Chromosomal Rearrangements

Chapter 13—Chromosomal Rearrangements and Changes in Chromosome Number Reshape Eukaryote Genomes

Fill in the Blank

|1. |Events that reshape genomes by reorganizing the DNA sequences within one or more chromosomes are known as ____________________. |
|Ans: |rearrangements |
| |Difficulty: 2 |

|2. |____________________, duplications, inversions, translocations, and movements of transposable elements are all types of |
| |rearrangements of chromosomal material.
|
|Ans: |Deletions |
| |Difficulty: 2 |

|3. |Very large deletions are visible at the relatively low resolution of a __________________, showing up as the loss of one or more|
| |bands from a chromosome. |
|Ans: |karyotype |
| |Difficulty: 2 |

|4. |Changes in gene ____________________, the number of times a given gene is present in the cell nucleus, can create a genetic |
| |imbalance. |
|Ans: |dosage |
| |Difficulty: 3 |

|5. |An unpaired bulge of one member of a homologous pair of chromosomes during prophase of meiosis I is known as a |
| |____________________. |
|Ans: |deletion loop |
| |Difficulty: 3 |

|6. |A recessive mutation in the mouse that prevents homozygous animals from walking in a straight line is known as the |
| |____________________ gene. |
|Ans: |shaker-1 |
| |Difficulty: 3 |

|7. |When repeats of a region lie adjacent to each other they are called ____________________ duplications. |
|Ans: |tandem |
| |Difficulty: 2 |

|8. |Inversions that include the centromere are termed ____________________. |
|Ans: |pericentric |
| |Difficulty: 2 |

|9.
|Inversions that exclude the centromere are termed ____________________. |
|Ans: |paracentric |
| |Difficulty: 2 |

|10. |A crossover product that lacks a centromere is known as a(n) ____________________fragment. |
|Ans: |acentric |
| |Difficulty: 2 |

|11. |The type of large-scale mutation in which parts of two different chromosomes trade places is a ____________________
|
| |translocation.
|
|Ans: |reciprocal |
| |Difficulty: 2 |

|12. |____________________ is the enzyme that catalyzes transposition. |
|Ans: |Transposase |
| |Difficulty: 2 |

|13. |Organisms with four copies of a particular chromosome (2n+2) are ________________. |
|Ans: |tetrasomic |
| |Difficulty: 3
|

|14. |Down syndrome is also known as ____________________. |
|Ans: |trisomy 21 |
| |Difficulty: 2 |

|15. |If a tetraploid derives all of its chromosome sets from the same species, we call this kind of polyploid a(n) |
| |____________________. |
|Ans: |autopolyploid |
| |Difficulty: 3 |

Multiple Choice

|16. |Which of the following are considered chromosomal rearrangements? |
|A) |inversions |
|B) |duplications |
|C) |deletions |
|D) |translocations |
|E) |all of the above |
| |Ans: E |
| |Difficulty: 1 |

|17. |Which of the following removes material from the genome? |
|A) |inversions |
|B) |duplications |
|C) |deletions |
|D) |translocations |
|E) |none of the above |
| |Ans: C |
| |Difficulty: 1 |

|18. |Which of the following adds material to the genome? |
|A) |inversions |
|B) |duplications |
|C) |deletions |
|D) |translocations |
|E) |none of the above |
| |Ans: B |
| |Difficulty: 1 |

|19. |The type of chromosomal rearrangement which reorganizes the DNA sequence within one chromosome is known as a(n): |
|A) |inversion |
|B) |duplication |
|C) |deletion |
|D) |translocation |
|E) |none of the above |
| |Ans: A |
| |Difficulty: 2 |

|20. |In general, which of the following usually has a greater chance of lethality than the others? |
|A) |inversion |
|B) |duplication |
|C) |deletion |
|D) |translocation |
|E) |all have an equal chance |
| |Ans: C |
| |Difficulty: 2 |

|21. |Sometimes a piece of one chromosome attaches to another chromosome. This is known as a(n): |
|A) |inversion |
|B) |duplication |
|C) |deletion |
|D) |translocation |
|E) |none of the above |
| |Ans: D |
| |Difficulty: 1 |

