MCAT Biological and Biochemical Foundations of Living Systems Quiz: 1b Dna Replication Repair

What this quiz covers

This quiz focuses on 1b Dna Replication Repair, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Biological and Biochemical Foundations of Living Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Accumulation of length changes in repeats due to failure to correct small misalignments during copying (correct answer)
2. Inability to reseal breaks between DNA fragments, producing many short linear DNA pieces
3. Increased C→T transitions at adjacent pyrimidines after UV exposure due to failure to remove bulky lesions
4. Failure to separate newly synthesized strands because covalent links between strands are not removed

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.

Question 2

1. It separates the two parental DNA strands to allow copying of each template
2. It seals adjacent DNA segments by forming phosphodiester bonds in the sugar–phosphate backbone (correct answer)
3. It selects the correct incoming nucleotide by monitoring base-pair geometry at the active site
4. 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.

Question 3

1. Removal of UV-induced bulky lesions by excising a short segment of the damaged strand (correct answer)
2. Direct reversal of methylated bases by transferring the methyl group to an enzyme
3. Correction of replication mismatches by recognizing the newly synthesized strand via nicks
4. 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.

Question 4

1. Mismatch correction that removes short patches containing mispaired bases after replication
2. End-joining of broken DNA ends without requiring extensive sequence matching, which can introduce small indels (correct answer)
3. Excision of bulky lesions followed by gap filling using the undamaged strand as a template
4. 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.

Question 5

1. It separates the two DNA strands ahead of synthesis to allow copying of the template (correct answer)
2. It removes RNA primers by degrading RNA and replacing it with DNA
3. It seals nicks between adjacent DNA fragments after synthesis is complete
4. It adds methyl groups to the newly synthesized strand to mark it for mismatch correction

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.

Question 6

1. Cutting the damaged strand on both sides of the lesion, removing a short segment, and resynthesizing using the intact strand (correct answer)
2. Removing a single damaged base by cleaving the base–sugar bond without removing neighboring nucleotides
3. Joining broken DNA ends directly without using sequence similarity
4. Correcting mismatches by excising the older strand based on chemical marks

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.

Question 7

1. The oxidized guanine can mispair during copying, leading to insertion of an incorrect base and fixation as a transversion (correct answer)
2. The oxidized guanine prevents sealing of backbone nicks, causing deletions rather than base substitutions
3. The oxidized guanine blocks strand separation at origins, preventing replication initiation and reducing mutations
4. The oxidized guanine is corrected by removing a short segment around bulky lesions, so loss of base removal has no effect

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.

Question 8

1. DNA synthesis resumes only after a double-strand break is joined by end-joining enzymes
2. DNA synthesis restarts by adding nucleotides to the $5'$ end, bypassing the need for a $3'$ end
3. DNA synthesis requires complete removal of the template strand before extension can continue
4. DNA synthesis can restart by extending an existing $3'$ end once the required nucleotide becomes available (correct answer)

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.

Question 9

1. Defective unwinding at replication origins, causing failure to initiate replication rather than translocations
2. Defective removal of uracil from DNA, causing primarily G:C→A:T transitions without rearrangements
3. Defective removal of UV-induced bulky lesions, causing primarily adjacent-pyrimidine substitutions
4. Defective accurate template-based double-strand break repair, shifting repair toward end-joining and increasing rearrangements (correct answer)

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.

Question 10

1. Correction of mismatched bases and small insertion–deletion loops that arise from strand slippage during DNA copying (correct answer)
2. Removal of UV-induced covalent links between adjacent pyrimidines by cutting out a short DNA segment and resynthesizing it
3. Joining of two broken DNA ends by aligning short homologous regions and trimming overhangs before ligation
4. Resolution of double-strand breaks using an intact sister DNA molecule as a template for accurate restoration

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.

