CHAPTER 6DNA REPLICATION, REPAIR, AND RECOMBINATION

2004 Garland Science Publishing

DNA Replication

6-1 DNA replication is considered semiconservative because(a) after many rounds of DNA replication, the original DNA double helix is still

intact.(b) each daughter DNA molecule consists of two new strands copied from the parent

DNA molecule.(c) each daughter DNA molecule consists of one strand from the parent DNA

molecule and one new strand.(d) new DNA strands must be copied from a DNA template.(e) an RNA primer must be used to initiate synthesis of the DNA strand.

6-2 If the genome of the bacterium E. coli requires about 20 minutes to replicate itself, how can the genome of the fruit fly Drosophila be replicated in only three minutes?(a) The Drosophila genome is smaller than the E. coli genome.(b) Eucaryotic DNA polymerase synthesizes DNA at a much faster rate than

procaryotic DNA polymerase.(c) The nuclear membrane keeps the Drosophila DNA concentrated in one place in

the cell, which increases the rate of polymerization.(d) Drosophila DNA contains more origins of replication than E. coli DNA.(e) Eucaryotes have more than one kind of DNA polymerase.

6-3 Answer the following questions about DNA replication.A. On a DNA strand that is being synthesized, which end is growing—the 3 end, the

5 end, or both ends? Explain your answer.B. On a DNA strand that is being used as a template, where is the copying occurring

relative to the replication origin—3 of the origin, 5, or both?

6-4 If DNA strands were paired in a parallel rather than antiparallel fashion, how would the replication of the DNA differ from that of normal double-stranded DNA?(a) Replication would not be semiconservative.(b) Replication origins would not be required.(c) The replication fork would not be asymmetrical.(d) The polymerase used would not be self-correcting.(e) Both new strands would be synthesized discontinuously.

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6-5 On Figure Q6-5 of a replication bubble:

Figure Q6-5

A. Indicate where the origin of replication was located (use O).B. Label the leading-strand template and the lagging-strand template of the right-

hand fork [R] as X and Y, respectively.C. Indicate by arrows the direction in which the newly made DNA strands (indicated

by dark lines) were synthesized.D. Number the Okazaki fragments on each strand 1, 2, and 3 in the order in which

they were synthesized. E. Indicate where the most recent DNA synthesis has occurred (use S).F. Indicate the direction of movement of the replication forks with arrows.

6-6 The lagging strand is synthesized discontinuously at the replication fork because(a) the lagging strand template is discontinuous.(b) DNA polymerase always falls off the template DNA every ten nucleotides or so.(c) DNA polymerase can polymerize nucleotides only in the 5-to-3 direction.(d) DNA polymerase removes the last few nucleotides synthesized whenever it stops.(e) None of the above

6-7 Is the following statement TRUE or FALSE?

When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand.

Explain your answer with a diagram illustrating sequential snapshots of the meeting of two adjacent replication forks.

6-8 Which one of the following statements about the newly synthesized strand of a human chromosome is correct?(a) It was synthesized from a single origin solely by continuous DNA synthesis.(b) It was synthesized from a single origin by a mixture of continuous and

discontinuous DNA synthesis.(c) It was synthesized from multiple origins solely by discontinuous DNA synthesis.(d) It was synthesized from multiple origins by a mixture of continuous and

discontinuous DNA synthesis.(e) It was synthesized from multiple origins by either continuous or discontinuous

DNA synthesis, depending on which specific daughter chromosome is being examined.

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6-9 You have discovered an “Exo–” mutant form of DNA polymerase in which the 3-to-5 exonuclease function has been destroyed but the ability to join nucleotides together is unchanged. Which of the following properties do you expect the mutant polymerase to have?(a) It will polymerize in both the 5-to-3 direction and the 3-to-5 direction.(b) It will polymerize more slowly than the normal Exo+ polymerase.(c) To replicate the same amount of DNA, it will hydrolyze fewer

deoxyribonucleotides than will the normal Exo+ polymerase.(d) It will fall off the template more frequently than the normal Exo+ polymerase.(e) It will introduce fewer mutations into new strands than the normal Exo+

polymerase.