|22. |Sometimes a part of the genome moves from chromosome to chromosome. This is known generally as a(n): |
|A) |inversion |
|B) |duplication |
|C) |deletion |
|D) |translocation |
|E) |transposable element |
| |Ans: E |
| |Difficulty: 2 |

|23. |Rearrangements and changes in chromosome number may affect gene activity or gene transmission by altering the |
| |________________________ of certain genes in a cell. |
|A) |position |
|B) |order |
|C) |number |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 2 |

|24. |Karyotypes generally remain constant within a species because: |
|A) |rearrangements occur frequently. |
|B) |changes in chromosome number occur infrequently. |
|C) |genetic instabilities produced by genomic changes usually are at a selective disadvantage. |
|D) |genetic imbalances are often at a selective advantage. |
| |Ans: C |
| |Difficulty: 2 |

|25. |Despite selection against chromosomal variations: |
|A) |related species almost always have the same karyotype. |
|B) |related species almost always have a different karyotype. |
|C) |closely related species diverge by many chromosomal rearrangements. |
|D) |distantly related species diverge by only a few chromosomal rearrangements. |
| |Ans: B |
| |Difficulty: 2 |

|26. |In higher organisms, using genetic analysis is usually difficult to distinguish small deletions in one gene from: |
|A) |heterozygotes. |
|B) |small duplications. |
|C) |monosomies. |
|D) |point mutations. |
| |Ans: D |
| |Difficulty: 1 |

|27. |For an organism to survive a deletion of more than a few genes, it must carry a nondeleted homolog of the deleted chromosome. |
| |This is known as: |
|A) |a deletion heterozygote. |
|B) |a deletion homozygote. |
|C) |dosage compensation. |
|D) |a triplolethal chromosome. |
| |Ans: A |
| |Difficulty: 2 |

|28. |Individuals born heterozygotes for certain deletions have a greatly increased risk of losing both copies of certain genes and |
| |developing cancer. One such disease is: |
|A) |triplolethal. |
|B) |scarlet eyes. |
|C) |retinoblastoma. |
|D) |cataracts. |
| |Ans: C |
| |Difficulty: 1 |

|29. |During the pairing of homologs in prophase of meiosis I, the region of a normal, nondeleted chromosome that has nothing with |
| |which to recombine forms a so-called: |
|A) |inversion loop. |
|B) |deletion heterozygote. |
|C) |crossover suppressor. |
|D) |deletion loop. |
| |Ans: D |
| |Difficulty: 2 |

|30. |Using Drosophila polytene chromosomes and small deletions, geneticists have been able to: |
|A) |map the shaker-1 gene in Drosophila. |
|B) |assign genes to regions of one or two polytene chromosome bands. |
|C) |assign genes to regions of 100kb or less of DNA. |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 2 |

|31. |Which of the following molecular techniques could a scientist use to help locate genes on cloned fragments of DNA with deletion |
| |mutants? |
|A) | In situ hybridization |
|B) |Crossover analysis |
|C) |Southern blot analysis |
|D) |all of the above |
|E) |both a and c |
| |Ans: E |
| |Difficulty: 3 |

|32. |Duplications arise by: |
|A) |chromosomal breakage and faulty repair. |
|B) |unequal crossing over. |
|C) |errors in replication. |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 1 |

|33. |During the pairing of homologs in prophase of meiosis I, the region of a chromosome bearing extra copies of a particular |
| |chromosomal region that has nothing with which to recombine forms a so-called: |
|A) |inversion loop. |
|B) |deletion heterozygote. |
|C) |duplication loop. |
|D) |deletion loop. |
| |Ans: C |
| |Difficulty: 2 |

|34. |An inversion may result from: |
|A) |a half-circle rotation of a chromosomal region following two double-strand breaks in a chromosome's DNA. |
|B) |the action of a transposable element. |
|C) |a crossover between DNA sequences present in two positions on the same chromosome in inverted orientation. |
|D) |all of the above |
|E) |none of the above |
| |Ans: D |
| |Difficulty: 2 |

