AP Biology: Unit 6, DNA Replication (Drill 23)

Question 1. Which model of DNA replication is uniquely supported by both the Generation 1 and Generation 2 results of the Meselson-Stahl experiment taken together?

• A) Semi-conservative replication: each daughter molecule contains one parental strand and one new strand, predicting all intermediate-density molecules in Generation 1 and a 1:1 ratio of intermediate to light molecules in Generation 2. ✓
• B) Conservative replication: the parental helix is preserved intact and a fully new double helix is synthesized, predicting two bands (heavy and light) in Generation 1 and two bands (heavy and light) in Generation 2.
• C) Dispersive replication: parental and new DNA are randomly interspersed throughout every strand, predicting a single intermediate band in Generation 1 that should shift progressively lighter in Generation 2 without producing a distinct light band.
• D) Bidirectional replication: replication forks moving in both directions produce molecules of varying density depending on how far replication proceeded.

Explanation: Correct answer: A. Semi-conservative replication is the only model consistent with both results. In Generation 1, each parental 15N-15N molecule unwinds and each strand serves as a template, producing two hybrid 15N-14N molecules -- all at intermediate density. In Generation 2, each hybrid molecule replicates: one strand (15N) templates a new 14N strand (producing another hybrid molecule), and the other strand (14N) templates a new 14N strand (producing a light molecule). The predicted result is a 1:1 ratio of intermediate to light molecules -- exactly two bands, as observed. (B) the conservative model predicts a heavy band and a light band in Generation 1 (the original helix preserved, new helix fully light) -- not the single intermediate band observed. It is ruled out by the Generation 1 result. (C) the dispersive model predicts a single intermediate band in Generation 1, consistent with that observation, but predicts that in Generation 2 all molecules would shift progressively lighter with no distinct light band -- it cannot produce the separate light band observed in Generation 2. (D) bidirectionality describes the mechanics of fork movement and does not make any prediction about density-gradient banding patterns.

Question 2. On the lagging strand during DNA replication, synthesis is discontinuous, producing Okazaki fragments. Which sequence of molecular events correctly describes how a gap between two Okazaki fragments is resolved?

• A) Helicase removes the RNA primer; DNA polymerase III fills the gap; DNA ligase seals the nick.
• B) Primase extends the existing RNA primer directly into DNA across the gap, and DNA ligase then seals the remaining nick between the two Okazaki fragments.
• C) DNA polymerase I removes the RNA primer of the upstream Okazaki fragment while simultaneously filling the gap with DNA; DNA ligase then seals the remaining nick between the new DNA and the downstream Okazaki fragment. ✓
• D) DNA polymerase III removes the RNA primer and replaces it with DNA; topoisomerase seals the nick.

Explanation: Correct answer: C. The process of resolving the gap at the 5-prime end of an Okazaki fragment involves two distinct enzymatic activities: (1) DNA polymerase I has 5'->3' exonuclease activity that removes the RNA primer of the upstream fragment nucleotide by nucleotide, and simultaneously fills the resulting gap with DNA using its 5'->3' polymerase activity (nick translation). (2) After the primer is removed and replaced, a single-stranded nick remains between the 3-prime end of the newly synthesized DNA and the 5-prime end of the next Okazaki fragment; DNA ligase seals this nick by forming a phosphodiester bond. (A) is incorrect; helicase unwinds the double helix at the replication fork -- it does not remove RNA primers. (B) is incorrect; primase only synthesizes RNA primers de novo; it does not extend primers into DNA, and ligase alone cannot fill a gap. (D) is incorrect; DNA polymerase III cannot remove RNA primers (it lacks the 5'->3' exonuclease activity of DNA polymerase I); topoisomerase relieves supercoiling, not nicks between fragments.

Question 3. A mutation eliminates the proofreading (3'->5' exonuclease) activity of DNA polymerase III. What would be the most likely consequence for DNA replication fidelity?

• A) Replication would stop entirely because DNA polymerase III cannot extend a strand without proofreading.
• B) Replication would proceed substantially faster overall because the proofreading exonuclease step is the rate-limiting bottleneck during strand elongation.
• C) Replication would complete but the error rate would increase substantially, because misincorporated nucleotides that would normally be detected and removed would remain in the new strand. ✓
• D) Replication fidelity would be unchanged because mismatch repair systems in the cell would compensate for the lost proofreading.

