molecular bio: test 1 chapter 6

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.

(d) Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

The sliding clamp complex encircles the DNA template and binds to DNA polymerase. This helps the polymerase synthesize much longer stretches of DNA without dissociating. While the loading of the clamp only occurs once on the leading strand, it must happen each time a new Okazaki fragment is made on the lagging strand. How does the cell expedite this process?

The cell employs an additional protein in order to make the constant reloading of the sliding clamp on the lagging strand much more efficient. The protein, called the clamp loader, harnesses energy from ATP hydrolysis to lock a sliding clamp complex around the DNA for every successive round of DNA synthesis.

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication. ______ primase ______ single-strand binding protein ______ sliding clamp ______ RNA primers ______ leading strand ______ lagging strand ______ Okazaki fragments ______ DNA helicase ______ DNA ligase

___3___ primase ___2___ single-strand binding protein ___3___ sliding clamp ___3___ RNA primers ___1___ leading strand ___2___ lagging strand ___2___ Okazaki fragments ___3___ DNA helicase ___2___ DNA ligase

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

(a) Because helicase unwinds the two DNA template strands, replication of both strands depends upon the activity of helicase at the time of initiation.

The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process. Which choice below describes the known sequence of events in this process? (a) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA ligase. (b) DNA repair polymerase simultaneously removes bases ahead of it and polymerizes the correct sequence behind it as it moves along the template. DNA ligase seals the nicks in the repaired strand. (c) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by an exonuclease, and the gap is repaired by DNA ligase. (d) A nick in the DNA is recognized, DNA repair proteins switch out the wrong base and insert the correct base, and DNA ligase seals the nick.

(a) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA ligase.

Homologous recombination is an important mechanism in which organisms use a "backup" copy of the DNA as a template to fix double-strand breaks without loss of genetic information. Which of the following is not necessary for homologous recombination to occur? (a) 3′ DNA strand overhangs (b) 5′ DNA strand overhangs (c) a long stretch of sequence similarity (d) nucleases

(b) 5′ DNA strand overhangs

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. Which of the following statements is true with respect to this in vitro replication system? (a) There will be only one leading strand and one lagging strand produced using this template. (b) The leading and lagging strands compose one half of each newly synthesized DNA strand. (c) The DNA replication machinery can assemble at multiple places on this plasmid. (d) One daughter DNA molecule will be slightly shorter than the other.

(b) Leading and lagging strands are synthesized bidirectionally from the replication origin, and are joined together by DNA ligase where the two replication forks meet at the termination site. Choice (a) is not correct, because this answer implies that the replication fork is not bidirectional and that replication continues around the plasmid until the process makes it back to the origin of replication. Choice (c) is incorrect because the origin is a specialized sequence where initiator proteins bind and open the DNA so that the DNA replication machinery can assemble. Choice (d) is incorrect because the daughter DNA molecules will be same size as the original plasmid (and each other).

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

(b) Okazaki fragment synthesis

Which of the following statements about sequence proofreading during DNA replication is false? (a) The exonuclease activity is in a different domain of the DNA polymerase. (b) The exonuclease activity cleaves DNA in the 5′-to-3′ direction. (c) The DNA proofreading activity occurs concomitantly with strand elongation. (d) If an incorrect base is added, it is "unpaired" before removal.

(b) The exonuclease activity cleaves DNA in the 5′-to-3′ direction.

Which of the following statements is not an accurate statement about thymine dimers? (a) Thymine dimers can cause the DNA replication machinery to stall. (b) Thymine dimers are covalent links between thymidines on opposite DNA strands. (c) Prolonged exposure to sunlight causes thymine dimers to form. (d) Repair proteins recognize thymine dimers as a distortion in the DNA backbone.

(b) Thymine dimers are covalent links between thymidines on opposite DNA strands.

Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the adenosine in the sequence TCAT is depurinated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication? (a) TCGT (b) TAT (c) TCT (d) TGTT

(c) TCT

In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes. During which process does swapping occur? (a) DNA replication (b) DNA repair (c) meiosis (d) transposition

(c) meiosis

The process of DNA replication requires that each of the parental DNA strands be used as a ___________________ to produce a duplicate of the opposing strand. (a) catalyst (b) competitor (c) template (d) copy

(c) template

Select the option that best completes the following statement: Nonhomologous end joining is a process by which a double-stranded DNA end is joined ___________________. (a) to a similar stretch of sequence on the complementary chromosome. (b) after repairing any mismatches. (c) to the nearest available double-stranded DNA end. (d) after filling in any lost nucleotides, helping to maintain the integrity of the DNA sequence.

(c) to the nearest available double-stranded DNA end.

How does the total number of replication origins in bacterial cells compare with the number of origins in human cells? (a) 1 versus 100 (b) 5 versus 500 (c) 10 versus 1000 (d) 1 versus 10,000

(d) 1 versus 10,000

Even though DNA polymerase has a proofreading function, it still introduces errors in the newly synthesized strand at a rate of 1 per 107 nucleotides. To what degree does the mismatch repair system decrease the error rate arising from DNA replication? (a) 2-fold (b) 5-fold (c) 10-fold (d) 100-fold

(d) 100-fold

The events listed below are all necessary for homologous recombination to occur properly: A. Holliday junction cut and ligated B. strand invasion C. DNA synthesis D. DNA ligation E. double-strand break F. nucleases create uneven strands Which of the following is the correct order of events during homologous recombination? (a) E, B, F, D, C, A (b) B, E, F, D, C, A (c) C, E, F, B, D, A (d) E, F, B, C, D, A

(d) E, F, B, C, D, A

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. You decide to use different bacterial strains (each having one protein of the replication machinery mutated) in order to examine the role of individual proteins in the normal process of DNA replication. What part of the DNA replication process would be most directly affected if a strain of bacteria lacking primase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

Choice (a) is the best answer because DNA synthesis cannot begin without the initial primers. Choice (b) is a good answer because lagging-strand synthesis requires continual use of RNA primers for discontinuous replication to occur.

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.

Choice (c) is the correct answer. Choices (a) and (b) are false. Although choice (d) is a correct statement, it is not the reason that DNA replication is called semiconservative.

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. What part of the DNA replication process would be most directly affected if a strain of bacteria lacking DNA ligase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

(d) lagging-strand completion

How many replication forks are formed when an origin of replication is opened? (a) 1 (b) 2 (c) 3 (d) 4

(b) 2

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. What part of the DNA replication process would be most directly affected if a strain of bacteria lacking the exonuclease activity of DNA polymerase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

(d) lagging-strand completion

Human beings with the inherited disease xeroderma pigmentosum have serious problems with lesions on their skin and often develop skin cancer with repeated exposure to sunlight. What type of DNA damage is not being recognized in the cells of these individuals? (a) chemical damage (b) X-ray irradiation damage (c) mismatched bases (d) ultraviolet irradiation damage

(d) ultraviolet irradiation damage

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. The repair polymerase is the enzyme that proofreads the newly synthesized strands to ensure the accuracy of DNA replication. B. There is a single enzyme that degrades the RNA primers and lays down the corresponding DNA sequence behind it. C. DNA ligase is required to seal the sugar-phosphate backbone between all the DNA fragments on the lagging strand. D. The repair polymerase does not require the aid of the sliding clamp, because it is only synthesizing DNA over very short stretches.

A. False. The repair polymerase is used to fill in the spaces left vacant after the RNA primers are degraded. B. False. This is a two-step process that requires two different enzymes. First, a nuclease removes the RNA primers. Then, the repair polymerase fills in the complementary DNA sequence. C. True. D. True.

Meselson and Stahl grew cells in media that contained different isotopes of nitrogen (15N and 14N) so that the DNA molecules produced from these different isotopes could be distinguished by mass. A. Explain how "light" DNA was separated from "heavy" DNA in the Meselson and Stahl experiments. B. Describe the three existing models for DNA replication when these studies were begun, and explain how one of them was ruled out definitively by the experiment you described for part A. C. What experimental result eliminated the dispersive model of DNA replication?

