Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank

Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

CHAPTER 5: DNA REPLICATION, REPAIR, AND RECOMBINATION

© Garland Science 2015

 

1.1.         Which of the following is correct regarding the mutation rate of genomic DNA in different organisms?

1. Human cells have a much higher mutation rate compared to bacteria when the rate is normalized to a single round of replication over the same length of DNA.
2. Mutation rates limit the number of essential genes in an organism’s genome.
3. Mutations in the somatic cells cannot be lethal.
4. Even if the mutation rate was 10 times higher than its current value, germ-cell stability in humans would not have been affected.
5. All of the above.

 

1.2.         The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time.

1. Less than 0.001
2. About 0.02
3. One or two
4. About 10
5. About 100

 

1.3.         On average, errors occur in DNA synthesis only once in every ten billion nucleotides incorporated. Which of the following does NOT contribute to this high fidelity of DNA synthesis?

1. Complementary base-pairing between the nucleotides
2. “Tightening” of the DNA polymerase enzyme around its active site to ensure correct pairing before monomer incorporation
3. Exonucleolytic proofreading by the 3′-to-5′ exonuclease activity of the enzyme to correct mispairing even after monomer incorporation
4. A strand-directed mismatch repair system that detects and resolves mismatches soon after DNA replication
5. All of the above mechanisms DO contribute to the fidelity.

 

1.4.         The nuclear DNA polymerases in human cells …

1. polymerize about 1000 nucleotides per second during DNA replication in vivo.
2. are incapable of 3′-to-5′ exonuclease activity.
3. are capable of 3′-to-5′ DNA polymerase activity.
4. have a single active site that is used for both polymerization and editing.
5. are unable to initiate polymerization de novo (i.e. in the absence of a primer).

 

1.5.         What is the main source of the free energy for the mechanical work performed by DNA helicases during DNA replication in our cells?

1. The hydrogen-bonding energy in the DNA double helix
2. Thermal energy in the nucleus
3. ATP hydrolysis by the helicase
4. The energy of SSB binding to single-stranded DNA
5. ATP hydrolysis by DNA topoisomerases

 

1.6.         During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. These primers …

1. are made up of DNA.
2. generally have a higher number of mutations compared to their neighboring DNA.
3. are made more frequently in the leading strand than the lagging strand.
4. are joined to the neighboring DNA by DNA ligase.
5. provide a 3′-phosphate group for the DNA polymerases to extend.

 

1.7.         DNA ligases are used in both DNA replication and repair to seal breaks in the DNA. But DNA damage can result in single- or double-strand breaks that are not normal ligase substrates. These need to be processed first before a ligase can act on them. One of the enzymes that is recruited to some of such breaks is called PNK. It has two separate activities on the DNA, both of which can help provide a canonical ligase substrate. Which of the following activities would you expect PNK to have in this context?

1. 5′ kinase (phosphorylation of a free 5′-OH group) and 3′ kinase
2. 5′ phosphatase (dephosphorylation to create a free 5′-OH group) and 3′ phosphatase
3. 3′ kinase and 3′ phosphatase
4. 5′ phosphatase and 3′ kinase
5. 5′ kinase and 3′ phosphatase

 

1.8.         Fill in the gap in the following paragraph using what you know about the activities of the proteins involved in DNA replication.

“Mitochondrial DNA replication requires a set of proteins similar to those used for the replication of the nuclear genome. However, mitochondria lack a dedicated DNA … and use the mitochondrial RNA polymerase instead.”

 

1.9.         During DNA replication, the single-strand DNA-binding (SSB) proteins …

1. are generally found more on the leading strand than the lagging strand.
2. bind cooperatively to single-stranded DNA and cover the bases to prevent base-pairing.
3. prevent the folding of the single-stranded DNA.
4. bind cooperatively to short hairpin helices that readily form in the single-stranded DNA.
5. All of the above.

 

1.10.       This protein is present at every replication fork and prevents DNA polymerase from dissociating, but does not impede the rapid movement of the enzyme. Which of the following is true regarding this protein?

1. It self-assembles onto DNA at the replication fork.
2. It is assembled on DNA as soon as DNA polymerase runs into a double-strand region of DNA.
3. Its assembly normally follows the synthesis of a new primer by the DNA primase.
4. It disassembles from DNA as soon as DNA polymerase runs into a double-strand region.
5. All of the above.

