DNA Replication - AP Biology

1. A biologist adds a drug that specifically inhibits helicase in rapidly dividing cells. In untreated cells, helicase separates the parental DNA strands at replication origins, allowing DNA polymerase to copy each template strand using complementary base pairing. After drug treatment, the biologist measures incorporation of labeled deoxyribonucleotides into DNA and finds a large decrease. Which explanation best accounts for the decreased nucleotide incorporation in treated cells?
   Answer: Inhibiting helicase prevents strand separation, reducing access to single-stranded templates for DNA polymerase. This question assesses the skill of analyzing DNA replication, investigating helicase's role in fork initiation and progression. Inhibiting helicase prevents parental strand separation, limiting single-stranded template availability for DNA polymerase to incorporate nucleotides via base pairing, thus decreasing labeled deoxyribonucleotide uptake. Without unwinding, replication cannot proceed beyond origins, as polymerase requires exposed templates. This explains the large reduction in incorporation compared to untreated cells where helicase enables continuous synthesis. A tempting distractor is choice B, claiming it increases primer synthesis and stalls polymerase, but this inverts the effect, misconceiving inhibition as overactivating priming when it actually blocks access upstream. To analyze inhibitor effects, identify the targeted enzyme's function and link it to downstream replication steps like nucleotide addition.
2. DNA replication uses complementary base pairing and requires short RNA primers that provide a free 3′-OH for DNA polymerase to extend. A drug prevents primase from synthesizing RNA primers but does not disrupt helicase activity. Which outcome is most likely at replication origins?
   Answer: Neither strand can be extended efficiently because DNA polymerase lacks a primer 3′ end. This question tests understanding of primer requirements in DNA replication. DNA polymerase cannot initiate synthesis de novo - it requires a primer with a 3'-OH group to begin adding nucleotides. Primase normally synthesizes short RNA primers that provide these 3' ends on both leading and lagging strands. Without primase activity, DNA polymerase has no 3' end to extend on either strand, effectively blocking all DNA synthesis at origins. Choice B incorrectly assumes the leading strand doesn't need primers, but even continuous synthesis requires an initial primer. The critical concept is that DNA polymerase absolutely requires a pre-existing 3'-OH group to function.
3. In a bacterial cell, replication begins at a single origin and proceeds bidirectionally. As the helix unwinds, DNA polymerase extends new DNA only by adding nucleotides to the 3′ end, using complementary pairing to the template strand. One new strand is synthesized continuously toward each fork, and the other is synthesized discontinuously as short DNA fragments. A mutation reduces the ability of DNA polymerase to proofread newly added bases, but does not affect unwinding or primer formation. Which outcome is most likely after several rounds of replication?
   Answer: More base-substitution mutations accumulate because mismatches persist during elongation. This question examines DNA replication fidelity and the consequences of impaired proofreading function. DNA polymerase normally has 3'→5' exonuclease activity that removes incorrectly paired bases immediately after incorporation, maintaining high fidelity during replication. When this proofreading function is reduced, mismatched base pairs that would normally be corrected persist in the newly synthesized DNA, leading to base-substitution mutations after subsequent rounds of replication. Choice B incorrectly suggests replication stops, but proofreading is not required for polymerase to continue synthesis—it only affects accuracy. The critical insight is that proofreading prevents mutations during elongation, not that it's required for replication to proceed.
4. During replication, each parental strand serves as a template for synthesis of a complementary strand, producing two double helices. A researcher provides a nucleotide pool lacking thymine (T) but containing adenine (A), cytosine (C), and guanine (G). Which outcome is most likely during DNA synthesis?
   Answer: DNA synthesis stalls when adenine is encountered on the template strand. This question explores the consequences of missing essential nucleotides during replication. DNA synthesis requires all four deoxyribonucleotides (dATP, dTTP, dCTP, dGTP) to create complementary strands. When thymine is absent, DNA polymerase cannot add the required nucleotide opposite adenine on the template strand, causing synthesis to stall at every A position. The polymerase cannot substitute uracil (choice A) because it specifically incorporates deoxyribonucleotides, not ribonucleotides. The key principle is that DNA polymerase has strict substrate specificity and cannot proceed without the correct nucleotide.
5. During S phase, a cell copies DNA by separating the two parental strands; each exposed base serves as a template for adding complementary nucleotides (A–T, C–G). DNA polymerase extends new DNA only by adding nucleotides to a free 3′ OH, so synthesis proceeds 5′→3′. At a replication fork, one new strand is synthesized continuously in the same direction as fork movement, while the other is synthesized discontinuously as short fragments that are later joined. If a base is mismatched, the shape of the helix is distorted, increasing the likelihood it is removed and replaced. Which outcome is most likely if DNA polymerase cannot add nucleotides to a 3′ end?
