DNA Replication and Repair (1B)

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MCAT Biological and Biochemical Foundations of Living Systems › DNA Replication and Repair (1B)

Questions 1 - 10
1

In a yeast model, a single gene knockout produces normal DNA synthesis rates but causes frequent small insertions and deletions specifically within short repeated sequences (e.g., AAAAAA). The phenotype is strongest after many rounds of cell division and is reduced when a second gene that promotes recombination is overexpressed. Based on the scenario, which outcome is most consistent with loss of the primary pathway that corrects small loops formed during copying of repetitive DNA?

Accumulation of length changes in repeats due to failure to correct small misalignments during copying

Failure to separate newly synthesized strands because covalent links between strands are not removed

Inability to reseal breaks between DNA fragments, producing many short linear DNA pieces

Increased C→T transitions at adjacent pyrimidines after UV exposure due to failure to remove bulky lesions

Explanation

This question tests knowledge of DNA replication and repair. Mismatch repair (MMR) corrects small insertions, deletions, or misalignments during replication, particularly in repetitive sequences, preventing microsatellite instability. The yeast knockout leads to indels in repeats, worsened by cell divisions and mitigated by recombination, indicating a MMR defect allowing loop accumulation. Choice A is consistent as it describes MMR's role in correcting replication slippage in repeats, matching the phenotype. Choice B distracts by suggesting a ligase issue causing short fragments, but the scenario shows normal synthesis rates without fragmentation. In similar questions, assess if errors accumulate in repeats over divisions, pointing to MMR. Confirm if recombination modulation affects the phenotype, distinguishing from other repair like NER or ligation.

2

In a model organism study, embryos carrying a temperature-sensitive mutation in DNA ligase develop normally at $25^\circ\text{C}$ but show extensive chromosome fragmentation and cell-cycle arrest at $37^\circ\text{C}$. DNA sequencing does not show a marked increase in base substitutions, but microscopy reveals many persistent single-strand breaks. Which statement best explains the role of DNA ligase in DNA replication that is most consistent with these findings?

It seals adjacent DNA segments by forming phosphodiester bonds in the sugar–phosphate backbone

It separates the two parental DNA strands to allow copying of each template

It selects the correct incoming nucleotide by monitoring base-pair geometry at the active site

It removes mismatched bases by excising a short patch surrounding the error

Explanation

This question tests understanding of DNA replication and repair, specifically the role of DNA ligase in sealing nicks. DNA ligase catalyzes the formation of phosphodiester bonds between adjacent DNA segments, joining Okazaki fragments on the lagging strand and sealing any remaining nicks after repair processes. The temperature-sensitive mutation causes ligase to lose function at 37°C, resulting in persistent single-strand breaks (nicks) that appear as chromosome fragmentation under microscopy. The absence of increased base substitutions indicates that polymerase fidelity and mismatch repair are intact - the problem is specifically in sealing the sugar-phosphate backbone. Choice A describes helicase function, and choice C describes polymerase selectivity, neither of which would cause the observed fragmentation pattern. To identify ligase defects, look for persistent nicks and fragmentation without increased point mutations.

3

A bacterial strain is exposed to ultraviolet light, then allowed to recover in the dark. Compared with wild-type cells, a mutant strain shows a large increase in C→T substitutions at sites where two adjacent pyrimidines occur, even though overall survival is similar. The mutant is known to be defective in a pathway that normally removes bulky lesions by cutting out a short stretch of the damaged strand and resynthesizing it. Based on the scenario, which process would be most disrupted in the mutant?

Removal of UV-induced bulky lesions by excising a short segment of the damaged strand

Direct reversal of methylated bases by transferring the methyl group to an enzyme

Correction of replication mismatches by recognizing the newly synthesized strand via nicks

Relief of supercoiling ahead of the replication machinery by transient strand breakage

Explanation

This question tests knowledge of DNA replication and repair. Nucleotide excision repair (NER) removes bulky lesions like UV-induced pyrimidine dimers by excising a short segment of the damaged strand and resynthesizing it using the intact strand as a template. In the scenario, the mutant defective in NER shows increased C→T substitutions at dipyrimidine sites after UV exposure, indicating unrepaired dimers lead to mutations during replication. Choice A is consistent because NER's role in removing such bulky lesions prevents these specific mutations, and its absence disrupts this process. Choice D is a distractor as it describes topoisomerase function in relieving supercoiling, which does not directly address UV damage repair or the mutation pattern observed. For similar questions, identify if the damage type (e.g., bulky vs. small) matches the repair pathway. Also, check if the mutation signature, like C→T at pyrimidines, points to UV-specific lesions rather than general replication issues.

