‐eyed locus (w) moves the w+ gene from one end of the X chromosome to a location nearer to the centromere. Flies with a w+ or w+ /w genotype are normally expected to have red eyes. However, in flies with the inversion, the eyes are mottled red and white. Explain why. Answer: The mottled phenotype is caused by a position effect. During the inversion, the w+ gene is moved from a euchromatic region on the X to a region near the heterochromatin at the centromere. Sometimes genes in this region are not transcribed because the heterochromatin remains condensed throughout the cell cycle. The eye cells in which the w+ gene has become inactivated produce white spots in the eyes. The cells in which the gene is not inactivated produce red spots.
Skill: Conceptual understanding
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40) What are the gametic consequences of a lack of cytokinesis following the first or second meiotic division? Answer: The result will be a change in the number of complete set of chromosomes. If the individual is a diploid organism, and cytokinesis fails to occurs at meiosis I, half the gametes will have no sets of chromosomes and half will have two sets of chromosomes. If cytokinesis fails at meiosis II, half the gametes will have one set of chromosomes (normal condition), one‐fourth of them will have two sets, and one‐fourth of them will have no sets.
Skill: Analytical reasoning
41) How does a Philadelphia chromosome occur, and what condition does it cause? Answer: Ninety percent of patients with the fatal cancer chronic myelogenous leukemia (CML cancer) have a chromosomal mutation in their myeloblasts caused by a reciprocal translocation event between chromosome 22 and chromosome 9. The defective chromosome 22 is called the Philadelphia chromosome.
Skill: Factual recall
42) How can the banding patterns on Drosophila polytene chromosomes be used to aid deletion mapping of genes on the chromosomes? Answer: The unexpected expression of a recessive trait caused by pseudodominance (the loss of a dominant allele) can be correlated with changes in band number and physical appearance of the polytene chromosomes. For example, in a polytene X bearing wild‐type alleles, deletions in certain regions can be visually observed. These can be correlated with the appearance of pseudodominant phenotypes for those alleles. The alleles can then be mapped to those missing regions on the chromosome.
Skill: Conceptual understanding
43) The diagram below shows a region near the center of a long arm of a chromosome. Assume that in the diagram below, every vertical hatch mark is 1 map unit from the next vertical hatch mark. The dashed double line represents an inversion. If you crossed this heterozygous individual with a normal homozygous recessive individual, and produced 1000 live offspring, how many would you expect to be Aa dd?
Answer: Recombination within this region would be lethal, producing no live offspring, so the distance between A and D will be effectively changed from 6 map units to 2 map units. As a result, 2% recombination is expected. Half of the recombinants will be Aadd, the other half aaDd. Therefore, we can expect the number of Aadd individuals to be 1% of 1000, or 10 individuals.
Skill: Problem-solving, conceptual understanding
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Chapter 16 Variations in Chromosome Structure and Number 143
44) What are all the possible gametic genotypes produced by a trisomic individual that is genotype A/A + /B for the trisomic chromosome? Answer: Three types of haploid gametes may be produced: A, A+ , and B; and three types of disomic gametes may be produced: A/A + , A/B, and A+ /B.
Skill: Problem-solving
45) How is hybrid sterility overcome in allopolyploid plants? Answer: Typically, allopolyploids are produced when two different species hybridize. The resulting offspring then contains one set of chromosomes from each parental species and is sterile because the sets are not homologous and cannot pair at meiosis to produce viable gametes. However, if both chromosome sets double by means of division error (nondisjunction), each diploid set can function normally during meiosis, and N1 +N2 gametes will be produced. Fusion of two such gametes will produce an allotetraploid offspring.
Skill: Conceptual understanding
46) A woman with normal vision, whose father has sex‐linked color blindness, marries a man with normal vision. They have a son who has Klinefelter syndrome and is color blind. Describe precisely what kind of event could caused have his Klinefelter syndrome. Answer: Nondisjunction must have occurred in the mother during meiosis II, when sister chromatids normally separate. With Klinefelter syndrome, the boy has an XXY karyotype. Since both of his X chromosomes contain the allele for color blindness, he likely has two copies of one homolog of his motherʹs X chromosome containing the allele for color blindness.
Skill: Problem-solving, conceptual understanding
47) Why do polyploid individuals with an even number of chromosome sets have a higher likelihood of fertility than those with an odd number of chromosome sets? Answer: Individuals with an even number of chromosome sets have the potential for homologs to be segregated equally during meiosis and are likely to produce balanced gametes. Polyploids with an odd number of chromosome sets always have an unpaired chromosome for each type and thus produce many unbalanced gametes.
Skill: Conceptual understanding
48) During meiosis in a reciprocal translocation heterozygote, what are the three ways that chromosomes may segregate at anaphase I, and what are the consequences for the resulting gametes? Answer: In alternate segregation, alternate centromeres (from nonhomologous chromosomes) migrate to the same pole. This results in normal gametes, each having a complete set of genes (one half with all normal chromosomes and one half with translocated chromosomes). In adjacent‐1 segregation, adjacent nonhomologous centromeres migrate to the same pole. This results in gametes with either duplications or deletions of genes, which are usually inviable. In adjacent‐2 segregation, adjacent homologous centromeres migrate to the same pole. This also results in inviable gametes with duplications or deletions.
Skill: Conceptual understanding
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49) A paracentric inversion heterozygote contains a normal chromosome with the long‐arm gene order GENE and an inverted chromosome with the long‐arm gene order GNEE. They form an inversion loop to pair during prophase I of meiosis. If a single crossover occurs between the G and the E segments of the loop, what will be the gene orders in the resulting recombinant dicentric chromosome and acentric fragment? Assume crossing over occurs at the four‐strand stage involving two nonsister chromatids. Answer: The dicentric chromosome gene order will become GENG, and the gene order in the acentric fragment will become EENE.
Skill: Problem-solving
50) During meiosis, what are the gametic consequences of a paracentric inversion with a single crossover event, compared to a pericentric inversion with a single crossover? Answer: The paracentric inversion results in two viable gametes: one with a normal chromosome and one with an inversion chromosome. Two gametes will contain fragments of a broken dicentric chromosome and will be inviable (since the acentric fragment is lost). The pericentric inversion also results in two viable gametes–one containing a normal chromosome and one containing an inversion chromosome–and two inviable gametes containing different deletion/duplication products.
Skill: Conceptual understanding
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages
MATCHING QUESTIONS Please select the best match for each term. 1) Operator
Skill: Factual recall
A) Molecule capable of down‐regulating protein synthesis B) Adjacent genes serving one function that are transcribed and regulated together C) Regulatory region to which repressor protein binds D) Regulatory molecule that leads to gene expression E) General class of molecules that help control regulation of expression 2) A 3) D 4) E 5) B
2) Repressor protein
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3) Inducer
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4) Effector molecule
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5) Operon
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Answers:
1) C
MULTIPLE-CHOICE QUESTIONS 6) The order of the structural genes in the lac operon is A) 5ʹ‐lacY‐lacZ‐lacA‐3ʹ. B) 3ʹ‐lacZ‐lacY‐lacA‐5ʹ. C) 5ʹ‐lacZ‐lacY‐lacA‐3ʹ. D) 3ʹ‐lacY‐lacA‐lacZ‐5ʹ. E) 5ʹ‐lacA‐lacY‐lacZ‐3ʹ. Answer: C
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7) In the lac operon in E. coli, a nonsense mutation in the ________ gene will result in a loss of β‐galactosidase activity in the cell. A) lacZ B) lacI C) lacY D) Both A and B E) All of these Answer: A
Skill: Factual recall
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8) The inducer of the lac operon is A) galactose. B) glucose. C) allolactose. D) repressor protein. E) permease. Answer: C
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9) Attenuation of the trp operon fails to occur when A) a lack of tryptophan causes stalling of ribosomes on the leader sequence of the RNA. B) an RNA stem‐loop forms that allows transcription to continue. C) an RNA stem‐loop that causes transcription to terminate fails to form. D) the cell has enough tryptophan. E) the operator is mutated. Answer: D
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10) To inhibit the transcription of operon genes, the lacI gene product binds to A) the operator. B) the promoter. C) the repressor. D) the inducer. E) β‐galactosidase. Answer: A
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11) A lacOC mutation is A) constitutive. B) cis‐dominant. C) incompletely dominant. D) codominant E) recessive. Answer: A
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12) During transcription of the trp operon, pairing of regions 2 and 3 in the leader peptide mRNA creates a(n) ________ signal. A) continuous B) termination C) antitermination D) induction E) repression Answer: C
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 147
13) What is the underlying reason that a lambda phage would need to switch from the lysogenic state to the lytic state? A) To have its genome replicated B) To kill its host cell C) To make more of the Cro protein D) To be more efficient in its use of resources E) To escape an endangered host cell Answer: E
Skill: Conceptual understanding
14) Which of the following mutations is dominant over wild‐type lacI+ in a partial diploid experimental cell? A) lacAB) lacOC) lacID) lacIS E) C and Donly Answer: D
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15) Which positive regulatory protein is required for the production of phage coat protein in lambda? A) CI B) Cro C) N D) Q E) None of these Answer: D
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16) How many structural genes are under the control of the trp operon promoter? A) 1 B) 2 C) 3 D) 4 E) 5 Answer: E
Skill: Factual recall
17) In terms of regulation of gene expression, tryptophan and allolactose are examples of ________ molecules. A) repressor B) effector C) inducer D) operator E) terminator Answer: B
Skill: Factual recall
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18) Which of the following can be a transdominant mutation in partial diploid cells? A) lacZB) PlacIC) lacIS D) lacYE) lacIAnswer: C
Skill: Factual recall
19) The interaction between allolactose and the repressor protein results in a shape change known as a(n) A) isomer. B) mutation. C) enantiomer. D) allosteric shift. E) transformant. Answer: D
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20) Operons can best be defined as A) operational mutations affecting gene expression. B) a set of genes under common regulatory control. C) dominant alleles responsible for enzyme synthesis. D) a balance between repressor and inducer molecules. E) a common mode of gene expression in bacteria. Answer: B
Skill: Factual recall
21) The ________ pathway in lambda phage leads to integration of viral DNA with that of the host, while the ________ pathway leads to induction of progeny viruses. A) lytic, lysogenic B) invasive, replicative C) repressive, inductive D) lysogenic, lytic E) replicative, lysogenic Answer: D
Skill: Factual recall
22) In bacteriophage lambda, the establishment of lysogeny requires the protein product of which gene? A) N B) cII C) cIII D) B and C only E) A, B, and C Answer: D
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 149
23) To maintain lysogeny in lambda, which of the following needs to occur? A) Binding of repressor protein to OL B) Binding of repressor protein to OR C) Inactivation of PL D) Inactivation of PR E) All of these Answer: B
Skill: Factual recall
24) The genetic switch for the infection pathways of lambda phage is controlled by two regulatory proteins: ________ and ________. A) lambda repressor, Cro B) lambda repressor, permease C) excisionase, integrase D) recA, Cro E) lyase, lysogenase Answer: A
Skill: Factual recall
25) Attenuation in trp gene regulation is accomplished by A) reducing permeability of tryptophan, the effector molecule. B) blocking expression of trp mRNA. C) formation of secondary mRNA structures in the leader region. D) binding with RNA polymerase. E) repression of the trp operator. Answer: C
Skill: Factual recall
TRUE-FALSE QUESTIONS 26) Genes required for maintaining basic cell structure, growth, and division are housekeeping genes and have regulated expression. Answer: FALSE Explanation: They are constitutive.
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27) The structural genes in the lac operon are transcribed into a single polycistronic mRNA molecule. Answer: TRUE
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28) Mutants that are lacIS do not produce lac enzymes, regardless of the presence or absence of lactose. Answer: TRUE
Skill: Factual recall
29) A constitutive gene is a gene that is occasionally expressed. Answer: FALSE Explanation: By definition, a constitutive gene is continually expressed.
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30) The enzyme transacetylase is responsible for uptake of lactose. Answer: FALSE Explanation: Permease, encoded by the lacY gene, plays this role.
Skill: Factual recall
31) Mutations of lacOC lead to attenuated expression of lac structural genes. Answer: FALSE Explanation: Mutations of lacOC lead to constitutive expression.
Skill: Factual recall
32) lacIS encodes superrepressors, repressors that always bind to the operator and prevent gene transcription. Answer: TRUE
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33) The purpose of regulation of the trp operon in bacteria is to ensure the cell can digest tryptophan if it is present and the cell is short on energy. Answer: FALSE Explanation: The purpose of regulation of the trp operon is only to make the gene products necessary for the synthesis of tryptophan if there is not enough tryptophan present in the cell, as it is an essential amino acid.
Skill: Conceptual understanding
34) The use of CAP‐ cAMP complex as a positive regulator is specific to the lac operon. Answer: FALSE Explanation: CAP protein and its need for cAMP are part of a general catobolite repression mechanism. Bacterial cells that have plenty of their preferred carbon source (glucose) limit the expression of genes used for alternative, non‐preferred energy sources.
Skill: Factual recall
35) The tryptophan codons of the attenuator sequence help speed up transcription of downstream trp genes at high concentrations of tryptophan. Answer: FALSE Explanation: These codons are readily read by trp‐tRNA at high tryptophan concentrations, leading to the attenuator formation that terminates transcription.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) How are partial diploid E. coli constructs made? Answer: The partial diploids are a special strain of bacteria that contain an engineered plasmid (the F factor) with a set of lac operon genes. This second set makes the cell ʺdiploidʺ with respect to these genes. Partial diploids are strains of bacteria that contain an Fʹ plasmid that bears a few chromosomal genes. These chromosomal genes in the Fʹ plasmid do not necessarily have to be lac operon genes.