6-10 Replication of DNA requires a primer to initiate DNA synthesis because(a) DNA polymerase can add its first nucleotide only to an RNA chain.(b) DNA polymerase can add a nucleotide only to a base-paired nucleotide with a

free 3 end.(c) DNA polymerase can polymerize nucleotides only in the 5-to-3 direction.(d) DNA polymerase can polymerize DNA only in short fragments.(e) DNA polymerase has a 3-to-5 exonuclease activity.

6-11 Indicate whether each of the following statements is correct or incorrect. Explain your answers.(a) Primase is less accurate than DNA polymerase at copying a DNA template.(b) The RNA primer remains as a permanent part of the new DNA molecule.(c) Replication of the leading strand does not require primase.(d) Longer primers are required to synthesize longer DNA fragments.(e) Primase can join ribonucleotides to create an RNA strand, using a single-stranded

DNA as a template, without the need for its own primer.

6-12 A. You are studying a strain of bacteria that carries a temperature-sensitive mutation in one of the genes required for DNA replication. The bacteria grow normally at the lower temperature, but when the temperature is raised they die. When you analyze the remains of the bacterial cells grown at the higher temperature you find evidence of partly replicated DNA. When the strands of this DNA are separated by heating, numerous single-stranded DNA molecules around 1000 nucleotides long are found. Which of the proteins listed below are most likely to be impaired in these mutant bacteria? Explain your answer.

B. Next to the proteins listed below, write the number (1, 2, 3, and so on) that corresponds to the order in which the proteins function during the synthesis of a new stretch of DNA.

DNA ligase PrimaseDNA polymerase Repair polymeraseHelicase RNA nucleaseInitiator proteins Single-stranded binding protein

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6-13 Which of the following proteins or protein complexes are most abundant near the replication fork? Why?(a) Single-strand binding protein(b) Sliding clamp(c) DNA polymerase(d) Helicase(e) Primase

6-14 A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but fails to replicate. Where do you think the block to replication arises? Choose the protein or protein complex below that is most likely responsible for the failure to replicate bacterial DNA. Give an explanation for your answer.(a) Primase(b) Helicase(c) DNA polymerase(d) Sliding clamp protein(e) Initiator proteins

6-15 Indicate whether the following statements about plasmids are TRUE or FALSE.(a) Replication of plasmid DNA is independent of a replication origin.(b) Maintenance of plasmid over the course of several cell divisions requires

telomerase activity.(c) DNA replication in plasmids is bidirectional.(d) Plasmids are experimentally useful for introducing specific DNA sequences into

yeasts and bacteria.(e) Plasmids were used to identify the replication origins from human DNA.

6-16 Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and thus not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

DNA Repair

6-17 A pregnant mouse is exposed to high levels of a chemical. Many of the mice in her litter are deformed, but when they are interbred with each other, all their offspring are normal. Which TWO of the following statements could explain these results?(a) In the deformed mice, somatic cells but not germ cells were mutated.(b) The original mouse’s germ cells were mutated.(c) In the deformed mice, germ cells but not somatic cells were mutated.(d) The toxic chemical affects development but is not mutagenic.(e) The original mouse was defective in DNA repair.

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6-18 During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine this error was not corrected and has no effect on the ability of the progeny to grow and reproduce.A. After this original bacterium divides once, what proportion of its progeny would

you expect to contain the mutation? B. What proportion of its progeny would you expect to contain the mutation after

three more rounds of DNA replication and cell division?

6-19 Mismatch repair of DNA(a) is carried out solely by the replicating DNA polymerase.(b) involves cleavage of the DNA backbone to excise a stretch of single-stranded

DNA containing a mispaired base.(c) preferentially repairs the leading strand to match the lagging strand.(d) makes replication 100,000 times more accurate.(e) can occur only on newly replicated DNA.

