Genetics Quiz: Dna Replication
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Dna ReplicationQuestion 1 of 20

In a variation of the Meselson-Stahl experiment, bacteria are grown in ¹⁵N, then transferred to ¹⁴N for one generation. The resulting hybrid DNA is isolated. If this hybrid DNA is denatured into single strands and then analyzed by density-gradient centrifugation, what result would provide definitive proof for the semiconservative model over the dispersive model?

Two distinct bands of equal intensity, one at ¹⁵N density and one at ¹⁴N density.
A single band of density corresponding to pure ¹⁴N.
A single band of density intermediate between ¹⁵N and ¹⁴N.
Two distinct bands of unequal intensity, with the ¹⁴N band being more intense than the ¹⁵N band.
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Genetics Quiz

Genetics Quiz: Dna Replication

Practice Dna Replication in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 Genetics.

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.

All questions

Question 1

In a variation of the Meselson-Stahl experiment, bacteria are grown in ¹⁵N, then transferred to ¹⁴N for one generation. The resulting hybrid DNA is isolated. If this hybrid DNA is denatured into single strands and then analyzed by density-gradient centrifugation, what result would provide definitive proof for the semiconservative model over the dispersive model?

  1. Two distinct bands of equal intensity, one at ¹⁵N density and one at ¹⁴N density. (correct answer)
  2. A single band of density corresponding to pure ¹⁴N.
  3. A single band of density intermediate between ¹⁵N and ¹⁴N.
  4. Two distinct bands of unequal intensity, with the ¹⁴N band being more intense than the ¹⁵N band.

Explanation: When analyzing DNA replication models, the key insight is understanding what happens to individual DNA strands after denaturation. In the Meselson-Stahl experiment setup, you're comparing two competing models: semiconservative (where each new DNA molecule contains one original strand and one new strand) versus dispersive (where new DNA contains mixed fragments of old and new material throughout each strand). After one generation in ¹⁴N following growth in ¹⁵N, both models produce hybrid DNA molecules with intermediate density. However, when you denature this hybrid DNA into single strands, the models predict different outcomes. In semiconservative replication, each hybrid DNA molecule contains one completely ¹⁵N-labeled strand (the original) and one completely ¹⁴N-labeled strand (the newly synthesized). Upon denaturation, you get equal numbers of heavy and light single strands, producing two distinct bands of equal intensity at the respective densities. This matches answer choice A. Choice B is incorrect because some strands must still contain ¹⁵N from the original DNA. Choice C represents what you'd see before denaturation, when strands are still paired as hybrid molecules. Choice D suggests unequal band intensities, which contradicts the fact that semiconservative replication produces exactly one old strand and one new strand per DNA molecule. In contrast, the dispersive model would predict single strands with mixed ¹⁵N/¹⁴N content, appearing as a single intermediate band even after denaturation. Study tip: Remember that denaturation separates the "history" of each individual strand, revealing whether replication preserves entire strands (semiconservative) or mixes old and new material within strands (dispersive).

Question 2

Replication of linear eukaryotic chromosomes faces the 'end-replication problem', which is not encountered with circular prokaryotic chromosomes. Which statement most accurately describes this problem and its enzymatic solution?

  1. DNA ligase cannot join the final Okazaki fragment to the end of the chromosome, a gap that is filled by telomerase.
  2. Helicase cannot unwind the very end of the chromosome, so telomerase must synthesize this region without a template.
  3. The final RNA primer on the lagging strand is removed, leaving a 3' overhang on the template that standard polymerases cannot fill, a problem solved by telomerase. (correct answer)
  4. Topoisomerase cannot relieve supercoiling at the chromosome ends, leading to DNA breaks that are repaired by telomerase.