Question 11

1. It removes torsional strain by cutting and rejoining DNA strands ahead of the copying site.
2. It provides a free 3' hydroxyl group by making a short RNA segment that the DNA-copying enzyme can extend. (correct answer)
3. It joins adjacent DNA fragments by forming phosphodiester bonds after RNA primers have been removed.
4. It identifies the newly synthesized strand by methylation patterns and removes mismatches after replication is complete.

Explanation: This question tests understanding of primase function in DNA replication initiation. DNA polymerases cannot initiate synthesis de novo - they require a pre-existing 3'-OH group to extend. Primase solves this problem by synthesizing short RNA primers (8-12 nucleotides) complementary to the DNA template, providing the 3'-OH group that DNA polymerase needs to begin synthesis. The scenario clearly shows that DNA synthesis fails without these RNA primers but proceeds when they're added exogenously, confirming primase's essential role. Option A describes topoisomerase function in relieving torsional strain, option C describes ligase function in joining fragments, and option D describes aspects of mismatch repair involving strand discrimination. The inability to start synthesis on bare template combined with the ability to extend existing primers is the key signature of primase requirement. For similar questions, recognize that primase is uniquely required because it's the only enzyme in replication that can begin nucleic acid synthesis without a primer.

Question 12

1. DNA ligase unwinds the duplex ahead of the replication machinery to prevent fragment formation
2. DNA ligase removes RNA primers to create a 3′ end that can be extended by polymerase
3. DNA ligase synthesizes DNA across the gap by adding nucleotides to the 5′ end of the downstream fragment
4. DNA ligase forms a phosphodiester bond between adjacent fragments after the gap has been filled, sealing the remaining nick (correct answer)

Explanation: This question tests knowledge of DNA replication and repair, focusing on the resolution of discontinuities in the lagging strand. During replication, the lagging strand is synthesized in short Okazaki fragments, which require primer removal, gap filling by polymerase, and sealing by an enzyme to form a continuous strand. In this scenario, the polymerase variant causes more frequent release, leading to increased gaps, but incubation allows resolution into a continuous chromosome. The correct answer is consistent because DNA ligase seals the nicks between filled fragments via phosphodiester bond formation, completing the strand. A distractor like choice B does not fit because DNA ligase does not remove RNA primers; that is the role of other enzymes like RNase H or polymerase-associated activities. For similar questions, confirm if the issue involves joining fragments post-synthesis and recall ligase's specific role in nick sealing. Also, differentiate ligase from polymerases, which fill gaps but do not seal them.

Question 13

1. Joining adjacent DNA fragments by forming phosphodiester bonds after primer removal
2. Initiation of DNA synthesis by creating short RNA segments that provide a 3′-OH for extension (correct answer)
3. Relief of torsional strain ahead of the fork by transient DNA breakage and resealing
4. Removal of bulky lesions by excising a short single-stranded DNA segment and filling the gap

Explanation: This question tests knowledge of DNA replication and repair, specifically the role of primase in initiation. Primase synthesizes short RNA primers that provide 3′-OH ends for DNA polymerase to extend, essential for starting new strands, especially on the lagging side. In this scenario, inhibiting primase reduces label incorporation at new origins but allows brief continuation from active forks. The correct answer is consistent because blocking primer synthesis disrupts initiation at new sites while elongation proceeds until primers are needed again. A distractor like choice A does not fit because it describes ligase, which acts post-synthesis and would not selectively affect new starts. For similar questions, check if the defect targets initiation versus elongation and recall primase's primer role. Also, distinguish primase from other enzymes like helicase or polymerase in replication dynamics.

