Microbiology 225 Exam Review

BIO 225 – Exam 2 Review Sheet

Chapter 9
1. Know the anatomy of the structures that make up the prokaryotic cell.
Know their functions.
Know any clinical significance each structure might have and if it is a target for antibiotics. (On separate sheet)
2. Know the differences between Gram positive and Gram negative cell walls. * Gram positive cell wall * In addition to many layers of peptidoglycan, the cell wall of Gram-positive bacterials cells also contain: * Teichoic acid * There are two forms * Wall teichoic acids – go part way through the wall. * Lipoteichoic acids – go completely through the wall and link to the plasma membrane * Both forms protrude above the wall, which gives the bacterial cell a negative charge * M protein * This is a virulence factor. * It protrudes from the cell wall. * It is required for infection. * It is highly susceptible to mutations. * Mycolic acid. * This is found in the Mycobacterium species. * It consists of a waxy lipid incorporated into the cell wall. * It makes cells extremely resistant to environmental stress. * It acts as a barrier against antibiotics and host defenses * CLINICAL SIGNIFICANCE OF THE BACTERIAL CELL WALL: Gram-positive bacteria * Teichoic acid can cause colonization of the nasal epithelium. * Lipoteichoic acid causes an inflammatory response.

* Gram-negative bacteria cell wall * Gram-negative bacteria only have a thin peptidoglycan layer. * Gram-negative bacteria only have a thin peptidoglycan layer. * The LPS layer is composed of lipids, proteins, and polysaccharides. * Lipoprotein molecules fasten the outer membrane to the peptidoglycan layer * The outer membrane of Gram-negative bacteria has a unique outer layer. * It is composed of lipopolysaccharides instead of the standard phospholipid molecules. * It serves as a major barrier to the outside world for the Gram-negative cell. * It contains specialized proteins called porin proteins: * These form a channel through the outer layer. * This channel is responsible for passage of molecules and ions into and out of the Gram-negative cell * CLINICAL SIGNIFICANCE OF THE BACTERIAL CELL WALL: Gram-Negative Bacteria * The gram-negative cell wall is a complex structure involved in infection in many ways. * The outer layer is a barrier to antiseptics, disinfectants, and antibiotics. * Porin proteins exclude large molecules. * The outer layer functions as endotoxin, with two parts having clinical relevance * Lipid A * Anchors the lipopolysaccharide part of the outer layer * Is released when the cell dies. * O polysaccharide * Carbohydrate chains that are part of the outer layer * Variable from one bacterial species to another * Recognized by the adaptive immune response * Sometimes used as a diagnostic marker - E. coli O157: H7 is designated by O polysaccharide 157

3. Know the special features of the cell wall such as M proteins, Mycolic acid and the feature of the outer membrane for the Gram negative cells. See above in red: 3. Know the terms for the number and arrangement of prokaryotic flagella * flagella consists * Filament- made of molecules of flagellin protein./ gives the flagellum a hollow core/ only seen in bacilli/ It is flexible * Hook - The hook links the flexible filaments to the basal body * Basal body- composed of paired ring structures that anchor the flagella to the cell * two different arrangements * Gram-positive – uses only one pair of ringed structures fastened to the plasma membrane * Gram-negative – uses two pairs of ringed structures: one is fastened to the plasma membrane and one to the outer layer * FLAGELLA CONFIGURATIONS * Monotrichous – one flagellum located at the end of the cell * Amphitrichous – two flagella, one at each end of the cell * Lophotrichous – two or more flagella located at the same end of the cell * Peritrichous – flagella surround the entire cell