Explanation: The end-replication problem arises on the lagging strand of linear chromosomes. When the RNA primer at the very 5' end of the newly synthesized lagging strand is removed, there is no pre-existing 3'-OH group for DNA polymerase to use to fill the gap. This results in a 3' overhang on the parental template strand and a shortening of the daughter chromosome with each replication cycle. Telomerase solves this by acting as a reverse transcriptase; it carries its own RNA template and extends the 3' overhang of the parental strand. This extended template then allows primase and polymerase to synthesize the missing portion of the lagging strand, preventing chromosome shortening. Distractors A, B, and D misidentify the cause of the problem or the function of telomerase.

Question 3

In E. coli, both DNA Polymerase I and DNA Polymerase III share 5'→3' polymerase and 3'→5' exonuclease activities. However, only DNA Polymerase I is capable of completing the synthesis of the lagging strand. Which unique enzymatic activity of DNA Polymerase I is essential for this specific role?

  1. Helicase activity to unwind the RNA primer.
  2. Reverse transcriptase activity to read the RNA primer.
  3. 5'→3' exonuclease activity to remove the RNA primer. (correct answer)
  4. Enhanced processivity to fill long gaps between fragments.

Explanation: The maturation of the lagging strand requires the removal of the RNA primers that initiated each Okazaki fragment. DNA Polymerase I has a unique 5'→3' exonuclease activity that allows it to excise nucleotides from the 5' end of a strand, such as the RNA primer. As it removes the RNA, its 5'→3' polymerase activity fills the gap with DNA. DNA Polymerase III lacks this 5'→3' exonuclease activity. Distractor A is incorrect; helicase, not polymerase, unwinds DNA. Distractor B is incorrect; reverse transcriptase synthesizes DNA from an RNA template, which is not what's happening here. Distractor D is incorrect; DNA Polymerase III has much higher processivity than DNA Polymerase I.

Question 4

The maturation of an Okazaki fragment requires its covalent linkage to the preceding fragment on the lagging strand. Which set of enzymatic activities is required, in order, to process the junction between two adjacent fragments after they have been synthesized by DNA Polymerase III?

  1. 5'→3' exonuclease, 5'→3' polymerase, DNA ligase (correct answer)
  2. 3'→5' exonuclease, 5'→3' polymerase, DNA ligase
  3. Primase, 5'→3' polymerase, DNA ligase
  4. Helicase, 5'→3' exonuclease, DNA ligase

Explanation: After DNA Polymerase III synthesizes an Okazaki fragment and runs into the primer of the previous fragment, three activities are needed to join them.

  1. 5'→3' exonuclease: The RNA primer of the preceding fragment must be removed. This is done by DNA Polymerase I's 5'→3' exonuclease activity.
  2. 5'→3' polymerase: As the primer is removed, the resulting gap is filled with DNA nucleotides by the 5'→3' polymerase activity of DNA Polymerase I.
  3. DNA ligase: Once the gap is filled, a single-strand break (a 'nick') remains in the sugar-phosphate backbone. DNA ligase seals this nick by forming a phosphodiester bond. Distractor B incorrectly lists 3'→5' exonuclease, which is for proofreading, not primer removal. Distractor C incorrectly includes Primase, which initiates fragments, but does not join them. Distractor D incorrectly includes Helicase, which unwinds DNA at the fork.

Question 5

A 10,000 bp segment of a bacterial chromosome is replicated by a single replication fork. Assuming Okazaki fragments are, on average, 2,000 nucleotides long, what is the total number of RNA primers required to replicate this entire segment?

  1. 1
  2. 5
  3. 6 (correct answer)
  4. 10

Explanation: Replication of a DNA segment involves both a leading and a lagging strand.

  1. Leading Strand: Synthesis is continuous. It requires only one RNA primer to initiate synthesis for the entire 10,000 nucleotide segment.
  2. Lagging Strand: Synthesis is discontinuous, occurring in Okazaki fragments. The length of the lagging strand to be synthesized is 10,000 nucleotides. With a fragment size of 2,000 nucleotides, the number of fragments is 10,000 / 2,000 = 5. Each Okazaki fragment requires its own RNA primer. So, 5 primers are needed for the lagging strand.
  3. Total: The total number of primers is the sum of primers for both strands: 1 (leading) + 5 (lagging) = 6 primers. Distractor A considers only the leading strand. Distractor B considers only the lagging strand. Distractor D incorrectly assumes 5 primers are needed for both strands (5 + 5 = 10).

