AP Biology Quiz: Dna Replication

What this quiz covers

This quiz focuses on Dna Replication, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

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. A mismatch escaped proofreading on one new strand, and the other duplex copied correctly. (correct answer)
2. Replication is conservative, so only one chromatid is newly synthesized and mutates.
3. A mutation can occur only in parental DNA, not in newly synthesized DNA strands.
4. Transcription introduced a permanent base change into one chromatid during S phase.
5. Complementary base pairing forces both chromatids to acquire identical mutations.

Explanation: 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.

Question 2

1. Templates re-form base pairs, reducing access for polymerase and slowing replication. (correct answer)
2. Polymerase switches to transcribing mRNA because single strands resemble RNA templates.
3. Replication speed increases because strands reanneal and guide polymerase more efficiently.
4. Replication becomes semiconservative only when binding proteins are absent.
5. Nucleotides pair randomly because binding proteins determine A–T and C–G specificity.

Explanation: 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.

Question 3

1. Failure to join lagging-strand fragments prevents formation of a continuous daughter strand. (correct answer)
2. Increased transcription causes RNA fragments to accumulate at replication forks.
3. Leading-strand synthesis requires fragments because polymerase works only 3′→5′.
4. Complementary base pairing is disrupted, preventing any nucleotide incorporation.
5. Chromosomes cannot condense, so replication forks cannot move along DNA.

Explanation: This question evaluates the skill of analyzing DNA replication, focusing on the discontinuous synthesis of the lagging strand and fragment joining. Accumulation of unjoined short DNA fragments near replication forks indicates a failure in the ligation process, which normally connects Okazaki fragments to form a continuous lagging strand. Since polymerase synthesizes these fragments in the 5′→3′ direction away from the fork, without joining, the daughter strand remains discontinuous despite proper base pairing and template usage. This disruption specifically affects the lagging strand, as the leading strand is synthesized continuously without needing ligation. Choice B is a tempting distractor, suggesting increased transcription and RNA accumulation, based on the misconception that replication issues directly interfere with RNA synthesis rather than DNA-specific processes. When troubleshooting replication observations, identify which strand and enzyme are implicated by matching symptoms to replication mechanics.

Question 4

1. All DNA strands are heavy because both strands in each molecule are newly synthesized.
2. All DNA molecules are half heavy because each contains one heavy new strand and one light parental strand. (correct answer)
3. Half the DNA molecules are fully heavy and half are fully light due to chromosome assortment.
4. No DNA strands are heavy because isotopes cannot be incorporated into nucleotides.
5. Only the leading strands are heavy because lagging strands are made of RNA fragments.

Explanation: 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.

Question 5

1. DNA replication, because uracil is required to pair with adenine in DNA.
2. Transcription, because RNA synthesis requires uracil-containing nucleotides. (correct answer)
3. DNA replication, because thymine-containing nucleotides cannot be incorporated.
4. Translation, because uracil is incorporated into polypeptides during elongation.
5. Chromosome segregation, because uracil controls spindle fiber attachment.

Explanation: This question evaluates the skill of analyzing DNA replication in comparison to transcription, highlighting nucleotide differences. The drug prevents uracil incorporation, directly inhibiting transcription because RNA synthesis requires uracil to pair with adenine on the DNA template, using ribonucleotides A, U, C, G. In contrast, DNA replication uses thymine instead of uracil in deoxyribonucleotides, so it proceeds normally without uracil. Transcription builds single-stranded RNA from one DNA template, while replication copies both strands with DNA nucleotides added 5′→3′. Choice A is a tempting distractor, claiming DNA replication needs uracil for adenine pairing, based on the misconception of confusing RNA and DNA base requirements. When distinguishing central dogma processes, note the specific nucleotides and products involved in each.

Question 6

1. Replication proceeds normally, but RNA nucleotides replace DNA nucleotides in both strands.
2. New DNA strands cannot elongate, so replication stalls at the fork despite template availability. (correct answer)
3. The lagging strand becomes continuous because fragments no longer require joining.
4. Leading-strand synthesis reverses direction and proceeds 3′→5′ to maintain base pairing.
5. Double-stranded DNA separates permanently because complementary bases no longer hydrogen-bond.

Explanation: 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.

Question 7

1. DNA polymerase cannot initiate synthesis de novo, leaving an unreplicated terminal region after primer removal. (correct answer)
2. Helicase cannot unwind chromosome ends, so replication stops early and deletes terminal genes.
3. Complementary base pairing fails at chromosome ends because there are no neighboring bases to stabilize pairing.
4. Chromosome ends are replicated by transcription, which produces shorter RNA copies each cell division.
5. DNA polymerase synthesizes the lagging strand 3′→5′, causing terminal nucleotides to be skipped.

Explanation: 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.