|35. |Inversions may be hard to detect because they: |
|A) |never visibly change chromosome banding patterns. |
|B) |increase recombination in heterozygotes. |
|C) |do not usually cause an abnormal phenotype. |
|D) |normally are removed immediately in natural populations. |
| |Ans: C |
| |Difficulty: 3 |

|36. |Which of the following does not happen when an intragenic inversion occurs? |
|A) |One part of the gene is relocated to a distant region of the chromosome. |
|B) |One part of the gene stays at its original site. |
|C) |Homozygotes for the inversion do not survive. |
|D) |The gene's function is not disrupted. |
| |Ans: D |
| |Difficulty: 3 |

|37. |When a crossover occurs within the inversion loop of a pericentric inversion each recombinant chromatid will have: |
|A) |a single centromere. |
|B) |a duplication of one region. |
|C) |a deletion different from the one of duplication. |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 2 |

|38. |Robertsonian translocations result from which of the following? |
|A) |Breaks at or near the centromeres of two acrocentric chromosomes followed by the reciprocal exchange of broken parts. |
|B) |A part of one chromosome becomes attached to a non-homologous chromosome. |
|C) |Unequal crossing over during meiosis. |
|D) |The fusion of two small chromosomes end-to-end such that a double centromere occurs. |
| |Ans: A |
| |Difficulty: 2 |

|39. |Which of the following does not usually show a problem during meiosis? |
|A) |translocation heterozygotes |
|B) |translocation homozygotes |
|C) |paracentric inversion |
|D) |pericentric inversion |
| |Ans: B |
| |Difficulty: 1 |

|40. |Of the following segregation patterns, which one is most likely to result in a normal zygote? |
|A) |alternate |
|B) |adjacent-1 |
|C) |adjacent-2 |
|D) |nondisjunction |
| |Ans: A |
| |Difficulty: 2 |

|41. |The condition of semisterility is most closely associated with: |
|A) |chromosomal duplications. |
|B) |pericentric inversions. |
|C) |translocation heterozygotes. |
|D) |translocation homozygotes. |
| |Ans: C |
| |Difficulty: 2 |

|42. |Translocations can help: |
|A) |determine linkage groups. |
|B) |aid in the diagnosis and treatment of certain cancers. |
|C) |map important genes. |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 1 |

|43. |Down Syndrome can result from: |
|A) |three copies of chromosome 21. |
|B) |a translocation of a part of chromosome 21. |
|C) |a reciprocal translocation between any two autosomes. |
|D) |a and b |
|E) |a, b, and c |
| |Ans: D |
| |Difficulty: 2 |

|44. |Which of the following do translocations and inversions not have in common? |
|A) |don't alter the amount of DNA in the genome |
|B) |ability to alter gene function |
|C) |use of inversion loops during crossing over |
|D) |catalysts of speciation |
| |Ans: C |
| |Difficulty: 2 |

|45. |A transposition is considered a cytologically invisible sequence rearrangement. With which of the following does it share this |
| |property? |
|A) |small deletion |
|B) |large duplication |
|C) |inversion |
|D) |translocation |
| |Ans: A |
| |Difficulty: 2 |

|46. |Barbara McClintock is most closely associated with which of the following? |
|A) |The initial discovery of genetic transposition. |
|B) |The discovery of transposable elements in corn. |
|C) |The mutation rate in translocation heterozygotes. |
|D) |The demonstration of the presence of transposable elements in polytene chromosomes. |
| |Ans: B |
| |Difficulty: 1 |

|47. |Transposable elements have many things in common. Which of the following is not a usual characteristic of them? |
|A) |Typically smaller than 50 bp. |
|B) |May be present in a genome from one to thousands of times. |
|C) |Are found only in a select group of organisms. |
|D) |Need not be sequences that do something for the organism. |
| |Ans: A |
| |Difficulty: 3 |

|48. |Retroposons and retro-viruses have structural parallels. Which of the following also shares structural parallels with them? |
|A) |tRNA |
|B) |DS-DNA |
|C) |rRNA |
|D) |mRNA |
| |Ans: D |
| |Difficulty: 2 |