 

1.11.       At the replication fork, the template for the lagging strand is thought to loop around. This looping would allow the lagging-strand polymerase to move along with the rest of the replication fork instead of in the opposite direction. The single-strand part of the loop is bound by the single-strand DNA-binding (SSB) proteins. As each Okazaki fragment is synthesized toward completion, how does the size of the loop change? What about the size of the SSB-bound part of the loop?

1. Increases; increases.
2. Increases; decreases.
3. Decreases; increases.
4. Decreases; decreases.
5. Decreases; does not change.

 

1.12.       The Dam methylase is responsible for methylating the adenine base in GATC sequences in Escherichia coli. Imagine two E. coli strains, one without any active Dam methylase, and the other with a hyperactive version of the enzyme that operates faster than the wild-type enzyme. Which of these strains would you expect to show a “mutator” phenotype?

1. Both of the strains
2. Neither of them
3. Only the first strain
4. Only the second strain

 

1.13.       In the following schematic drawing, two DNA molecules are shown before and after the action of a protein that is also involved in the process of DNA replication. What is this protein called?

 

1. DNA ligase
2. DNA helicase
3. DNA polymerase I
4. DNA topoisomerase I
5. DNA topoisomerase II

 

1.14.       What do the enzymes topoisomerase I and topoisomerase II have in common?

1. They both have nuclease activity.
2. They both create double-strand DNA breaks.
3. They both require ATP hydrolysis for their function.
4. They both can create winding (tension) in an initially relaxed DNA molecule.
5. All of the above.

 

1.15.       In Escherichia coli, replication of DNA can occur throughout the cell cycle while the cell is also actively transcribing its genes. This means collisions between replication forks and RNA polymerases are inevitable. Depending on the orientation of the genes, collisions can be rear-end (when both machines are traveling in the same direction) or head-on (when they are traveling in opposite directions). In the following paragraph, match each of the letters (A to D) to one appropriate number below. Do not use a number more than once. Your answer would be a four-digit number composed of digits 1 to 5 only, e.g. 1253.

“Typically, in a rear-end collision, the (A) of RNA polymerase collides with the (B) in the replication fork. In contrast, in a head-on collision, the (C) of RNA polymerase hits the (D) in the fork.”

1. front edge (of RNA polymerase)
2. rear edge (of RNA polymerase)
3. DNA helicase
4. leading-strand DNA polymerase
5. lagging-strand DNA polymerase

 

1.16.       Which of the following features is common between the replication origins in Escherichia coli and Saccharomyces cerevisiae?

1. They both normally exist in one copy per genome.
2. Both are specified by DNA sequences of tens of thousands of nucleotide pairs.
3. Both contain sequences that attract initiator proteins, as well as stretches of DNA rich in A-T base pairs.
4. Both contain GATC repeats that are methylated to prevent the inappropriate “firing” of the origin.
5. All of the above.

 

1.17.       You have found a strain of Escherichia coli that has an unusually short doubling time of only 15 minutes, despite the fact that its complete DNA replication should take almost 35 minutes. You also find that there is only one replication origin on its chromosome from which two forks originate, just like the normal process described for E. coli. However, you discover that the origin of replication in this strain has a significantly shorter “refractory period,” resulting in the reactivation of the origin before the previous round of replication is over. Based on this model, if you examine the chromosomes of this strain (under conditions of fast growth), how many replication forks would you expect to observe per chromosome on average?

1. Two, just like the wild-type strain
2. Four
3. Six
4. Eight
5. Ten

 

1.18.       The telomerase enzyme in human cells …

1. has an RNA component.
2. extends the telomeres by its RNA polymerase activity.
3. polymerizes the telomeric DNA sequences without using any template.
4. removes telomeric DNA from the ends of the chromosomes.
5. creates the “end-replication” problem.

 

1.19.       If this protein complex does not function normally, the ends of the eukaryotic chromosomes would activate the cell’s DNA damage response, causing chromosomal fusions and other genomic anomalies. What is this protein complex called?