Skill: Factual recall
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 151
37) Distinguish between cis‐ and trans‐dominance. Answer: Cis‐dominant genes affect upstream or downstream genes on the same chromosome, while trans‐dominant genes can affect expression of genes located on another chromosome.
Skill: Factual recall
38) For each of the following lac operon mutations, state whether transcription of β‐galactosidase will increase, decrease, or be unaffected. a. Placb. lacY+ c. lacId. lacOe. lacAAnswer: (a) decrease. (b) initially unaffected. (c) increase. (d) increase. (e) unaffected.
Skill: Problem-solving
39) If a cell has the partial diploid genotype
lacI+ lacO+ lacZ- lacY+ _______________________ lacI- l acO+ lacZ+ lacY- ,
how will its gene expression in the presence of allolactose? Answer: The lacI+ gene is dominant over lacI- , and the operator is functional. Similarly, the wild‐type (+) versions of the lacZ and lacY genes are dominant. So, this construct should function normally in the presence of allolactose.
Skill: Problem-solving
40) If a cell has the partial diploid genotype
lacI+ lacO- lacZ+ lacY+ ______________________ lacI- lacO- lacZ+ lacY- ,
how will its gene expression function in the presence of allolactose? Answer: In this construct, lacI+ gene is dominant over lacI- , which means repressor protein is produced. Both copies of the operator are defective, however, and the repressor cannot bind to it. The structural genes will therefore be constitutively produced.
Skill: Problem-solving
41) Briefly define what is meant by negative and positive control of the lac operon. Answer: In negative control, the repressor protein blocks lac gene transcription in the absence of the inducer. In positive control, the catabolite activator protein complexed with cAMP binds to the CAP site in the lac promoter, facilitating the transcription of lac genes.
Skill: Conceptual understanding
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42) Compare the action of repressor proteins in the lac operon and the trp operon. Answer: In both cases, the active repressor binds to the operator and prevents transcription of the operon. In the lac operon, the repressor is inactivated by allolactose, an isomer of lactose that functions as an inducer. Allolactose is only made when lactose is present; therefore, the operon makes the gene products for lactose catabolism only when lactose is present. On the other hand, the trp repressor is active only when bound to tryptophan. Thus, the gene product for the biosynthesis of tryptophan is not made when sufficient quantities of the amino acid are already present.
Skill: Factual recall
43) Why do lacOC mutants express in cis‐dominant and not in transdominant fashion? Answer: Because in this mutant, the repressor cannot bind to the lac operator, preventing the downstream lac structural genes from being transcribed. Thus this mutant has no bearing on the functionality of another lac operon in a partial diploid construct.
Skill: Conceptual understanding
44) What selective pressures are likely to have led to the evolution of regulatory systems like operons? Why not simply express these genes continuously? Answer: Continual production of unnecessary enzymes is energetically costly; presumably, this is the main selective pressure to evolve regulatory mechanisms that express genes only when needed.
Skill: Conceptual understanding
45) What effect would mutational damage of the following genes or gene regions have on the induction of an integrated lambda phage virus? a. damage to the cI gene b. damage to the cro gene c. damage to the xis gene Answer: (a) The cI gene encodes the lambda repressor, elimination of which permits transcription from the PL and PR promoters, leading to the lytic pathway. (b) Eliminating the Cro protein also permits transcription from the PL and PR promoters. (c) Damage to the enzyme excisionase would circumvent induction because this enzyme is necessary for excision from the bacterial chromosome.
Skill: Problem-solving
46) At the molecular level, how do proteins like repressors bind to DNA? How can binding affinities be increased or decreased? Answer: Protein‐nucleic acid interaction is mediated by hydrogen bonding, not covalent bonding. DNA is a negatively charged molecule, so binding occurs through positively charged amino acid residues in the protein. Binding affinity can be increased or decreased through amino acid substitutions that affect charge or shape.
Skill: Conceptual understanding
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Chapter 17 Regulation of Gene Expression in Bacteria and Bacteriophages 153
47) Give an overview of the mechanism by which ultraviolet radiation induces the lytic pathway in lambda phage. Answer: The DNA damage that accompanies UV irradiation may include damage to the recA gene, causing it to code for a modified protein which stimulates the lambda repressor polypeptides to cleave, inactivating them. Lack of repressor allows RNA polymerase to bind to the PR promoter, transcribing cro. Cro protein reduces synthesis of cII, which regulates lambda repressor synthesis. Eventually, enough Q proteins accumulate from the transcription from the PR promoter to initiate the lytic pathway.
Skill: Factual recall
48) Should viruses like lambda phage be considered living entities? Or are they more properly viewed as parasitic fragments of DNA? Answer: The answer depends on the definition of life; viruses are not alive under the cellular definition of life, which holds that living organisms are composed of one or more cells. They do, however, evolve by natural selection, the main criterion for life as suggested by geneticist H. J. Muller.
Skill: Conceptual understanding
49) Contrast negative (repressor) and positive (CAP) control in the lac operon. Answer: Under negative control, transcription of the structural genes is shut down by a repressor protein until inactivated by the inducer (allolactose). Under positive control, transcription of the structural genes is shut down by the inducer (CAP).
Skill: Conceptual understanding
50) Microbial genomics is a rapidly growing field. Of great interest to geneticists is the search for new operons as new microbial genome sequences become available. What kind of ʺsignatureʺ or identifying feature would you look for as you searched new genomes for candidate operons? Answer: Operons are sets of genes under common regulatory control, with promoter and operator regions situated adjacent to structural gene regions. Those regions that are conserved (i.e., occur with similar structure) across multiple genomes are good candidate operons for research and can often be identified via computer‐based searches.
Skill: Problem-solving
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Chapter 18 Regulation of Gene Expression in Eukaryotes
MATCHING QUESTIONS Please select the best match for each term. 1) Alternative polyadenylation
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A) Controls transcription initiation B) Produces more than one protein from one gene C) A method of targeting proteins for degradation D) Gene regulation dependant on parent of origin E) A method for targeting RNAs for degradation
2) Genomic imprinting
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3) Enhancer
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4) Ubiquitination
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5) siRNA
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Answers:
1) B
2) D
3) A
4) C
5) E
MULTIPLE-CHOICE QUESTIONS 6) A similarity between regulation of the lac operon in bacteria and the regulation of galactose utilization in yeast is that both pathways A) transcribe all the genes necessary as one polycistronic mRNA. B) use feedback inhibition to stop making their sugars. C) use a modified version of their sugars as inducers. D) are subject to catabolite repression. E) C and D only Answer: E
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7) When a promoter element is bound by a positive regulatory protein, the result is A) activation of replication. B) activation of transcription. C) activation of translation. D) repression of replication. E) repression of transcription. Answer: B
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154
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Chapter 18 Regulation of Gene Expression in Eukaryotes 155
8) Unlike bacterial operons, eukaryotic operons A) are all positively controlled. B) are all negatively controlled. C) are restricted to certain eukaryotic kingdoms. D) are exceedingly rare. E) None of these Answer: D
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9) Hormones can be considered A) inducers. B) repressors. C) transcription regulators. D) translation regulators. E) enzymes. Answer: C
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10) Unlike prokaryotes, most gene regulation in eukaryotes A) takes place at the transcriptional level. B) is controlled at the level of transcript processing. C) is controlled by inhibitory proteins. D) is based on posttranslational modification. E) None of these Answer: B
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11) Which of the following secondary structures may be a DNA‐binding domain? A) Zinc finger B) Leucine zipper C) Helix‐turn‐helix D) A and B only E) A, B, and C Answer: E
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12) In eukaryotic cells, steroid hormones A) bind to cytoplasmic receptors, are transported to the nucleus, bind DNA, and regulate gene expression. B) bind to cell surface receptors and send signals that result in regulation of genes in the nucleus. C) act as enhancers and bind to an assortment of activators and repressors. D) methylate DNA to regulate gene expression. E) cause maternal effect. Answer: A
Skill: Factual recall
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13) RNA silencing involves A) miRNAs. B) siRNAs. C) double‐stranded RNA. D) a protein called Dicer. E) All of these Answer: E
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14) HREs are A) hormone receptors. B) DNA sequences in promoter regions to which steroid hormone receptors bind. C) proteins that interact with histones. D) elements that lead to histone deacetylation. E) None of these Answer: B
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15) Chromosome‐level gene repression involves A) supercoiling. B) physical blockage of gene regions. C) chromatin formation. D) nucleosomes. E) All of these Answer: E
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16) Change in the DNA‐histone complex that can increase or decrease transcriptional activity is termed A) induction. B) gene regulation. C) chromatin remodeling. D) nucleosome formation. E) None of these Answer: C
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17) The protein class(es) involved in the activation of transcription is (are) A) general transcription factors (GTFs). B) transactivators. C) coactivators. D) A and B only E) A, B, and C Answer: E
Skill: Factual recall
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Chapter 18 Regulation of Gene Expression in Eukaryotes 157
18) Control of transcription initiation of eukaryotic protein‐coding genes involves A) enhancers. B) promoters. C) repressors. D) activators. E) All of these Answer: E
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19) Activators in eukaryotes A) bind to enhancers. B) may act as monomers C) may act as homodimers. D) may act as heterodimers with a DNA binding subunit and an activation subunit. E) All of these Answer: E
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20) The half‐life of a protein is directly related to A) what kind of cell it is produced in. B) its tertiary structure. C) whether it is encoded by a constitutively expressed gene. D) its N‐terminal amino acid residue. E) the number of stabilizing cross‐linkages. Answer: D
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21) Removal of the 5ʹ G‐cap of mRNA transcripts A) prevents transcript transport. B) leads to degradation by exonucleases. C) extends mRNA half‐life. D) leads to polyadelylation. E) reduces mRNA half‐life. Answer: B
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22) mRNA regulation can be achieved at the level of A) transport. B) processing. C) translation. D) life span (degradation). E) All of these Answer: E
Skill: Factual recall
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23) With respect to genetics, URS stands for A) uracil response system. B) upstream regulatory sequence. C) upstream repressor‐binding sequence. D) unregulated repression series. E) uniform response system. Answer: B
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24) Proteins that are encoded by the same gene but differ in structure and function are the product of alternative splicing in pre‐mRNA processing. Such protein variants are termed A) isoteins. B) heteromeric proteins. C) isozymes. D) protein isoforms. E) allelomorphs. Answer: D
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25) Heterochromatin is associated with A) differential gene expression. B) gene silencing. C) mRNA translation. D) heterozygotes. E) A, B, and C only Answer: B
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TRUE-FALSE QUESTIONS 26) Transcriptionally active genes show lower levels of DNA methylation compared with transcriptionally inactive genes. Answer: TRUE
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27) Gene silencing in eukaryotes is often achieved through chromatin structure. Answer: TRUE
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28) Eukaryotic repression generally occurs via physical blockage of the promoter. Answer: FALSE Explanation: In eukaryotes, repressors generally interfere with activators, preventing them from initiating transcription.
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29) Longer poly(A) tails on mRNAs are associated with less translational activity. Answer: FALSE Explanation: Shorter poly(A) tails are associated with less translational activity.
Skill: Factual recall
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Chapter 18 Regulation of Gene Expression in Eukaryotes 159
30) Recombinases mediate transcript processing from pre‐mRNA to mature mRNA. Answer: FALSE Explanation: The spliceosome mediates transcript processing.
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31) Steriod hormones are effector molecules. Answer: TRUE
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32) Steroid and peptide hormone receptors are typically cytoplasmic. Answer: FALSE Explanation: Steroid hormone receptors are cytoplasmic, while peptide hormone receptors are on the cell surface, in the cell membrane.
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33) Imprinting is implicated in Prader‐Willi and Angelman syndromes. Answer: TRUE
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34) Gene silencing is a form of regulation controlled by repressor proteins. Answer: FALSE Explanation: Gene silencing is a form of expression regulation either based on packaging (heterochromatin), which usually affects large numbers of genes at once, or by RNAi, which is often gene specific.
Skill: Factual recall
35) Noncoding RNAs called mRNAs are present in most eukaryote genomes and are involved in negatively regulating gene expression through degradation of mRNAs with complementary sequences. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Describe how siRNA functions in regulation of expression and how this can be used as a tool in genetic research. Answer: ʺSmall interferingʺ RNA can complex with mRNA with complementary base sequences, preventing transcription. This makes siRNAs valuable tools, as they can be synthesized to experimentally block the expression of genes in specific tissue types with very precise timing, helping geneticists determine gene function and interaction.
Skill: Conceptual understanding
37) Some genes that are expressed in the fetal stage are never expressed again in that individualʹs life. What do you think is the most common mechanism for preventing the expression of such genes? Answer: Such genes are permanently silenced, typically via heterochromatic packaging. This is more efficient than relying on repression.
Skill: Conceptual understanding
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38) What essential difference between the growth and reproductive strategies of prokaryotes and multicellular eukaryotes explains why the latter have far more sophisticated regulatory mechanisms? Answer: An important point made in this chapter is that while prokaryotes grow and divide, multicellular eukaryotes develop and differentiate. Tissue differentiation and the homeostatic maintenance of the multicellular body require a diversity of flexible, precise regulatory mechanisms.