Question 6

A mutant strain of E. coli possesses a temperature-sensitive DNA ligase, which is functional at 30°C but inactive at 42°C. If a culture of this strain is shifted from 30°C to 42°C during active DNA replication, which of the following molecular structures would be expected to accumulate?

  1. Short RNA-DNA hybrid fragments corresponding to the lagging strand.
  2. Unreplicated parental DNA due to the stalling of the replication fork.
  3. A fully replicated leading strand and a completely single-stranded lagging strand template.
  4. Numerous short, double-stranded DNA segments on the lagging strand template, separated by nicks. (correct answer)

Explanation: DNA ligase is responsible for sealing the final phosphodiester bond (the 'nick') between adjacent Okazaki fragments after the RNA primer has been removed and the gap filled with DNA. If DNA ligase is inactive, DNA Polymerase III will synthesize Okazaki fragments, and DNA Polymerase I will remove the RNA primers and fill the gaps with DNA. However, the final covalent link between the 3' end of one fragment and the 5' end of the next will not be formed. This results in the accumulation of many fully synthesized but unjoined DNA fragments on the lagging strand. Distractor A is incorrect because DNA Polymerase I, which removes the RNA primers, is still functional. Distractor B is incorrect as helicase and polymerases are functional, so replication will proceed until the lack of ligation causes downstream problems, but accumulation of fragments is the immediate effect. Distractor C is an oversimplification; lagging strand synthesis will initiate and proceed, but fragments won't be joined.

Question 7

An E. coli strain harbors a mutation in the polA gene that inactivates the 5'→3' exonuclease domain of DNA Polymerase I, while leaving its other functions intact. Which of the following phenotypes is expected in this mutant?

  1. A significant increase in the overall mutation rate due to failed proofreading.
  2. Replication forks will stall immediately after initiation.
  3. The bacteria will be unable to replicate their DNA due to an inability to synthesize Okazaki fragments.
  4. Daughter DNA strands will contain segments of RNA covalently linked to DNA. (correct answer)

Explanation: The 5'→3' exonuclease domain of DNA Polymerase I is specifically responsible for removing the RNA primers of Okazaki fragments. If this domain is inactive, the primers cannot be excised. DNA Polymerase I's polymerase activity can still fill any available gaps, and DNA ligase may seal nicks, but the RNA segments will remain incorporated within the lagging strand. This results in a final product where the daughter DNA contains covalently linked ribonucleotides. Distractor A is incorrect because proofreading is handled by the 3'→5' exonuclease domain, which is unaffected. Distractor B and C are incorrect because the main replicative enzyme, DNA Polymerase III, is normal, so replication and Okazaki fragment synthesis will proceed.

Question 8

The function of single-strand binding proteins (SSBPs) is to bind to the unwound parental DNA strands. Which of the following would be the most direct consequence of non-functional SSBPs in a bacterial cell?

  1. The DNA helix would be unable to be unwound by helicase at the origin of replication.
  2. The separated template strands would be prone to re-annealing and to degradation by cellular nucleases. (correct answer)
  3. RNA primers would fail to anneal to the template strands, preventing the initiation of synthesis.
  4. Supercoiling ahead of the replication fork would increase, leading to a stall in replication.

Explanation: Single-strand binding proteins coat the unwound DNA strands behind the helicase. They have two primary functions: 1) to prevent the complementary strands from immediately re-forming a double helix (re-annealing), and 2) to protect the vulnerable single-stranded DNA from being broken down by nucleases. Without functional SSBPs, the replication bubble would be unstable and the template DNA could be damaged, severely impairing or halting replication. Distractor A is incorrect; helicase activity is separate from SSBP function. Distractor C is incorrect; primer annealing depends on base pairing, not directly on SSBPs. Distractor D describes the consequence of non-functional topoisomerase/gyrase.