Question 14

1. Direct reversal of modified bases by transferring a methyl group from DNA to the repair enzyme
2. Removal of a short stretch of nucleotides surrounding a helix-distorting lesion followed by resynthesis using the intact strand (correct answer)
3. Correction of replication mismatches by excising a region of the parental strand containing the mispaired base
4. Repair of double-strand breaks by aligning homologous chromosomes and exchanging large DNA segments

Explanation: This question tests knowledge of DNA replication and repair, particularly nucleotide excision repair (NER) for bulky DNA lesions. NER recognizes helix-distorting damage, excises a short oligonucleotide segment containing the lesion, and resynthesizes the gap using the intact complementary strand as a template. In this scenario, the chemical induces bulky modifications, and the mutation disrupts the endonuclease that cuts around the lesion, leading to unrepaired distortions that cause double-strand breaks during replication in S phase. The correct answer is consistent because disrupting NER would prevent removal of these lesions, allowing them to stall replication forks and increase breaks. A distractor like choice A does not fit because it describes direct reversal, such as alkyltransferase activity, which handles small modifications like methylation without excision. To evaluate similar questions, identify if the damage is helix-distorting and if repair involves segment excision rather than base-specific reversal. Additionally, note that NER deficiency often leads to replication-associated breaks and sensitivity to UV or chemicals causing bulky adducts.

Question 15

1. Repair of double-strand breaks by aligning homologous chromosomes and exchanging large DNA segments
2. Sealing of sugar–phosphate backbone nicks after removal of RNA primers during DNA synthesis
3. Direct reversal of UV-induced covalent links between adjacent pyrimidines on the same strand
4. Correction of mismatched bases on the newly synthesized strand using strand-specific cues shortly after replication (correct answer)

Explanation: This question tests understanding of DNA replication and repair mechanisms, specifically post-replicative mismatch repair. The mismatch repair system corrects base-pairing errors that escape polymerase proofreading by recognizing distortions in the DNA helix and removing the incorrect base from the newly synthesized strand. The scenario describes a mutation affecting mismatch repair, as evidenced by increased substitutions without increased double-strand breaks, indicating the primary defect is in correcting replication errors rather than managing DNA damage. The mismatch repair system uses strand-specific cues (like hemimethylation patterns in bacteria) to distinguish the new strand from the template strand and correct errors on the new strand. Choice B describes ligase function, which would cause strand breaks if defective, not substitutions. To identify mismatch repair defects, look for increased point mutations without other types of DNA damage.

Question 16

1. It prevents lesions by reversing chemical modifications of bases without cutting the DNA backbone
2. It increases accuracy by removing incorrectly paired nucleotides immediately after incorporation
3. It reduces replication stalling by inserting nucleotides across from non-informative lesions, at the cost of lower accuracy (correct answer)
4. It restores plasmid copy number by recombining the plasmid with homologous chromosomal DNA during mitosis

Explanation: This question tests understanding of DNA replication and repair, specifically translesion synthesis by damage-tolerant polymerases. When replicative polymerases encounter non-instructive lesions like abasic sites, they stall because they cannot determine which base to insert. Specialized translesion polymerases can bypass these lesions by inserting nucleotides (often adenine by the 'A-rule') opposite the damage, allowing replication to continue at the cost of increased mutations. In Strain 1 with the translesion polymerase, replication proceeds past the abasic site but with increased errors; in Strain 2 without it, replication stalls at the lesion. This trade-off between replication completion and accuracy is characteristic of translesion synthesis. Choice B describes proofreading function, which would not help with abasic sites that lack pairing information. When analyzing translesion synthesis, expect continued replication with increased mutations near lesions.

Question 17

1. More frequent removal of the newly added nucleotide at the mismatch site before synthesis continues
2. Increased retention of an incorrect newly added nucleotide, allowing extension past the mismatch and fixation after the next round (correct answer)
3. Failure to join adjacent DNA fragments, leaving a nick that is later converted into a substitution
4. Preferential replacement of a damaged base by copying information from the opposite strand without changing replication accuracy

Explanation: This question tests understanding of DNA replication and repair, specifically the role of polymerase proofreading in maintaining replication fidelity. DNA polymerases have two key activities: a polymerase domain that adds nucleotides and a 3'-to-5' exonuclease domain that removes incorrectly paired bases immediately after incorporation. In this scenario, the editing site mutation eliminates the exonuclease activity while leaving polymerase activity intact. When the exonuclease cannot remove the mismatched T opposite G, the polymerase continues synthesis beyond the mismatch, and after the next round of replication, the mismatch becomes fixed as a permanent G→A substitution (since T pairs with A). Choice A is incorrect because it describes increased editing activity, not loss of editing. This demonstrates that polymerase proofreading prevents mutations by removing mismatches before they can be extended and fixed in subsequent rounds.