Skill: Conceptual understanding
39) Describe the use of the DNA‐degrading enzyme DNase I in experiments that explore the effect of chromatin on gene inactivation. Answer: DNA from cells with certain transcriptionally active vs. inactive genes can be isolated and treated with DNase I; then the condition of the genes are assessed with a Southern blot. Inactive, sequestered genes should remain intact, while active, exposed genes should have been degraded by the enzyme.
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40) Explain how combinatorial gene regulation permits regulation of a large diversity of genes with relatively few regulatory proteins. Answer: Although the number of different regulatory proteins is limited, in combination they can interact with a far greater number of genes with specificity.
Skill: Conceptual understanding
41) Embryonic development is orchestrated by a cascading series of genes, the expression of many of which are controlled in time and space by transcription factors and effector molecules of various kinds. But what controls the expression of the first regulatory genes to be transcribed in the newly fertilized zygote? Answer: The egg contains stored mRNAs as well as transcription factors derived from the mother, and both of these orchestrate the expression of the first regulatory genes.
Skill: Conceptual understanding
42) What would happen developmentally if a zygote with XY sex chromosome karyotype had a defective testosterone receptor? Answer: We learned in this chapter that testosterone is necessary for initiating the male developmental pathway; an XY individual lacking this receptor would develop as a female.
Skill: Conceptual understanding
43) What are enhancers? Answer: Enhancers are DNA sequences that typically have binding sites for several different activators and repressors. They are located somewhere in the vicinity of the gene, much farther away than promoter sequences, sometimes tens of kilobases away. They allow complex and nuanced control of gene expression depending on a number of different conditions such as developmental stage, tissue type, and response to extracellular signals.
Skill: Conceptual understanding
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Chapter 18 Regulation of Gene Expression in Eukaryotes 161
44) Plants produce steroids called phytosterols, which have various functions in plant physiology. Some phytosterols are thought to have a defensive function, however, specifically in repelling insect herbivores. How might plant‐derived steroids help protect plants from insect attack? Answer: In some cases, the plant‐derived steroids are sufficiently similar to insect steroid hormones that they can wreak havoc with insect physiological functioning and development when ingested. Debilitating or killing insect herbivores reduces tissue loss to the plant.
Skill: Conceptual understanding
45) Insect growth and metamorphosis is controlled by the steroid hormones ecdysone and juvenile hormone (JH). In insects like butterflies and moths that have a distinct larval (juvenile) and adult stage, the relative amount of JH determines whether the insect molts to the next juvenile stage or switches to become an adult. Can you think of a way to use this hormonal system to control caterpillar agricultural pests? Answer: The best approach would be to somehow suppress JH early in caterpillar development, forcing them to mature early. This would reduce the duration of the most destructive stage (the leaf‐eating caterpillar). The opposite treatment (extending the caterpillar stage) would have dire consequences, because the quantity of leaf biomass these insects consume increases geometrically with each larval stage.
Skill: Conceptual understanding
46) Interference RNA technology is one of the most exciting new developments in genetics in recent years. In April 2004, Cancer Research UK and the Netherlands Cancer Institute announced creation of a 24,000‐molecule iRNA library, designed to inactivate about 8,000 human genes. One hope for this library is its application in combating cancer. How might iRNA be implemented in cancer treatment? Answer: Cancers for which the genetic basis is understood might be treated by preventing gene expression through iRNA‐mediated gene silencing. Use of this technology may depend on the type of cancer and its localization.
Skill: Conceptual understanding
47) In genetic imprinting, expression depends on the parent of origin of a given allele. How might imprinting affect the expression of a hypothetical autosomal disease that expresses in dominant fashion and recessive fashion, and how might imprinting be detected? Answer: In individuals heterozygous for the dominant allele, expression of a disease phenotype depends on which allele is silenced–and thus which parent donated the disease allele. This will not matter in homozygous recessive individuals with respect to recessively expressing disease alleles. In the former case, imprinting could be detected through analysis of enough pedigrees with known male vs. female heterozygotes and known progeny phenotypes. This would reveal the influence of parent of origin on allele expression.
Skill: Problem-solving
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48) Evolutionary biologists suspect that genetic imprinting evolved through parent‐offspring conflict. What kinds of ʺconflict of interestʺ in progeny development can you imagine between mammalian parents? Answer: Fetal development requires enormous energy investment, and in many species the degree of pre‐ and postnatal investment directly influences future maternal reproduction. From the point of view of the motherʹs evolutionary fitness, there should be a balance between present and future investment. In terms of the fatherʹs evolutionary fitness, however, his unborn offspring should take as much energy from the mother as possible to safeguard his present investment, since his future investments may not necessarily be with the same mother. Imprinting may have evolved as a result of conflict between the maternal and paternal genomes. (See Reik and Walter. 2001. Nat Genet. 27:255‐256; PMID 11242103.)
Skill: Problem-solving
49) What are some advantages and disadvantages of arranging gene clusters into operons? Answer: Operons can be advantageous in that regulation of expression can be controlled by a single regulatory region, with the whole set up‐ or down‐regulated. In prokaryotes, induced operons can respond to needs quickly by transcribing a single polycistronic mRNA. Some of these features can be disadvantageous as well; control by a single regulatory region also means greater vulnerability to mutation. A mutant operator or promoter, for example, might render all genes in the operon dysfunctional.
Skill: Conceptual understanding
50) Several common protein structural motifs are now known to be involved in DNA recognition and binding. What are these, and how would researchers use genomic data to help figure out if new candidate genes encode such DNA‐binding proteins? Answer: DNA‐binding structural motifs of proteins include zinc and leucine zippers and the helix‐turn‐helix (HTH) motif. Researchers can electronically scan the sequence of new genes for the particular nucleotide sequence underlying such motifs, thereby identifying them as candidate DNA recognition or binding proteins.
Skill: Factual recall, conceptual understanding
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Chapter 19 Genetic Analysis of Development
MATCHING QUESTIONS Please select the best match for each term. 1) Differentiation
Skill: Factual recall, conceptual understanding
A) Genes that determine the number of segments in Drosophila B) Process through which vaccines increase production of specific antibodies C) The specialization of cells during development D) Formation of limbs and organs during development E) Tissue patches in larvae that will develop into adult structures
2) Morphogenesis
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3) Imaginal discs
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4) Clonal selection
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5) Pair-rule genes
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Answers:
1) C
2) D
3) E
4) B
5) A
MULTIPLE-CHOICE QUESTIONS 6) ________ cells may differentiate into any cell or tissue type. A) Pluripotent B) Omnipotent C) Totipotent D) Hyperpotent E) Potent Answer: C
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7) Which of the following cell types has the lowest developmental potential? A) Oocyte B) Red blood cell C) Embryonic stem cell D) Zygote E) All have equal developmental potential. Answer: B
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8) The developmental fate of a cell is established by the process termed A) development. B) differentiation. C) determination. D) specialization. E) A, B, and C only Answer: C
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9) ________ is the process of cell or tissue change in development. A) Transformation B) Morphogenesis C) Development D) Change E) None of these Answer: B
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10) Which of the following correctly expresses the order of development? A) Fertilization, development, maturity B) Induction, determination, differentiation C) Induction, differentiation, determination D) Fertilization, differentiation, determination E) None of these Answer: B
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11) Which of the following statements is true? A) Sex determination in fruit flies depends on the X : autosome ratio. B) Mammalian sex determination can be influenced by temperature. C) Mammals experience random Y inactivation. D) The testis‐determining factor gene is X‐linked. E) Dosage compensation occurs in the same way in most animals. Answer: A
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12) The X‐inactive specific transcripts gene is designated A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: E
Skill: Factual recall
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Chapter 19 Genetic Analysis of Development 165
13) The best model organism for studying cell fate from fertilized egg through adult is A) Arabidopsis thaliana. B) Caenorhabditis elegans. C) Drosophila melanogaster. D) Mus musculus. E) Saccharomyces cerevisiae. Answer: B
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14) Which of the following is not true of X‐inactivation? A) It is necessary for dosage compensation. B) It is the cause of calico color in cats. C) The mechanism includes transcription of the XIST gene and chromatin remodeling. D) The XIST RNA is non‐coding. E) The master regulatory gene for X‐inactivation is sxl or sex‐lethal. Answer: E
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15) In Drosophila, sex determination is carried out by A) a cascade of alternative splicing. B) the SRY gene. C) histone modification. D) the DMRT1 gene. E) All of these Answer: A
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16) Antibody genes A) contain both constant and variable regions. B) are formed by cutting and splicing DNA in B cells during development. C) are made from two heavy and two light chains, each of which has a variable region. D) each recognize one antigen and are the only antibody made by their B cell. E) All of these Answer: E
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17) The X‐inactivation center gene is A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: B
Skill: Factual recall
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18) SRY is A) a single‐region Y gene. B) a Y‐inactivation center gene. C) the sex‐determining region of the Y chromosome. D) a Y‐controlling element. E) None of these Answer: C
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19) The X‐controlling element gene is A) SRY. B) XIC. C) Xce. D) MATa. E) Xist. Answer: C
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20) Homeobox genes A) encode proteins with a conserved sequence of amino acids. B) are important in development. C) are conserved in multicellular animals. D) are transcription factors. E) All of these Answer: E
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21) Mammalian sex determination is controlled by A) XIST. B) SRY. C) XIC. D) All of these E) None of these Answer: A
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22) The central function of X-inactivation is A) silencing imprinted genes. B) enhancing the expression of genes on the chromosome left active. C) dosage compensation. D) silencing deleterious alleles. E) None of these Answer: C
Skill: Factual recall
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Chapter 19 Genetic Analysis of Development 167
The following scenario applies to the questions below. In the snail Limnaea peregra, shell coiling is determined by the nuclear alleles S (dextral) and s (sinistral). This gene is a maternal effect gene that influences egg formation, which in turn influences whether the snail that hatches from the egg has a shell that coils to the right (dextral) or left (sinistral). A female sinistral snail is mated with a sinistral male snail. The genotypes of both are unknown. All the F 1 progeny are dextral. When the members of the F 1 generation are crossed, they yield an F 2 generation that is dextral and sinistral.
4 4 3 1
23) What was the genotype of the mother of the female snail? A) SS B) Ss C) ss D) Ss or ss E) SS or Ss Answer: C
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24) What were the genotypes of the sinistral snails? A) ss female and Ss male B) Ss female and Ss male C) SS female and ss male D) Ss female and ss male E) ss female and SS male Answer: B
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25) The ________ is a conserved 180‐bp sequence found in homeotic genes. A) TATA box B) bicoid box C) Hox box D) homeobox E) Pribnow box Answer: D
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TRUE-FALSE QUESTIONS 26) The developmental trajectory of the cells of an embryo can be tracked and diagrammatically expressed as a developmental atlas. Answer: FALSE Explanation: The correct term for the diagrammatical expression of this information is fate map.
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27) Determination occurs via induction. Answer: TRUE
Skill: Factual recall
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28) Antibodies elicit the production of antigens. Answer: FALSE Explanation: Antigens elicit antibody production.
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29) Development occurs through the activation and inactivation of a series of specific genes in a precise sequence. Answer: TRUE
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30) Morphogens are substances that help control development. Answer: TRUE
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31) Antennapedia and bithorax belong to the class of mutants referred to as atavistic mutants. Answer: FALSE Explanation: Both antennapedia and bithorax are homeotic mutants.
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32) Retention of the full genomic structure during cell differentiation is termed DNA constancy. Answer: TRUE
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33) The major classes of antibody molecules are called antigen‐globulins. Answer: FALSE Explanation: Antibody molecules are immunoglobulins, of which there are five classes in humans.
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34) T and B cells are so named for the types of antigens they attack. Answer: FALSE Explanation: They are named for the site of production: T cells are produced in the thymus, B cells in the bone marrow.
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35) No human cells lose all of their DNA during differentiation. Answer: FALSE Explanation: Red blood cells (erythrocytes) are anucleate; that is, they have no nucleus and thus no DNA.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Why are the changes associated with development irreversible? Answer: Development is a programmed sequence of cellular‐level phenotypic events, a cascade in which activation of certain genes in turn sets into motion the activation of other genes. The changes set in motion involve morphology and physiology –phenotypic changes that are too complex to reverse.
Skill: Analytical reasoning
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Chapter 19 Genetic Analysis of Development 169
37) Vascular plant cells generally have the ability to differentiate into various types of tissue–a fact long known to horticulturists who propagate plants by cuttings. This means that plants do not display the strict separation of germline and somatic cells that animals do. Might this be expected to accelerate or decelerate evolution in plants relative to animals? Answer: Since somatic cell lines can become germline in plants (as when vegetative cells give rise to germ‐bearing flower tissue), somatic mutations can become heritable. Opportunities for the generation and propagation of heritable genetic variants would be greater in plants as a result. One consequence of this could be more rapid evolution in some cases, but a host of other factors play a role in setting evolutionary rates.
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38) If cell and tissue differentiation is controlled genetically, how can environmental factors play a role in determining the final phenotype? Answer: Environmental factors may influence the timing and degree of expression of developmental genes, thereby influencing the phenotype.
Skill: Analytical reasoning
39) The use of embryonic stem cells in federally funded research in the United States is a hotly debated topic. What are the arguments for and against use of such stem cell lines in research and treatment of disease? Answer: The arguments in favor of this research primarily focus on the research and therapeutic gains that may be made through their use–in particular, the promise of using stem cells to treat diseases that involve deterioration of certain cell types. Arguments opposed to this research largely stem from opposition to one source of stem cells: human embryos.