Question 9

Fluoroquinolone antibiotics, such as ciprofloxacin, function by inhibiting bacterial DNA gyrase. If a susceptible bacterial culture in logarithmic growth phase is treated with ciprofloxacin, what is the most immediate consequence at the molecular level of DNA replication?

  1. Separation of the two parental DNA strands at the origin of replication is blocked.
  2. Positive supercoils accumulate ahead of the replication fork, causing it to stall. (correct answer)
  3. Newly synthesized Okazaki fragments on the lagging strand fail to be joined.
  4. RNA primers are not removed from the newly synthesized DNA strands.

Explanation: DNA gyrase, a type of topoisomerase II, is responsible for introducing negative supercoils into DNA, which relieves the torsional stress (positive supercoils) that builds up ahead of the replication fork as helicase unwinds the double helix. Inhibition of DNA gyrase prevents the removal of these positive supercoils. The resulting strain on the DNA molecule will eventually halt the progression of the helicase and, consequently, the entire replication fork. Distractor A describes a failure of initiator proteins or helicase. Distractor C describes the effect of a DNA ligase inhibitor. Distractor D describes an effect of inhibiting DNA Polymerase I's 5'→3' exonuclease activity.

Question 10

A researcher performs an experiment modeled after Meselson and Stahl. They grow E. coli for many generations in a medium containing a heavy nitrogen isotope (¹⁵N). The culture is then transferred to a medium containing a different, lighter isotope (¹⁴N) and allowed to complete one round of replication. Finally, the cells are transferred to a third medium containing only the lightest nitrogen isotope (¹³N) for a second round of replication. After isolating DNA and using density-gradient centrifugation, what is the expected distribution of DNA molecules?

  1. A single band of intermediate density, between ¹⁴N/¹⁴N and ¹³N/¹³N.
  2. Two distinct bands of equal intensity, one corresponding to ¹⁵N/¹³N density and one to ¹⁴N/¹³N density. (correct answer)
  3. Three distinct bands corresponding to ¹⁵N/¹⁴N, ¹⁵N/¹³N, and ¹⁴N/¹³N densities.
  4. Two distinct bands of unequal intensity, with the ¹⁴N/¹³N band being three times more intense than the ¹⁵N/¹³N band.

Explanation: This problem requires tracking DNA strands through two rounds of semiconservative replication with changing isotopic labels.

  1. Start: All DNA is ¹⁵N/¹⁵N (heavy).
  2. Generation 1 (in ¹⁴N): The ¹⁵N strands separate, and each serves as a template for a new ¹⁴N strand. All resulting DNA molecules are ¹⁵N/¹⁴N hybrids of intermediate density.
  3. Generation 2 (in ¹³N): The hybrid ¹⁵N/¹⁴N molecules from generation 1 unwind. The ¹⁵N template strand is replicated using ¹³N, creating a ¹⁵N/¹³N molecule. The ¹⁴N template strand is replicated using ¹³N, creating a ¹⁴N/¹³N molecule. This results in a 1:1 ratio of two distinct types of molecules: ¹⁵N/¹³N and ¹⁴N/¹³N. These would appear as two separate bands of equal intensity on a density gradient. Distractor A is incorrect because two different molecular species are formed. Distractor C incorrectly assumes some ¹⁵N/¹⁴N hybrids remain. Distractor D incorrectly calculates the ratio of the resulting molecules.

Question 11

The 'processivity' of a DNA polymerase refers to its ability to catalyze consecutive polymerization reactions without releasing its template. This property is critical for efficient genome replication. What is the primary basis for the high processivity of the main replicative polymerase in E. coli, DNA Polymerase III?