A. The DNA samples collected were placed into centrifuge tubes containing cesium chloride. After high-speed centrifugation for 2 days, the heavy and light DNA products were separated by density. B. The three models were conservative, semiconservative, and dispersive. The conservative model suggested a mechanism by which the original parental strands stayed together after replication and the daughter duplex was made entirely of newly synthesized DNA. The semiconservative model proposed that the two DNA duplexes produced during replication were hybrid molecules, each having one of the parental strands and one of the newly synthesized strands. The dispersive model predicted that the new DNA duplexes each contained segments of parental and daughter strands all along the molecule. The conservative model was ruled out by the density-gradient experiments. C. The dispersive model was ruled out by using heat to denature the DNA duplexes and then comparing the densities of the single-stranded DNA. If the dispersive model had been correct, individual strands should have had an intermediate density. However, this was not the case; only heavy strands and light strands were observed, which convincingly supported the semiconservative model for DNA replication.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. Homologous recombination cannot occur in prokaryotic cells, because they are haploid, and therefore have no extra copy of the chromosome to use as a template for repair. B. The first step in repair requires a nuclease to remove a stretch of base pairs from the 5′ end of each strand at the site of the break. C. The 3′ overhang "invades" the homologous DNA duplex, which can be used as a primer for the repair DNA polymerase. D. The DNA template used to repair the broken strand is the homologous chromosome inherited from the other parent.

A. False. Homologous recombination also occurs in prokaryotic cells, and typically occurs very shortly after DNA replication, when the newly replicated duplexes are in close proximity. B. True. C. True. D. False. Although it is called homologous recombination, this is not a process that depends on the proximity of parental homologs. When used as a mechanism for DNA repair, homologous recombination uses the sister chromatids in an undamaged, newly replicated (homologous) DNA helix as a template.

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 Table Q6-55, where a "yes" indicates that the mutant is more sensitive than a normal fly, and blanks indicate normal sensitivity. Table Q6-55 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?

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 the 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.

The synthesis of DNA in living systems occurs in the 5′-to-3′ direction. However, scientists synthesize short DNA sequences needed for their experiments on an instrument dedicated to this task. A. The chemical synthesis of DNA by this instrument proceeds in the 3′-to-5′ direction. Draw a diagram to show how this is possible and explain the process. B. Although 3′-to-5′ synthesis of DNA is chemically possible, it does not occur in living systems. Why not?

A. The actual chemical reaction in DNA synthesis is the same regardless of whether going in the 5′-to-3′ or in the 3′-to-5′ direction. The most important distinction between these two options is that if DNA is synthesized in the 3′-to-5′ direction, the 5′ end of the elongating strand, rather than the 3′ end, will have a nucleoside triphosphate. B. DNA synthesis from 3′ to 5′ does not allow proofreading. If the last nucleotide added is mispaired and is removed, the last nucleotide on the growing strand is a nucleoside monophosphate and the nucleotide coming in only has a hydroxyl group on the 3′ end. Thus, there is no favorable hydrolysis reaction to drive the addition of new nucleotides.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. Primase is needed to initiate DNA replication on both the leading strand and the lagging strand. B. The sliding clamp is loaded once on each DNA strand, where it remains associated until replication is complete. C. Telomerase is a DNA polymerase that carries its own RNA molecule to use as a primer at the end of the lagging strand. D. Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication.

A. True B. False. Although the sliding clamp is only loaded once on the leading strand, the lagging strand needs to unload the clamp once the polymerase reaches the RNA primer from the previous segment and then reload it where a new primer has been synthesized. C. True. D. False. Primase does not have a proofreading function, nor does it need one because the RNA primers are not a permanent part of the DNA. The primers are removed, and a DNA polymerase that does have a proofreading function fills in the remaining gaps.

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? It is possible that the individuals with the cancer have _______________________. (a) inherited a cancer-causing gene that suffered a mutation 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.

Choice (c) is the correct 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, so choice (a) is incorrect. A defect in DNA synthesis or nucleotide biosynthesis would probably be lethal, so choices (b) and (d) are incorrect.