4

A clinician-scientist studies a tumor with a loss-of-function mutation in a gene required for accurate repair of double-strand breaks using an intact homologous DNA sequence as a template. The tumor cells show frequent small insertions and deletions at break sites and are unusually sensitive to inhibitors of a single-strand break repair enzyme. Based on the mechanism, which repair pathway is most likely upregulated to compensate for the lost function, leading to the observed insertions/deletions?

Excision of bulky lesions followed by gap filling using the undamaged strand as a template

End-joining of broken DNA ends without requiring extensive sequence matching, which can introduce small indels

Mismatch correction that removes short patches containing mispaired bases after replication

Direct reversal of alkylated bases by transferring the alkyl group to a repair protein

Explanation

This question tests understanding of DNA repair pathways, specifically the relationship between homologous recombination and non-homologous end joining (NHEJ). When cells lose homologous recombination repair (which uses a homologous template for accurate repair), they become dependent on NHEJ, which directly ligates broken ends without a template. NHEJ often introduces small insertions or deletions at the junction because the ends may require processing before ligation, and no template ensures accuracy. The tumor's sensitivity to single-strand break repair inhibitors suggests these cells rely heavily on preventing double-strand breaks from forming. Choice A describes mismatch repair (for replication errors, not breaks), and choice C describes nucleotide excision repair (for bulky lesions, not breaks). To identify compensatory repair mechanisms, look for the error signatures characteristic of each pathway - NHEJ produces indels at break sites.

5

A plasmid replication system is assembled with purified proteins. When a specific ATP-dependent enzyme is omitted, replication initiates normally but stalls after short stretches, and the DNA remains largely double-stranded near the growing region. Adding the enzyme restores long products. Which statement best explains the role of the omitted enzyme during replication?

It adds methyl groups to the newly synthesized strand to mark it for mismatch correction

It seals nicks between adjacent DNA fragments after synthesis is complete

It separates the two DNA strands ahead of synthesis to allow copying of the template

It removes RNA primers by degrading RNA and replacing it with DNA

Explanation

This question tests knowledge of DNA replication and repair. Helicases unwind the DNA double helix during replication, providing single-stranded templates for polymerase activity. Omitting the helicase causes replication to stall after short stretches, with DNA remaining double-stranded, restored by adding it back. Choice A is consistent as helicase's unwinding is essential for progression beyond initial synthesis. Choice C distracts by describing ligase, which seals nicks post-synthesis but would not cause stalling or double-stranded persistence. In similar questions, identify stalling with intact helices as helicase defects. Verify if adding the enzyme resumes long synthesis, distinguishing from primer or ligase issues.

6

A bacterial culture is treated with a compound that causes adjacent bases on the same strand to become covalently linked, distorting the DNA helix. Wild-type cells remove the damage efficiently, but a mutant lacking the relevant repair pathway accumulates replication stalls and mutations clustered near the lesions. Which repair strategy is most consistent with removal of a helix-distorting lesion of this type?

Cutting the damaged strand on both sides of the lesion, removing a short segment, and resynthesizing using the intact strand

Joining broken DNA ends directly without using sequence similarity

Correcting mismatches by excising the older strand based on chemical marks

Removing a single damaged base by cleaving the base–sugar bond without removing neighboring nucleotides

Explanation

This question tests knowledge of DNA replication and repair. Nucleotide excision repair (NER) removes helix-distorting lesions like intrastrand crosslinks by dual incisions and resynthesis. The mutant accumulates stalls and mutations near lesions from failed removal. Choice A is consistent as NER excises such distortions. Choice D distracts by misdescribing MMR strand targeting. For similar questions, match distortions to NER. Check if stalls cluster near damage, not general mismatches.

7

A researcher sequences clones derived from a bacterial population and finds a high frequency of G:C→T:A transversions that increase after oxidative stress. The mutant strain lacks an enzyme that normally removes an oxidized form of guanine from DNA before replication. Based on the scenario, which mechanism best explains the observed transversions when the oxidized guanine is not removed?