4. Know the types of passive and active membrane transport mechanisms. * Passive transport * two types of passive transport * Simple diffusion * does not require ATP. * It is based on the development of concentration gradients. * Solutes move from regions of higher concentration to regions of lower concentration. * The higher the concentration gradient between two regions, the faster diffusion occurs. * Diffusion slows down as equilibrium is reached. * Simple diffusion only occurs with: * Lipid soluble molecules * Molecules small enough to pass through the membrane. * * Facilitated diffusion * does not require ATP. * Molecules are brought across the plasma membrane by carrier proteins called permease proteins. * Permeases achieve this by changing their three-dimensional shape. * Molecules too large to fit into a permease are chopped into smaller pieces by enzymes secreted by the cell. * Active transport. * requires ATP. * Solutes are carried either into or out of a cell against the concentration gradient. * It is the most common form of membrane transport. * three types of active transport: * Efflux pumping * Efflux pumping involves proteins called the super family of transporters. * It employs a revolving door mechanism - pumps bring in certain molecules and expel others at the same time. * The energy source is the proton motive force of electron transport. * It is efficient because two functions (intake and output) occur at the same time * ABC transport systems * ABC transport systems are very complex and involve several proteins. * The molecule to be transported binds to a protein on the outside of the plasma membrane. * It is handed off to a complex of proteins located in the plasma membrane. * These proteins then transport the molecule into the cytoplasm. * Group translocation * Group translocation is unique to bacteria. * It is very energy expensive and uses PEP instead of ATP. * It helps make sure molecules stay inside the cell. * An enzyme attaches a phosphate to the molecule, preventing the molecule from leaving the cell

5. Know the osmotic solutions. * Osmosis- water chases the concentration of solutes. * If the solute concentration is greater outside the cell: * It allows water to leave the cell and results in plasmolysis (cell shrinks)- * hypertonic solution will undergo plasmolysis * If the solute concentration is greater inside the cell: * It allows water to enter the cell and results in osmotic lysis (cell bursts).- * hypotonic solution will undergo osmotic lysis * an isotonic solution, the concentration is the same inside and outside the cell

6. Know the process of sporulation and germination. * Process of sporulation: * The first step is replication. * The second step is the sequestration of a copy of the chromosome. * This chromosome copy is surrounded by a septum * The third step is formation of the forespore. * Large amounts of peptidoglycan are deposited round the forespore. * In the last step, the rest of the original cell deteriorates and degrades. * Bacterial genetic information is protected inside the endospores. * Process of Germination- * Germination of the endospore back into a vegetative cell occurs when the environmental stress has subsided. * The endospore accepts water molecules, swells, and cracks. * The water activates metabolism and the cell begins to grow

Chapter 10
1. Know the terms generation and generation time. * Generation- Each division of bacteria * Generation time- The time between divisions

2. Know the physical requirements for microbial growth. * Temperature * Bacteria can be separated according to temperature ranges in which they grow best. * Psychrophiles – grow at cold temperatures * Mesophiles – grow at moderate temperatures * Thermophiles – grow at high temperatures * The minimum growth temperature is the lowest temperature at which an organism grows. * The maximum growth temperature is the highest temperature at which an organism grows. * The optimum growth temperature is the temperature at which the highest rate of growth occurs * Temperature affects growth * Increased temperature breaks chemical bonds. * This causes changes in the three dimensional structure. * These changes can inhibit or destroy the ability for the molecules to function properly. * pH * Bacteria grow in a wide range of pH values. * Most bacteria prefer the neutral pH of 7.0. * Some bacteria are acidophiles that grow at extremely low pH values. * Helicobacter pylori (causes stomach and duodenal ulcers) grows at a low pH value but is not an acidophile * pH can negatively affect protein structure. * An excess of hydrogen ions causes bonds to break. * This changes three-dimensional structure. * Changes in three-dimensional structure destroy protein function. * Destruction of protein function can be a lethal event * Osmotic pressure * Osmotic pressure is the pressure exerted on bacteria by their environment. * One of the major agents exerting such pressure is water. * Osmotic pressure can inhibit bacterial growth. * High salt concentrations can be used to preserve food (cure meats). * Cause a hypertonic environment and plasmolysis. * This is an imperfect way to preserve food because some bacteria are halophilic and thrive in high salt concentrations. * Halophilic organisms can be divided into: * Obligate – requiring a high salt concentration * Extreme – requiring very high levels of salt * Facultative – can grow either with or without high salt levels