Skill: Conceptual understanding
40) A maternal effect mutation bicoid (bcd) is recessive. The bcd allele makes no Bicoid protein. In the absence of the Bicoid protein, product embryogenesis is not completed and eggs fail to hatch into viable larva. Consider a cross between a female who is heterozygous ( bcd+ /bcd) and a homozygous mutant male (bcd/bcd). a. How is it possible for a male homozygous for the mutation to exist? b. Predict the outcome (normal and/or failed embryogenesis) in the F 1 and F2 generations resulting from this cross. Answer: (a) The bcd/bcd male must have had a heterozygous mother. (b) The first generation offspring will all hatch normally, although half of them will be bcd/bcd. In the F2 generation, half the eggs will fail to produce larva. An F 1 female who is bcd/bcd will produce no live offspring, but the bcd+ /bcd females will produce all live offspring.
Skill: Problem-solving
41) Recount the evidence that Dolly the sheep was indeed the product of cloning, or cell fusion. Answer: Dollyʹs phenotype matched that of her nuclear mother, rather than that of the recipient mother that carried her to term. Genetic analysis (DNA fingerprinting) confirmed that Dolly matched the genetic signature of the donorʹs somatic cell type rather than the recipient motherʹs.
Skill: Factual recall
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42) Can you think of a specific example in which multiple genome datasets would be useful in the study of morphogenesis? Answer: Datasets for phenotypically variable, or plastic, organisms would be especially valuable. A good candidate group for study would be domesticated dogs, characterized by many and varied breeds.
Skill: Conceptual understanding
43) The cells of triploid animals, like certain salamanders, are individually larger, yet their overall body size is no different from that of diploids. Pose a developmental hypothesis for the control of morphogenesis in such animals. Answer: The increase in cell size with no concomitant increase in body size means that the triploids produce fewer cells for the same body size. Intercellular signals are important in morphogenesis, so perhaps size and shape of tissues and organs is sensed and communicated by chemical signals generated, say, when certain cell types come in contact. A certain size or dimension would be reached sooner with larger cells, triggering the next set of developmental genes and preserving overall body dimensions.
Skill: Problem-solving
44) Contrast the mechanisms of dosage compensation in humans with those found in insects like Drosophila. Answer: The mechanisms of dosage compensation found in humans and fruit flies represent two different solutions to the same problem. In both cases, X-linked genes are present in twice the copy number in XX females as in XY or XO males, potentially creating a dosage imbalance. In mammals like humans, this is solved through X-inactivation, while in insects like Drosophila it is solved through X-hyperactivation. Both act to restore X-expression parity between females and males.
Skill: Conceptual understanding
45) Mutations like bithorax, in which a second pair of wings develops in the place of halteres, give clues to the evolutionary origin of differentiated structures. What inferences might you draw from bithorax, or antennapedia? Answer: These otherwise different structures have a similar genetic basis, suggesting that they are likely to have a common origin (gene duplication and divergence, for example).
Skill: Analytical reasoning
46) In what ways does the cloned cat Cc demonstrate the limitations of cloning technology in obtaining desired phenotypes? Answer: Cc differs in several respects from her somatic cell‐donor mother despite being genetically identical. Her coat color differs, as does her shape and personality. These differences stem from various complicating factors, including patterns of X‐inactivation and environmental contributions to certain quantitative traits.
Skill: Factual recall
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Chapter 19 Genetic Analysis of Development 171
47) Cloned animals are typically not entirely genetically identical to their cloned parent. Why? Answer: Cloned animals are usually made by transplanting a somatic cell nucleus into an enucleated oocyte from a donor. Unless the donor cell is from the same maternal lineage as the somatic cell, the mitochondria and the mitochondrial DNA will be that of the egg donor, not the clone parent.
Skill: Problem-solving
48) Is there a relationship between the problems encountered cloning mammals and the rarity of polyploidy in this group? Answer: Possibly so; mammals, unlike some animal groups, appear to be extremely sensitive to gene dosage and even small perturbations in developmental pathways.
Skill: Analytical reasoning
49) Your body is likely to have existing antibodies that will counter antigens you have yet to contact. How is this accomplished? Answer: Differential splicing of numerous gene elements (V, J, C, K, and D segments) underlies the tremendous antibody diversity of animals. With distinct molecules ranging to 108 , chances are good that at least one of these molecules will recognize a novel antigen.
Skill: Factual recall
50) As of 2008, human cloning was banned in the United States. What are the arguments for maintaining such a ban? Answer: Human cloning is a hugely complex ethical issue. For the time being, the extreme inefficiency and high developmental error rate associated with mammalian cloning are sufficient to justify a ban.
Skill: Conceptual understanding
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Chapter 20 Genetics of Cancer
MATCHING QUESTIONS Please select the best match for each term. 1) CDK
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A) Genes that produce abnormally active cell growth proteins B) Phosphorylate proteins important to the cell cycle C) Normal growth-promoting genes D) Proteins that can stop the cell cycle E) Are made and degraded at specific points in the cell cycle 2) E 3) A 4) C 5) D
2) Cyclins
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3) Oncogenes
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4) Proto-oncogenes
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5) Tumor suppressors
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Answers:
1) B
MULTIPLE-CHOICE QUESTIONS 6) Hereditary retinoblastoma is an autosomal dominant hereditary cancer. Cells from retinal tumors in a child who has this disease possess A) evidence of non‐disjunction in all autosomes. B) one non‐functional copy of the retinoblastoma gene. C) two non‐functional copies of the retinoblastoma gene. D) trisomy for the retinblastoma gene. E) a dominant mutation in the retinoblastoma gene. Answer: C
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7) Apoptosis is A) rapid cell division. B) caused by proto‐oncogenes. C) the change in shape of a cell when it becomes cancerous. D) programmed cell death. E) None of these Answer: D
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172
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Chapter 20 Genetics of Cancer 173
8) Protein kinases are enzymes that normally catalyze the A) degradation of cellular proteins. B) phosphorylation of cellular proteins. C) formation of peptide bonds in proteins. D) phosphorylation of ADP. E) synthesis of kinins. Answer: B
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9) A proto‐oncogene that gains function due to amplification, translocation to a more transcriptionally active area of the genome, or a mutation that causes either an increase in function or the suppression of down‐regulation, is A) a pseudo‐oncogene. B) recessive. C) dominant. D) an oncogene. E) C and D only Answer: E
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10) Metastatic cancer is A) terminally differentiated. B) malignant. C) invasive. D) B and C only E) A, B, and C Answer: D
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11) Retroviruses are responsible for which of the following conditions? A) Rous sarcoma B) AIDS C) Feline leukemia D) Mouse mammary tumors E) All of these Answer: E
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12) Mutations in a normal growth‐stimulating gene are most likely to have no effect or to cause A) the loss of responsiveness to growth stimulation in the cell with the mutant gene. B) the cell with the mutant gene to become cancerous. C) excessive growth in the cells surrounding the one with the mutant gene. D) a mutator effect in the cell with the mutant gene. E) apoptosis of the cell with the mutant gene. Answer: A
Skill: Conceptual understanding
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13) Small DNA tumor viruses such as SV40 and HPV have viral oncogenes that are not homologs of host cell genes. Some of these novel viral oncogenes make proteins that exert their tumorigenic effect by A) binding to and inactivating pRB. B) making Gag, Pol and Env proteins more active. C) causing cellular proto‐oncogenes to mutate into oncogenes. D) causing apoptosis in target tissues. E) inactivating the immune system. Answer: A
Skill: Conceptual understanding
14) ________ play a pivotal role in programmed cell death. A) Cyclins B) Telomeres C) Proto‐oncogenes D) Oncogenes E) Telomerases Answer: A
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15) In sporadic (nonhereditary) cancers, mutations occur in ________ cells. A) germ B) somatic C) undifferentiated D) mammary E) T Answer: B
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16) Cancer is best defined as A) immune dysfunction. B) uncontrolled and abnormal cell division. C) viral malignancy. D) regulated differentiation of tissue. E) cellular deregulation. Answer: B
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17) Agents that can induce cancer through mutagenesis are called A) destabilizers. B) cancerogens. C) mutagens. D) carcinogens. E) None of these Answer: D
Skill: Factual recall
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Chapter 20 Genetics of Cancer 175
18) In retroviruses, the pol gene product is a(n) A) DNA polymerase. B) reverse transcriptase. C) integrase. D) RNA polymerase. E) recombinase. Answer: B
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19) Cancer may arise as a result of A) mutations associated with viral infection. B) mutations associated with chemical mutagens. C) mutations associated with radiation. D) spontaneous mutations. E) All of these Answer: E
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20) v‐onc genes are found in A) vital cells. B) viral cells. C) cancer‐causing retroviruses. D) very rapidly growing cells. E) bacteriophages. Answer: C
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21) Malignant breast, colon, and lung cancers typically have A) mutated tumor suppressor genes. B) active oncogenes. C) chromosomal abnormalities. D) activated telomerase. E) All of these Answer: E
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22) Which of the following statements is not true? A) Certain types of virus can cause cancer. B) Melanomas are commonly caused by ionizing radiation. C) Neoplasia is another term for cancerous cells. D) About half the people with cancer have a mutation in a p53 gene. E) Most cancers are hereditary. Answer: E
Skill: Factual recall
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23) Which of the following is not likely to be a tumor suppressor? A) A growth factor receptor B) A checkpoint protein C) A DNA repair enzyme D) An apoptosis‐promoting factor E) The negative regulator of a CDK Answer: A
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24) Cell proliferation in culture is normally limited by A) cytophagy. B) quorum sensing. C) contact inhibition. D) competition. E) nutrient limitation. Answer: C
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25) In normal cells, p53 protein acts as A) a signal transducer. B) a tumor suppressor. C) a proto‐oncogene activator. D) a GTPase inhibitor. E) None of these Answer: E
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TRUE-FALSE QUESTIONS 26) Ionizing radiation is emitted at low levels by many natural objects, including some rocks and gases. Answer: TRUE
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27) The products of tumor suppressor genes stimulate cell proliferation, while the products of proto‐oncogenes inhibit cell proliferation. Answer: FALSE Explanation: The statement is correct the other way around: proto‐oncogene products stimulate cell proliferation, while those of tumor suppressor genes inhibit cell proliferation.
Skill: Factual recall
28) The nucleic acid of retroviral provirus is composed of RNA. Answer: FALSE Explanation: Although viral nucleic acid is composed of RNA, that of the provirus is composed of DNA.
Skill: Factual recall
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Chapter 20 Genetics of Cancer 177
29) Some carcinogens act on the genome directly, while others are converted to mutagenic substances by the cellsʹs enzymes. Answer: TRUE
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30) Terminally differentiated cells are noncancerous, normal cells. Answer: TRUE
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31) The process of relaying a growth‐stimulatory or growth‐inhibitory signal in response to an extracellular factor binding at the cell surface is called intercellular signaling. Answer: FALSE Explanation: This process is called signal transduction.
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32) Wild‐type mutator genes produce substances that induce point mutations and chromosomal rearrangements. Answer: FALSE Explanation: Wild‐type mutator gene products function in DNA replication and repair.
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33) Inherited mutant tumor suppressors produce a dominant susceptibility to cancer, but the mutant alleles are actually recessive in the cancer. Answer: TRUE
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34) The only genes necessary for the retroviral lifecycle are gag and pol. Answer: FALSE Explanation: In addition to these, the env gene is also necessary.
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35) Most cancer deaths in the United States are caused by radiation‐induced cancer. Answer: FALSE Explanation: Random mutations due to metabolic processes and chemical carcinogens, including natural and synthetic chemicals, are responsible for most cancer deaths in the United States.
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) List the ways in which proto‐oncogenes may be converted to oncogenes. Answer: The mechanisms include point mutations, deletions, and gene amplification.
Skill: Analytical reasoning
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37) The terms proto‐oncogene and oncogene may give the misleading impression that these are ʺgenes for cancerʺ that are waiting to be turned on. What are these genes, in terms of normal function, and what name might have been given to them to reflect this function rather than their role in cancer? Answer: Proto‐oncogene products play a role in cell growth or the cell cycle. Examples discussed in this chapter include growth factors, protein kinases, and membrane‐associated G proteins. Had they not been discovered in defective form as oncogenes, they would be named more precisely for their function.
Skill: Analytical reasoning
38) Programmed cell death, or apoptosis, is a necessary and useful property of cells. Why is this so? Answer: It is useful for multicellular organisms to have the ability to have some cells selectively self‐destruct. Cells that have become irreparably genetically damaged, are infected by viruses, or have completed their developmental role may pose a danger to the organism if they live.
Skill: Conceptual understanding
39) How does telomerase, which is not a cause of cancer, nonetheless play a role in its development? Answer: The enzyme telomerase helps cancerous cells continue to divide by lengthening chromosomal telomeres.
Skill: Factual recall
40) Describe Knudsonʹs two‐mutation model: When and where are the mutations posited to occur? Answer: In Knudsonʹs model, one mutation is inherited and the second occurs in a somatic cell. The inherited mutation would be found in all cells, requiring only one more mutation in one of them to lead to cancer.
Skill: Factual recall
41) Reverse transcriptase, the enzyme responsible for copying viral RNA into cDNA prior to integration into the host genome, lacks 3ʹ‐to‐5ʹ exonuclease ʺproofreadingʺ ability. How is this advantageous to organisms that depend on such enzymes? Answer: Retroviruses depend on reverse transcriptase. Part of the successful infection and proliferation strategy of retroviruses like HIV is their high mutability. The errors made by the enzyme are useful in generating variable progeny viruses that can evade the hostʹs immune system.