The oxidized guanine blocks strand separation at origins, preventing replication initiation and reducing mutations

The oxidized guanine prevents sealing of backbone nicks, causing deletions rather than base substitutions

The oxidized guanine is corrected by removing a short segment around bulky lesions, so loss of base removal has no effect

The oxidized guanine can mispair during copying, leading to insertion of an incorrect base and fixation as a transversion

Explanation

This question tests knowledge of DNA replication and repair mechanisms, specifically how base modifications can lead to mutations if not corrected. The biological principle involved is that oxidative damage to DNA, such as the formation of 8-oxoguanine (8-oxoG), can cause mispairing during replication, leading to transversion mutations like G:C to T:A. In this scenario, the mutant bacterial strain lacks the enzyme (likely MutM or FPG) that removes 8-oxoG before replication, resulting in increased transversions after oxidative stress as observed in the sequenced clones. The correct answer, choice A, is consistent because unrepaired 8-oxoG pairs with adenine instead of cytosine during replication, fixing the transversion upon subsequent DNA synthesis. A distractor like choice B does not fit, as oxidized guanine primarily causes base substitutions via mispairing rather than deletions from unsealed nicks, which relate more to ligase deficiencies. To check similar questions, verify if the mutation type matches the expected outcome of the damaged base's mispairing preference. Additionally, confirm that the repair pathway mentioned aligns with the specific lesion, such as base excision repair for oxidized bases rather than nucleotide excision for bulky adducts.

8

A plasmid is replicated in vitro with a limited supply of one nucleotide (dGTP). The reaction produces many truncated products that terminate at positions where G would be added. When the missing nucleotide is supplied later, extension resumes from the existing ends without needing a new start. Which statement is most consistent with how DNA synthesis proceeds under these conditions?

DNA synthesis requires complete removal of the template strand before extension can continue

DNA synthesis resumes only after a double-strand break is joined by end-joining enzymes

DNA synthesis restarts by adding nucleotides to the $5'$ end, bypassing the need for a $3'$ end

DNA synthesis can restart by extending an existing $3'$ end once the required nucleotide becomes available

Explanation

This question tests knowledge of DNA replication and repair. DNA polymerases extend from 3' ends and can resume upon nucleotide availability, without new initiation. Truncated products extend when dGTP is added, indicating restart from existing ends. Choice D is consistent as it describes extension from 3' ends post-pause. Choice B distracts by suggesting 5' addition, which does not occur. For similar questions, note resumption without new starts. Check if pauses are nucleotide-specific, ruling out breaks.

9

A lab compares two repair-defective mammalian cell lines after exposure to ionizing radiation. Line 1 shows many large chromosomal translocations; line 2 shows fewer translocations but prolonged cell-cycle arrest and reliance on an intact sister chromatid for recovery. Based on the scenario, which pairing of defective pathway and expected phenotype is most consistent for Line 1?

Defective removal of UV-induced bulky lesions, causing primarily adjacent-pyrimidine substitutions

Defective accurate template-based double-strand break repair, shifting repair toward end-joining and increasing rearrangements

Defective unwinding at replication origins, causing failure to initiate replication rather than translocations

Defective removal of uracil from DNA, causing primarily G:C→A:T transitions without rearrangements

Explanation

This question tests knowledge of DNA replication and repair. Homologous recombination (HR) accurately repairs double-strand breaks using sister chromatids, preventing rearrangements. Line 1's translocations indicate HR defect, shifting to error-prone alternatives. Choice D is consistent as HR loss increases rearrangements post-radiation. Choice B distracts by focusing on uracil repair causing transitions, not rearrangements. For similar questions, associate translocations with HR failure. Check reliance on sister chromatids, distinguishing from base repairs.

10

A bacterial strain was engineered with a point mutation in a gene required for correcting single-base mismatches that remain after DNA copying. When grown for many generations, this strain accumulates short repeat-length changes (e.g., +1 or −1 base) within simple sequence repeats, while showing relatively fewer large deletions. Which replication/repair process would be most directly disrupted by this mutation?

Removal of UV-induced covalent links between adjacent pyrimidines by cutting out a short DNA segment and resynthesizing it

Joining of two broken DNA ends by aligning short homologous regions and trimming overhangs before ligation

Resolution of double-strand breaks using an intact sister DNA molecule as a template for accurate restoration

Correction of mismatched bases and small insertion–deletion loops that arise from strand slippage during DNA copying

Explanation

This question tests knowledge of DNA mismatch repair and its role in preventing replication errors. Mismatch repair (MMR) is a post-replicative process that recognizes and corrects base-base mismatches and small insertion-deletion loops that escape DNA polymerase proofreading. The bacterial strain described has a mutation affecting single-base mismatch correction, which is the hallmark of MMR deficiency. The accumulation of repeat-length changes in simple sequence repeats is characteristic of MMR defects because DNA polymerase frequently slips on repetitive sequences during replication, creating small loops that MMR normally corrects. Option B describes nucleotide excision repair for UV damage, option C describes non-homologous end joining, and option D describes homologous recombination repair - none of which specifically target replication errors in simple repeats. The relatively fewer large deletions indicate that other repair pathways remain intact. For similar questions, recognize that MMR deficiency specifically leads to microsatellite instability and increased point mutations rather than large-scale genomic changes.

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