3. Know the chemical requirements for microbial growth. Know the terms for the microbial groups as characterized by their growth requirements. * chemical requirements * * Carbon- * Bacteria are classified based on ways that they acquire carbon. * Chemoheterotrophs – obtain carbon by breaking down other carbon molecules * Chemoautotrophs – obtain carbon from CO2. * Nitrogen- * is involved in protein synthesis. * It is an integral part of amino acid structure. * It is part of the structure of nucleic acids. * It can be obtained in a variety of ways: * Decomposition of existing proteins * From ammonium ions found in organic materials * Nitrogen fixation * Sulfur- * Bacteria must have sulfur to make amino acids and some vitamins. * Sulfur is obtained from decomposition. * Sulfur can be procured in the sulfate ions (SO42-) and from H2S * Phosphorus * is essential for the synthesis of nucleic acids, AMP, ADP, and ATP. * It is a major component for the development of the plasma membrane. * Bacteria obtain phosphorus by cleaving ATP or ADP or from phosphate ions * Organic growth factors and trace elements: * Bacteria use growth factors such as vitamin B1, B2, and B6. * Bacteria cannot synthesize these growth factors so they must be obtained from the environment. * Bacteria also require potassium, magnesium, and calcium as enzyme co-factors. * Some bacteria also require trace elements such as iron, copper, molybdenum, and zinc
4. Know the terms for the microbial groups based on their oxygen requirements. Know the techniques used to culture anaerobic organisms. * Many bacteria do not require oxygen for growth. * Some die in the presence of oxygen. * This is due to the production of the superoxide free radical form of oxygen. * This form is unstable and steals electrons from other molecules

* two types of bacteria that grow in the presence of oxygen: * Aerobes – require oxygen for growth * Facultative anaerobes – can grow with or without oxygen. * Both types produce an enzyme called superoxide dismutase that converts free radical oxygen to molecular oxygen and peroxide * Peroxide is also poisonous. * Bacteria produce two enzymes to deal with peroxide: * Catalase – converts peroxide to water and oxygen * Peroxidase – converts peroxide to water.

* There are three major categories of bacteria based on oxygen use: * Obligate aerobes – require oxygen for growth * Obligate anaerobes – cannot survive in the presence of oxygen * Facultative anaerobes – can grow with or without oxygen.

* There are two additional smaller categories of bacteria based on oxygen use: * Aerotolerant – grows in oxygen but does not use it in metabolism * Microaerophile – requires only low levels of oxygen for growth.

* Know the techniques used to culture anaerobic organisms * The first method uses the medium sodium thioglycolate, which forms an oxygen gradient during growth * Aerobic organisms grow at the top. * Anaerobic organisms grow at the bottom. * Facultative anaerobes grow throughout the medium * The second method for growing anaerobic organisms is in a GasPakTM jar * This incubation container provides an oxygen-free environment. * Only obligate and facultative anaerobes can grow via this method.

5. Know the types of culture media. * Chemically defined media * Chemically defined media are those in which the chemical composition is precisely known. * They are used for the laboratory analysis of compounds produced by specified bacteria.

* Complex media * Complex media contain not only numerous ingredients of known chemical composition but also digested proteins and extracts derived from plants or meats. * The exact chemical composition of these digests and extracts is not known. * often called a nutrient and are available in two forms: * Nutrient agar (solid) * Nutrient broth (liquid) * Growth media (Selective/Differiential) * can be used to identify pathogens in several ways. * All use selective media and differential media * A selective medium is one that contains ingredients that prohibit the growth of some organisms while fostering the growth of others. * A differential medium is one that contains ingredients that can differentiate between organisms. * Many selective media are also differential media

* Blood agar media identify production of hemolysins. * There are three types of hemolysins: alpha, beta, and gamma. * EMB (Eosin methylene blue) * MAC (MacConkey) * MSA (mannitol-salt)
6. Know the phases of the bacterial growth curve and what happens to the cells in each phase. * Lag phase – bacteria are adjusting to their environment * This varies depending on the organisms and the environment. * Log phase – the number of bacteria doubles exponentially * There is a constant minimum generation time. * This phase lasts only as long as a suitable level of nutrients is available. * Bacteria are the most metabolically active and most susceptible to antibiotics. * Stationary phase – the number of cells dividing is equal to the number dying * It is caused by a decreasing availability of nutrients.