  1. An intrinsic helicase subunit that clears the path for synthesis, preventing dissociation.
  2. Its 3'→5' exonuclease activity, which corrects errors and stabilizes the enzyme on the DNA.
  3. A high affinity for the RNA primer, which anchors the enzyme firmly at the start of synthesis.
  4. A ring-shaped sliding clamp subunit (β-clamp) that encircles the DNA and tethers the polymerase to the template. (correct answer)

Explanation: DNA replication requires polymerases to add thousands of nucleotides continuously without "falling off" the template strand. This property, called processivity, is what distinguishes the main replicative enzymes from repair polymerases that only add a few nucleotides at a time. DNA Polymerase III achieves its remarkable processivity through the β-clamp (beta clamp), a ring-shaped protein complex that completely encircles the double-stranded DNA like a sliding donut. This clamp physically tethers Pol III to the DNA template, allowing it to slide along while remaining attached, enabling the addition of thousands of nucleotides in a single binding event. The correct answer is D. Let's examine why the other options don't explain Pol III's high processivity: Option A is incorrect because Pol III doesn't contain an intrinsic helicase subunit—helicases like DnaB work separately at the replication fork. Option B misunderstands the role of 3'→5' exonuclease activity, which provides proofreading capability but doesn't significantly contribute to processivity; in fact, this activity can temporarily stall synthesis during error correction. Option C is wrong because while Pol III does interact with RNA primers, this interaction isn't particularly high-affinity, and the primer only provides the initial 3'-OH group for synthesis to begin—it doesn't maintain the enzyme's attachment throughout elongation. When studying DNA replication, remember that processivity and proofreading are separate functions. The sliding clamp mechanism is a key evolutionary solution that appears in all domains of life (PCNA in eukaryotes, β-clamp in bacteria) precisely because keeping polymerases attached is crucial for efficient genome duplication.

Question 12

A novel antiviral drug is found to be a potent inhibitor of the primase enzyme in a virus with a dsDNA genome. If this drug is administered to an infected cell culture during viral replication, what would be the observed state of the viral DNA molecules?

  1. Parental strands will be separated by helicase, but no synthesis of new complementary strands will occur. (correct answer)
  2. Only the leading strand will be synthesized, resulting in molecules that are partially double-stranded and partially single-stranded.
  3. Replication will be completed, but the daughter strands will consist of small, unlinked DNA fragments.
  4. The viral DNA will remain in its condensed, supercoiled state, and no unwinding will occur.

Explanation: Primase synthesizes the short RNA primers that are absolutely required for DNA polymerase to initiate DNA synthesis. DNA polymerases cannot start synthesis de novo; they can only add nucleotides to a pre-existing 3'-OH group. Without the action of primase, no primers can be made. Consequently, even if helicase unwinds the parental DNA, DNA polymerase will be unable to synthesize any new DNA on either the leading or the lagging strand. Distractor B is incorrect because even the leading strand requires one initial primer to get started. Distractor C describes the effect of a DNA ligase inhibitor. Distractor D describes the effect of an inhibitor of initiator proteins or topoisomerase.

Question 13

An E. coli strain harbors a mutation in the polA gene that inactivates the 5'→3' exonuclease domain of DNA Polymerase I, while leaving its other functions intact. Which of the following phenotypes is expected in this mutant?

  1. A significant increase in the overall mutation rate due to failed proofreading.
  2. Replication forks will stall immediately after initiation.
  3. The bacteria will be unable to replicate their DNA due to an inability to synthesize Okazaki fragments.
  4. Daughter DNA strands will contain segments of RNA covalently linked to DNA. (correct answer)

Explanation: The 5'→3' exonuclease domain of DNA Polymerase I is specifically responsible for removing the RNA primers of Okazaki fragments. If this domain is inactive, the primers cannot be excised. DNA Polymerase I's polymerase activity can still fill any available gaps, and DNA ligase may seal nicks, but the RNA segments will remain incorporated within the lagging strand. This results in a final product where the daughter DNA contains covalently linked ribonucleotides. Distractor A is incorrect because proofreading is handled by the 3'→5' exonuclease domain, which is unaffected. Distractor B and C are incorrect because the main replicative enzyme, DNA Polymerase III, is normal, so replication and Okazaki fragment synthesis will proceed.