Researchers have isolated a mutant strain of E. coli that carries a temperature- sensitive variant of the enzyme DNA ligase. At the permissive temperature, the mutant cells grow just as well as the wild-type cells. At the nonpermissive temperature, all of the cells in the culture tube die within 2 hours. DNA from mutant cells grown at the nonpermissive temperature for 30 minutes is compared with the DNA isolated from cells grown at the permissive temperature. The results are shown in Figure Q6-33, where DNA molecules have been separated by size by means of electrophoresis (P, permissive; NP, nonpermissive). Explain the appearance of a distinct band with a size of 200 base pairs (bp) in the sample collected at the nonpermissive temperature. Figure Q6-33

DNA ligase has an important role in DNA replication. After Okazaki fragments are synthesized, they must be ligated (covalently connected) to each other so that they finally form one continuous strand. At the nonpermissive temperature this does not happen, and although there may be a range of fragments, the notable band at 200 base pairs is the typical size of an individual Okazaki fragment.

Figure Q6-15 shows a replication bubble. Figure Q6-15 A. On the figure, 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 as 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.

See Figure A6-15.

Use the components in the list below to label the diagram of a replication fork in Figure Q6-32. A. DNA polymerase B. single-strand binding protein C. Okazaki fragment D. primase E. sliding clamp F. RNA primer G. DNA helicase

See Figure A6-32.

During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine that this error was not corrected and that it has no effect on the ability of the progeny to grow and reproduce. A. After this original bacterium has divided 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?

a. One-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. One-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, whereas the other bacteria will give rise to normal bacteria. The proportion of progeny containing the mutation will therefore remain at 25%.

Beside the distortion in the DNA backbone caused by a mismatched base pair, what additional mark is there on eukaryotic DNA to indicate which strand needs to be repaired? (a) a nick in the template strand (b) a chemical modification of the new strand (c) a nick in the new strand (d) a sequence gap in the new strand

c

6-27 Which diagram accurately represents the directionality of DNA strands at one side of a replication fork?

d

Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the cytosine in the sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication? (a) TTAT (b) TUA T (c) TGA T (d) TAAT

(a) TTAT

DNA polymerase catalyzes the joining of a nucleotide to a growing DNA strand. What prevents this enzyme from catalyzing the reverse reaction? (a) hydrolysis of pyrophosphate (PPi) to inorganic phosphate (Pi) + Pi (b) release of PPi from the nucleotide (c) hybridization of the new strand to the template (d) loss of ATP as an energy source

(a) hydrolysis of pyrophosphate (PPi) to inorganic phosphate (Pi) + Pi

Nonhomologous end joining can result in all but which of the following? (a) the recovery of lost nucleotides on a damaged DNA strand (b) the interruption of gene expression (c) loss of nucleotides at the site of repair (d) translocations of DNA fragments to an entirely different chromosome

(a) the recovery of lost nucleotides on a damaged DNA strand

In somatic cells, if a base is mismatched in one new daughter strand during DNA replication, and is not repaired, what fraction of the DNA duplexes will have a permanent change in the DNA sequence after the second round of DNA replication? (a) 1/2 (b) 1/4 (c) 1/8 (d) 1/16

(b) 1/4

The DNA duplex consists of two long covalent polymers wrapped around each other many times over their entire length. The separation of the DNA strands for replication causes the strands to be "overwound" in front of the replication fork. How does the cell relieve the torsional stress created along the DNA duplex during replication? (a) Nothing needs to be done because the two strands will be separated after replication is complete. (b) Topoisomerases break the covalent bonds of the backbone allowing the local unwinding of DNA ahead of the replication fork. (c) Helicase unwinds the DNA and rewinds it after replication is complete. (d) DNA repair enzymes remove torsional stress as they replace incorrectly paired bases.

(b) Topoisomerases break the covalent bonds of the backbone allowing the local unwinding of DNA ahead of the replication fork

Sickle-cell anemia is an example of an inherited disease. Individuals with this disorder have misshapen (sickle-shaped) red blood cells caused by a change in the sequence of the β-globin gene. What is the nature of the change? (a) chromosome loss (b) base-pair change (c) gene duplication (d) base-pair insertion

(b) base-pair change

The classic experiments conducted by Meselson and Stahl demonstrated that DNA replication is accomplished by employing a ________________ mechanism. (a) continuous (b) semiconservative (c) dispersive (d) conservative

(b) semiconservative

DNA polymerases are processive, which means that they remain tightly associated with the template strand while moving rapidly and adding nucleotides to the growing daughter strand. Which piece of the replication machinery accounts for this characteristic? (a) helicase (b) sliding clamp (c) single-strand binding protein (d) primase