6-20 Which of the following DNA repair processes accurately restores genetic information only immediately after the DNA has been replicated? Explain your answer.(a) Repair of deamination(b) Repair of depurination(c) Mismatch repair(d) Repair of pyrimidine dimers

6-21 Several members of the same family were diagnosed with the same kind of cancer when they were unusually young. Which one of the following is the most likely explanation for this phenomenon? Possibly, the individuals with the cancer have(a) inherited a cancer-causing gene that was mutated in an ancestor’s somatic cells.(b) inherited a mutation in a gene required for DNA synthesis.(c) inherited a mutation in a gene required for mismatch repair.(d) inherited a mutation in a gene required for the synthesis of purine nucleotides.(e) independently accumulated multiple random mutations over a period of years

leading to cancer.

6-22 If uncorrected, deamination of cytosine in DNA is most likely to lead to(a) substitution of an AT base pair for a CG base pair.(b) deletion of the altered CG base pair from the DNA.(c) conversion of the DNA into RNA.(d) generation of a thymine dimer.(e) None of the above.

6-23 Which of the following compounds is likely to be the most mutagenic?(a) One that depurinates DNA.(b) One that replaces adenine with guanine during DNA replication.(c) One that nicks the sugar-phosphate backbone.(d) One that causes thymidine dimers.(e) One that crosslinks together the two strands of the double helix.

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6-24 You have made a collection of mutant fruit flies that are defective in various aspects of DNA repair. You test each mutant for its hypersensitivity to three DNA-damaging agents: sunlight, nitrous acid (which causes deamination of cytosine), and formic acid (which causes depurination). The results are summarized in Figure Q6-24, where a “yes” indicates that the mutant is more sensitive than a normal fly and blanks indicate normal sensitivity.

Figure Q6-24

A. Which mutant is most likely to be defective in the DNA repair polymerase?B. What aspect of repair is most likely to be affected in the other mutants?

6-25 You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. What is the most likely explanation for this?(a) Coding sequences are repaired more efficiently.(b) Coding sequences are replicated more accurately.(c) Coding sequences are packaged more tightly in the chromosomes to protect them

from DNA damage.(d) Mutations in coding sequences are more likely to be deleterious to the organism

than mutations in noncoding sequences.(e) DNA repair enzymes preferentially repair the newly replicated strand to match the

original strand.

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DNA Recombination

6-26 Homologous recombination is initiated by double-strand breaks (DSBs) in a chromosome. DSBs arise from DNA damage caused by harmful chemicals or by radiation (for example, x-rays). During meiosis, the specialized cell division that produces gametes (sperm and eggs) for sexual reproduction, the cells intentionally cause DSBs in order to stimulate crossover homologous recombination. If there is not at least one occurrence of crossing-over within each pair of homologous chromosomes during meiosis, those non-crossover chromosomes will segregate randomly during division.

Figure Q6-26

A. Consider the copy of chromosome 3 that you received from your mother. Is it identical to the chromosome 3 that she received from her mother (her maternal chromosome) or identical to the chromosome 3 she received from her father (her paternal chromosome) or neither? Explain.

B. Starting with the representation in Figure Q6-26 of the double stranded maternal and paternal chromosomes found in your mother, draw two possible chromosomes you may have received from your mother.

C. What does this indicate about your resemblance to your grandfather and grandmother?

6-27 Identify each statement below as TRUE or FALSE. Explain your answers.A. Site-specific recombination can repair sites of damaged DNA.B. Mobile genetic elements comprise nearly half of the human genome.C. Viruses probably evolved from intracellular mobile genetic elements.D. Genes that contain instructions for making motor proteins are called mobile

genetic elements.E. Homologous recombination results in the accumulation of mobile genetic

elements.