   Answer: New DNA strands cannot elongate, so replication stalls at the fork despite template availability. This question assesses the skill of analyzing DNA replication processes, particularly the role of DNA polymerase in strand elongation. If DNA polymerase cannot add nucleotides to a 3′ end, it loses its ability to extend new DNA strands in the required 5′→3′ direction, halting the incorporation of complementary bases despite available templates. This leads to stalled replication forks because both leading and lagging strand synthesis rely on 3′ end addition, preventing any elongation of daughter strands. Consequently, replication cannot proceed normally, resulting in incomplete DNA copying during S phase. A tempting distractor is choice A, which suggests RNA nucleotides replace DNA ones, stemming from the misconception that replication could switch to RNA-based synthesis without polymerase's directional constraint. To analyze similar replication defects, always trace the impact on directionality and enzyme requirements step by step.
6. DNA replication uses each parental strand as a template, with new nucleotides added by DNA polymerase according to base-pairing rules (A–T, C–G). DNA polymerase extends from a free 3′ OH, synthesizing 5′→3′. A proofreading activity can remove an incorrectly paired nucleotide when a mismatch distorts the helix, allowing the correct nucleotide to be inserted. In a mutant cell line, proofreading is absent but polymerase still adds nucleotides normally. Which outcome is most likely in the replicated DNA of the mutant cells?
   Answer: A higher mutation rate because mismatched bases are less likely to be removed during synthesis. This question tests the skill of analyzing DNA replication fidelity, emphasizing the proofreading mechanism's role in error correction. Without proofreading, mismatched nucleotides that distort the helix are not removed, leading to a higher incorporation of incorrect bases during 5′→3′ synthesis. Although base pairing guides nucleotide selection, occasional mismatches occur, and lacking removal increases the likelihood they persist in the daughter strands. This results in an elevated mutation rate in the replicated DNA of mutant cells compared to normal cells with functional proofreading. Choice C is a common distractor, claiming no change because base pairing alone ensures accuracy, overlooking the misconception that replication is error-free without additional correction mechanisms like proofreading. For evaluating replication mutations, consider both initial pairing accuracy and post-synthesis error correction processes.
7. A short DNA segment is shown on one parental strand as 3′-T A C G G A-5′. During replication, the complementary daughter strand is synthesized by adding DNA nucleotides according to base pairing, producing a strand antiparallel to the template. DNA polymerase adds nucleotides to a 3′ end, so the new strand is written 5′→3′. Which sequence represents the newly synthesized complementary strand (5′→3′)?
   Answer: 5′-A T G C C T-3′ This question evaluates the skill of analyzing DNA replication by determining complementary strand sequences based on base pairing and directionality. The newly synthesized strand must be antiparallel to the template 3′-TAC GGA-5′, resulting in 5′-ATG CCT-3′ through complementary pairing where T pairs with A, A with T, C with G, G with C, G with C, and A with T. DNA polymerase adds nucleotides to the 3′ end, ensuring the new strand is written in the 5′→3′ orientation as it elongates. This sequence accurately reflects the replication logic of matching bases while maintaining antiparallel structure. Choice C, 5′-TAC GGA-3′, is a tempting distractor as it mirrors the template without reversal, stemming from the misconception of ignoring antiparallel orientation in strand synthesis. To solve sequence problems, always reverse the template direction and apply base pairing rules systematically.
8. DNA replication relies on complementary base pairing and the requirement that DNA polymerase extend from a 3′ end. At one replication fork, a student labels the template strand running 3′→5′ toward the fork as Template X and the opposite template running 5′→3′ toward the fork as Template Y. DNA polymerase synthesizes one strand continuously and the other discontinuously in fragments that are later joined. Which statement best accounts for which template produces the discontinuous (lagging) strand at this fork?
   Answer: Template Y produces the lagging strand because synthesis must proceed 5′→3′ away from the fork. This question assesses understanding of leading versus lagging strand synthesis based on template orientation. DNA polymerase can only synthesize in the 5'→3' direction, which means it reads the template 3'→5'. At a replication fork, the template running 3'→5' toward the fork (Template X) allows continuous synthesis toward the fork (leading strand), while the template running 5'→3' toward the fork (Template Y) requires discontinuous synthesis away from the fork in short fragments (lagging strand). Choice B incorrectly states that polymerase reads templates 5'→3', when it actually reads 3'→5' while synthesizing 5'→3'. The strategy is to match the directionality constraint of DNA polymerase with the orientation of each template strand relative to fork movement.