Explanation: Correct answer: C. DNA polymerase III has two distinct activities: 5'->3' polymerase (synthesis) and 3'->5' exonuclease (proofreading). Proofreading allows the enzyme to detect and excise a misincorporated nucleotide immediately after it is added, before proceeding. Without this activity, the synthesis activity continues, but errors are no longer removed in real time. The result is a substantially elevated mutation rate. Estimates suggest proofreading reduces the intrinsic error rate of DNA polymerase by 100-fold. (A) is incorrect; the 3'->5' exonuclease activity is separate from the polymerase activity -- losing proofreading does not stop synthesis. (B) is partially plausible in that proofreading does slow elongation, but the effect on rate would be modest relative to its effect on fidelity; the primary consequence is elevated mutation rate, not speed. (D) is incorrect; mismatch repair operates post-replication and does compensate for some errors, but it is not a perfect substitute for proofreading. The two systems are additive, not redundant -- losing proofreading increases errors even if mismatch repair is intact.

Question 4. DNA replication must occur before mitosis and meiosis. Which of the following accurately describes a key difference between how replication errors affect outcomes in mitosis versus meiosis?

• A) Replication errors that occur before mitosis result in two daughter cells with identical mutations; replication errors before meiosis affect only one of the four resulting gametes.
• B) Mitosis corrects all replication errors using its spindle assembly checkpoint proteins, whereas meiosis lacks any comparable proofreading mechanism.
• C) Replication errors before meiosis produce genetic variation, which is beneficial; replication errors before mitosis produce cancer, which is always harmful.
• D) Replication errors that occur before mitosis can only affect somatic cells and are not heritable; replication errors that occur before meiosis in germline cells can be passed to offspring and are heritable. ✓

Explanation: Correct answer: D. The most fundamental difference is heritability. Somatic cells (produced by mitosis) are not transmitted to offspring; mutations in somatic cells affect only the individual organism (and potentially contribute to cancer or other somatic conditions, but do not cross generations). Germline cells (produced in the gonads by meiosis and its preceding mitotic divisions) give rise to gametes; a replication error before meiosis can be incorporated into a gamete and, if that gamete participates in fertilization, will be present in every cell of the offspring. This is the basis of inherited mutations. (A) is partially correct that a mitotic replication error produces two daughter cells with the mutation, but it incorrectly says meiotic errors affect "only one of four gametes" -- the distribution depends on which strand carries the error and when the error occurred relative to the meiotic divisions. (B) is incorrect; spindle checkpoint proteins check chromosome attachment to the spindle, not DNA sequence errors; neither mitosis nor meiosis "corrects" replication errors in the sense described. (C) is incorrect in its absolute framing: replication errors before meiosis are mutations, not automatically beneficial; and mitotic mutations, while they can contribute to cancer, are not "always harmful."

Question 5. In eukaryotic DNA replication, multiple origins of replication fire simultaneously along each chromosome. Why is this feature essential for timely replication of eukaryotic genomes?

• A) Multiple origins are required because eukaryotic DNA polymerase can only synthesize in the 3'->5' direction, and multiple enzymes working simultaneously overcome this limitation.
• B) Eukaryotic chromosomes are much longer than prokaryotic chromosomes, and DNA polymerase moves at a finite rate (roughly 50-100 bp/s in eukaryotes, slower than in prokaryotes). Replicating a single large chromosome from one origin would take far longer than the cell cycle allows; multiple simultaneous origins reduce the effective replication time proportionally. ✓
• C) Eukaryotic chromosomes are circular, so replication must proceed from multiple origins at once to complete copying the entire circle before the cell divides.
• D) Multiple origins are needed because each origin can only replicate a single gene, and eukaryotic chromosomes contain thousands of genes.

Explanation: Correct answer: B. Human chromosomes can be hundreds of millions of base pairs long. DNA polymerase in eukaryotes operates at approximately 50-100 bp/s (slower than in prokaryotes). Replicating a single large human chromosome (e.g., chromosome 1, ~250 Mb) from a single origin at 50-100 bp/s would take on the order of millions of seconds -- far longer than the S phase of the human cell cycle (~6-8 hours). With thousands of origins firing simultaneously, the effective replication distance per origin is greatly reduced, allowing entire chromosomes to be replicated within hours. (A) is incorrect; DNA polymerase synthesizes in the 5'->3' direction, not 3'->5'. Multiple origins do not address directionality; that issue is handled by the lagging/leading strand mechanism at each replication fork. (C) is incorrect; eukaryotic chromosomes are linear, not circular. Prokaryotic chromosomes are typically circular and usually have a single origin. (D) is incorrect; one origin can replicate an entire chromosome given sufficient time; the number of origins is not determined by gene count but by the need to complete replication within the S phase window.