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6-28 Which of the following DNA sequences are commonly carried on mobile genetic elements? You may choose more than one option.(a) Transposase gene(b) Holliday junction(c) Recognition site for transposase(d) Antibiotic resistance gene(e) Replication origin

6-29 Retrotransposons(a) are found only in eucaryotes.(b) can move by either the cut-and-paste mechanism or by a mechanism requiring an

RNA intermediate.(c) include the transposable elements LINE-1, Alu, and Tn10.(d) can move only if they encode a reverse transcriptase.(e) were multiplied to high copy numbers in a common ancestor of all mammals.

6-30 Describe the likely consequence of introducing high levels of reverse transcriptase into a human embryo.

6-31 Some retrotransposons and retroviruses integrate preferentially into regions of the chromosome that are (a) packaged in euchromatin and (b) located outside the coding regions of genes that contain information for making a protein. Why might these mobile genetic elements have evolved this strategy?

6-32 All viruses(a) have single-stranded genomes.(b) lyse the cells they infect.(c) encode all of the enzymes needed to replicate themselves.(d) contain both nucleic acid and protein.(e) have the same size genomes.

6-33 The enzymes reverse transcriptase and DNA polymerase both synthesize DNA. Which of the following statements about them are TRUE?(a) Reverse transcriptase uses only an RNA template. DNA polymerase uses only a

DNA template.(b) Reverse transcriptase can use either an RNA or a DNA template. DNA

polymerase uses only a DNA template.(c) DNA polymerase is used only by cells. Reverse transcriptase is used only by

viruses.(d) DNA polymerase uses deoxynucleotides. Reverse transcriptase uses

ribonucleotides.(e) Reverse transcriptase and DNA polymerase both contain a 3-to-5 exonuclease

activity.

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6-34 Once a retrovirus has integrated into the genome of a host cell, why would it not be possible to eradicate the virus by treating the infected cell with reverse transcriptase inhibitors?

6-35 Why do retroviruses need to package reverse transcriptase molecules into their virus particles even though they carry the gene for reverse transcriptase in their genomes?

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Answers

6-1 Choice (c) is the answer. Choices (a) and (b) are false. Although choices (d) and (e) are correct statements, they are not the reasons that DNA replication is called semiconservative.

6-2 Choice (d) is the answer. Bacteria have one origin of replication and Drosophila has many. Choice (a) is incorrect because the Drosophila genome is bigger than the E. coli genome. Choice (b) is incorrect, as eucaryotic polymerases are not faster than procaryotic polymerases. Choices (c) and (e) are technically correct statements, but are not relevant to the question.

6-3 A. The 3 end. DNA polymerase can add nucleotides only to the 3-OH end of a nucleic acid chain.

B. Both, due to the bidirectional nature of chromosomal replication.

6-4 (c) The antiparallel nature of the strands of normal DNA, combined with the 5-to-3 activity of the DNA polymerase, precludes the continuous synthesis of both strands and thus requires that the replication fork be asymmetrical. If the strands were parallel, both new strands could be synthesized continuously.

6-5 See Figure A6-5.

Figure A6-5

6-6 (a) False(b) False(c) True(d) False(e) False

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6-7 True. See Figure A6-7 for an illustration of the meeting of two adjacent replication forks.

Figure A6-7

6-8 (d) Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin: The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis; the fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.

6-9 Choice (c) is the answer. An Exo— polymerase will be unable to proofread and thus will hydrolyze fewer nucleotides than one that can proofread, because it cannot remove an incorrect nucleotide and try again to add the correct nucleotide (thus choice (e) is false). Choice (a) is unlikely because it postulates an entirely new function for the defective polymerase. Choice (b) is incorrect because if the rate of polymerization changes, it would become faster not slower, as a consequence of the Exo— polymerase lacking a proofreading function that might cause it to stop frequently to check its products and remedy errors. Choice (d) is unlikely, because the ability to stay on the template is due to the association of polymerase with additional proteins.