9. Replication requires unwinding of the double helix and stabilization of exposed single-stranded templates so they do not reanneal. A mutation prevents single-strand binding proteins from attaching to separated DNA, but polymerase function is normal. Which outcome is most likely at the replication fork?
   Answer: Templates re-form base pairs, reducing access for polymerase and slowing replication. This question examines the role of single-strand binding proteins (SSBs) in DNA replication. SSBs normally coat exposed single-stranded DNA after helicase unwinds the double helix, preventing the complementary strands from re-annealing through hydrogen bonding. Without SSBs, the separated template strands can spontaneously re-form base pairs, creating a physical barrier that blocks DNA polymerase access and slows replication fork progression. Choice C incorrectly suggests reannealing would help, but reformed double-stranded regions actually impede polymerase movement. The key insight is that maintaining single-stranded templates is essential for polymerase accessibility.
10. In semiconservative DNA replication, each parental strand serves as a template for synthesis of a complementary strand. DNA polymerase adds nucleotides to the 3′ end, generating new DNA 5′→3′. The two new DNA molecules each contain one parental strand paired with one newly synthesized strand. A scientist labels newly synthesized DNA by providing nucleotides containing a heavy isotope for one S phase only, then switching back to normal nucleotides. Which pattern is most likely after one round of replication in the presence of heavy nucleotides?
    Answer: All DNA molecules are half heavy because each contains one heavy new strand and one light parental strand. This question tests the skill of analyzing semiconservative DNA replication through isotope labeling experiments. After one round with heavy nucleotides, each daughter DNA molecule consists of one light parental strand and one heavy newly synthesized strand, resulting in all molecules being hybrid or half heavy. This pattern arises because replication uses each parental strand as a template, incorporating heavy nucleotides into the new strands via 5′→3′ synthesis. The semiconservative model predicts this uniform hybrid density rather than a mix of fully heavy or light molecules. Choice C is a tempting distractor, suggesting half fully heavy and half fully light due to assortment, reflecting the misconception of conservative replication where parental strands stay together. For interpreting labeling results, recall the semiconservative mechanism and track parental versus new strands across generations.
11. At a replication fork, single-strand DNA-binding proteins (SSBs) bind to exposed parental DNA strands after helicase separates them. DNA polymerase then uses each parental strand as a template, extending new DNA by complementary base pairing. In a mutant extract lacking functional SSBs, the parental strands frequently re-form hydrogen bonds with each other soon after unwinding. Which outcome is most likely in the mutant extract compared with a normal extract?
    Answer: Replication fork progression will slow because reannealing reduces the availability of single-stranded templates. This question assesses the skill of analyzing DNA replication, addressing the function of single-strand binding proteins in maintaining template availability. Without SSBs, parental strands reanneal after unwinding, reducing single-stranded template exposure and slowing replication fork progression as polymerase has less access to templates. This reannealing disrupts the normal process where SSBs stabilize separated strands, allowing continuous complementary synthesis. Consequently, the mutant extract exhibits delayed replication compared to normal, where SSBs prevent such interference. A tempting distractor is choice B, suggesting accuracy increases due to forced base pairing, but this misconceives reannealing as aiding new strand fidelity when it actually hinders synthesis by reforming parental duplexes. When evaluating replication mutants, consider how each component stabilizes the fork and predict effects on speed versus accuracy.
12. At a replication fork, DNA polymerase synthesizes new DNA 5′→3′ by extending from a 3′ end, using each parental strand as a template. Because the templates are antiparallel, one new strand is synthesized continuously and the other discontinuously as fragments. A mutation causes DNA polymerase to add nucleotides only very slowly but does not affect strand separation. Which outcome is most likely at replication forks in the mutant cells?
    Answer: Single-stranded DNA accumulates because unwinding continues while synthesis lags behind. This question assesses the skill of analyzing DNA replication dynamics when polymerase activity is slowed. A mutation reducing polymerase speed causes synthesis to lag behind strand separation by helicase, leading to accumulation of exposed single-stranded DNA at the replication fork. Although unwinding proceeds normally, the slowed 5′→3′ addition on both leading and lagging strands cannot keep pace, resulting in persistent ssDNA regions. This imbalance specifically affects the coordination between unwinding and synthesis without altering base pairing or strand continuity directly. Choice A is a tempting distractor, suggesting faster fork movement with unchanged helicase, stemming from the misconception that polymerase speed directly accelerates rather than limits fork progression. To analyze enzyme mutations in replication, compare the rates of coupled processes like unwinding and synthesis.