1. Telomerase
2. T-loop
3. ORC
4. Shelterin
5. RecA

 

1.20.       Which of the following schematic drawings better depicts the end of mammalian chromosomal DNA?

 

 

 

 

 

 

1.21.       Which of the following spontaneous lesions in DNA occurs most frequently in a mammalian cell?

1. Depurination
2. Cytosine deamination
3. Guanine oxidation
4. Guanine alkylation
5. Depyrimidination

 

1.22.       DNA glycosylases constitute an enzyme family found in all three domains of life. They can …

1. add sugar moieties to DNA.
2. remove sugar moieties from DNA.
3. add a purine or pyrimidine base to DNA.
4. remove a purine or pyrimidine base from DNA.
5. remove a nucleotide from DNA.

1.23.       Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. These backup polymerases …

1. lack 3′-to-5′ exonucleolytic proofreading activity.
2. are replaced by the replicative polymerases after adding only a few nucleotides.
3. can create mutations even on undamaged DNA.
4. may recognize specific DNA damage and add the appropriate nucleotide to restore the original sequence.
5. All of the above.

 

1.24.       What are the products of deamination of cytosine and 5-methyl cytosine, respectively?

1. Thymine and uracil
2. Thymine in both cases
3. Uracil and thymine
4. Uracil in both cases
5. Xanthine and hypoxanthine

 

1.25.       Which of the following repair pathways can accurately repair a double-strand break?

1. Base excision repair
2. Nucleotide excision repair
3. Direct chemical reversal
4. Homologous recombination
5. Nonhomologous end joining

 

1.26.       This protein folds into a doughnut shape that can encircle DNA. It can load on the DNA only when the DNA is broken in both strands, so that the DNA can thread through the hole in the protein. Which of the following proteins do you think matches this description?

1. PCNA, the sliding clamp for DNA polymerases at the replication forks
2. Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining
3. MCM, the helicase critical for the initiation and elongation of replication
4. Topoisomerase II, which can create or relax superhelical tension in DNA
5. RecA/Rad51, which carries out strand invasion in homologous recombination

 

1.27.       In contrast to vertebrates, there is very little DNA methylation in the genomes of invertebrates such as Drosophila melanogaster and Caenorhabditis elegans. Indicate whether you expect each of the following statements to be true (T) or false (F) regarding 5′-CG-3′ dinucleotide sequences in the genome. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT.

(  )  On average, approximately one dinucleotide out of every 16 in the human genome is a CG dinucleotide.

(  )  On average, approximately one dinucleotide out of every 256 in the D. melanogaster genome is a CG dinucleotide.

(  )  The proportion of CG dinucleotides in the human genome is more than that of C. elegans.

(  )  The proportion of CG dinucleotides in the human genome is more than that expected by chance.

 

1.28.       In which phases of the eukaryotic cell cycle does homologous recombination often occur to repair DNA damage?

1. G₁ and S phases
2. S and G₂ phases
3. G₂ and M phases
4. M and G₁ phases
5. G₁ and G₂ phases

 

1.29.       Which of the following is NOT correct regarding homologous recombination and its regulation?

1. Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome.
2. Repair of double-strand breaks by homologous recombination is favored during or soon after DNA replication.
3. Homologous recombination can rescue broken or stalled replication forks in S phase.
4. Excessive use of homologous recombination by human cells can lead to cancer.
5. Low usage of homologous recombination by human cells can lead to cancer.

 

1.30.       In the following schematic drawing of a Holliday junction that undergoes branch migration, cutting at which combination of the sites a to d would generate a crossover?

1. a and b
2. a and c
3. a and d
4. b and c
5. b and d

 

1.31.       In meiosis, a crossover in one position is thought to inhibit crossing-over in the neighboring regions. This regulatory mechanism …

1. results in a very uneven distribution of crossover points along each chromosome.
2. ensures that even small chromosomes undergo at least one crossover.
3. controls how the Holliday junctions are resolved.
4. All of the above.

 

1.32.       What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics?

1. DNA-only transposons
2. Retroviral-like retrotransposons
3. Nonretroviral retrotransposons
4. Site-specific recombinases

 

1.33.       DNA-only transposons …

1. can be recognized by the presence of short inverted repeats at each end.
2. often encode a transposase that mediates the transposition process.
3. leave double-strand breaks in the donor chromosome.
4. can move by a cut-and-paste mechanism.
5. All of the above.

1.34.       Which of the following is true regarding retroviral-like retrotransposons?

1. They encode both a reverse transcriptase and an RNA polymerase.
2. They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA.
3. Their genomic RNA can be translated to produce viral coat proteins.
4. They leave double-strand breaks in the original donor DNA.
5. The Alu element in our genome is an example of retroviral-like retrotransposons.