* Death phase (logarithmic decline phase) -a continuous decline in the number of dividing cells * It is caused by the exhaustion of the nutrient supply and by a build-up of metabolic waste
7. Know the direct and indirect methods to count cells and whether they give a total or viable cell count. (pg. 199) * There are two ways to measure bacterial growth: * Direct methods of measuring bacterial growth include: * Direct cell counts ( determines total # of cells) * Viable cell counts (viable) * Plate counts (total) (used if sample has at least 100 cells/ml) * Most probable number (not precise- statistical estimation) * Membrane filtration- (used to count cells in dilute environment) * is used to look for water contamination * This method generates the fecal coliform count. * The filter pores are small enough to exclude bacteria. * Filters are placed on media plates and incubated. * The number of contaminating bacteria can then be counted. * Indirect methods of measuring bacterial growth include: * Turbidity – most often used * Total weight * Chemical constituents –use chemical means * Measuring cell products-must be correlated with cell number
8. Know the clinical implications of microbial growth. * Many bacteria are fastidious in a laboratory setting. * Some bacteria cannot be grown in a laboratory setting. * Some bacteria have stringent nutritional requirements. * All of these things can affect diagnosis and treatment of infection * Bacterial growth requirements can lead to missed identification of pathogens and wrong diagnoses * This can be caused by improper handling of clinical specimens. * It can also be cause by improper culturing. * Specific standard procedures for collecting specimens are intended to limit these problems

Chapter 11
1. Know the structure of DNA. * DNA: * is a double helix made of two strands of nucleotides. * Each nucleotide is made of a phosphate group, the sugar deoxyribose and one of four nitrogenous bases. * In DNA these bases are adenine (A), thymine (T), guanine (G) and cytosine (C). * The bases are located on the inside of the molecule and are held together with hydrogen bonds. * The base A always pairs with T, and the base G always pairs with C. * Since these bases are specifically paired, any mismatched pairs are chemically unstable. * The strands are antiparallel and run in opposite directions. * remember that the two strands of nucleotides that make up the DNA molecule are antiparallel.

2. Know the process of DNA replication. * a) The DNA double helix is unwound by helicase and topoisomerase. Topoisomerase unwinds the supercoils and helicase unwinds and separates the helix. * b) An RNA primer is put into place at the primer/template junction using enzyme Primase. This RNA primer gives the DNA polymerase something to use (like starting material) to synthesize the new DNA strand (daughter strand). * c) DNA Polymerase constructs the new strand in the 5’-3’ direction with new building blocks (nucleotides) added on to the 3’ end of the growing strand. Because DNA polymerase can only work in one direction, the leading strand is synthesized continuously while the lagging strand is synthesized in pieces. The pieces are called Okazaki fragments and will be put together later. * d) RNAse H removes the RNA primers and DNA polymerase fills in any spaces between Okazaki fragments. * e) The Okazaki fragments are covalently bonded together by DNA ligase.

3. Know the structure of RNA. * RNA is a single strand of nucleotides. * Each nucleotide is made of a phosphate group, the sugar ribose and one of four nitrogenous bases. * In RNA these bases are adenine (A), uracil (U), guanine (G), and cytosine (C). * While RNA is single stranded, the different types of RNA can interact with each other through base pairing. * When the bases pair A binds with U and G binds with C. * There are three types of RNA: * Messenger RNA (mRNA) – is a copy of one gene in the DNA * Transfer RNA (tRNA) – carries the amino acids needed to produce a protein * Ribosomal RNA (rRNA) – combines with proteins to produce a ribosome