13. A DNA segment contains the template sequence 3′-A C G T A A-5′ at a replication fork. DNA polymerase adds nucleotides to the 3′ end of the growing strand using complementary base pairing. Which sequence is produced on the new strand (written 5′→3′)?
    Answer: 5′-T G C A T T-3′ This question tests application of complementary base pairing rules during DNA synthesis. Given the template 3'-ACGTAA-5', DNA polymerase reads it 3' to 5' and synthesizes the new strand 5' to 3' using complementary base pairing (A pairs with T, C with G). Reading the template from 3' to 5': A→T, C→G, G→C, T→A, A→T, A→T, producing 5'-TGCATT-3'. Choice C incorrectly uses RNA nucleotides (uracil), while choice D has the wrong polarity. The strategy is to carefully track template reading direction and apply base pairing rules systematically.
14. During S phase, a eukaryotic replication fork opens as helicase separates the two parental strands. Each parental strand serves as a template, and DNA polymerase adds nucleotides only to a free 3′ end, forming phosphodiester bonds as complementary base pairs (A–T, G–C) align. One strand is synthesized continuously in the same direction as fork movement, while the other is synthesized in short segments that are later joined. A researcher briefly inhibits the enzyme that seals nicks between adjacent DNA segments, then removes the inhibitor and allows replication to finish. Which outcome is most likely in the newly synthesized DNA?
    Answer: The lagging strand contains persistent single-strand breaks between short DNA segments. This question tests understanding of DNA replication mechanics, specifically the role of DNA ligase in joining Okazaki fragments. During replication, the leading strand is synthesized continuously in the 5'→3' direction toward the fork, while the lagging strand must be synthesized discontinuously as short Okazaki fragments away from the fork due to DNA polymerase's directional constraint. When DNA ligase (the enzyme that seals nicks between adjacent DNA segments) is temporarily inhibited, these Okazaki fragments cannot be joined together, leaving persistent single-strand breaks in the lagging strand even after replication completes. Choice B incorrectly assumes helicase function is affected, but the inhibitor only targets ligase, not helicase. The key strategy is to identify which replication component is affected by the inhibitor and predict the specific consequence for that strand.
15. In a model system, a DNA molecule is replicated in the presence of a nucleotide analog that can base-pair like thymine but lacks a 3′-OH group. DNA polymerase incorporates the analog occasionally during synthesis on both leading and lagging strands. Incorporation does not prevent base pairing, but it prevents addition of any subsequent nucleotide to that strand end. Which outcome is most likely when the analog is incorporated into a growing DNA strand?
    Answer: Elongation stops at that point because no further phosphodiester bonds can form at the 3′ end. This question examines the requirement for a 3'-OH group in DNA synthesis. DNA polymerase catalyzes phosphodiester bond formation by attacking the α-phosphate of an incoming nucleotide with the 3'-OH of the growing strand. When a nucleotide analog lacking a 3'-OH is incorporated, it acts as a chain terminator because no subsequent nucleotide can be added to that strand—there's no 3'-OH to form the next phosphodiester bond. Choice C incorrectly claims only the 5' end is needed for extension, but DNA polymerase absolutely requires a free 3'-OH for elongation. The key concept is that the 3'-OH group is essential for the chemistry of DNA synthesis, making its absence a complete block to further elongation.
16. In a cell-free replication system, DNA polymerase extends a primer on a DNA template using complementary base pairing. The system is supplied with abundant dATP, dTTP, dCTP, and dGTP. In one reaction, dTTP is replaced with a nucleotide analog that can base-pair with adenine but lacks a 3′ OH group. After initiation, synthesis begins but stops after the analog is incorporated. Which explanation best accounts for the premature termination of DNA synthesis?
    Answer: The analog lacks a 3′ OH, preventing formation of the next phosphodiester bond at the growing strand end. This question assesses the skill of analyzing DNA replication, examining how nucleotide structure affects chain elongation. The analog, lacking a 3' OH, halts synthesis after incorporation because DNA polymerase requires a 3' OH to form the phosphodiester bond with the next nucleotide's 5' phosphate. Although the analog base-pairs correctly with adenine, its structural defect prevents further extension at the growing strand's end. This chain termination occurs specifically when the analog is added, explaining the premature stop despite abundant normal nucleotides. A tempting distractor is choice A, suggesting it prevents helicase unwinding, but this misconceives the analog's role, as it affects elongation post-incorporation, not initial strand separation. For analog-based questions, identify which replication step the modification targets and trace its impact on the process.