 

1.35.       Which of the following is NOT correct regarding long and short interspersed nuclear elements?

1. Each of them encodes a reverse transcriptase.
2. They rely on the cellular transcription machinery to produce their RNA transcript.
3. They use one of the strands in the target DNA as a primer to synthesize their DNA.
4. Together, they make up about 40% of our genome.
5. They can move into new regions of genome that do not have any homology with their DNA.

 

1.36.       Phase variation helps protect the bacterium Salmonella typhimurium against the immune system of its host by switching the orientation of a certain promoter. This process …

1. is carried out through a DNA transposition mechanism.
2. is irreversible.
3. can often result in the excision of the promoter from the chromosome altogether.
4. is mediated by enzymes that form transient covalent bonds with the DNA.

 

1.37.       A replication fork is shown schematically below. The strand labeled A is called the … strand.

 

1.38.       The sliding clamp and the DNA helicase that function at the replication fork both have three-dimensional structures resembling a ring with a central hole through which DNA is threaded. Which of these proteins, the clamp (C) or the helicase (H), do you think has a wider hole in its structure? Write down C or H as your answer.

 

1.39.       Indicate true (T) and false (F) statements below regarding the initiation of replication in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

(  )  Tens of thousands of replication origins are used each time a cell in our body replicates its DNA.

(  )  Different cells in our body use different sets of replication origins.

(  )  Both replication forks in a replication bubble are normally active in replication.

(  )  Gene expression and chromatin structure can affect the choice of the origins to use as well as the order in which they are activated.

 

1.40.       What combination of the following events normally prevents the origins of replication in the yeast Saccharomyces cerevisiae from “firing” more than once during the cell cycle? Your answer is a two-letter string composed of letters A to E only, e.g. AB. Order the letters in your answer alphabetically.

(A) The helicase-loading proteins Cdc6 and Cdt1 are only active in S phase.

(B) The helicase Mcm1 can be delivered to the origin recognition complex (ORC) only in S phase.

(C) The ORC can only become active (by dephosphorylation) in G₁ phase.

(D) The helicase can only become active (by phosphorylation) in S phase.

(E) The ORC can bind to the origin only in S phase.

 

1.41.       Examples of two general types of DNA damage are shown in the following drawing. Which type of damage (1 or 2) is more common in our cells?

 

1.42.       In cells that are exposed to sunlight, ultraviolet (UV) light can result in covalent linkage between two adjacent DNA bases. If not repaired in time, such dimers can lead to inheritable mutations. Consider the sequence 5′-GGTATGATCATTATAA-3′ in the chromosome of a cell that is exposed to intense sunlight. How many possible dimers can form in the region of DNA double helix corresponding to this sequence?

 

1.43.       Indicate whether each of the following DNA lesions is typically repaired via the base excision (B) or nucleotide excision (N) repair pathway? Your answer would be a four-letter string composed of letters B and N only, e.g. BBBB.

1. Deaminated cytosines
2. Depurinated residues
3. Thymine dimers
4. Bulky guanine adducts

 

1.44.       Indicate true (T) and false (F) descriptions below regarding the nucleotide excision repair pathway. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  It involves recognition of distortions in the DNA double helix rather than specific base changes.

(  )  It involves endonucleolytic cleavage and helicase-mediated strand removal.

(  )  It involves cleavage by the AP endonuclease.

(  )  It is coupled to the DNA transcription machinery of the cell.

 

1.45.       Fill in the blank in the following paragraph. Do NOT use abbreviations.

 

“In human cells, the predominant pathway to repair double-strand breaks is …, in which the broken ends are simply rejoined with the concomitant loss of a few nucleotides. This leaves scars at the breakage sites. This pathway can potentially create chromosome translocations.”

 

3.46.       Sort the following steps in the order that they normally happen during the process of repairing double-strand breaks by homologous recombination. Your answer would be a six-letter string composed of letters A to F only, e.g. DEFABC.

(A) Ligation

(B) DNA synthesis using undamaged DNA as the template

(C) DNA synthesis using original DNA as the template

(D) Release of the invading strand

(E) Strand invasion

(F) Nuclease digestion (resection)

 

1.47.       The RecA/Rad51 protein carries out strand exchange during homologous recombination. Indicate true (T) and false (F) statements about this process below. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  RecA hydrolyzes ATP upon binding to the invading DNA strand.