4. Know the terms codon, anticodon, mRNA, tRNA, and rRNA. * Codon- genetic code employs three letter combinations, based on a four letter alphabet (A, T, C, G). * Three rules govern arrangement * Codons are always read in one direction. * The message is translated in a fixed reading frame. * There is no overlap or gap in the code. * Anti-codon-a triplet of nucleotides in transfer RNA that is complementary to the codon in messenger RNA which specifies the amino acid * Messenger RNA (mRNA) – is a copy of one gene in the DNA * Transfer RNA (tRNA) – carries the amino acids needed to produce a protein * Ribosomal RNA (rRNA) – combines with proteins to produce a ribosome
5. Know the processes of transcription and translation. * Transcription: Transcription is the copying of the DNA code onto an RNA molecule in the nucleus. Transcription involves three steps – initiation, elongation and termination. * Initiation: * a. DNA unwinds and one strand is used as the template strand (the strand to which the mRNA nucleotides will bind). * b. The RNA polymerase binds to the promoter region of the DNA/ gene to be copied. So transcription begins at the promoter. * Elongation: * c. Using base pairing rules the RNA polymerase puts new RNA nucleotides in place creating the new RNA strand in a 5’ to 3’ direction. * Termination: * d. When the RNA polymerase reaches the terminator sequence on the DNA (like a stop light) the RNA polymerase falls off, the RNA transcript is released, and the DNA recoils and remains unchanged. So transcription ends at the termintor(Remember we only made a quick copy of the genes we needed from the DNA. Our goal is to leave the DNA unchanged so we can transcribe again in the future when necessary.) * In eukaryotic cells, the RNA transcript is usually not complete and needs some modification (processing) before it leaves the nucleus. “Unnecessary” pieces called introns are removed using enzymes and then the rest of the transcript is linked back together. * RNA is not processed in prokaryotic cells since it has no nucleus. Sometimes as the RNA is made from the DNA, ribosomes will already be attaching and translating it before the mRNA is finished. This is called simultaneous transcription and translation * Translation is the converting (translating) the RNA code to a protein molecule at the ribosome * Initiation * a. Translation begins on the mRNA at the start codon AUG (in the P site). At the start codon the small ribosomal subunit, the first tRNA (with the corresponding anticodon UAC), and the second ribosomal subunit all come together to form the initiation complex. * Elongation * b. The second tRNA fills the A site and the amino acid from the first tRNA (in the P site) links to the amino acid in the A site using a peptide bond. * c. The ribosome moves forward to the next codon. The first tRNA in the P site is moved to the E site, the second tRNA in the A site is moved to the P site (it has the growing protein still attached), and the A site is now open and available again. * d. The third tRNA comes into the A site delivering its appropriate amino acid for the protein chain. * e. The amino acids from the second tRNA (in the P site) are transferred to the amino acid at the A site using a peptide bond. * f. The ribosome moves forward again to the next codon. This causes the first tRNA to leave from the E site, the second tRNA moves from the P site to the E site, and the third tRNA moves from the A to the P site opening up the A site again for the fourth tRNA arrival. * Termination * g. This process continues until the ribosome reaches the STOP codon. At the stop codon the ribosome comes apart, the tRNA’s are released, the mRNA is released, and the new protein chain is released as well.
6. Know the types of mutations.
7. Know the basic structure of an operon and the two types of operons and how they function. * An operon is a group of genes that are regulated together. (Regulated genes can be shut “turned off” or can be “turned on”.) * Operons contain a promoter region, an operator region, and structural genes which code for your product. * Often there is regulatory gene (i) that codes for a regulatory protein that helps control expression of the operon.

* the two types of operons: * inducible operon –turns the operon on (ex. LAC operon) * t he binding of the effectors molecule to the repressor greatly reduces the affinity of the repressor for the operator, the repressor is released and transcription proceeds * repressible operon- turns the operon off (ex. TRP operon)
With repressible systems, the binding of the effector molecule to the repressor greatly increases the affinity of repressor for the operator and the repressor binds and stops transcription. Thus, for the trp operon , the addition of tryptophan (the effector molecule) to the E. coli environment shuts off the system because the repressors binds at the operator.
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8. Know the types of mutations and how DNA damage can be repaired. * Mutations are changes in the base sequence in the DNA of a gene. There are two general categories of mutations. Base substitution mutations and frameshift mutations. * Base substitution mutations occur when, during DNA replication, one base is substituted for another. This alters the DNA code and would affect one codon. Sometimes these have no effect, sometimes they have a bad effect, and sometimes they can even be beneficial. * There are three types of base substitution mutations we see; silent mutations, missense mutations and nonsense mutations. * In silent mutations the base substitution changes the DNA code which changes the RNA transcript. However, because of the redundancy of the genetic code (many codons for 1 amino acid) sometimes a change in a codon does not translate to a change in the amino acid. As a result, the protein sequence remains unchanged so the product suffers no change. The mutation is there but it is “silent” because the protein produced remains the same so no visible change is seen. * Missense mutations are base substitutions that change the DNA code, change the RNA transcript, change the codons, and result in an incorrect amino acid placed in the protein structure. This will not be silent because the amino acid sequence will change which may result in the protein function changing as well. * Nonsense mutations are base substitutions that change the DNA code, change the RNA transcript, and change a normal codon to a stop codon. These are almost always devastating because they stop protein synthesis prematurely so the protein is not completely made. * Frameshift mutations are when the order of the codon reading is disrupted due to an insertion or a deletion of one or more bases. In the example below the black lines show the normal reading frame for the first two amino acids AGA and TGC in the form of DNA. When transcribed to mRNA, these codons would code for the amino acids Serine and Threonine. However, if the first A is deleted, then the reading frame shifts one base to the right and the new codons read are GAT and GCG which will not produce the same amino acids (they will produce leucine and arginine when transcribed and transplated). This will result in a completely different protein that is most likely not functional.