6-10 Choice (b) is the answer. Choices (a) and (d) are false statements. Choices (c) and (e) are true but are not the reason DNA polymerase requires a primer.

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6-11 (a) Correct. Primase lacks a 3-to-5 exonuclease activity and thus cannot proofread the nucleotide chain it makes.

(b) Incorrect. A nuclease removes the RNA primer after it provides its 3-OH to prime the synthesis of DNA. Then enzymes involved in repair synthesis and ligation join the newly synthesized stretches of DNA to make a continuous strand containing only DNA.

(c) Incorrect. The leading strand requires primase to initiate DNA synthesis, although fewer RNA primers will be needed on the leading strand than the lagging strand.

(d) Incorrect. The length of the primer does not correlate with the length of the DNA strand that is synthesized.

(e) Correct. Like other RNA polymerases, primase does not require a primer to initiate synthesis of the nucleotide chain.

6-12 A. The 1000-nucleotide fragments that accumulate in the mutant are likely to be Okazaki fragments. Thus, the mutant is likely to be defective in the function of RNA nuclease, repair polymerase, or DNA ligase. These proteins are required to remove the RNA primer, fill in the gap, and stitch together the Okazaki fragments, respectively. We could test which enzyme was defective by determining if the fragments contained a short stretch of RNA at the 5 end (nuclease defective) or if the fragments annealed to the template leaving gaps (repair polymerase defective).

B. DNA ligase—8DNA polymerase—5Helicase—2Initiator proteins—1Primase—4Repair polymerase—7RNA nuclease—6Single-stranded binding protein—3

6-13 (a) Single-strand binding protein is required to coat all single-stranded regions of DNA that form at a replication fork; this requires many molecules of single-strand binding protein at each fork. Only one or two molecules of each of the other proteins or protein complexes are required at each replication fork.

6-14 Choice (e) is the answer. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices (a) through (d) can act on any DNA regardless of its sequence. In contrast, the initiator proteins recognize specific DNA sequences at the origins of replication. These sequences differ between bacteria and yeast.

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6-15 (a) False(b) False(c) True(d) True(e) False

6-16 In the absence of telomerase, the lifespan of a cell and its progeny cells is limited. With each round of DNA replication, the length of telomeric DNA will shrink, until finally all the telomeric DNA will disappear. Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage or crucial genetic information might be lost. Cells whose DNA lacks telomeres will stop dividing or die. However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication.

6-17 Choice (a) or (d) is correct. Choice (b) cannot account for these results since a mutation in the original mouse’s germ cells would have no effect on the fetuses she was already carrying. Neither can choice (c), as mutations in the germ cells of the fetuses while in utero would have had no effect on their development, but might have led to mutant mice among their offspring. If the original mouse were defective in DNA repair (choice (e)), this would increase the number of mutations caused by the chemical but cannot explain the observed effects on her offspring and their offspring.

6-18 A. Half or 50%. DNA replication in the original bacterium will create two new DNA molecules, one of which will now carry a mismatched C-T base pair. So one daughter cell of that cell division will carry a completely normal DNA molecule; the other cell will have the molecule with the mutation mispaired to a correct nucleotide.

B. A quarter or 25%. At the next round of DNA replication and cell division, the bacterium carrying the mismatched C-T will produce and pass on one normal DNA molecule from the undamaged strand containing the T and one mutant DNA molecule with a fully mutant C-G base pair. So at this stage, one out of the four progeny of the original bacterium is mutant. Subsequent cell divisions of these mutant bacteria will give rise only to mutant bacteria, while the other bacteria will give rise to normal bacteria. The proportion of progeny containing the mutation will, therefore, remain at 25%.

6-19 Choice (b) is the answer. Choice (a) is incorrect, as mismatch repair requires specialized repair proteins that act after the DNA is replicated or at other times during the life of a cell (thus choice (e) is incorrect). Choice (c) is incorrect because mismatch repair repairs the newly replicated leading strand or the lagging strand to match its template strand. Choice (d) is incorrect, since mismatch repair makes replication approximately 100 (not 100,000) times more accurate.