Skill: Analytical reasoning
42) When oncologists (scientists who study cancer) refer to ʺtransformation,ʺ they mean something different from what the word ʺtransformationʺ signifies when used by biotechnologists. Explain the difference between the two uses of the term in genetics. Answer: In terms of cancer, cells that lose their growth control and no longer behave like normal cells are referred to as transformed. In the context of bacterial genetics and gene manipulation, however, as was discussed in Chapter 9 and Chapter 15, transformation refers to a living cell picking up naked DNA from outside the cell and incorporating it into its genome.
Skill: Conceptual understanding
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Chapter 20 Genetics of Cancer 179
43) People with the genetic disease xeroderma pigmentosum have compromised DNA repair mechanisms, making them extremely sensitive to genetic damage from sunlight. Without taking measures to minimize exposure, these individuals typically succumb to skin cancer in childhood or early adulthood–a demonstration of the mutagenicity of ultraviolet radiation. Yet, deliberate exposure in the form of sunbathing is very popular with some segments of the population. Why do you think this is so, and how might public health professionals raise awareness of the danger in which these people place themselves? Answer: The reasons for the popularity of sunbathing in some groups are probably numerous and complex, but generally relate to the wish to project a culturally conditioned image of health and vigor. This is ironic in that the practice actually causes genetic damage–decreasing health and vigor. Health professionals might try to combat this practice with advertising campaigns highlighting the effects of genetic damage, perhaps along the lines of the campaigns against cigarette smoking.
Skill: Analytical reasoning
44) The Bcl-2 gene product in its normal form can be activated to prevent cells from undergoing apoptosis (programmed cell death) in specific circumstances. It is a carefully regulated gene, being ʺturned onʺ only under specific circumstances. A null mutation in Bcl-2 leads to excessive cell loss. Some Bcl-2 mutation can cause a ʺgain of functionʺ in which the protein is active all the time. a. In the terms used in this chapter, what would the normal version be called? b. What would the gain‐of‐function mutation be called? c. How would you expect the gain‐of‐function mutation to be involved in cancer? Answer: (a) A proto‐oncogene. (b) An oncogene. (c) The gain‐of‐function mutation could cause abnormal cells to fail to commit suicide when their genome has been irreparably damaged. As a result, these cells could mutate into cancer cells.
Skill: Conceptual understanding
45) BRCA1 is a gene involved in repair of double‐stranded DNA breaks. Mutant forms of this gene are linked to a substantial proportion of familial breast cancers. A woman who inherits a certain allele of the gene has about a 60‐80% chance of getting breast cancer, as well as an elevated chance of getting ovarian cancer, in her lifetime. a. Why donʹt a higher percentage of women with the mutation get breast cancer? b. How would the breast cancer inheritance caused by BRCA1 be described in Mendelian terms? Answer: (a) BRCA1 is a tumor suppressor gene. Tumor suppressor genes generally conform to Knudsonʹs ʺtwo‐hitʺ mutational model of cancer. For a cell to become cancerous as a result of BRCA1 mutation both copies of the gene need to have become mutant and non‐functional. If a woman is born with one defective copy her odds of ending up with cells with two defective copies are greatly increased over that of a person whose cells started out with two good copies, but the chance is still less than 100%. (b) In Mendelian terms breast cancer caused by a hereditary mutation in the BRCA1 gene would appear to be dominant but with partial penetrance.
Skill: Conceptual understanding, analytical reasoning
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46) Li-Fraumeni syndrome is a rare inherited disease that greatly increases a personʹs risk of developing several types of cancer. This disease stems from a mutation in the TP53 tumor suppressor gene (Nichols et al. 2001. Cancer Epidemiol Biomarkers Prev 10:83‐87). In most cases, approximately 95% of the mutations can be detected by sequence analysis of exons 4 through 9. If you isolated and analyzed DNA from a Li -Fraumeni patient, what genotype should you find in DNA isolated from a. normal tissue? b. malignant tissue? Answer: (a) The DNA from normal tissue should reveal one loss‐of‐function TP53 mutation (TP53‐/TP53+). (b) Malignant tissue should show loss of both alleles (TP53-/TP53-).
Skill: Problem-solving
47) At the molecular level, what exactly are the mutagenic effects of UVA and UVB radiation? Answer: As discussed in Chapter 7, this radiation primarily causes point mutations by inducing tautomeric shifts and thymine dimers.
Skill: Factual recall
48) Suppose you are studying a rare spontaneous cancer in cats, which you suspect is retrovirus‐induced. What kinds of retroviral insertion could lead to cancer, and what kind of genetic signature would you look for to distinguish between them? Answer: One possible scenario is stimulation of a proto‐oncogene through insertional mutagenesis, inserting near and overexpressing a proto‐oncogene under the regulatory control of viral control elements. Another scenario is infection with a transducing retrovirus, in which expression of the v-onc gene induces tumor formation. The genetic signature of the two scenarios differs in the presence of v-onc. Isolating DNA from tumor tissue and sequencing the viral insertional site (perhaps first identified through viral genes like gag, pol, and env) would reveal whether v-onc or c-onc genes are present. Most proto‐oncogenes contain introns that are not present in the corresponding v-onc.
Skill: Conceptual understanding
49) What kinds of cancers might eventually be treated with RNAi, which you learned about in Chapter 18? Answer: siRNAs can be engineered to silence or down‐regulate genes by binding with their mRNA. A potential application may thus be treating cancers that stem from the expression or overexpression of a gene product; for example, overproduction of growth factors or protein kinases by an oncogene.
Skill: Analytical reasoning
50) What is metastasis, and why is metastatic cancer the most difficult cancer to treat? Answer: Metastasis is the spreading of cancerous cells to several locations of the body via the bloodstream. The cancerous cells are thus no longer at a point source, which might be treatable locally with surgery or radiation. Metastatic cancers must be treated systemically, as with chemotherapy or radiation.
Skill: Conceptual understanding
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Chapter 21 Population Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) Favorable alleles of genes can cause selection for an allele of an unrelated linked gene, a phenomenon known as
Skill: Factual recall
A) gene flow. B) overdominance. C) genetic hitchhiking. D) postzygotic isolation. E) genetic drift.
2) Sickle‐cell heterozygotes have a selective advantage, exhibiting
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3) Immigration of individuals with different allele frequencies into a population causes
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4) Hybrid sterility is an example of
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5) Very small, isolated populations are subject to
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Answers:
1) C
2) B
3) A
4) D
5) E
MULTIPLE-CHOICE QUESTIONS 6) In the Hardy‐Weinberg model, the ideal population is A) very large. B) randomly mating. C) free of mutations. D) non‐migrating. E) All of these Answer: E
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7) If there are two alleles of a gene, B and b, and the frequency of the B allele (p) is 0.90, the frequency of the b allele (q) is A) 0.81. B) 0.30. C) 0.10. D) 0.09. E) 0.01. Answer: C
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8) For alleles of genes in Hardy‐Weinberg equilibrium, the frequency of heterozygotes is represented as A) p. B) q. C) p 2 . D) q2 . E) 2pq. Answer: E
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9) Which of the following methods is least useful for assessing levels of genetic variation in populations? A) DNA sequencing B) VNTRs C) Protein electrophoresis D) Phenotypic observation E) RFLP analysis Answer: D
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10) If the genotypic frequencies for an allele in a population at Hardy‐Weinberg equilibrium are 0.635 homozygous dominants, 0.430 heterozygotes, and 0.125 homozygous recessives, what is the frequency of the recessive allele? A) 0.35 B) 0.65 C) 0.125 D) 0.875 E) 0.01 Answer: A
Skill: Problem-solving
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Chapter 21 Population Genetics 183
11) Allele frequencies can change across the geographic range of a species according to climate, geological features, or direction in the range. For instance, a certain allele frequency might go up along with the elevation or down as you go south in the speciesʹ range. This systematic change in frequency is referred to as A) genetic drift. B) q2 . C) directional selection. D) a cline. E) heterosis. Answer: D
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12) Which of the following is not a form of genetic drift? A) Founder effect B) Nonrandom mating C) Population bottleneck D) Heterozygote advantage E) All of these are forms of genetic drift. Answer: D
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13) Genetic drift may lead to A) loss of genetic variation in a population. B) a decrease in the mutation rate of a particular gene. C) increased heterozygosity at many loci. D) Hardy‐Weinberg equilibrium. E) All of these Answer: A
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The following scenario applies to the questions below. For a gene A, a geneticist studying a population of butterflies found the following genotypes: 108 AA butterflies 144 Aa butterflies 48 aa butterflies 14) What is the observed genotypic frequency of Aa individuals? A) 0.144 B) 0.72 C) 0.40 D) 0.48 E) 1.44 Answer: D
Skill: Problem-solving
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15) What is the frequency of the a allele in the butterfly population? A) 0.16 B) 0.04 C) 0.40 D) 0.64 E) 0.40 Answer: C
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16) Synonymous mutations are found much more frequently than nonsynonymous mutations because A) mutations do not occur randomly B) the genetic code is redundant. C) of genetic drift. D) there is a selective advantage in looking like another species. E) changes in the amino acid sequence of proteins are likely to be harmful. Answer: E
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17) ________ is the ultimate source of genetic variation in populations. A) Natural selection B) Genetic drift C) Gene flow D) Mutation E) Recombination Answer: D
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18) In a population of frogs, RR homozygotes produce an average of 5 offspring that survive to reproduce. Rr frogs produce 4, and rr frogs produce 1. The fitness of the three genotypes w11, w12, and w22 are respectively A) 5, 4, and 1. B) 0.5, 0.4, and 0.1. C) 1.0., 0.8, and 0.2. D) 0.5, 1.0, and 0.1. E) None of these Answer: C
Skill: Problem-solving
19) In reciprocal hybrid crosses, the heterogametic sex is typically sterile while the homogametic sex is not. This phenomenon is referred to as A) hybrid incompatibility. B) postzygotic isolation. C) Haldaneʹs rule. D) male sterility. E) None of these Answer: C
Skill: Factual recall
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Chapter 21 Population Genetics 185
20) A population genetic study of a certain insect revealed four alleles for the GPI (glucose‐phosphate isomerase) locus. How many different genotypic classes should be found? A) 4 B) 8 C) 10 D) 16 E) 20 Answer: C
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21) A sizable population was analyzed for SNPs in a certain DNA region. Four possible variants were found, with the following frequencies: SNP 1: GGTCTAGGA; frequency = 0.91 SNP 2: GGTGTAGGA; frequency = 0.03 SNP 3: GGTATAGGA; frequency = 0.03 SNP 4: GGTTTAGGA; frequency = 0.03 If the population is in Hardy‐Weinberg equilibrium, what is the total frequency of heterozygotes at this SNP locus? A) 0.828 B) 0.009 C) 0.170 D) 0.055 E) 0.0018 Answer: C
Skill: Problem-solving
22) Heterozygotes for the sickle‐cell allele show higher fitness than either homozygote genotype in malarial regions. This is an example of A) epistasis. B) pleiotropy. C) founder effect. D) overdominance. E) codominance. Answer: D
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23) For which of the following traits is human mating expected to be nonrandom? A) Blood type B) Isocitrate dehydrogenase enzyme C) Skin color D) Presence or absence of a widowʹs peak E) Fingerprint loci Answer: C
Skill: Factual recall
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24) Twenty loci are screened for genetic variation in a common caterpillar species. Four loci are found to have two or more alleles. The proportion of polymorphic loci for this population is thus A) 0.10. B) 0.20. C) 0.30. D) 0.40. E) 0.50. Answer: B
Skill: Problem-solving
25) Although bees carry pollen from tulip flowers to the blossoms on cherry trees, hybrid seeds are not produced because of A) gametic isolation. B) temporal isolation. C) ecological isolation. D) mechanical isolation. E) All of these Answer: A
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TRUE-FALSE QUESTIONS 26) For two populations to remain genetically homogenized, a significant proportion of the population of each must be exchanged at least every other generation. Answer: FALSE Explanation: Remarkably little genetic exchange is necessary to keep two populations from diverging in allele frequencies by genetic drift. Theory suggests that as little as one migrant every other generation is sufficient to maintain similar allele frequencies.
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27) When the population mean for a given measured trait is observed to decline over the course of several generations, it is an example of disruptive selection. Answer: FALSE Explanation: A positive or negative change in the population mean value-i.e., in one or the other direction-is defined as directional selection.
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28) The Hardy‐Weinberg relationship cannot be used to compute allele frequencies when one or more alleles are recessive. Answer: FALSE Explanation: The mode of expression of alleles has no bearing on whether the Hardy‐Weinberg relationship can be used to compute their frequency.
Skill: Factual recall
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Chapter 21 Population Genetics 187
29) Two closely related insect species are active at different times of day, with one diurnal and the other nocturnal. This is an example of spatial isolation. Answer: FALSE Explanation: This is an example of temporal isolation.
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30) A selection coefficient of 1 means zero fitness. Answer: TRUE
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31) Survival is the single most important process in evolution by natural selection. Answer: FALSE Explanation: Reproduction is the primary evolutionary currency. Survival clearly has a bearing on this, but individuals that survive will have zero evolutionary fitness if they do not reproduce.