Question 14

A mutant strain of E. coli possesses a temperature-sensitive DNA ligase, which is functional at 30°C but inactive at 42°C. If a culture of this strain is shifted from 30°C to 42°C during active DNA replication, which of the following molecular structures would be expected to accumulate?

  1. Short RNA-DNA hybrid fragments corresponding to the lagging strand.
  2. Unreplicated parental DNA due to the stalling of the replication fork.
  3. A fully replicated leading strand and a completely single-stranded lagging strand template.
  4. Numerous short, double-stranded DNA segments on the lagging strand template, separated by nicks. (correct answer)

Explanation: DNA ligase is responsible for sealing the final phosphodiester bond (the 'nick') between adjacent Okazaki fragments after the RNA primer has been removed and the gap filled with DNA. If DNA ligase is inactive, DNA Polymerase III will synthesize Okazaki fragments, and DNA Polymerase I will remove the RNA primers and fill the gaps with DNA. However, the final covalent link between the 3' end of one fragment and the 5' end of the next will not be formed. This results in the accumulation of many fully synthesized but unjoined DNA fragments on the lagging strand. Distractor A is incorrect because DNA Polymerase I, which removes the RNA primers, is still functional. Distractor B is incorrect as helicase and polymerases are functional, so replication will proceed until the lack of ligation causes downstream problems, but accumulation of fragments is the immediate effect. Distractor C is an oversimplification; lagging strand synthesis will initiate and proceed, but fragments won't be joined.

Question 15

A 12,000 base pair linear viral genome with a 40% G-C content is replicated in vitro. The reaction mixture contains all necessary enzymes, a non-radioactive template genome, and dNTPs where only the dATP is supplied with a radioactive phosphorus atom in the alpha position (α-³²P). After exactly one round of semiconservative replication, how many radioactive phosphorus atoms will have been incorporated into the two resulting daughter DNA molecules?

  1. 3,600
  2. 4,800
  3. 7,200 (correct answer)
  4. 12,000

Explanation: This problem requires calculating the number of specific nucleotides incorporated during replication.

  1. Determine the total number of A-T base pairs: The genome is 12,000 base pairs (bp) long. The G-C content is 40%, which means the A-T content is 100% - 40% = 60%.
  2. Calculate the number of thymine (T) bases in the template: The number of A-T pairs is 60% of the total base pairs: 12,000 bp * 0.60 = 7,200 A-T pairs. In a double-stranded DNA molecule, the number of adenine bases equals the number of thymine bases. Therefore, the original template molecule contains 7,200 thymine bases.
  3. Determine the number of dATPs incorporated: During semiconservative replication, two new strands are synthesized using the original strands as templates. An adenine (A) is incorporated into a new strand for every thymine (T) present on the template strand. Since there are 7,200 thymine bases in the template DNA, 7,200 radioactively labeled dATP molecules will be incorporated into the two new strands. Distractor A (3,600) might result from incorrectly dividing the number of A-T bases by two. Distractor B (4,800) corresponds to the number of G-C pairs, a misreading of the question. Distractor D (12,000) is the total length of one new strand, not the count of a specific base.

Question 16

A single bacterium with a ¹⁵N-labeled chromosome is placed in a ¹⁴N medium. It is allowed to divide for exactly three generations. What fraction of the total bacterial chromosomes present in the population will contain any of the original ¹⁵N-labeled DNA?

  1. 1/2
  2. 1/4 (correct answer)
  3. 1/8
  4. 1/16

Explanation: This is a classic semiconservative replication problem. The single original chromosome has two ¹⁵N strands. These two strands will be conserved throughout the subsequent generations.