(  )  RecA forces the invading strand into a conformation that fully mimics the geometry of a long DNA double helix.

(  )  Sampling of the homologous duplex by the invading strand is likely to occur in triplet nucleotide blocks.

(  )  RecA-bound, invading, single-stranded DNA binds and destabilizes the homologous duplex to allow the sampling of its sequence by base-pairing.

 

1.48.       Indicate true (T) and false (F) statements below regarding the use of homologous recombination in meiosis. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF.

(  )  Meiotic recombination starts with a double-strand break caused by errors in DNA replication.

(  )  Meiotic recombination occurs preferentially between DNA from maternal and paternal chromosome pairs.

(  )  Holliday junctions can form during meiotic recombination, sometimes in pairs.

(  )  During meiotic recombination in human cells, the majority of the invading strands are released, leading to no crossover.

 

1.49.       Indicate whether each of the following mobile elements in Drosophila is a DNA-only transposon (D), a retroviral-like retrotransposon (R), or a nonretroviral retrotransposon (N). Your answer would be a four-letter string composed of letters D, R, and N only, e.g. RRRR.

(  )  P elements; these have inverted terminal repeats and move with the help of a transposase enzyme.

(  )  Copia elements; these have directly repeated long terminal repeats and move with the help of a reverse transcriptase and an integrase.

(  )  F elements; these are long interspersed nuclear elements (LINEs).

(  )  Mariner elements; these move via a DNA intermediate.

 

1.50.       Consider three types of mobile genetic elements that are found in our genome: the DNA-only transposons (D), the retroviral-like retrotransposons (R), and the nonretroviral retrotransposons (N). Which type appears to still be active and move in our genome, accounting for a detectable fraction of human mutations? Write down your answer as D, R, or N.

 

1.51.       As shown in the following drawing, a researcher has engineered three pairs of LoxP sites (for conservative site-specific recombination) in a region that contains three reporter genes coding for red, yellow, or cyan fluorescent proteins, respectively. Each type of LoxP sequence (shown as a black, gray, or white arrowhead) is specific, meaning it does not recombine with the other types of LoxP sequences. Upon Cre recombinase activation, depending on which recombination event occurs first (which we assume is random), a number of possible combinations of reporters can remain in the final DNA. For each of the following combinations, indicate whether it can (C) or cannot (N) result from this recombination scheme. Do not consider the re-integration of excised DNA, which happens very rarely. Your answer would be a six-letter string composed of letters C and N only, e.g. CCCCNN.

 

 

(  )  Red and yellow

(  )  Red only

(  )  Yellow only

(  )  Cyan only

(  )  Yellow and cyan

(  )  Red and cyan

 

 

Answers:

1. Answer: B

Difficulty: 2

Section: The Maintenance of DNA Sequences

Feedback: The mutation rate per nucleotide pair per replication cycle in humans is similar to that of Escherichia coli, and both are extremely low, less than one in a billion. Nevertheless, current mutation rates are thought to limit the number of essential genes in an organism to about 30,000. The stability of the germ-line genome depends on mutation rates, as does somatic-cell stability. Cancers typically arise from somatic mutations.

2. Answer: B

Difficulty: 3

Section: The Maintenance of DNA Sequences

Feedback: About 15 cell generations pass in 10 hours of growth. Thus, the expected average number of mutations is equal to:

(15 generations) × (3 × 10^–10 mutations/generation/nucleotide) × (4.5 × 10⁶ nucleotides/genome)

= ~2 × 10^–2 mutations per genome.

This is equivalent to about 98% of the bacterial cells having zero mutations and the remaining 2% having only one mutation each. Please note that the number of mutations in different cells in a given culture will vary widely around the average calculated above.

3. Answer: E

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: Base-pairing is the basis of all DNA replication and repair. The other three mechanisms each increase accuracy by about 100-fold.

4. Answer: E

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: Eukaryotic DNA polymerases are not as fast as their bacterial counterparts, presumably due to the difficulty in replicating through nucleosomes. All known DNA polymerases polymerize DNA from 5′ to 3′ and require a free 3′ end provided by a primer. The self-correcting polymerases have an additional 3′-to-5′ exonuclease activity that takes place at a distinct site in the enzyme.