17. A mutation alters a DNA polymerase so that it frequently inserts guanine opposite thymine during replication. The polymerase still extends DNA in the 5′→3′ direction and does not remove mismatches. After replication and one additional round of replication in a normal cell, some daughter DNA molecules contain a stable base-pair change at the original site. Which outcome is most likely for the sequence at that position in molecules where the mismatch became fixed?
    Answer: A G–C base pair will replace the original A–T base pair after replication copies the mismatch. This question assesses the skill of analyzing DNA replication, exploring the consequences of insertion errors without correction. The mismatch, such as T-G instead of T-A, becomes fixed after another replication round, resulting in a G-C pair in some daughter molecules where the erroneous G templates a C. This occurs because the uncorrected mismatch leads to one strand carrying the wrong base, which then directs complementary pairing in the next cycle, permanently changing the sequence. Over two rounds, half the molecules may retain the original A-T, but the mutated ones stabilize as G-C. A tempting distractor is choice D, claiming a T-G pair persists indefinitely, but this ignores how replication resolves mismatches into stable pairs, misconceiving the process as tolerant of perpetual mismatches. To predict mutation outcomes, simulate two replication cycles, tracking base pairing from the error point.
18. After replication, each daughter DNA molecule contains one parental strand and one newly synthesized complementary strand. A researcher observes a point mutation present in only one of the two sister chromatids immediately after S phase. Which explanation best accounts for this observation?
    Answer: A mismatch escaped proofreading on one new strand, and the other duplex copied correctly. This question explores how mutations arise during semiconservative replication. Each sister chromatid contains one parental strand and one newly synthesized strand. If a replication error occurs on one new strand and escapes proofreading, only that chromatid will carry the mutation, while the other chromatid (with a different new strand) remains normal. This explains why mutations can affect just one sister chromatid. Choice B incorrectly invokes conservative replication, which doesn't occur in normal cells. The key concept is that each chromatid's new strand is synthesized independently, allowing errors to affect them differently.
19. A student examines replication of a linear eukaryotic chromosome end. DNA polymerase requires a primer and can extend only from an existing 3′ end, synthesizing DNA 5′→3′. After the final RNA primer on the lagging strand is removed, there is no upstream 3′ OH available for DNA polymerase to fill in the remaining gap at the extreme 5′ end of the new strand. Over many cell divisions, this leads to progressive shortening of chromosome ends. Which explanation best accounts for the shortening based on replication mechanism?
    Answer: DNA polymerase cannot initiate synthesis de novo, leaving an unreplicated terminal region after primer removal. This question assesses the skill of analyzing DNA replication, specifically the end-replication problem in linear chromosomes. Shortening occurs because DNA polymerase cannot fill the gap at the 5' end of the new lagging strand after RNA primer removal, as it requires a primer and an existing 3' OH for extension, leaving an unreplicated terminal region. This progressive loss happens over divisions since each cycle removes the terminal primer without replacement. The mechanism stems from the 5' to 3' synthesis direction and primer necessity, unique to linear ends. A tempting distractor is choice E, stating polymerase synthesizes lagging strand 3' to 5', but this reverses directionality, misconceiving that all synthesis is 5' to 3' regardless of strand. For telomere-related issues, map the final primer position and identify why the gap persists post-removal.
20. In DNA replication, the parental strands are antiparallel, and DNA polymerase extends from primers by adding nucleotides to the 3′ end. A student claims both new strands are synthesized continuously because polymerase can follow the fork on both templates. Which explanation best accounts for why one strand is synthesized discontinuously?
    Answer: One template is oriented so synthesis requires repeated priming to maintain 5′→3′ extension. This question addresses why DNA replication produces leading and lagging strands. Because DNA strands are antiparallel and polymerase only synthesizes 5' to 3', one template (3' to 5' oriented toward the fork) allows continuous synthesis, while the other template (5' to 3' oriented toward the fork) requires discontinuous synthesis. The lagging strand must be made in fragments because its template orientation means synthesis proceeds away from the advancing fork, requiring repeated priming events. Choice B incorrectly attributes fragments to base composition rather than template orientation. Understanding antiparallel structure and polymerase directionality explains asymmetric synthesis patterns.

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