|49. |Which of the following is a possible effect that a transposable element may have on a gene? |
|A) |Shift the reading frame. |
|B) |Diminish the efficiency of splicing. |
|C) |Provide a transcription stop signal. |
|D) |all of the above |
| |Ans: D |
| |Difficulty: 2 |

|50. |Which of the following is not an aneuploidy? |
|A) |monosomy |
|B) |tetraploid |
|C) |trisomy |
|D) |tetrasomy |
| |Ans: B |
| |Difficulty: 1 |

|51. |The most common human aneuploidy is trisomy 21, Down syndrome. All of the effects listed below may be seen in this syndrome |
| |except: |
|A) |death always by age 25. |
|B) |mental retardation. |
|C) |skeletal abnormalities. |
|D) |heart defects. |
|E) |increased susceptibility to infection. |
| |Ans: A |
| |Difficulty: 2 |

|52. |Which of the following sex chromosome aneuploidies is not usually seen in live births? |
|A) |XO |
|B) |XXY |
|C) |YO |
|D) |XXX |
|E) |None of the above |
| |Ans: C |
| |Difficulty: 2 |

|53. |Turner syndrome, XO, is a sex chromosome aneuploidy. Of the effects listed below, which one is not usually seen in this |
| |syndrome? |
|A) |unusually short stature |
|B) |infertility |
|C) |skeletal abnormalities |
|D) |unusually long limbs |
| |Ans: D |
| |Difficulty: 2 |

|54. |In Drosophila, a gynandromorph, which is composed of equal parts male and female tissue, results from: |
|A) |an XX female losing one X chromosome during the first mitotic division after fertilization. |
|B) |an egg carrying an X chromosome fertilized by a Y-carrying sperm. |
|C) |a normal egg fertilized by both an X-carrying sperm and a Y-carrying sperm. |
|D) |the fusion of a female embryo with a male embryo. |
| |Ans: A |
| |Difficulty: 3 |

|55. |Which of the following is not an example of a euploid condition? |
|A) |triploidy |
|B) |diploidy |
|C) |Down syndrome |
|D) |tetraploidy |
| |Ans: C |
| |Difficulty: 1 |

|56. |Triploid organisms usually result from: |
|A) |the union of haploid and diploid gametes. |
|B) |unequal disjunction during embryogenesis. |
|C) |propagation of fused cell lines. |
|D) |fusion of three gametes simultaneously. |
| |Ans: A |
| |Difficulty: 2 |

|57. |During mitosis, if the chromosomes in a diploid tissue fail to separate after replication, the resulting daughter cells will be:|
|A) |monoploid. |
|B) |tetrasomic. |
|C) |triploid. |
|D) |tetraploid. |
| |Ans: D |
| |Difficulty: 2 |

|58. |Hybrids in which the chromosome sets come from two distinct, though related, species are known as: |
|A) |autopolyploids. |
|B) |allopolyploids. |
|C) |amphiploids. |
|D) |bivalents. |
| |Ans: B |
| |Difficulty: 2 |

|59. |The genus Triticale is a new genus of the various allopolyploid hybrids between wheat and rye. Some of the members of this genus|
| |show agricultural promise because: |
|A) |wheat has a high yield. |
|B) |rye adapts well to unfavorable environments. |
|C) |wheat has a high level of protein. |
|D) |rye has a high level of lysine. |
|E) |all of the above |
| |Ans: E |
| |Difficulty: 2 |

|60. |Which of the following rarely, if ever, results in a positive force for evolution? |
|A) |polyploidy |
|B) |allopolyploidy |
|C) |trisomy |
|D) |amphidiploidy |
| |Ans: C |
| |Difficulty: 2 |