5. Answer: C

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: The helicase at the replication fork binds and hydrolyses ATP to move on the DNA in a stepwise fashion.

6. Answer: B

Difficulty: 1

Section: DNA Replication Mechanisms

Feedback: The DNA primase enzyme syntheses RNA primers (with free 3′-OH groups) that are extended and later replaced by DNA that is more accurately synthesized. This happens more frequently in the lagging strand.

7. Answer: E

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The DNA ligase reaction normally requires a free 3′-OH group and a 5′-phosphate group on the two DNA ends to proceed. If a break has created 3′-phosphate and 5′-OH ends instead, for example, an enzyme such as PNK can dephosphorylate the 3′ end in the upstream molecule, and phosphorylate the 5′ end in the downstream molecule, to create the canonical ends.

8. Answer: primase

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: Like the RNA polymerases involved in transcription, primases do not require priming and can synthesize RNA molecules that can be used as primers for DNA synthesis by the replicative DNA polymerases. RNA polymerase is used to create RNA primers for mitochondrial DNA replication.

9. Answer: C

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: The SSB proteins bind cooperatively to the regions of exposed single-stranded DNA (which are routinely found in the lagging strand) and prevent the formation of hairpin helices without blocking the base-pairing potential.

10. Answer: C

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: The sliding clamp is normally loaded (more frequently on the lagging strand) on the primer–template duplex by the clamp loader, a process that requires ATP hydrolysis by the loader. As soon as the polymerase runs into a double-strand region downstream, the clamp releases the polymerase but is not itself immediately disassembled from DNA.

11. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The double-strand region of the loop represents the growing Okazaki fragment and gradually increases in size. Additionally, as each Okazaki fragment is growing, the fork is also traveling, exposing more single-stranded DNA in the loop. The loop size resets to zero upon initiating the synthesis of the next Okazaki fragment.

12. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: An inactive or hyperactive Dam methylase can interfere with strand-directed mismatch repair, as both make it harder to distinguish between the newly replicated and old DNA strands. Both situations are expected to increase the overall mutation rate, giving rise to a mutator phenotype.

13. Answer: E

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: This untangling of intertwined DNA is mediated by the enzyme topoisomerase II. This type of reaction helps solve the winding problem during replication.

14. Answer: A

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: Unlike topoisomerase I, which can only relieve the tension in DNA through introducing a nick (single-strand break) in one of the DNA strands, topoisomerase II can actively introduce or relieve tension by creating double-strand breaks (that remain tightly associated with the enzyme) using energy from ATP hydrolysis.

15. Answer: 2413

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The replication fork in E. coli can progress almost 20 times faster than the RNA polymerase, meaning that rear-end collisions are typically between the rear edge of the RNA polymerase and the front edge of the leading-strand DNA polymerase that is traveling (faster) in the same direction on the same template strand. In a head-on collision, however, the front edge of RNA polymerase hits the DNA helicase that is traveling on the same (lagging) strand in the opposite direction. Note that the polymerases translocate with respect to their template in the 3′-to-5′ direction, while the helicase in the fork translocates in the 5′-to-3′ direction.

16. Answer: C

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The yeast origins of replication are similar to the bacterial origins in that they are A-T-rich and have particular sequences that attract replication initiator proteins. In higher eukaryotes, in contrast, the determinants of the replication origins are probably less sequence-specific. E. coli and S. cerevisiae both have relatively short origins compared to higher organisms.

17. Answer: C

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Even though each round of replication takes 35 minutes, the “firing” interval can be adjusted independently, and even go out of control. In this strain, possibly due to mutations in the methylation pathway, the origins can become activated every 15 minutes, resulting in an average of six forks (three bubbles) per chromosome, the newest of which have started about 30 minutes after the oldest ones.

18. Answer: A

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Telomerase uses its RNA component as a template to polymerize DNA sequences at the chromosome ends to solve the “end-replication” problem.

19. Answer: D

Difficulty: 1

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: With the help of the t-loop structure, shelterin “hides” the telomere ends from the cell’s double-strand break detector and repair systems.

20. Answer: D

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The telomeres have 3′ overhangs and are folded in structures called t-loops, which help protect the chromosome ends.

21. Answer: A

Difficulty: 1

Section: DNA Repair

Feedback: Depurination (that is, the hydrolytic removal of purine bases from DNA) is by far the most common endogenous DNA lesion in our cells. Please refer to Table 5–3 for details.