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32) According to the Hardy‐Weinberg principle, at equilibrium the allele frequencies are dependent on the genotypic frequencies. Answer: FALSE Explanation: According to the Hardy‐Weinberg principle, the genotypic frequencies of a trait follow from the frequencies of the alleles for that trait.
Skill: Factual recall
33) The frequency of a recessive X‐linked allele is equal to the phenotypic frequency of males exhibiting that allele. Answer: TRUE
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34) The genetic variation between human ʺraces,ʺ such as Africans and Europeans, is greater than the variation within a single race. Answer: FALSE Explanation: There is less variation between the human ʺracesʺ than within a single race, showing that our concept of race has little genetic validity.
Skill: Factual recall
35) A gene is known to have three alleles: p, q, and r. The frequency of p is estimated at 0.70. Thus, the frequencies or q and r must be 0.15 each. Answer: FALSE Explanation: The three frequencies can assume any value as long as they all sum to 1.0. Moreover, for a three‐allele gene, knowing the frequency of just one allele is not sufficient to determine the frequencies of the other two alleles.
Skill: Factual recall
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188 Test Bank for iGenetics
SHORT ANSWER QUESTIONS AND PROBLEMS 36) In light of the fact that the assumptions of the Hardy‐Weinberg principle are unrealistic (no mutation, no selection, etc.), how can the principle be useful to population geneticists? Answer: In one respect, the Hardy‐Weinberg principle is a null hypothesis: It provides insight into expected allele and genotypic frequencies, and when real populations are observed to deviate substantially from Hardy‐Weinberg expectations, the reasons can be investigated in the form of alternative hypotheses. Also, it turns out that real populations are often robust in deviations from Hardy‐Weinberg assumptions, making the principle useful in estimating the frequencies of alleles and genotypes in populations.
Skill: Conceptual understanding
37) The deciduous forest of eastern North America has experienced dramatic changes over the past two to three centuries, transitioning from a nearly continuous forested area to increasingly patchy areas of forest broken up by farming and development. Some species are more sensitive to severe habitat fragmentation than others. What are some characteristics that might make a species more susceptible to the effects of habitat and population fragmentation? Answer: Habitat fragmentation leads to ʺislandsʺ of habitat, introducing the potential for inbreeding and genetic drift. More sensitive species might be those that, for whatever reason, are not highly mobile, and thus tend to become marooned in subpopulations that may experience drift effects that carry them below the minimum viable population size. Highly mobile species–certain birds, for example–would be better able to maintain gene flow between the habitat islands. A related characteristic of sensitive species might be dependence on a specialized resource that is itself increasingly scarce as a result of habitat fragmentation. Generalists may have an easier time finding the resources they need in whatever habitat island they find themselves.
Skill: Conceptual understanding
38) Observed genotypic frequencies in populations rarely match Hardy‐Weinberg expectations exactly. Why is this so, and how do scientists determine if the frequencies they observe depart from the Hardy‐Weinberg equilibrium to a biologically meaningful degree? Answer: There will always be chance deviations in observed data owing to sampling error or the random element of the mating process. Statistical testing is used to compute the probability of observing a given level of discrepancy between observed and expected values. The chi‐square test is most often applied to these kinds of genetic data.
Skill: Conceptual understanding
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Chapter 21 Population Genetics 189
39) With respect to a gene with two alleles, for each generation of complete inbreeding the proportion of heterozygotes is expected to be reduced by while the proportion of each homozygous class is expected to increase in frequency by . Explain why this is so. Answer: The three genotypes for our two‐allele gene are AA, aa, and Aa. Complete inbreeding for AA and aa will yield the same genotypes. Each cross of Aa × Aa, however, will yield AA, Aa, and aa in a 1:2:1 ratio [ : : ]. Thus, the heterozygotes are reduced by
1 1 , while each homozygous class increases by . 2 4 1 1 1 4 2 4 1 4 1 2
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40) Populations that suffer significant reductions in number may experience two population-genetic consequences that together can hasten their decline. Which two processes are likely to act in concert in small populations, and what is their effect? Answer: The two processes are inbreeding and genetic drift. As populations decrease in size, inbreeding is expected to increase. This in turn may further accelerate population decline by increasing the expression of deleterious recessive alleles, decreasing survivorship or fecundity. Genetic drift leads to further loss of genetic variation, accelerating inbreeding.
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41) Show algebraically that under Hardy‐Weinberg equilibrium, allele frequencies will not change from one generation to another. Answer: The demonstration is simple. Following allele p, start with the Hardy‐Weinberg genotypic frequencies p 2 + 2pq + q2 . The proportion of genotypes contributing p to the next generation is: (p 2 + pq) / (p 2 + 2pq + q2 ). p is transmitted via genotype p 2 and one‐half the frequency of the 2pq individuals, or p 2 + pq. Rewriting p 2 as p(1 - q) enables us reduce this to simply p as follows: p 2 + pq = p(1 - q) + pq = p - pq + pq = p. This means that the allele is transmitted to the next generation at frequency p, unchanged.
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42) Two neighboring Caribbean islands harbor iguana populations. One island (island A) has a total iguana population of 800, while the other (island B) has a population less than half that size, numbering 250. On island A, the frequency of an allele p is estimated at 0.75, while this same allele is present at a frequency of 0.90 on island B. During a hurricane tracking across the islands, a group of about 100 iguanas get rafted from island A to island B. What effect does this have on the frequency of allele p on island B? Answer: The migrant subpopulation is 100/800, or 0.125. This can be taken as m, the migration coefficient. The frequency of p in island A is 0.75; call this p A . p B is given as 0.9. Using the relationship Δp = m(p A - p B), Δp = 0.125(0.75 - 0.9) = 0.125(-0.15) = -0.019. The frequency of p on island B thus declines by 0.019, to 0.9 - 0.019 = 0.881.
Skill: Problem-solving
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190 Test Bank for iGenetics
43) Red‐green color blindness is an X‐linked trait. In a population genetics study, 1,000 people (500 men and 500 women) were tested for this trait, and 35 men were found to be color blind. Use this information to compute the frequency of the allele for color blindness and the wild‐type allele in this population, and estimate the expected number of carrier females. Answer: 35/1,000 = 0.035 = q; this is the frequency of the trait in this population. Since p = (1 q) = (1 - 0.035) = 0.965, you can use p and q to compute 2pq at 2(0.035)(0.965) = 0.068; this is the frequency of heterozygotes, which can only be female. This gives an estimated 33 female carriers in this population.
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44) The northern elephant seal, Mirounga angustirostris, suffered a significant population bottleneck in the late nineteenth century, when hunting reduced their population size to as few as 20 individuals. It has since rebounded to several tens of thousands. The related southern elephant seal, M. leonine, was not hunted as intensively. What prediction can you make about the relative levels of heterozygosity in populations of these two species? Answer: Much genetic variation would have been lost in the northern elephant seal population as a result of such a severe bottleneck, so heterozygosity will be very low. The southern elephant seal, in contrast, should still have significant genetic variation and correspondingly higher heterozygosity.
Skill: Conceptual understanding
45) Two populations experience equally severe bottlenecks, reducing each to one‐tenth of its original size. One population is in the bottleneck for one generation, and the other is in the bottleneck for five generations. You assess the heterozygosity of both populations after they have been restored to their prebottleneck size. What do you expect to find, and why? Answer: The population that was bottlenecked for five generations is likely to have lost much more genetic variation than the one bottlenecked for only one generation. This is because each generation in a bottleneck state is equivalent to a separate bout of genetic drift, losing variants by chance each time.
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46) In the early twentieth century, evolutionary biologists argued whether there was enough genetic variation in populations to explain the diversity of life by natural selection. With the advent of protein electrophoresis and DNA‐level analysis, the problem turned on its head: the neutralist school arose from the idea that there was too much genetic variation to be maintained by natural selection. What are some of the arguments and observations neutralists cited to support the idea that much variation is neither favored nor disfavored by natural selection? Answer: More genetic variation is expected in noncoding regions, such as intergenic areas and introns, and has indeed been found to be the case. Also, owing to redundancy in the genetic code, the third codon position can often be variable and still specify the same amino acid. More variation might therefore be expected at third base‐pair positions in codons, which has also been found to be the case.
Skill: Conceptual understanding
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Chapter 21 Population Genetics 191
47) Why do we expect that postzygotic isolation will evolutionarily precede prezygotic isolation in the speciation process? Answer: Postzygotic isolation is fitness‐reducing in that mating effort is wasted on offspring that have low survivorship or fecundity. Natural selection should act to reduce this waste by favoring, for example, mate recognition signals or spatial or temporal partitioning that prevent such matings in the first place.
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48) Some recessive mutations can be exceedingly debilitating or lethal when expressed in homozygotes. If their effects are so severe, why doesnʹt natural selection simply purge such alleles from the population completely? Answer: Natural selection does act on these alleles when they are expressed, but for the most part, when alleles are rare they occur in heterozygotes and are thus masked. Another reason selection cannot eliminate such alleles completely is that they arise continually by mutation. Finally, any recessive alleles that are expressed late in life will be shielded from natural selection, since they will not affect an individualʹs chances of producing viable offspring.
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49) This chapter discussed speciation by genetic divergence following geographic isolation, which is expected to lead to reduced gene flow, a process known as allopatric speciation. A more controversial form of speciation is the genetic divergence of populations without physical isolation, a process known as sympatric speciation. Can you envision a mechanism or process that would permit two coexisting populations of the same species to begin to diverge without being isolated from one another? Answer: Sympatric speciation is thought to be uncommon but almost certainly has occurred. In one general scenario, a form of isolation ʺin place,ʺ or in sympatry, may occur if certain males or females begin (by drift, perhaps) to prefer different resources–say, a special site for mating or laying eggs. If their mates follow in this ʺimprintingʺ process, the two preference populations may begin to experience reduced gene flow between them, which if maintained long enough may bring about a degree of genetic divergence that would be selectively reinforced should some of the individuals from the two populations ever hybridize.
Skill: Analytical reasoning
50) Remote oceanic islands are characterized as having disharmonic biota, meaning the number and relative proportions of taxa living on the island differ significantly from the number and relative proportions of taxa on the nearest continental mainland area. Each island system is disharmonic in its own unique way. Why? Answer: This pattern stems from the random nature of the ancestral colonization of those islands. Successfully colonizing a remote island is rare, but in the vastness of time such events occur regularly. Precisely what groups make it is a matter of chance, but some have a far greater likelihood than others. Flying animals, for example, are more likely colonists than large or sensitive terrestrial animals (and in fact no native mammals or amphibians are found on the remotest islands). Since the colonization process is random, each island system experiences a unique colonization history. Thus, each will have its own complement of taxa from continental areas, which, depending on the timing of ancestral colonization and the available niches, will spread and diversify to different degrees on each.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics
MATCHING QUESTIONS Please select the best match for each term. 1) VG/V P
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A) Narrow‐sense heritability B) Genetic variance C) Determines whether differences in means are significant D) Variance E) Broad‐sense heritability
2) V A /V P
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3) V A + V D + V I
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4) s2
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5) ANOVA
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Answers:
1) E
2) A
3) B
4) D
5) C
MULTIPLE-CHOICE QUESTIONS 6) Which of the following is a source of genetic variation in a population caused by epistatic interactions between genes? A) Additive genetic variance B) Dominance variance C) General environmental effects D) Interaction variance E) Maternal effect Answer: D
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7) Quantitative traits typically have multiple genetic influences in which alleles of a number of genes contribute individual amounts to the final phenotype without classical dominance or epistasis. The term used for the observed variation that results from this is A) V A . B) V D . C) V E. D) V G. E) V I. Answer: A
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192
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Chapter 22 Quantitative Genetics 193
8) A female cat that is small due to nutritional deficiencies experienced in early life will give birth to smaller kittens than a genetically identical cat that is larger in size due to better nutrition in early life, even if both female cats were bred to the same male cat and both had good nutrition during pregnancy. The term that is used in measuring this type of variation is A) V P . B) V E. C) V EG. D) V ES. E) V EM. Answer: E
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9) The distribution of continuous traits can best be described as A) linear. B) bell‐shaped. C) hyperbolic. D) logarithmic. E) bimodal. Answer: B
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10) Ornamental hydrangeas are well known to produce either blue or magenta flowers, depending on soil pH. This may be a good example of a discrete trait influenced by A) genetics. B) environment. C) multiple genes. D) A and B only E) A, B, and C Answer: B
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11) Which of the following statements about heritability is true? A) Broad‐sense heritability defines the complete genetic basis of a trait. B) Heritability is fixed for a given trait. C) Traits shared by members of the same family do not necessarily have high heritability. D) Heritability is the measure of the proportion of an individualʹs phenotype that is due to genetics. E) If a population displays little variation in a given trait, then that trait has low heritability Answer: C
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194 Test Bank for iGenetics
12) When graphing a distribution of phenotypes for a continuous trait, a broad curve implies A) little variation in the phenotypes. B) a large standard deviation. C) a small variance. D) a large correlation coefficient. E) a zero regression line slope. Answer: B
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13) The proportion of phenotypic variance that results from additive genetic variance is known as ________ heritability. A) broad‐sense B) narrow‐sense C) multifactorial D) generic E) genetic Answer: B
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14) A broad‐sense heritability of 0.8 for a particular trait indicates that A) phenotypic variation stems from environmental influences. B) the trait is 80% heritable. C) genes and environment play equal roles in shaping the trait. D) genetic differences among individuals account for much of the phenotypic variation for that trait. E) None of these Answer: D
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15) A Mendelian cross for a three‐gene trait is expected to yield ________ F 2 genotypes. A) 4 B) 16 C) 27 D) 64 E) 265 Answer: C
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16) In parent‐offspring regression, the average value of the parents is termed the A) mean parental value. B) parental average. C) midparent value. D) median parental value. E) modal parental value. Answer: C
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Chapter 22 Quantitative Genetics 195
17) A ________ trait is typically influenced by one or two genes. A) quantitative B) continuous C) polygenic D) multifactorial E) discrete Answer: E
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18) In maize, some alleles at the seed coat locus influence our ability to visually determine the endosperm color phenotype. Specifically, an opaque seed coat prevents us from seeing the endosperm within, while we can readily see the endosperm in seeds with transparent coats. The opaque coat color allele thus appears to mask the expression of endosperm color alleles, providing an example of A) codominance. B) pleiotropy. C) polygenic inheritance. D) epistasis. E) maternal effect. Answer: D
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19) In the general regression equation y = mx + b, variables m and b represent ________ and ________, respectively. A) the x‐intercept, the y‐intercept B) the regression line slope, the y‐intercept C) the y‐intercept, the regression line slope D) the y‐intercept, the x‐intercept E) narrow‐sense heritability, broad‐sense heritability Answer: B
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20) Assume that the length of a type of cucumber at maturity is controlled by three genes ( A, B, and C), each of which has two alleles. The A, B, and C alleles each add 2 inches of cucumber growth, while the a, b, and c alleles add only 1 inch. A true‐breeding 6‐inch strain is bred to a true‐breeding 12‐inch strain. All the F 1 s produce 9‐inch cucumbers. What sizes would you expect to see in the mature F 2 s? A) Discrete lengths of 6, 9, and 12 inches B) Discrete lengths of 6 and 12 inches C) Discrete lengths of 9 inches D) Discrete lengths of 6, 7, 8, 9, 10, 11, and 12 inches E) Lengths ranging continuously from 6 to 12 inches Answer: D
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21) Analysis of specific genetic contributions to quantitative traits has been done by A) altering the environment of individuals and observing phenotype alterations. B) completing sequencing of specific genes for various species. C) measuring V G by adding V A , V D , and V I. D) conducting ANOVA on genetic markers and phenotypes to identify linkage to QTLs. E) measuring specific environmental effects, such as sunlight, on phenotype. Answer: D
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22) Which of the following problems might arise in a statistical study as a result of a population sample of inadequate size? A) A non‐representative sample B) Sampling error C) A non‐random sample D) A sample drawn from a biased subset of data points E) All of these Answer: E
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23) Heritability can only be measured for A) traits with phenotypic variance. B) traits with genotypic variance. C) discrete traits. D) polygenic traits. E) None of these Answer: A
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24) Populations of the snail Cepaea nemoralis exhibit three color morphs, a good example of a ________ trait. A) continuous B) discontinuous C) quantitative D) polygenic E) B, C, and D only Answer: E
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25) A quantitative trait influenced by four loci, each with two alleles, would be expected to show ________ possible genotypes. A) 81 B) 16 C) 8 D) 256 E) None of these Answer: A
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Chapter 22 Quantitative Genetics 197
TRUE-FALSE QUESTIONS 26) If a populations shows very little phenotypic variance for a trait, then that trait is probably controlled principally by the environment rather than genetics. Answer: FALSE Explanation: A population could be homozygous for alleles that control a trait, or have genetic variations that do not appreciably affect phenotype for that trait, yet the trait itself might still be influenced entirely by genes. As an example, the human trait of possessing two arms and two legs is determined genetically and is quite constant.