Matching

| |

Match the following descriptions with the terms below
a. inversion
b. duplication
c. deletion
d. translocation
e. transposable element

|61. |A piece of genetic material that moves from place to place in the genome. |
|Ans: |e |
| |Difficulty: 2 |

|62. |A change in the genome whereby new material is added to the genome. |
|Ans: |b |
| |Difficulty: 1 |

|63. |A change in the genetic material where a DNA sequence changes direction. |
|Ans: |a |
| |Difficulty: 1 |

|64. |A decrease of genetic material in the genome. |
|Ans: |c |
| |Difficulty: 1 |

|65. |A piece of chromosome attaches to another chromosome. |
|Ans: |d |
| |Difficulty: 2 |

| |

Match the following descriptions with the terms below a. retroposon b. transposon c. transposable element d. transposase

|66. |Any DNA segment that moves about in the genome. |
|Ans: |c |
| |Difficulty: 3 |

|67. |Moves in the genome with the aid of an RNA intermediate. |
|Ans: |a |
| |Difficulty: 2 |

|68. |Moves DNA directly. |
|Ans: |b |
| |Difficulty: 3 |

|69. |An enzyme that catalyzes a transposition event. |
|Ans: |d |
| |Difficulty: 1 |

True or False

|70. |When comparing mouse and human Giemsa-stained karyotypes, we see no conservation of banding patterns. |
| |Ans: True |
| |Difficulty: 2 |

|71. |Karyotypes generally remain constant within a species because rearrangements and changes in chromosome number occur |
| |infrequently. |
| |Ans: False |
| |Difficulty: 3 |

|72. |Changes in chromosome number include aneuploidy, monoploidy, polyploidy, and duplications. |
| |Ans: False |
| |Difficulty: 2 |

|73. |Deletion may arise from errors in replication, from faulty meiotic or mitotic recombination, and from exposure to X-rays. |
| |Ans: True |
| |Difficulty: 2 |

|74. |Homozygosity for a deletion is often, but not always, lethal. |
| |Ans: True |
| |Difficulty: 2 |

|75. |Recessive mutations can often be covered by deletions in heterozygotes. |
| |Ans: True |
| |Difficulty: 2 |

|76. |Most duplications have no obvious phenotypic consequences and can be detected only by cytological or molecular means. |
| |Ans: False |
| |Difficulty: 2 |

|77. |Duplication of chromosomal segments rarely has an effect on the evolution of genomes. |
| |Ans: True |
| |Difficulty: 2 |

|78. |Crossing-over within an inversion loop produces aberrant recombinant chromatids. |
| |Ans: False |
| |Difficulty: 3 |

|79. |Reciprocal translocations are usually phenotypically abnormal because they have neither lost nor gained genetic material. |
| |Ans: True |
| |Difficulty: 1 |

|80. |A hallmark of transposons is that their ends are inverted repeats of each other. |
| |Ans: False |
| |Difficulty: 3 |

|81. |The mouse genome has high synteny with the human genome since about 170 DNA blocks are simply rearranged between the two |
| |genomes. |
| |Ans: True |
| |Difficulty: 2 |

|82. |Euploid cells contain only incomplete sets of chromosomes. |
| |Ans: False |
| |Difficulty: 2 |

|83. |Down syndrome is an example of triploidy. |
| |Ans: False |
| |Difficulty: 2 |

|84. |Genetic imbalance results from polyploidy. |
| |Ans: False |
| |Difficulty: 1 |

|85. |An acentric fragment is an inversion cross-over product lacking a centromere. |
| |Ans: True |
| |Difficulty: 1 |

Short Answer

|86. |Explain how data from the linkage groups of the mouse can be used as a resource for assessing human linkage groups. |
|Ans: |Because virtually all genes cloned from the mouse genome are conserved in the human genome and vice versa, it is |
| |possible to construct linkage maps for the two genomes from the same set of markers. Comparisons of the mouse and human|
| |linkage groups allow one to see a picture somewhere between complete correspondence and unrelatedness. Genes closely |
| |linked in the mouse tend to be closely linked in humans, but genes that are less tightly linked in one species tend not|
| |to be linked at all in the other. This shows that even though mice and humans diverged about 65 million years ago, the |
| |DNA sequences in many regions are very similar. |
| |Difficulty: 4 |