* how DNA damage can be repaired * Photoreactivation is a process of DNA repair that fixes damage caused by UV light. Photolyases are enzymes that use the energy from the sun to break the thymine dimer and therefore repair the mutation damage.

* base excision - Repair enzymes cut and pull out incorrect bases and replace them with the correct one

* nucleoside excision- repair enzymes cut out incorrect DNA sequences and then repair the removed area with the correct DNA sequence
9. Know the methods of genetic recombination. * There are four ways in which genetic recombination can occur: * Transposition – within the same cell * Transformation – between cells * Conjugation – between cells * Transduction – between cells. * Genetic recombination involves one cell (donor cell) giving some of its DNA to a recipient cell which often integrates the donor DNA with its own. Once the recipient integrates the donor DNA we call it a “recombinant” because it has a mixture of DNA. The recipient is usually changed in some way due to the additional DNA it has acquired from the donor. There are three ways that bacteria exchange genetic information; transformation, conjugation, and transduction. * Transformation * Transformation involves the transfer of genetic material between cells. * It involves naked DNA. * This DNA is taken up by a bacterial cell and recombines with genes of that cell. * The recipient cell must be competent. * Must be able to take up large molecules such as pieces of DNA. * Some bacteria are naturally competent, whereas others can become competent after chemical treatment. * Only a small amount of DNA is actually taken up * EXAMPLE: * The picture in your book that shows Frederick Griffith’s experiment with S. pneumoniae is a great example of transformation. He was studying 2 strains of S. pneumonia in an attempt to make a vaccine. The R strain had no capsule (and was not deadly to the mice) and the S strain which was encapsulated (and was very deadly to the mice). Griffith also tried to heat kill the S strain which disrupts the capsule and makes the S strain not deadly. However, when he combined the R strain (not deadly) with the heat killed S strain (not deadly) in a mouse the subject died. Why did both of strains, which are not deadly alone, kill the mouse in combination? The answer was transformation. The heat killed S strain must have released its DNA into the area and the R strain bacteria picked it up and incorporated it into its own DNA. This caused the previously harmless R strain to now make capsules and this kills the mouse.