22. Answer: D

Difficulty: 1

Section: DNA Repair

Feedback: These enzymes can recognize altered bases in DNA and catalyze their removal by hydrolyzing the glycosidic bond between the base and the deoxyribose sugar.

23. Answer: E

Difficulty: 1

Section: DNA Repair

Feedback: The translesion polymerases come in different flavors and some of them are very efficient in repairing specific damage, but they generally have a low fidelity as they cannot proofread. The cell thus limits their usage to very few polymerization steps.

24. Answer: C

Difficulty: 2

Section: DNA Repair

Feedback: The use of thymine instead of uracil by DNA has the advantage that deamination of cytosine (to uracil) can be readily detected as an abnormal base in DNA. However, deamination of 5-methyl cytosine (a modification that is found in vertebrate DNA) can produce thymine and lead to mutations at a higher rate.

25. Answer: D

Difficulty: 1

Section: DNA Repair

Feedback: Compared to the faithful homologous recombination pathway to repair double-strand breaks, nonhomologous end joining is a “quick and dirty” solution that often results in mutations at the site of repair.

26. Answer: B

Difficulty: 3

Section: DNA Repair

Feedback: The Ku dimer forms a hole that accommodates a DNA double helix. In addition, its structure reflects the fact that it should only bind to free DNA ends (for example, from double-strand breaks) in a fashion similar to threading a needle.

27. Answer: FFFF

Difficulty: 3

Section: DNA Repair

Feedback: In a random DNA sequence, one out of every 16 dinucleotides is a CG. However, these dinucleotides are found at a much lower frequency in the human genome, which is probably related to the fact that cytosine methylation at these sites (which has a role in transcription regulation) can be followed by deamination to cause a C-to-T substitution. Consistent with this notion, the CG frequency in D. melanogaster and C. elegans is close to the expected value.

28. Answer: B

Difficulty: 2

Section: Homologous Recombination

Feedback: Homologous recombination is used to repair DNA damage during and shortly after the S phase of the cell cycle, when a sister chromatid is available for faithful repairs.

29. Answer: A

Difficulty: 2

Section: Homologous Recombination

Feedback: If a homologous chromosome is used (instead of a sister chromatid) to repair a double-strand break, one of the parental alleles of the affected gene can be replaced by the other allele, leading to loss of heterozygosity.

30. Answer: E

Difficulty: 3

Section: Homologous Recombination

Feedback: Cutting at b and d would result in a crossover. Cutting at a and c would only lead to a heteroduplex (no crossover). The other combinations are not normally chosen.

31. Answer: B

Difficulty: 2

Section: Homologous Recombination

Feedback: The poorly understood regulatory mechanism of crossover control ensures a roughly even distribution of crossover points along chromosomes. It also ensures that each chromosome undergoes at least one crossover every meiosis. For many organisms, roughly two crossovers per chromosome occur, one on each arm.

32. Answer: A

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: DNA-only transposons that carry genes encoding antibiotic-inactivating enzymes are largely responsible for the spread of antibiotic resistance in bacterial strains.

33. Answer: E

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: DNA-only transposons often encode a transposase to move by a cut-and-paste mechanism that results in a broken donor chromosome. Short inverted repeats of DNA sequence are found at the ends of these transposons.

34. Answer: B

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Retroviral-like retrotransposons resemble retroviruses in their replication, but lack the ability to produce a protein coat. They have directly repeated long terminal repeats at each end and are transcribed by cellular RNA polymerases. The transcript is then used as a template by a reverse transcriptase (encoded by the element) to make a double-stranded DNA copy that is then integrated into a new site on the chromosome using an integrase (also encoded by the element). This mechanism keeps the original copy unchanged. Alu elements are nonretroviral retrotransposons.

35. Answer: A

Difficulty: 2

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Most short interspersed nuclear elements (SINEs) do not encode any proteins and spread by pirating enzymes encoded by other transposons, such as long interspersed nuclear elements (LINEs).

36. Answer: D

Difficulty: 2   

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Phase variation in Salmonella is brought about by a conservative site-specific recombinase encoded by the bacterium. The recombinases that catalyze conservative site-specific recombination form transient, high-energy covalent bonds with the DNA and use this energy to complete the DNA rearrangements, an action reminiscent of topoisomerases.