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27) Narrow‐sense heritability is a more useful measure than broad‐sense heritability in agricultural selective breeding. Answer: TRUE
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28) The mean value of a quantitative trait in F 1 offspring is generally smaller than either of the values of the parentals. Answer: FALSE Explanation: The F1 value will be intermediate between the two parentals.
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29) Narrow‐sense heritability for a trait can be obtained from a regression of midparent value on offspring value. Answer: TRUE
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30) If members of the same family display the same phenotypic traits, these traits have high heritability. Answer: FALSE Explanation: Even if the phenotypic traits of members of the same family are correlated, they are not necessarily influenced by genetics. The common familial environment may lead to any such correlation.
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31) Continuous traits are often influenced by multiple simple (Mendelian) genes. Answer: TRUE
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32) Evolution is defined as genetic change in populations over time. Answer: TRUE
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33) A gene that influences several traits is termed pleiotropic. Answer: TRUE
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34) As the number of genes influencing a quantitative trait increases, the range of phenotypic variation is expected to increase. Answer: FALSE Explanation: The range of phenotypic variation may have little to do with the number of genes, although the number of observable phenotypes is expected to increase.
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35) In descriptive statistics, the median is defined as the value exactly in the center of a distribution, with half the values lying above and half lying below. Answer: TRUE
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SHORT ANSWER QUESTIONS AND PROBLEMS 36) Continuously varying traits typically show a bell‐shaped frequency distribution. Explain why this is the case, and why intermediate phenotypes are more frequent than phenotypes at either end of the distribution. Answer: The greater frequency of intermediate phenotypes is a consequence of a greater frequency of intermediate genotypes stemming from allelic combinatorials. Even a single two-allele gene possesses twice as many ways to make a heterozygote ( Aa and aA), the number of which increase exponentially as the number of genes increase.
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37) Distinguish between polygenic and multifactorial traits. Answer: Polygenic traits are quantitative traits, shaped by multiple genes. Multifactorial traits are shaped both by multiple genes and environmental factors.
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38) Explain why narrow‐sense heritability is a more accurate indicator of responsiveness to selection than is broad‐sense heritability. Answer: Narrow‐sense heritability is defined in terms of additive genetic variance, meaning that each allele of each contributing gene has a phenotypic effect. Broad-sense heritability is based on genetic variance in general. Since selective breeding is based on phenotypes, insofar as additive genetic variance is a more precise reflection of genotype underlying phenotype, this measure will help in making better selective breeding decisions.
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39) The correlation coefficient is a measure of the strength of association between two traits. Correlation, however, does not indicate causation. Think of one or two variables (not necessarily phenotypes) that are correlated and are causally linked, and one or two variables that may be correlated but are not likely to be causally linked. Answer: Traits like possession of a Y chromosome and possession of testes or a heavy beard are causally linked, owing to the testis‐determining factor gene on the Y chromosome. Blonde hair and blue eyes are often associated but are not necessarily causally linked.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics 199
40) Define, in your own words, the biological meaning of heritability. Answer: Heritability is often imprecisely defined as a measure of the genetic component of a trait. More precisely, it is a statistical measure that quantifies the influence of genes on phenotypic variance with respect to a given trait.
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41) A botanist studying a plant with geographically separated and morphologically distinct populations wants to know the degree to which the differences between the populations are attributable to genetic factors. Outline a simple experiment she can perform to address this question. Answer: Two classic experimental approaches would be useful here. The first is reciprocal transplant, which involves switching the environments of the plants, for example, by creating test plots in each area and planting seeds or seedlings collected from the other site. The second would be to create a common garden, in which seeds or seedlings of plants from each population are reared under identical conditions, for example, side by side in a greenhouse. Comparison of key morphological traits across the populations will provide a way to estimate heritability of the traits.
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42) If you measured the narrow‐sense heritability of a crop plant trait at 0.25, what would that value mean in terms of the ease of selecting the trait in a selective breeding program? Answer: This heritability value is rather low but does indicate that the observed phenotypic variation is influenced by genetics. The trait is selectable, albeit perhaps slowly.
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43) A regression analysis of two traits yields a slope of 0. Explain the biological meaning of this value. Answer: Phenotypic variation in the traits is not correlated.
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44) The broad‐sense heritability of a trait is estimated at 0.8. If the total phenotypic variance is estimated at 32.20, what is the genetic variance? Answer: V G = 25.76; broad‐sense heritability = V G / V P
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45) Human fingerprints are a good example of a multifactorial trait. What evidence can you cite to suggest that fingerprints are partially influenced by genes? How about evidence suggesting that they are partially influenced by environmental factors? Answer: Observations suggesting there is a genetic component to fingerprint patterns include consistent population differences in the frequency of basic patterns (whorls, arches, loops). However, even identical twins have unique fingerprints, indicating that they are also determined by environmental factors.
Skill: Analytical reasoning
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200 Test Bank for iGenetics
46) In a controlled cross for a quantitative trait, 600 F 2 offspring are obtained, 10 of which exhibit the phenotype of one of the parentals. How many genes are estimated to control this trait based on these results? Answer: 1/(4 n) gives the fraction of F2 offspring resembling either parental for a polygenic trait in which n is the number of genes involved. 10/600 is approximately corresponding with a value of n = 3 genes.
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1 64
,
47) In Mendelʹs day, ʺparticulateʺ inheritance would have made less sense to people than blending inheritance. Why? Answer: At the time Mendel performed his experiments, it was unusual to focus on specific traits. If one looks at an entire organism, it is easy to see how offspring could appear to have been produced by blending the traits of each parent, which was the standard assumption among Mendelʹs contemporaries. Mendelʹs insight came from focusing on the inheritance of specific traits, leading him to uncover patterns of mathematical regularity in their occurrence in his controlled breeding experiments.
Skill: Analytical reasoning
48) The average adult height in the U.S. population has increased by several centimeters over the past two generations. Is this likely to reflect a change in the genetic makeup of the population, a change in environmental factors, or both? Explain. Answer: There has been no directional change in the genetic makeup of the U.S. population in this time interval. Rather, environmental factors–nutritional improvement, in particular-underlie this trend.
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49) Identical (monozygotic) and fraternal (dizygotic) twins provide a natural experiment for investigating the influence of genetics on traits. Outline a study approach using twins, assuming you could identify enough pairs, in which you are examining the relative contribution of genetics on blood cholesterol level. Answer: If there is a genetic predisposition to achieving a certain blood cholesterol level, monozygotic twins might be expected to show a stronger statistical association in cholesterol level than dizygotic twins. You would need to control for environment, however: ideally, your twins should be given a controlled diet for a specified period.
Skill: Analytical reasoning
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Chapter 22 Quantitative Genetics 201
50) An isolated population of wild turkeys has adult males that weigh an average of 18 pounds and females that weigh an average of 10 pounds. Another isolated population living 100 miles north has a population where adult males weigh 13 pounds and females 8 pounds on average. Both populations have a very low phenotypic variance. How could you determine genetic contributions to the weight differences between the populations? Answer: The phenotypic variation (V P) is dependent on genotypic variation (V G) plus environmental variation (V E) and gene‐environment interactions (2COV G,E + V G ×
E). To isolate V G members of both populations could be captured and bred in
captivity under identical circumstances. After a few generations, environmental influences, including V M, should have equalized. If the separate populations have achieved identical V P then V G contributions are limited. If the separate populations maintain a significant size difference, the populations could be crossed. If the F 1 s are intermediate in weight and the F 2 s are average but have a much wider variance, then quantitative genetic factors are more important and narrow‐sense heritability can be measured as H N = V A /V P. If, instead, the F 1 s have one parental type and the F2 s are of that type and of the other parental type, a single gene with a dominant and recessive phenotype could cause the size difference. A number of other possibilities exist depending on one or two other genes that might have incompletely dominant or additive alleles.
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3 4 1 4 2
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Chapter 23 Molecular Evolution
MATCHING QUESTIONS Please select the best match for each term. 1) Transition
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A) A mutation that alters the encoded amino acid B) A purine-to-pyrimidine mutation C) Nonfunctional evolutionary remnant of a gene D) An A‐to‐G or C-to‐T mutation E) A mutation that does not alter the encoded amino acid.