(b) sliding clamp

Initiator proteins bind to replication origins and disrupt hydrogen bonds between the two DNA strands being copied. Which of the factors below does not contribute to the relative ease of strand separation by initiator proteins? (a) replication origins are rich in A-T base pairs (b) the reaction can occur at room temperature (c) they only separate a few base pairs at a time (d) once opened, other proteins of the DNA replication machinery bind to the origin

(b) the reaction can occur at room temperature

The chromatin structure in eukaryotic cells is much more complicated than that observed in prokaryotic cells. This is thought to be the reason that DNA replication occurs much faster in prokaryotes. How much faster is it? (a) 2× (b) 5× (c) 10× (d) 100×

(c) 10×

Which of the following statements about the newly synthesized strand of a human chromosome is true? (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.

(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.

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) It will fall off the template more frequently than the normal Exo+ polymerase. (d) It will be more likely to generate mismatched base pairs.

(d) It will be more likely to generate mismatched base pairs.

Telomeres serve as caps at the ends of linear chromosomes. Which of the following is not true regarding the replication of telomeric sequences? (a) The lagging-strand telomeres are not completely replicated by DNA polymerase. (b) Telomeres are made of repeating sequences. (c) Additional repeated sequences are added to the template strand. (d) The leading strand doubles back on itself to form a primer for the lagging strand.

(d) The leading strand doubles back on itself to form a primer for the lagging strand.

Which of the following statements correctly explains what it means for DNA replication to be bidirectional? (a) The replication fork can open or close, depending on the conditions. (b) The DNA replication machinery can move in either direction on the template strand. (c) Replication-fork movement can switch directions when the fork converges on another replication fork. (d) The replication forks formed at the origin move in opposite directions.

(d) The replication forks formed at the origin move in opposite directions.

Recombination has occurred between the chromosome segments shown in Figure Q6-61. The genes A and B, and the recessive alleles a and b, are used as markers on the maternal and paternal chromosomes, respectively. After alignment and homologous recombination, the specific arrangements of A, B, a, and b have changed. Figure Q6-61 Which of the choices below correctly indicates the gene combination from the replication products of the maternal chromosome? (a) AB and aB (b) ab and Ab (c) AB and Ab (d) aB and Ab

(d) aB and Ab

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. When DNA is being replicated inside a cell, local heating occurs, allowing the two strands to separate. B. DNA replication origins are typically rich in G-C base pairs. C. Meselson and Stahl ruled out the dispersive model for DNA replication. D. DNA replication is a bidirectional process that is initiated at multiple locations along chromosomes in eukaryotic cells.

A. False. The two strands do need to separate for replication to occur, but this is accomplished by the binding of initiator proteins at the origin of replication. B. False. DNA replication origins are typically rich in A-T base pairs, which are held together by only two hydrogen bonds (instead of three for C-G base pairs), making it easier to separate the strands at these sites. C. True. D. True.

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 so as 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 noncrossover chromosomes will not segregate properly. Figure Q6-43 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 Q6-43 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?

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 A6-43. The correct 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 A6-43 C. As a result of extensive crossing-over, you resemble both your grandmother and your grandfather. If there were no crossing-over, you might have a much stronger resemblance to one than the other.

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?

A. The 3′ end. DNA polymerase can add nucleotides only to the 3′-OH end of a nucleic acid chain. B. Both, as a result of the bidirectional nature of chromosomal replication.

Indicate whether the following statements are true or false. If a statement is false, explain why it is false. A. Ionizing radiation and oxidative damage can cause DNA double-strand breaks. B. After damaged DNA has been repaired, nicks in the phosphate backbone are maintained as a way to identify the strand that was repaired. C. Depurination of DNA is a rare event that is caused by ultraviolet irradiation. D. Nonhomologous end joining is a mechanism that ensures that DNA double-strand breaks are repaired with a high degree of fidelity to the original DNA sequence.