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6-20 (c) Mismatches occur most often as a result of replication errors, in which case the erroneous base is found only on the newly synthesized strand. In most eucaryotic cells, the preferential repair of the erroneous base instead of its pairing partner is thought to require recognition of nicks in the sugar-phosphate backbone of the newly synthesized DNA strand. These nicks are sealed soon after replication, so mismatch repair must occur in the short interval between passage of the replication fork and sealing of the sugar-phosphate backbone in order to accurately restore the original sequence. When mismatch repair occurs at other times, it has an equal probability of restoring the original information or setting the mutation. For the other kinds of DNA damage (U-G mismatch, abasic site, pyrimidine dimer) it is always evident to determine which strand is damaged and which strand contains reliable original information.

6-21 Choice (c) is the answer. In fact, affected individuals in some families with a history of early-onset colon cancer have been found to carry mutations in mismatch repair genes. Mutations arising in somatic cells are not inherited (choice (a) is incorrect). A defect in DNA synthesis or nucleotide biosynthesis would likely be lethal (choices (b) and (d) are incorrect). Choice (e) describes the way that most cancers arise in people late in life, but it is extremely unlikely that several individuals in the same family spontaneously acquired similar random mutations leading to the early onset of the same kind of cancer.

6-22 (a)

6-23 Choice (e) is the answer. Altogether, the DNA repair pathways described in Chapter 6 can easily repair the DNA damage described in options (a)–(d), these types of damage occur on one strand of the double helix, and thus can be repaired using the intact genetic information encoded on the complementary strand. When the two strands are crosslinked and thus both strands are damaged, entirely different and probably less accurate pathways of repair are required.

6-24 A. Mr. Self-Destruct is more likely than the other mutants to be defective in the DNA repair polymerase because Mr. Self-Destruct is defective in repair of all three kinds of DNA damage. The repair pathways for all three kinds of damage are similar in the later steps, including a requirement for the DNA repair polymerase.

B. The other mutants are specific for a particular type of damage. Thus the mutations are likely to be in genes required for the first stage of repair, the recognition and excision of the damaged bases. Dracula and Mole are likely to be defective in the recognition or excision of thymidine dimers; Faust is likely to be defective in the recognition or excision of U-G mismatched base pairs; and Marguerite is likely to be defective in the recognition or excision of abasic sites.

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6-25 Choice(d) is the answer. Mutations—whether they arise by mistakes in replication or by damage to the DNA that remains unrepaired—tend to hit the DNA fairly randomly (choices (a) and (b) are false). However, if a mutation occurs in a protein-coding sequence rather than in the surrounding DNA, it is more likely to cause a deleterious change that kills or impairs the organism and thereby decreases the likelihood that the mutation will be passed on to future generations. Since skinks and Komodo dragons share a common lizard ancestor, differences in their genomes have arisen during their divergence from this ancestor. Mutations in noncoding sequences are more likely to have no effect on the functioning of the organism and thus frequently get passed along to progeny. Choice (c) is incorrect, as genes that are being expressed tend to be more loosely packaged than the noncoding DNA. Choice (e) is true, but has no bearing on the phenomenon described.

6-26 A. Neither. The copy of chromosome 3 you received from your mother is a hybrid of the ones she received from her mother and her father.

B. See Figure A6-26. The right answers include any chromosome in which a portion matches the information from the paternal chromosome and the remainder matches the information from the maternal chromosome.

Figure A6-26

C. Due to extensive crossing over, you resemble both your grandmother and your grandfather. If there were no crossing over, then you might have a much stronger resemblance to one than the other.

6-27 A. False. Homologous recombination, not site-specific recombination, is sometimes used to repair sites of damaged DNA. Site-specific recombination is used mostly in the mobilization of mobile genetic elements.