  • Generation 0: 1 cell, 1 chromosome (¹⁵N/¹⁵N). Total chromosomes = 1. Chromosomes with ¹⁵N = 1.
  • Generation 1: The cell divides into 2. The two original ¹⁵N strands serve as templates, creating 2 hybrid (¹⁵N/¹⁴N) chromosomes. Total chromosomes = 2. Chromosomes with ¹⁵N = 2.
  • Generation 2: The 2 cells divide into 4. The two ¹⁵N strands again serve as templates, creating 2 hybrid (¹⁵N/¹⁴N) chromosomes. The two ¹⁴N strands serve as templates for 2 fully light (¹⁴N/¹⁴N) chromosomes. Total chromosomes = 4. Chromosomes with ¹⁵N = 2.
  • Generation 3: The 4 cells divide into 8. The two ¹⁵N strands once again create 2 hybrid (¹⁵N/¹⁴N) chromosomes. The other six ¹⁴N strands create 6 fully light (¹⁴N/¹⁴N) chromosomes. Total chromosomes = 8. Chromosomes with ¹⁵N = 2. The fraction of chromosomes containing any original ¹⁵N DNA is 2 out of 8, which simplifies to 1/4. Distractor A (1/2) is the fraction after 2 generations. Distractor C (1/8) is the fraction of total DNA strands that are ¹⁵N (2 original strands out of 16 total strands), not the fraction of chromosomes.

Question 17

Eukaryotic chromosomes have multiple origins of replication, whereas the prokaryotic chromosome of E. coli has only one. What is the most fundamental reason for this difference?

  1. The vast size of eukaryotic genomes must be replicated within the limited time of the S phase, requiring a massively parallel approach. (correct answer)
  2. The presence of nucleosomes in eukaryotes impedes the replication fork, necessitating re-initiation at multiple sites.
  3. Eukaryotic DNA polymerases are much slower and less processive than prokaryotic polymerases, requiring many origins to compensate.
  4. Linear eukaryotic chromosomes require multiple origins to solve the end-replication problem, which circular chromosomes do not have.

Explanation: When you encounter questions about DNA replication differences between prokaryotes and eukaryotes, focus on the fundamental constraints each cell type faces during replication. The key issue is timing and scale. Eukaryotic genomes are enormous compared to prokaryotic genomes—human cells contain about 3 billion base pairs versus E. coli's 4.6 million. Yet eukaryotic cells must replicate their entire genome during S phase, typically within 6-8 hours. If eukaryotic chromosomes used just one origin of replication like E. coli, replication would take days or weeks to complete, making cell division impossible. Multiple origins (thousands per chromosome) allow simultaneous replication along each chromosome, dramatically reducing the time needed. This makes A correct—the massive parallel approach is essential to meet S phase timing constraints. B is incorrect because while nucleosomes do temporarily slow replication forks, they don't necessitate multiple origins—the forks can still proceed through chromatin with histone chaperones helping reassemble nucleosomes behind them. C misrepresents polymerase differences. Eukaryotic DNA polymerases aren't dramatically slower or less processive than prokaryotic ones—both replicate at roughly 50 nucleotides per second. D confuses different problems. Multiple origins aren't the solution to end-replication problems (telomeres and telomerase handle that issue). Linear chromosomes could theoretically use single origins if time weren't limiting. Study tip: Remember that eukaryotic replication strategies primarily solve timing problems created by genome size, not mechanical problems created by chromosome structure.

Question 18

A novel antiviral drug is found to be a potent inhibitor of the primase enzyme in a virus with a dsDNA genome. If this drug is administered to an infected cell culture during viral replication, what would be the observed state of the viral DNA molecules?

  1. Parental strands will be separated by helicase, but no synthesis of new complementary strands will occur. (correct answer)
  2. Only the leading strand will be synthesized, resulting in molecules that are partially double-stranded and partially single-stranded.
  3. Replication will be completed, but the daughter strands will consist of small, unlinked DNA fragments.
  4. The viral DNA will remain in its condensed, supercoiled state, and no unwinding will occur.