A. True. B. False. It is believed that the nicks are generated during DNA replication as a means of easy identification of the newly synthesized strand but are sealed by DNA ligase shortly after replication is completed. C. False. Depurination occurs constantly in our cells through spontaneous hydrolysis of the bond linking the DNA base to the deoxyribose sugar. D. False. Homologous recombination can repair double-strand breaks without any change in DNA sequence, but nonhomologous end joining always involves a loss of genetic information because the ends are degraded by nucleases before they can be ligated back together.

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 3 minutes? (a) The Drosophila genome is smaller than the E. coli genome. (b) Eukaryotic DNA polymerase synthesizes DNA at a much faster rate than prokaryotic 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.

Choice (d) is the correct 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, because eukaryotic polymerases are not faster than prokaryotic polymerases.

A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but fails to replicate with the yeast DNA. Where do you think the block to replication arises? Choose the protein or protein complex below that is most probably responsible for the failure to replicate bacterial DNA. Give an explanation for your answer. (a) primase (b) helicase (c) DNA polymerase (d) initiator proteins

Choice (d) is the correct answer. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices (a) to (c) 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.

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.

Choices (a) or (d) are correct. Choice (b) cannot account for these results because a mutation in the original mouse's germ cells would have no effect on the fetuses she was already carrying. Neither can choice (c), because mutations in the germ cells of the fetuses while in utero would have had no effect on their development, but they might have led to mutant mice among their offspring.

Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore 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.

In the absence of telomerase, the life-span 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 has disappeared. 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.

A mismatched base pair causes a distortion in the DNA backbone. If this were the only indication of an error in replication, the overall rate of mutation would be much higher. Explain why.

The distortion in the DNA backbone is insufficient information for the mismatch repair system to identify which base is incorrect and which was originally part of the chromosome when replication began. Without additional marks that identify the difference between the newly synthesized strand and the template strand, the repair would be corrected only 50% of the time by random chance. The error rate (and therefore the mutation rate) would still be less than in a system that lacked the mismatch repair enzymes (1 mistake per 107 base pairs), but greater than the error rate in a system that accurately identifies the newly synthesized strand (1 mistake per 109 base pairs).

Use your knowledge of how a new strand of DNA is synthesized to explain why DNA replication must occur in the 5′-to-3′ direction. In other words, what would be the consequences of 3′-to-5′ strand elongation?

There would be several detrimental consequences to 3′-to-5′ strand elongation. One of those most directly linked to the processes of DNA replication involves synthesis of the lagging strand. After the RNA primers are degraded, the DNA segments remaining will have 5′ ends with a single phosphate group. The incoming nucleotide will have a 3′-OH group. Without the energy provided by the release of PPi from the 5′ end, the process of elongation would no longer be energetically favorable.

The deamination of cytosine generates a uracil base. This is a naturally occurring nucleic acid base, and so does not represent a DNA lesion caused by damage due to chemicals or irradiation. Why is this base recognized as "foreign" and why is it important for cells to have a mechanism to recognize and remove uracil when it is found in the DNA duplex?

Uracil is an RNA base and it is recognized as a mutational lesion because, as it is formed from the deamination of cytosine, it will be paired with a guanine in the context of the DNA duplex. Uracil pairs by forming two hydrogen bonds, similar to thymine, and is thus a poor partner for guanine, which forms three hydrogen bonds with cytosine. The mismatch causes a distortion of the DNA backbone, allowing the repair machinery to recognize the uracil as a lesion. Because uracil pairs preferably with adenine (its partner in double-stranded RNA), the deamination of cytosine to uracil is highly mutagenic. If unrepaired, it can result in the transition of a C-G base pair to a T-A base pair.

You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. 6-16 In addition to the extracts and the plasmid DNA, are there any additional materials you should add to this in vitro replication system? Explain your answer.

You will probably add exogenous nucleoside triphosphates to serve as the building blocks needed to make new strands of DNA. Although these monomers will be present in the extracts, they will be present at lower concentrations than are normally found inside the cell. They may also be subject to hydrolysis, and the nucleoside diphosphates that are the products of this hydrolysis are not usable substrates for DNA replication. For both of these reasons, it is important to add excess nucleotides to the reaction mixture for efficient DNA replication to occur.

Which of the choices below represents the correct way to repair the mismatch shown in Figure Q6-41? Figure Q6-41

a