B. True. Forty-five percent of the human genome is comprised of mobile genetic elements.C. True. A good guess for how viruses evolved is that some mobile genetic elements

acquired genes encoding coat proteins and other proteins required for packaging and cellular escape of the nucleic acids of mobile genetic elements.

D. False. Cellular motor proteins are completely unrelated to mobile genetic elements.E. False. Site-specific recombination, not homologous recombination, is the primary

mechanism for the accumulation of mobile genetic elements. Homologous recombination does sometimes aid in the repair of DNA damage caused by the excision of a mobile genetic element from the chromosome, but it does not aid in the insertion of the element into a new location.

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6-28 Choices (a), (c), and (d) are correct. A Holliday junction is not a sequence, but a structural intermediate in homologous recombination (choice (b) is false). Mobile genetic elements often have no replication origin, since their movement and proliferation occurs by a type of repair DNA synthesis and their replication during cell division occurs by passage of a replication fork that originates elsewhere in the chromosome (choice (e) is false).

6-29 Choice (a) is the answer. Retrotransposons are found only in eucaryotes. Retrotransposons (by definition) move only through an RNA intermediate (choice (b) is false), and include LINE-1 and Alu sequences (but not the bacterial transposon Tn10, so choice (c) is false). Retrotransposons do not necessarily need to provide their own reverse transcriptase so long as there is an alternative source of reverse transcriptase in the cell (choice (d) is false). The LINE-1 and Alu retrotransposons that comprise about a third of the human genome are not identical in sequence or location to the retrotransposons in other mammals like mice (choice (e) is false).

6-30 The embryo would probably have severe developmental abnormalities or die. The huge numbers of retrotransposons littering the human genome are largely immobile due to the accumulation of disabling mutations. However, it is likely that at least a few of the millions of copies of the transposons still contain the sites necessary for retrotransposition, although they do not encode a functional reverse transcriptase. High levels of reverse transcriptase will probably cause many retrotransposition events. The resultant insertion of retrotransposons is likely to disable genes required for development or survival.

6-31 The most evolutionarily successful mobile genetic elements are those that are best at reproducing themselves. In order to increase the number of copies of a particular element, the element must meet two criteria: (1) it must not kill its host and (2) it must maximize its ability to continue reproducing. If an element inserts into the coding region of a gene, it might disable the gene and thereby confer a selective disadvantage in the reproduction or survival of its host. Thus, elements that devised a way to avoid insertion into coding regions probably were better able to increase their copy number throughout the human population. If an element inserts into a heterochromatic region of a chromosome, its genes may not be expressed and therefore it may become immobile. Elements that devised a way to direct insertion into euchromatin would be more likely to maintain mobility and thereby increase their copy number over time.

6-32 Choice (d) is the answer. All viruses contain both protein and nucleic acid. Viruses can have either double- or single-stranded genomes (choice (a)). Not all viruses lyse the cells they infect (choice (b)); for example, some bud out of the cell without killing it. Viruses can have as few as three genes or more than a hundred (choice (e)). No virus is able to replicate in the absence of a host cell (choice (c)).

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6-33 Choice (b) is the answer. Reverse transcriptase can use an RNA or DNA template (thus choice (a) is false). Choice (c) is false because DNA polymerase is often used by DNA viruses. Choice (d) is false because both enzymes polymerize deoxynucleotides. Choice (e) is false because reverse transcriptase does not have proofreading exonuclease activity.

6-34 Once the virus has integrated into the genome, it has no further need for reverse transcriptase. Therefore, an inhibitor of reverse transcriptase may be able to block infection of other cells by viruses that bud off the infected cell, but it will not be able to eradicate the integrated virus. If the virus integrates into the genome of a cell with the potential to divide, it will be faithfully propagated along with all the genomic DNA to all progeny of that cell.

6-35 Retroviruses carry their own reverse transcriptase with them, as they must produce a double-stranded DNA copy of their genome before their genes can be transcribed and expressed.

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