|87. |Explain the differences between chromosomal rearrangements and changes in chromosome number. Cite at least one example of each. |
|Ans: |Chromosomal rearrangements reorganize the DNA sequences within one or more chromosomes. Changes in chromosome number |
| |involve losses or gains of entire chromosomes or sets of chromosomes. (Student may cite as an example of |
| |rearrangements: deletion, duplication, inversion, translocation, and transposable elements. For changes in chromosome |
| |number student may cite an aneuploidy such as a monosomy or trisomy, monoploidy, or polyploidy.) |
| |Difficulty: 4 |

|88. |Describe how an inversion heterozygote can reduce the number of recombinant progeny. |
|Ans: |When inversion heterozygotes have chromosomes pair up during meiosis, an inversion loop is formed to allow the tightest|
| |possible alignment of homologous regions. This always produces aberrant recombinant chromatids. Two inversion cases are|
| |possible – pericentric and paracentric. In a pericentric crossover within the inversion loop each recombinant will |
| |carry a duplication of one region and a deletion of another. This abnormal dosage of some genes will result in abnormal|
| |gametes and if they fertilize normal gametes, zygotes may die because of genetic imbalance. In a paracentric crossover |
| |within the inversion loop the recombinant chromatids will be unbalanced in both gene dosage and centromere number. |
| |(Student may then explain how centromere number can result in genetically unbalanced gametes such as what acentric and |
| |dicentric chromatids would produce.) |
| |Difficulty: 4 |

|89. |Discuss the several effects that translocations and inversions have in common. |
|Ans: |Both translocations and inversions change genomic position without affecting the total amount of DNA. If a breakpoint |
| |of either one is within a gene, the gene function may be altered or lost. Both types may produce genetically imbalanced|
| |gametes that may negatively affect a zygote or developing embryo. (Student may explain at this point the differences |
| |between how the imbalanced gametes are produced.) Because both reduce viable progeny and heterozygotes, they may play a|
| |role in speciation and evolution. |
| |Difficulty: 4 |

|90. |Explain the possible effects that a transposable element may have on a gene. |
|Ans: |Insertion of a transposable element near or within a gene can affect gene expression and alter phenotype. For example, |
| |a B type hemophilia occurs after insertion of Alu into the gene encoding clotting factor IX. Secondly, the effect of |
| |insertion depends on what the element is and where the insertion point is. If insertion is into a protein-coding exon, |
| |the reading frame may shift or a stop codon may be introduced. Insertion into an intron may lower the efficiency of |
| |splicing, which may result in removal from the transcript that could lower production of a normal polypeptide. A stop |
| |signal could also affect genes downstream. Upstream insertion into a regulatory gene could affect gene function in |
| |various ways also. |
| |Difficulty: 4 |

|91. |Explain the mechanism by which aneuploidy occurs. |
|Ans: |Aneuploidy occurs because of meiotic nondisjunction either in meiosis I or meiosis II. In meiosis I if homologs do not |
| |separate all gametes produced will contain an error. Two of the gametes will contain both homologs and two will contain|
| |neither. When fertilization of a normal gamete occurs by either of these abnormal gametes, aneuploidy results. Half of |
| |the zygotes will be trisomic and half will be monosomic. Meiotic nondisjunction during meiosis II will produce two |
| |normal and two abnormal gametes. If fertilization occurs with either of the abnormal gametes, aneuploid zygotes are |
| |produced. |
| |Difficulty: 4 |

|92. |Discuss why triploid organisms are almost always sterile. |
|Ans: |(Student may explain how triploids occur.) Triploids are almost always sterile because meiosis produces mostly |
| |unbalanced gametes. During the first meiotic division in a triploid germ cell, three sets of chromosomes must segregate|
| |into two daughter cells. Most likely one daughter will end up with two chromosomes and the other will have only one of |
| |any one set of homologs. Some cells will have two of some chromosomes and the normal one of others. Many combinations |
| |of incorrect number of chromosomes will occur with very little chance of the normal amount. Most gametes will be |
| |aberrant and will have a reduced chance of producing viable offspring. |
| |Difficulty: 4 |