2) Transversion
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3) Synonymous
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4) Nonsynonymous
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5) Pseudogene
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Answers:
1) D
2) B
3) E
4) A
5) C
MULTIPLE-CHOICE QUESTIONS 6) Homologous proteins A) share a common ancestor. B) share a common amino acid sequence. C) are determined by alleles on homologous chromosomes. D) A and B only E) A, B, and C Answer: A
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7) Synonymous changes that arise within coding sequences of genes A) generally have no effect on the phenotype. B) are usually detrimental to fitness. C) may be eliminated by natural selection. D) occur more frequently than nonsynonymous changes. E) B and C only Answer: A
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202
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Chapter 23 Molecular Evolution 203
8) Indels are A) index deletions used to identify a species. B) parsimonious trees. C) transitions and transversions combined. D) small deletions or insertions that produce gaps in sequence alignments. E) the result of codon usage bias. Answer: D
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9) Higher substitution rates are seen in ________ than in ________. A) coding regions, noncoding regions B) 5ʹ flanking regions, 3ʹ flanking regions C) first codon positions, second codon positions D) introns, exons E) charged amino acid codons, noncharged amino acid codons Answer: D
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10) Nonsynonymous substitutions A) do not lead to amino acid change. B) lead to amino acid change. C) refer to substitutions in unrelated lineages. D) affect only certain amino acids. E) lead to transversions. Answer: B
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11) What is the number of possible unrooted trees for 5 taxa? A) 5 B) 10 C) 15 D) 20 E) 25 Answer: C
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12) Which of the following statements about molecular evolution is true? A) Homologous genes often evolve in the same way. B) Different regions of the same gene evolve at different rates. C) Genes in the same organism evolve at about the same rate. D) Genes in different organisms evolve at about the same rate. E) Nucleotide substitutions are nonrandom. Answer: B
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204 Test Bank for iGenetics
13) Which of the following does not provide molecular evidence for evolution? A) The 1,000‐fold differences in nonsynonymous substitutions between some genes B) The existence of pseudogenes C) The fact that high‐expression genes use amino acids that are less energetically costly than those used by low‐expression genes D) The fact that there is a lower rate of nucleotide substitutions in coding regions than in noncoding regions of genes E) All these are molecular evidence for evolution Answer: E
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14) With regards to the sequence alignment
G T C C A T C G G G T C G A T C — G,
which of the following statements is not true? A) There has been a base deletion. B) There has been a base substitution. C) There has been a gene conversion. D) There has been a transversion. E) There is evidence of an indel. Answer: C
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15) Which of the following mutations is considered a synonymous change? A) ACC to AUG B) CUU to AUU C) CUU to CUA D) CUU to CUA E) UUU to UUA Answer: D
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16) Codon bias suggests that A) some codons evolved earlier than others. B) not all nonsynonymous substitutions are neutral. C) some codons are selectively favored over others. D) small fitness effects can make a large difference over time. E) some tRNAs are more abundant than others. Answer: C
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Chapter 23 Molecular Evolution 205
17) From the standpoint of evolutionary fitness, it is expected that genes with a high rate of expression will make use of amino acids that are ________ than those amino acids used by genes with a low rate of expression. A) more common B) more energetically economical C) more energetically costly D) rarer E) None of these Answer: B
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18) Which of the following types of proteins evolves at the slowest rate? A) Peptide hormones B) Collagen C) Insulin D) Histones E) Lipoprotein Answer: D
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19) Which of the following factors does not interfere with the interpretation of evolutionary time using the molecular clock? A) Horizontal gene transfer B) Differences in generation time C) Mutations due to background cosmic radiation D) Different selection pressures on different taxonomic groups E) Differences in DNA repair efficiency in different taxonomic groups Answer: C
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20) The greatest genetic differences found in humans are A) found between individuals from different continents. B) found between individuals of different racial groups. C) found between island populations and mainland populations. D) found between individuals of African descent. E) found between individuals of European descent. Answer: D
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21) The average synonymous substitution rate in mammalian mitochondrial genes is ________ the average rate for nuclear genes. A) 2 times B) 5 times C) 10 times D) 50 times E) 100 times Answer: C
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206 Test Bank for iGenetics
22) In the unweighted pair group method (UPGMA) of determining evolutionary relationships, A) each set of taxa is connected by the shortest genetic distance. B) a constant rate of evolution across all lineages is assumed. C) a distance matrix is first constructed. D) A and B only E) A, B, and C Answer: E
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23) Reconstructions of the evolutionary relationships of a set of taxa are called A) classifications. B) phylogenetic trees. C) UPGMA networks. D) networks. E) evolutionary scenarios. Answer: B
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24) The differences in evolutionary rates of change revealed by relative rate tests may be attributable to A) differential strength of selection. B) generation time. C) average DNA repair efficiency. D) differential exposure to mutagens. E) All of these Answer: E
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25) _______ tend(s) to occur more frequently than ________. A) Synonymous substitutions, nonsynonymous substitutions B) Genetic divergence, genetic convergence C) Transitions, transversions D) A and B only E) A, B, and C Answer: E
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TRUE-FALSE QUESTIONS 26) The noncoding 5ʹ flanking region of genes is more constrained by selection than the noncoding 3ʹ flanking region. Answer: TRUE
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27) Many viruses are helped along in generating antibody‐evading capsid protein variations by replication error. Answer: TRUE
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Chapter 23 Molecular Evolution 207
28) Pseudogenes tend to have lower average substitution rates than functional genes. Answer: FALSE Explanation: Pseudogenes are nonfunctional and are therefore not subject to natural selection. Substitutions thus accumulate more rapidly than in functional genes.
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29) The counterpart to the use of mtDNA to reconstruct maternal lineages is the use of chloroplast DNA to reconstruct paternal lineages. Answer: FALSE Explanation: The patrilineally inherited Y chromosome is the male counterpart to mtDNA.
Skill: Factual recall
30) High molecular similarity always reflects evolutionary relatedness. Answer: FALSE Explanation: Molecular data are less susceptible to convergence than certain morphological traits, but it is possible that a given pair of similar sequences (particularly coding sequences) from unrelated groups may have converged owing to natural selection. Horizontal (or lateral) transfer also occurs.
Skill: Factual recall
31) Maximum confidence is a procedure for reconstructing phylogenetic trees invoking the fewest number of steps or mutations. Answer: FALSE Explanation: The procedure invoking the fewest steps or mutations is maximum parsimony because the shortest tree is also the most parsimonious in requiring the fewest evolutionary events to obtain it.
Skill: Factual recall
32) Nuclear autosomal DNA sequence is more abundant than mtDNA and is therefore preferable for use in reconstructing events like domestication or human migration. Answer: FALSE Explanation: Recombination and slower overall base substitution rates make nuclear DNA less useful for such purposes than mtDNA.
Skill: Factual recall
33) The fact that approximately 80% of the six codons that encode the amino acid leucine in yeast are UUG is an example of codon bias. Answer: TRUE
Skill: Factual recall
34) Genetic distance refers to the proportion of dissimilar nucleotides between two homologous sequences. Answer: TRUE
Skill: Factual recall
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208 Test Bank for iGenetics
35) Mitochondria and chloroplasts have genomes that have evolved from the nuclear genomes in eukaryotes. Answer: FALSE Explanation: Organellar genomes have many similarities to bacterial genomes and are thought to have arisen from the endosymbiosis of bacteria by the ancestors of modern eukaryotes.
Skill: Factual recall
SHORT ANSWER QUESTIONS AND PROBLEMS 36) What is the rationale behind the bootstrap procedure, in which trees are repeatedly reconstructed following resampling with replacement of the original data? Answer: Phylogenetic bootstrapping gives an estimate of the robustness of each node in an inferred tree by seeing to what extent the node is affected by random resampling of the data. A very weakly supported node would be more likely to ʺcollapseʺ with resampling (showing that not much of the ʺsignalʺ in the original data supported it), while a strongly supported node will be reconstructed time and again with resampling because it is based on support from a wider array of informative sites.
Skill: Analytical reasoning
37) Given the following aligned sequences Sequence 1 Sequence 2 Sequence 3 Sequence 4
ATA ATA ATC ATC
TGT TGT TGT TGG
ATA TTA ATA ATA
AAA AAG AAA AAG,
identify any transitions and transversions as well as any synonymous and nonsynonymous substitutions. Answer: First codon: A to C: Second codon: T to G: Third codon: A to T: Fourth codon: A to G: transversion, synonymous substitution (Ile - Ile) transversion, nonsynonymous substitution (Cys - Trp) transversion, nonsynonymous substitution (Ile - Leu) transition, synonymous substitution (Lys - Lys)
Skill: Problem-solving
38) What are the major weaknesses of the parsimony and distance matrix methods of phylogeny reconstruction? Answer: The main problem with parsimony is that, by definition, it assumes the most parsimonious (shortest or simplest) evolutionary pathway, while it is almost certainly true that this is not invariably the pathway taken. Still, as a guiding principle, parsimony has much to recommend it. Distance methods make no such assumptions about how evolution should work, but on the other hand assume that simple unweighted similarity and dissimilarity reflect evolutionary relatedness and that evolutionary rate is constant across lineages. We know that this is not true, owing to selective constraints, convergence, etc.
Skill: Conceptual understanding
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Chapter 23 Molecular Evolution 209
39) The McDonald‐Kreitman test is an important tool in molecular evolutionary analysis. What is this test, and what kinds of insights can it provide? Answer: The test compares synonymous and nonsynonymous sites in gene-coding regions. If the nonsynonymous : synonymous substitution ratio within species is the same as that between species, then the substitutions are likely to be neutral. If the ratios differ, natural selection is likely to have been responsible by favoring certain substitutions over others. Insights the test provides include identifying genes on which natural selection may be acting for further study, and testing for neutrality for population genetic analyses that assume or require neutrality.
Skill: Factual recall
40) What is codon bias? If synonymous substitutions give the same amino acid, why should there be a bias in which codons encode that amino acid? Answer: Codon bias is the disproportionate use of certain synonymous codons over others. Some synonymous codons pair with different tRNAs that carry the same amino acid. One explanation for this concerns efficiency in translation: While a mutation to a synonymous codon does not change the amino acid, it may change the tRNA used by a ribosome during translation. Over evolutionary time, natural selection may favor use of codons corresponding with the most readily available tRNA.
Skill: Conceptual understanding
41) In most genes, the rate of synonymous mutations is significantly higher than that of nonsynonymous mutation. The MHC genes exhibit the reverse, having over twice as many nonsynonymous as synonymous mutations. Explain this phenomenon. Answer: Synonymous mutations do not change the amino acid composition of a gene product. Nonsynonymous mutations do. Most changes in amino acids in a protein could have a deleterious effect and would be selected against. The MHC relies on genetic diversity to optimize immune responses both for the individual and for the species. Therefore, nonsynonymous mutations could provide a selective advantage.
Skill: Factual recall
42) In the 1960s and 1970s, organisms were classified into five kingdoms, one of which was prokaryotes. In the 1980s, organisms were reorganized into three domains, with the prokaryotes split into Archaea and Bacteria, and all of the other four kingdoms lumped together into Eukarya. Why did this happen? Answer: In the original classification, prokaryotes were separated from all other organisms because of cell structure. With the advent of DNA sequencing, gene sequence comparisons became possible. Carl Woese, Norm Pace, and others compared sequences of conserved genes such as ribosomal RNAs and discovered that prokaryotes fall into two distinct molecular classes that are as different from each other as either is from the eukaryotes.
Skill: Factual recall
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210 Test Bank for iGenetics
43) In the following alignment of 12 base pairs, which sites are phylogenetically informative? Sequence 1 Sequence 2 Sequence 3 Sequence 4
ATTGGTATAAAA ATTAGTATAAAA ATTGGTATCAAA ATTAGTATCAAA
Answer: Only the fourth and ninth sites show variation, and are thus potentially informative sites.
Skill: Conceptual understanding
44) It is not unusual for a phylogeny reconstruction to result in several equally robust phylogenetic trees that differ in their branching patterns. How can this be the case if evolution only took one true pathway? Answer: Phylogeny reconstruction is in part a statistical procedure, and the reconstructions depend on the amount and nature of the input data and the assumptions inherent in the particular reconstruction method (maximum parsimony, maximum likelihood, etc.). Given these constraints, it is expected that phylogenetic analysis of a given data set will, at times, produce multiple, equally well‐supported trees.
Skill: Analytical reasoning
45) One problem commonly experienced in phylogeny reconstruction arises when two or more of the groups being compared have accumulated far more genetic changes (mutations) than the others in the comparison. This situation often results in a phenomenon known as ʺlong branch attraction,ʺ in which the highly mutated lineages end up grouping together in a phylogenetic analysis even if they are distantly related. What do you think causes long branch attraction? How might it be compensated for? Answer: Since nucleotides can change only among four states ( A,C,T, and G), in lineages with disproportionately many base substitutions the odds of becoming convergently similar (or at least obscuring the underlying pattern of homology with their true nearest relative) increases greatly. This problem may be compensated for by using data from several different genes or regions of the genome.
Skill: Conceptual understanding
46) Distinguish between a gene tree and a species tree. Answer: Trees based on a single homologous gene reflect the evolutionary history of that gene, but its pattern of divergence may not exactly match that of the species carrying that gene. Single‐gene phylogenies are gene trees, while several such genes can be collectively analyzed to infer a more accurate picture of species‐level phylogeny, or species tree.
Skill: Conceptual understanding
47) In a molecular‐evolutionary analysis, you estimate the number of base substitutions per site (K) at 0.82. Given an estimated divergence time ( T) of 10 million years based on the fossil record, what is the substitution rate (r)? Answer: Using the relationship r = K/(2T) to compute r, the rate is: 0.82/2(1 × 10 7 ) = 0.82 × 10 -8 .
Skill: Problem-solving
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Chapter 23 Molecular Evolution 211
48) One of the lines of evidence supporting the theory that modern humans originated in Africa is that the modern‐day human population of Africa contains the greatest diversity of genetic variants of any human population. Why is this consistent with the idea that humans first evolved in Africa? Answer: The reasoning is that populations arising in and anciently occupying an area will have had more time to accumulate genetic variants. Derivative populations migrating elsewhere would experience a founder effect, which would lead to loss of some variation.
Skill: Analytical reasoning
49) Humans have an extremely high level of genetic similarity, with greater than 99% sequence identity across populations. Yet, many people appear very different phenotypically. How do you reconcile these observations? Answer: These observations are very easily reconciled. The phenotypic differences are actually genetically trivial. Differences in body shape, skin color, eye color, body shape, and so on, undoubtedly result from variation in a scant few coding loci. In addition, much phenotypic variation is caused by environmental factors, such as nutrition. The human genome consists of some three billion nucleotides, so the few loci that underlie phenotypic differences constitute so trivial a subset of these that geneticists generally do not accept the idea of a genetic basis for ʺrace.ʺ
Skill: Analytical reasoning
50) Describe briefly the mechanisms that are thought to give rise to gene duplication and multigene families. Answer: Gene duplication, arising from unequal crossing‐over, can give rise to tandem gene copies. Some of these may then evolve different but related functions. Transposition can also create duplicate copies of genes, but these are less likely to occur in tandem, typically inserting elsewhere in the genome.
Skill: Factual recall
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