* Conjugation * Conjugation involves the transfer of material between cells. * Conjugation requires direct contact between the donor and recipient cells. * Gram-positive cells stick to each other. * Gram-negative cells use pili as a conduit for DNA transfer. * DNA moves from the donor to recipient cell * There are several steps in conjugation: * The sex pilus of the donor cell recognizes specific receptors on the cell wall of recipient cell. * An enzyme in the donor cell causes the plasmid DNA to unwind. * One of the two single strands of plasmid DNA stays in the donor cell * The other moves across the plasmid into the recipient cell. * Both single strands are replicated. * After replication, the donor and the recipient contain identical plasmids * Conjugation can have several outcomes for the recipient cell: * The plasmid can remain as a plasmid. * The plasmid can become incorporated into the recipient cell chromosome. * When this happens, the recipient cell is then referred to as Hfr. * DNA from Hfr can be moved into a new recipient. * This replaces sections of the host chromosome * Another explanation: * Conjugation is the transfer of genetic information mediated by a plasmid. * A plasmid is a small piece of circular DNA found in bacteria that usually contains non-essential genes like antibiotic resistance or toxin production. Plasmids usually remain separate from the regular bacterial chromosome. * Conjugation requires the two cells (donor and recipient) to touch in order for the transfer to take place. Also, conjugation usually takes place between donor cells that have plasmids (F+) and recipient cells that do initially have the plasmids (F-). Your book gives examples of conjugation of the F plasmid in E. coli. The F (fertility) plasmid gives the cell the ability to make a sex pilus and allows conjugation with other cells. * The bacterium that has the F plasmid is called the F+ cell. This cell creates a sex pilus (plural is pili – we talked about these in chapter 9) and attaches to a cell that does not have the F plasmid (an F- cell). Conjugation allows the donor cell to copy the F plasmid and send a copy to the F- cell. Once the recipient receives the F plasmid it is converted from an F- to an F+ cell (because it has the plasmid) and it can now transfer it to other cells. * Sometimes the F plasmid integrates into the recipient chromosome and converts the recipient into an Hfr cell (high frequency of recombination) cell. The unique feature of Hfr cells is that when they begin conjugation and start transfer of the genetic material to a recipient cell, they not only transfer part of the plasmid but also part of their own DNA to the recipient cell. Therefore when an Hfr cell transfers DNA to a recipient the recipient gets a piece of the F plasmid and part of the donor cell DNA. Watch the animation on this process to get a clearer picture. * Transduction * Transduction involves the transfer of genetic material between cells. * It is a common event in both Gram-positive and Gram-negative bacteria. * It uses a bacterial virus (phage) for transfer * There are two forms of transduction: * Generalized – random * There are three phases to generalized transduction. * The original infected cell chromosome is cleaved into pieces. * Some of this bacterial DNA is incorporated into a newly made phage. * When these phages infect the next cell, original DNA recombines with host chromosome * Explanation of generalized transduction: * During viral replication inside the host cell, the bacteriophage takes over the host cell machinery the host DNA is often degraded too. When the new virus assembles its parts it usually only packages up new viral DNA into the new viral package. However, sometimes because the host DNA is degraded too the virus accidentally picks up a piece of bacterial DNA and puts it in the virus package. When the new viruses leave the host cell this virus with the bacterial DNA can infect a new bacterium and inject the previous host’s DNA into the new host bacteria. Now instead of a viral infection the new host has just been injected with previous host DNA which it can integrate with its own and form a recombinant. This is called generalized transduction.

* Specialized – specific * During specialized transduction: * Phage DNA incorporates into the host chromosome. * Phage DNA excises itself from the host chromosome. * Part of the host DNA is taken along. * Original host DNA is incorporated into the next host chromosome. * Specialized transduction is used in biotechnology * Transposition is caused by transposons. * Transposons move from one place on the chromosome to another. * They can move into or out of the chromosome. * They use cleavage and rejoining mechanisms * Transposition causes random rearrangements. * The results can be beneficial or detrimental. * Beneficial changes will be selected for and maintained. * They may be the reason for several human diseases.

Possible ESSAY questions are color coded throughout the documents.
Chapter 9 - you will be asked to describe the clinical significance of three prokaryotic cell structures (6 points)

Chapter 10 - you will need to know how to draw the bacterial growth curve and to describe what happens in each phase of the bacterial growth curve. (4 points) (pg. 196) * Lag phase – bacteria are adjusting to their environment * This varies depending on the organisms and the environment. * Log phase – the number of bacteria doubles exponentially * There is a constant minimum generation time. * This phase lasts only as long as a suitable level of nutrients is available. * Bacteria are the most metabolically active and most susceptible to antibiotics. * Stationary phase – the number of cells dividing is equal to the number dying * It is caused by a decreasing availability of nutrients.

* Death phase (logarithmic decline phase) -a continuous decline in the number of dividing cells * It is caused by the exhaustion of the nutrient supply and by a build-up of metabolic waste

Chapter 11 - you will need to be able to describe DNA replication, Transcription or Translation (I will pick one of these for you) (5 points)
Chapter 11 - you will need to be able to describe an operon and describe either an inducible or repressible operon (I will pick one for you) (3 points)
Chapter 11 - you will need to be able to describe Transposition, Transformation, Transduction or Conjugation (I will pick one of these for you) (2 points)