|93. |Discuss how deletions and duplications may contribute to evolution. |
|Ans: |General examples of how chromosomal rearrangements might contribute to evolution: |
| |Deletions – a small deletion that moves a coding sequence of one gene next to a promoter or other regulatory element of|
| |an adjacent gene may, rarely, allow expression of a protein at a novel time in development or in a novel tissue. If the|
| |new time or place of expression is advantageous to the organism, it might become established in the genome. |
| |Duplications – a duplication will provide at least two copies of a gene. If one copy maintains the original function, |
| |the other could conceivably acquire a new function that would probably be related to the original function. Many |
| |examples can be seen in higher plants and animals. (Students may also write about the evolutionary contributions of the|
| |other chromosomal rearrangements and might even mention the role of changes in chromosome number.) |
| |Difficulty: 4 |

|94. |Why do inversions act as cross-over suppressors? |
|Ans: |Inversions act as cross-over suppressors because only progeny that do not recombine within an inversion loop will |
| |survive. |
| |Difficulty: 4 |

|95. |What is a balancer chromosome? |
|Ans: |A balancer chromosome is a special chromosome often created by the use of X-rays for the purpose of genetic |
| |manipulation; these chromosomes often carry multiple, overlapping inversions that enable researchers to follow them |
| |through crosses, and a recessive lethal mutation that prevents the survival of homozygotes. |
| |Difficulty: 4 |

|96. |What is the difference between alternate and an adjacent-1 segregation or an adjacent-2 segregation pattern? |
|Ans: |An alternate segregation pattern results in balanced chromosomes while adjacent 1or 2 patterns yield chromosomes that |
| |are unbalanced. |
| |Difficulty: 4 |

Experimental Design and Interpretation of Data

|97. |We now know that several organisms have a high degree of synteny at the genomic level. You wish to test the hypothesis that the|
| |laboratory mouse and human share genomic similarities. What tests would you complete and given that we now know that the mouse |
| |and human genomes are highly syntenic, what results would you expect? |
|Ans: |Karyotype analysis can be used to test the hypothesis of genomic similarities however, only animals that have high |
| |homology will show similar banding patterns. Therefore, FISH (fluorescence in situ hybridization) would be a more |
| |useful technique to determine synteny. The mouse and human genomes are similar in that approximately 170 similar |
| |fragments an average length of about 18 Mb are simply rearranged (this is not visible in a karyotype). |
| |Difficulty: 4 |

|98. |You are mapping traits in your favorite organism but unbeknownst to you, your laboratory model organism contains a rare |
| |deletion. How will your mapping results be affected? |
|Ans: |The mapping distance will appear smaller than the actual physical distance in the wild-type organism. |
| |Difficulty: 4 |

|99. |You have discovered an altered phenotype and cloned the gene responsible. However, the gene you cloned appears to have an |
| |unusual sequence in it. In order to determine the chromosomal location of your new gene, you perform FISH, using only the |
| |unusual sequence, on several animals. To your surprise, the FISH results suggest that each animal contains the gene on a |
| |different chromosome. How would you interpret your results. |
|Ans: |The unusual sequence is a transposon and your “new” phenotype arose via the disruption of its gene by the transposon. |
| |Difficulty: 4 |

|100. |You are a master gardener and your favorite tomato plant is very sensitive to a pesticide called DEADBUG. You wish to make your|
| |special tomato plants resistant to the pesticide which you spray on other bushes in your garden. Using microbial techniques |
| |give sufficient and complete details of how you would do this (include ploidy status). |
|Ans: |Haploid pollen grains are cold treated and plated on agar plates. The resulting embryoids are treated with hormone in |
| |liquid culture and eventually grown as a monoploid plant. The plant is treated with a mutagen to induce mutations that |
| |can result in insensitivity to the pesticide. Somatic cells are removed from the treated plant and plated on agar |
| |containing DEADBUG. Only cells resistant to DEADBUG will grow. Again the embryoid is hormone treated and grown into a|
| |resistant monoploid plant. Treatment with colchicine will allow duplication of chromosomes without separation |
| |resulting in a normal diploid plant. |
| |Difficulty: 4 |