Explanation: Primase synthesizes the short RNA primers that are absolutely required for DNA polymerase to initiate DNA synthesis. DNA polymerases cannot start synthesis de novo; they can only add nucleotides to a pre-existing 3'-OH group. Without the action of primase, no primers can be made. Consequently, even if helicase unwinds the parental DNA, DNA polymerase will be unable to synthesize any new DNA on either the leading or the lagging strand. Distractor B is incorrect because even the leading strand requires one initial primer to get started. Distractor C describes the effect of a DNA ligase inhibitor. Distractor D describes the effect of an inhibitor of initiator proteins or topoisomerase.

Question 19

A single bacterium with a ¹⁵N-labeled chromosome is placed in a ¹⁴N medium. It is allowed to divide for exactly three generations. What fraction of the total bacterial chromosomes present in the population will contain any of the original ¹⁵N-labeled DNA?

  1. 1/2
  2. 1/4 (correct answer)
  3. 1/8
  4. 1/16

Explanation: This is a classic semiconservative replication problem. The single original chromosome has two ¹⁵N strands. These two strands will be conserved throughout the subsequent generations.

  • Generation 0: 1 cell, 1 chromosome (¹⁵N/¹⁵N). Total chromosomes = 1. Chromosomes with ¹⁵N = 1.
  • Generation 1: The cell divides into 2. The two original ¹⁵N strands serve as templates, creating 2 hybrid (¹⁵N/¹⁴N) chromosomes. Total chromosomes = 2. Chromosomes with ¹⁵N = 2.
  • Generation 2: The 2 cells divide into 4. The two ¹⁵N strands again serve as templates, creating 2 hybrid (¹⁵N/¹⁴N) chromosomes. The two ¹⁴N strands serve as templates for 2 fully light (¹⁴N/¹⁴N) chromosomes. Total chromosomes = 4. Chromosomes with ¹⁵N = 2.
  • Generation 3: The 4 cells divide into 8. The two ¹⁵N strands once again create 2 hybrid (¹⁵N/¹⁴N) chromosomes. The other six ¹⁴N strands create 6 fully light (¹⁴N/¹⁴N) chromosomes. Total chromosomes = 8. Chromosomes with ¹⁵N = 2. The fraction of chromosomes containing any original ¹⁵N DNA is 2 out of 8, which simplifies to 1/4. Distractor A (1/2) is the fraction after 2 generations. Distractor C (1/8) is the fraction of total DNA strands that are ¹⁵N (2 original strands out of 16 total strands), not the fraction of chromosomes.

Question 20

Fluoroquinolone antibiotics, such as ciprofloxacin, function by inhibiting bacterial DNA gyrase. If a susceptible bacterial culture in logarithmic growth phase is treated with ciprofloxacin, what is the most immediate consequence at the molecular level of DNA replication?

  1. Separation of the two parental DNA strands at the origin of replication is blocked.
  2. Positive supercoils accumulate ahead of the replication fork, causing it to stall. (correct answer)
  3. Newly synthesized Okazaki fragments on the lagging strand fail to be joined.
  4. RNA primers are not removed from the newly synthesized DNA strands.

Explanation: DNA gyrase, a type of topoisomerase II, is responsible for introducing negative supercoils into DNA, which relieves the torsional stress (positive supercoils) that builds up ahead of the replication fork as helicase unwinds the double helix. Inhibition of DNA gyrase prevents the removal of these positive supercoils. The resulting strain on the DNA molecule will eventually halt the progression of the helicase and, consequently, the entire replication fork. Distractor A describes a failure of initiator proteins or helicase. Distractor C describes the effect of a DNA ligase inhibitor. Distractor D describes an effect of inhibiting DNA Polymerase I's 5'→3' exonuclease activity.