Question 8

1. Decreased mutation rate because fewer proteins contact DNA during replication.
2. Increased mutation rate because mismatches persist and become fixed in later replication. (correct answer)
3. No mutation rate change because proofreading occurs only during transcription.
4. Only deletions increase because mismatch repair acts exclusively on missing bases.
5. Mutations occur only in RNA because DNA repair proteins do not affect DNA.

Explanation: This question tests understanding of post-replicative DNA repair mechanisms and their impact on mutation rates. Mismatch repair systems identify and correct base-pairing errors that escape DNA polymerase proofreading, providing an additional layer of replication fidelity. Without functional mismatch repair, these errors persist through subsequent cell divisions and become permanently fixed as mutations in daughter cells, leading to an increased mutation rate over time. Choice C incorrectly claims proofreading occurs only during transcription, confusing DNA replication accuracy mechanisms with RNA synthesis. The key insight is that multiple repair systems work together to maintain genome stability, and losing any one increases mutation accumulation.

Question 9

1. DNA polymerase initiates synthesis de novo because base pairing supplies the first 3′ end.
2. Leading-strand synthesis continues, but lagging-strand synthesis cannot occur.
3. Neither strand can be extended efficiently because DNA polymerase lacks a primer 3′ end. (correct answer)
4. Okazaki fragments become longer because primers are produced less frequently.
5. Replication switches to using uracil so primers are unnecessary.

Explanation: 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.

Question 10

1. Replication proceeds normally because antiparallel strands allow extension from either end.
2. Replication fails because extension requires adding nucleotides to a 3′ end, not a 5′ end. (correct answer)
3. Transcription increases because RNA polymerase replaces DNA polymerase at forks.
4. Replication becomes conservative, producing one entirely old and one entirely new duplex.
5. Replication continues but produces RNA–DNA hybrids across the genome.

Explanation: This question examines the directional constraints of DNA polymerase activity. DNA polymerase can only add nucleotides to the 3'-OH group of a growing strand, synthesizing in the 5' to 3' direction. A mutant polymerase that could only add to 5' ends would be unable to extend any primer because primers present 3' ends, not 5' ends. This would completely halt replication at all forks. Choice A is wrong because having antiparallel strands doesn't change the fundamental requirement for 3' end extension - both strands need synthesis in the 5' to 3' direction. Remember that DNA polymerase's directional constraint is absolute and cannot be bypassed.

Question 11

1. Fewer replication errors occur because exonuclease activity normally introduces mutations.
2. More replication errors persist because mismatches are less likely to be removed. (correct answer)
3. Replication switches from semiconservative to conservative because proofreading is missing.
4. Only the lagging strand accumulates errors because the leading strand lacks primers.
5. Errors decrease because complementary base pairing no longer constrains nucleotide choice.

Explanation: This question examines the consequences of losing DNA polymerase's proofreading function. The 3' to 5' exonuclease activity normally removes mismatched nucleotides immediately after incorporation, maintaining replication fidelity. Without this activity, mismatches that occur during base selection remain in the DNA, increasing the mutation rate in daughter molecules. Choice A backwards suggests exonuclease causes errors rather than correcting them. The critical understanding is that proofreading is a error-correction mechanism that reduces, not increases, replication mistakes.

Question 12

1. A higher frequency of base-substitution mutations remains in daughter DNA molecules. (correct answer)
2. RNA nucleotides are incorporated throughout both strands instead of DNA nucleotides.
3. Replication stops entirely because helicase cannot separate the parental strands.
4. The new strands are synthesized only in the 3′→5′ direction on both templates.
5. Both daughter molecules retain one old strand but no newly synthesized strand.

Explanation: This question tests understanding of DNA replication proofreading mechanisms. DNA polymerase normally has 3' to 5' exonuclease activity that removes mismatched bases during synthesis, maintaining high fidelity. When proofreading is disabled but base-pairing rules remain intact, polymerase still adds mostly correct nucleotides, but occasional mismatches cannot be removed. These uncorrected errors become permanent mutations in the daughter DNA molecules. Choice B is incorrect because DNA polymerase specifically adds deoxyribonucleotides, not RNA nucleotides, regardless of proofreading ability. The key strategy is recognizing that proofreading is a quality control mechanism separate from the basic synthesis machinery.

Question 13

1. New DNA strands incorporate uracil in place of thymine to maintain pairing with adenine.
2. DNA synthesis stalls when adenine is encountered on the template strand. (correct answer)
3. Replication proceeds normally because thymine is needed only during transcription.
4. Replication becomes error-free because missing thymine prevents mismatches.
5. Both strands are synthesized without templates because nucleotides self-assemble.

Explanation: 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.

Question 14

1. An A–T base pair will be restored because mismatches always revert during the next replication cycle.
2. A G–C base pair will replace the original A–T base pair after replication copies the mismatch. (correct answer)
3. A U–A base pair will form because thymine is replaced by uracil during DNA replication.
4. A T–G base pair will persist indefinitely because DNA polymerase cannot distinguish mismatches.
5. A C–T base pair will form because cytosine is complementary to thymine during replication.

Explanation: 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.