37. Answer: lagging

Difficulty: 2

Section: DNA Replication Mechanisms

Feedback: Strand A is extended 5′-to-3′ (from right to left) whereas the fork moves in the opposite direction; therefore, the strand should be polymerized discontinuously.

38. Answer: C

Difficulty: 3

Section: DNA Replication Mechanisms

Feedback: The central hole in the clamp can accommodate a double-stranded DNA molecule with extra room for smooth sliding. However, the helicase only allows a single strand of DNA to pass through its central hole.

39. Answer: TTTT

Difficulty: 2

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: Human cells have probably hundreds of thousands of potential origins, out of which about 40,000 are used each time a cell divides. Depending on gene expression and chromatin state, different sets of origins can be used in different cells of the body. Once an origin is activated, the resulting replication bubble expands on both sides of the origin.

40. Answer: CD

Difficulty: 3

Section: The Initiation and Completion of DNA Replication in Chromosomes

Feedback: The helicase is delivered to the ORC by active Cdc6 and Cdt1 proteins in G₁ phase, but is activated by phosphorylation only in S phase. At the same time, phosphorylation of ORC prevents its reactivation in S phase. ORC can be reactivated again later in G₁.

41. Answer: 2

Difficulty: 1

Section: DNA Repair

Feedback: Depurination (2) and deamination (1) are both common, although depurination is by far the most common endogenous DNA lesion in our cells.

42. Answer: 5

Difficulty: 3

Section: DNA Repair

Feedback: UV irradiation can cause two adjacent pyrimidine bases (T or C) to form these covalent dimers. Please note that the complementary strand should also be taken into account. The possible sites of damage are TC and TT doublets in the strand shown, and TT, TC, and CC doublets in the complementary strand.

43. Answer: BBNN

Difficulty: 2

Section: DNA Repair

Feedback: Please note that depurinated residues are repaired through only part of the base excision repair pathway.

44. Answer: TTFT

Difficulty: 1

Section: DNA Repair

Feedback: In nucleotide excision repair, a multienzyme complex recognizes DNA double-helix distortions caused by damage. This pathway involves cutting the strand containing the damage at two positions flanking the damage. A helicase is also recruited and catalyzes strand removal. This pathway is also linked to the transcription machinery.

45. Answer: nonhomologous end joining

Difficulty: 1    Section: DNA Repair

Feedback: In nonhomologous end joining, the broken ends are simply brought together and rejoined by DNA ligation, often concomitant with the loss of nucleotides at the site of joining.

46. Answer: FEBDCA

Difficulty: 2

Section: Homologous Recombination

Feedback: This order of events results in the accurate repair of double-strand breaks.

47. Answer: FFTT

Difficulty: 2

Section: Homologous Recombination

Feedback: RecA binds cooperatively to the invading strand, forcing blocks of three nucleotides into a conformation mimicking that of three nucleotides in a double helix (although, between adjacent triplets, the DNA backbone is untwisted and stretched out). This complex then binds and destabilizes the target homologous duplex, allowing the sampling to take place by base-pairing interactions. RecA hydrolyzes its bound ATP after the strand exchange is completed.

48. Answer: FTTT

Difficulty: 2

Section: Homologous Recombination

Feedback: Meiotic recombination requires an initial double-strand break. In contrast to the breaks resulting from DNA damage, this one is provided in a controlled manner by a specialized meiotic protein.

49. Answer: DRND

Difficulty: 2

Section: Homologous Recombination

Feedback: P elements and mariner elements are DNA-only transposons in different Drosophila species, copia is an abundant retroviral-like retrotransposon, and F elements are nonretroviral retrotransposons.

50. Answer: N

Difficulty: 1

Section: Transposition and Conservative Site-Specific Recombination

Feedback: Some nonretroviral retrotransposons (such as Alu elements) are still moving in our genome, accounting for perhaps two mutations out of every thousand new human mutations. Although our genome is also littered with relics of retroviral-like retrotransposons, none appear to be active today.

51. Answer: CNNCNC

Difficulty: 3

Section: Transposition and Conservative Site-Specific Recombination

Feedback: The arrangement of LoxP sites matters. For example, here, it does not allow the final combination (yellow + cyan) because deleting the red reporter requires the use of the white sequence, which also deletes the yellow reporter.