DNA Replication
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AP Biology › DNA Replication
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?
A mutation can occur only in parental DNA, not in newly synthesized DNA strands.
A mismatch escaped proofreading on one new strand, and the other duplex copied correctly.
Replication is conservative, so only one chromatid is newly synthesized and mutates.
Complementary base pairing forces both chromatids to acquire identical mutations.
Transcription introduced a permanent base change into one chromatid during S phase.
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.
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?
Templates re-form base pairs, reducing access for polymerase and slowing replication.
Replication speed increases because strands reanneal and guide polymerase more efficiently.
Nucleotides pair randomly because binding proteins determine A–T and C–G specificity.
Polymerase switches to transcribing mRNA because single strands resemble RNA templates.
Replication becomes semiconservative only when binding proteins are absent.
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.
A DNA molecule replicates when helicase separates strands and DNA polymerase adds complementary DNA nucleotides to each template strand. Polymerase adds nucleotides only to the 3′ end, producing new strands in the 5′→3′ direction. Because the two templates are antiparallel, one new strand is synthesized continuously toward the replication fork and the other in short fragments away from the fork that are later joined. A researcher observes many unjoined short DNA fragments accumulating near replication forks. Which explanation best accounts for the accumulation of these fragments?
Increased transcription causes RNA fragments to accumulate at replication forks.
Complementary base pairing is disrupted, preventing any nucleotide incorporation.
Failure to join lagging-strand fragments prevents formation of a continuous daughter strand.
Leading-strand synthesis requires fragments because polymerase works only 3′→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.
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?
Only the leading strands are heavy because lagging strands are made of RNA fragments.
No DNA strands are heavy because isotopes cannot be incorporated into nucleotides.
All DNA strands are heavy because both strands in each molecule are newly synthesized.
Half the DNA molecules are fully heavy and half are fully light due to chromosome assortment.
All DNA molecules are half heavy because each contains one heavy new strand and one light parental strand.
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.
A student compares DNA replication and transcription. In replication, both DNA strands are copied to produce two double-stranded DNA molecules, and DNA polymerase uses DNA nucleotides (A, T, C, G) added to a 3′ end. In transcription, one DNA strand serves as a template to build an RNA molecule using ribonucleotides (A, U, C, G). A cell is treated with a drug that prevents incorporation of uracil-containing nucleotides. Which process is most directly inhibited by the drug?
Chromosome segregation, because uracil controls spindle fiber attachment.
DNA replication, because thymine-containing nucleotides cannot be incorporated.
Translation, because uracil is incorporated into polypeptides during elongation.
DNA replication, because uracil is required to pair with adenine in DNA.
Transcription, because RNA synthesis requires uracil-containing nucleotides.
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.
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?
Leading-strand synthesis reverses direction and proceeds 3′→5′ to maintain base pairing.
New DNA strands cannot elongate, so replication stalls at the fork despite template availability.
The lagging strand becomes continuous because fragments no longer require joining.
Replication proceeds normally, but RNA nucleotides replace DNA nucleotides in both strands.
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.
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?
Complementary base pairing fails at chromosome ends because there are no neighboring bases to stabilize pairing.
Helicase cannot unwind chromosome ends, so replication stops early and deletes terminal genes.
DNA polymerase cannot initiate synthesis de novo, leaving an unreplicated terminal region after primer removal.
Chromosome ends are replicated by transcription, which produces shorter RNA copies each cell division.
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.
Replication accuracy depends on correct base pairing and repair of mismatches that escape initial nucleotide selection. After replication, some cells use mismatch repair to identify the newly synthesized strand by transient discontinuities and then replace the incorrect nucleotide. A toxin prevents mismatch repair proteins from binding DNA but does not affect replication speed. Which change is most likely observed over several cell divisions?
Decreased mutation rate because fewer proteins contact DNA during replication.
Only deletions increase because mismatch repair acts exclusively on missing bases.
Mutations occur only in RNA because DNA repair proteins do not affect DNA.
No mutation rate change because proofreading occurs only during transcription.
Increased mutation rate because mismatches persist and become fixed in later replication.
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.
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?
Neither strand can be extended efficiently because DNA polymerase lacks a primer 3′ end.
Okazaki fragments become longer because primers are produced less frequently.
Leading-strand synthesis continues, but lagging-strand synthesis cannot occur.
DNA polymerase initiates synthesis de novo because base pairing supplies the first 3′ end.
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.
At a replication fork, DNA polymerase extends a primer by adding deoxyribonucleotides to the 3′ end, creating a complementary strand antiparallel to the template. In a mutant cell, the polymerase can bind DNA but can add nucleotides only to a 5′ end. Which explanation best accounts for the effect on replication?
Replication becomes conservative, producing one entirely old and one entirely new duplex.
Transcription increases because RNA polymerase replaces DNA polymerase at forks.
Replication continues but produces RNA–DNA hybrids across the genome.
Replication proceeds normally because antiparallel strands allow extension from either end.
Replication fails because extension requires adding nucleotides to a 3′ end, not a 5′ end.
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.