Question 15

1. Both strands remain as single-stranded DNA because hydrogen bonds cannot reform.
2. The leading strand contains many unjoined fragments, while the lagging strand is continuous.
3. The lagging strand contains unjoined fragments, while the leading strand is mostly continuous. (correct answer)
4. Replication converts DNA into RNA because fragments cannot be joined as DNA.
5. Chromosomes replicate conservatively, leaving one duplex completely unchanged.

Explanation: This question tests understanding of continuous versus discontinuous DNA synthesis. The leading strand is synthesized continuously in one long piece following the replication fork, while the lagging strand is made in short Okazaki fragments that must be joined by DNA ligase. Without ligase activity, the leading strand remains mostly intact (except for the junction with its initial primer), but the lagging strand persists as multiple unjoined fragments. Choice B reverses this relationship, misunderstanding which strand requires fragment joining. Remember that ligase specifically affects the lagging strand's discontinuous synthesis pattern.

Question 16

1. Replication fork progression will slow because reannealing reduces the availability of single-stranded templates. (correct answer)
2. Replication accuracy will increase because reannealing forces correct base pairing in the new strands.
3. Okazaki fragments will become longer because polymerase can add nucleotides in both directions.
4. The leading strand will be synthesized as RNA because SSBs are required for DNA nucleotide selection.
5. The parental strands will be degraded because SSBs normally catalyze phosphodiester bond formation.

Explanation: 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.

Question 17

1. Without mismatch removal, incorrect bases remain paired and become fixed after the next replication cycle. (correct answer)
2. Without mismatch removal, helicase cannot unwind DNA efficiently, causing strand breaks that appear as mutations.
3. Without mismatch removal, DNA polymerase will synthesize RNA instead of DNA, increasing base substitutions.
4. Without mismatch removal, complementary base pairing rules change so A pairs with C more often than T.
5. Without mismatch removal, replication becomes conservative so the original strands accumulate copying errors.

Explanation: This question assesses the skill of analyzing DNA replication, emphasizing the importance of proofreading in maintaining fidelity. Without mismatch removal, incorrect bases inserted during synthesis are not excised, allowing them to persist in the new strand and become permanently incorporated as mutations after the next replication cycle when the mismatched base serves as a template. This increases point mutations because the polymerase, while still adding nucleotides via base complementarity, lacks the exonuclease activity to correct errors immediately after insertion. In extract 1, normal proofreading reduces errors, explaining the difference in mutation frequency between the extracts. A tempting distractor is choice B, which claims helicase cannot unwind DNA efficiently without mismatch removal, but this arises from the misconception that proofreading is linked to unwinding, whereas they are separate processes with proofreading occurring post-insertion. For similar questions, focus on distinguishing the roles of replication enzymes and how defects in one specifically impact fidelity versus progression.

Question 18

1. Replication forks move faster because helicase activity is unchanged.
2. Single-stranded DNA accumulates because unwinding continues while synthesis lags behind. (correct answer)
3. The lagging strand becomes continuous because slower synthesis prevents fragment formation.
4. RNA nucleotides replace DNA nucleotides because slow polymerase switches enzymes.
5. Base pairing rules change so A pairs with C and G pairs with T to speed synthesis.

Explanation: 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.

Question 19

1. The leading strand is synthesized continuously in the same direction as fork movement. (correct answer)
2. The leading strand is synthesized discontinuously because DNA is double stranded.
3. The leading strand is synthesized by ribosomes, which move away from forks.
4. The leading strand grows 3′→5′ to keep the two strands antiparallel.
5. The leading strand forms by pairing RNA bases with the DNA template.

Explanation: This question examines the skill of analyzing DNA replication, specifically the distinction between leading and lagging strand synthesis at replication forks. The student's label is incorrect because the leading strand is actually synthesized continuously in the 5′→3′ direction, aligning with the movement of the replication fork as it unwinds the DNA. This continuous synthesis occurs on the template strand oriented such that polymerase can extend toward the advancing fork without interruption. In contrast, the lagging strand is synthesized discontinuously away from the fork in short fragments due to the antiparallel nature of DNA strands. Choice B is a tempting distractor, stating the leading strand is discontinuous because DNA is double-stranded, reflecting the misconception that both strands must be fragmented regardless of template orientation. To correctly identify strand types, visualize the fork direction and apply the 5′→3′ synthesis rule to each template.

Question 20

1. Inhibiting helicase prevents strand separation, reducing access to single-stranded templates for DNA polymerase. (correct answer)
2. Inhibiting helicase increases primer synthesis, causing polymerase to stall at too many initiation sites.
3. Inhibiting helicase converts DNA replication into RNA replication, so labeled deoxyribonucleotides are excluded.
4. Inhibiting helicase strengthens phosphodiester bonds, preventing nucleotides from being added to the backbone.
5. Inhibiting helicase changes base-pairing rules, so nucleotides cannot hydrogen-bond to the template.

Explanation: 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.