Question 18

1. It identifies incorrect base pairs and replaces them using the original strand as a guide
2. It synthesizes short RNA segments that provide starting points for DNA synthesis
3. It covalently seals discontinuities between adjacent DNA fragments after synthesis is complete
4. It relieves torsional strain generated during strand separation by transiently cutting and rejoining DNA (correct answer)

Explanation: This question tests understanding of DNA replication machinery, specifically the role of topoisomerase in managing DNA topology. During replication, helicase unwinds the double helix, creating positive supercoiling (overwinding) ahead of the replication fork that would eventually halt replication if not resolved. Topoisomerase relieves this torsional strain by transiently cutting one or both DNA strands, allowing rotation, then resealing the break. Inhibiting topoisomerase causes accumulation of positive supercoils, slowing replication and potentially causing DNA breaks under mechanical stress. The observation of breaks in high-strain regions confirms topoisomerase's role in managing replication-induced torsion. Choice B describes primase function, and choice C describes ligase function, neither of which directly manage topological strain. When analyzing replication stress, consider how mechanical forces during DNA unwinding require topological management.

Question 19

1. Exchange of DNA segments between homologous chromosomes to restore missing genetic information
2. Addition of nucleotides to the $3'$ end of the new strand using the old strand as a template
3. Removal of RNA primers from newly synthesized DNA to prevent ribonucleotide retention
4. Recognition and excision of a short stretch of the damaged strand surrounding a bulky UV lesion (correct answer)

Explanation: This question tests understanding of DNA repair mechanisms, specifically nucleotide excision repair (NER) of UV-induced DNA damage. UV light causes covalent bonds between adjacent pyrimidines (often thymine dimers), creating bulky lesions that distort the DNA helix. NER recognizes these distortions and removes them by cutting out a short DNA segment containing the damage, then resynthesizing using the undamaged strand as template. The mutant strain's defect in this pathway leads to persistence of UV lesions, which during replication can cause C→T transitions when damaged cytosines are read as thymine. The first step of NER - recognition and excision of the damaged segment - is most directly impaired. Choice B describes normal polymerase function, not repair, and choice C describes primer removal, unrelated to UV damage. To identify NER defects, look for increased mutations at UV damage sites, particularly C→T transitions at dipyrimidine sequences.

Question 20

1. The nick can be converted into a double-strand break when the replication machinery encounters it, leading to incomplete products (correct answer)
2. The nick increases polymerase editing, causing excessive removal of correct nucleotides and slowing synthesis uniformly
3. The nick prevents base pairing, so all downstream bases are copied as RNA rather than DNA
4. The nick blocks ligase from sealing fragments, which directly increases base substitutions without affecting fragment length

Explanation: This question tests understanding of DNA replication and repair, specifically how single-strand breaks interact with the replication machinery. When a replication fork encounters a nick on the template strand, the fork can collapse, converting the single-strand break into a double-strand break as the replication machinery attempts to use the nicked strand as a template. This fork collapse results in truncated replication products because synthesis cannot continue past the break. The conversion of a nick to a double-strand break during replication is a well-established mechanism of replication fork collapse. Choice B incorrectly suggests nicks increase editing (they don't affect exonuclease activity), and choice C incorrectly proposes RNA synthesis (DNA polymerase doesn't switch to RNA synthesis at nicks). When analyzing replication problems, consider how pre-existing DNA damage can be converted to more severe lesions by the replication process itself.