IB Biology Quiz: Apply Dna Replication
20 questions · exam conditions
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Apply Dna ReplicationQuestion 1 of 20

Analysis of a failed DNA replication reaction reveals that the parent DNA is unwound and stabilized by single-strand binding proteins, and numerous short RNA sequences are attached to the template strands, but very little new DNA has been formed. Which enzyme was most likely defective or inhibited?

DNA helicase
Primase
DNA polymerase
DNA ligase
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IB Biology Quiz

IB Biology Quiz: Apply Dna Replication

Practice Apply Dna Replication in IB Biology 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 Apply Dna Replication, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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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.

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Question 1

Analysis of a failed DNA replication reaction reveals that the parent DNA is unwound and stabilized by single-strand binding proteins, and numerous short RNA sequences are attached to the template strands, but very little new DNA has been formed. Which enzyme was most likely defective or inhibited?

  1. DNA helicase
  2. Primase
  3. DNA polymerase (correct answer)
  4. DNA ligase
Explanation: The DNA is unwound, so helicase is functional. There are short RNA sequences, so primase is functional. The lack of new DNA synthesis following the primers points to a failure of DNA polymerase to elongate the strands. A defect in ligase would result in replicated fragments that are not joined, not a lack of replication itself.

Question 2

Which of the following is a key reason why eukaryotes have multiple origins of replication on their chromosomes, whereas prokaryotes typically have only one?

  1. Eukaryotic DNA polymerase works much more slowly than prokaryotic DNA polymerase.
  2. Eukaryotic chromosomes are linear, while prokaryotic chromosomes are circular.
  3. Eukaryotes have vastly larger genomes that must be replicated within a similar timeframe. (correct answer)
  4. The semi-conservative replication in eukaryotes is more complex than in prokaryotes.
Explanation: Eukaryotic genomes are orders of magnitude larger than prokaryotic genomes. To replicate all of this DNA during the S phase of the cell cycle, replication must begin at many points simultaneously. Multiple origins of replication dramatically shorten the total time required to copy the entire genome.

Question 3

A single replication bubble forms in a linear eukaryotic DNA molecule and replication proceeds bidirectionally. Considering the entire bubble, how many leading strands and lagging strands are being synthesized at any given moment?

  1. One leading strand and one lagging strand.
  2. One leading strand and two lagging strands.
  3. Two leading strands and one lagging strand.
  4. Two leading strands and two lagging strands. (correct answer)
Explanation: A replication bubble has two replication forks moving in opposite directions from the origin. At each fork, there is one continuously synthesized leading strand and one discontinuously synthesized lagging strand. Therefore, within the entire bubble, there are two forks, meaning two leading strands and two lagging strands are being synthesized simultaneously.

Question 4

A single DNA molecule undergoes four complete rounds of semi-conservative replication. What fraction of the total DNA strands (not molecules) in the final population will be composed of the original, template DNA?

  1. 1/4
  2. 1/8
  3. 1/16 (correct answer)
  4. 1/32
Explanation: The starting molecule has 2 original strands. After 4 rounds of replication, there will be 2⁴ = 16 DNA molecules. The total number of strands will be 16 molecules × 2 strands/molecule = 32 strands. The 2 original strands are conserved within this population. Therefore, the fraction of original strands is 2/32, which simplifies to 1/16.

Question 5

A new experimental drug is found to specifically inhibit the action of DNA ligase. Which of the following would be the most direct consequence for a cell undergoing DNA replication after being treated with this drug?

  1. The DNA double helix would fail to unwind at the replication fork, halting all synthesis.
  2. The sugar-phosphate backbone of the newly synthesized lagging strand would have breaks. (correct answer)
  3. RNA primers would not be removed from the newly synthesized DNA strands.
  4. DNA polymerase would be unable to add new deoxynucleotides to the growing strands.
Explanation: DNA ligase is responsible for joining the Okazaki fragments on the lagging strand by forming the final phosphodiester bond. If ligase is inhibited, the fragments will be synthesized but will remain as separate pieces, leaving breaks or 'nicks' in the sugar-phosphate backbone.

Question 6

If DNA replication were conservative instead of semi-conservative, what would be the expected composition of DNA molecules after two rounds of replication, starting from a single heavy (¹⁵N-¹⁵N) DNA molecule in a ¹⁴N medium?

  1. Two hybrid (¹⁵N-¹⁴N) molecules and two light (¹⁴N-¹⁴N) molecules.
  2. Four molecules, all of which are hybrid (¹⁵N-¹⁴N).
  3. One heavy (¹⁵N-¹⁵N) molecule and three light (¹⁴N-¹⁴N) molecules. (correct answer)
  4. One heavy (¹⁵N-¹⁵N) molecule and one light (¹⁴N-¹⁴N) molecule.
Explanation: In a conservative model, the original parent molecule remains intact. After one generation, there would be one original heavy molecule and one new light molecule. After the second generation, the original heavy molecule would template another new light molecule, and the light molecule from the first generation would template another light molecule. This results in one original heavy molecule and three new light molecules.

Question 7

A strain of bacteria possesses a mutation that inactivates the 3' to 5' exonuclease activity of its main DNA polymerase, but its 5' to 3' polymerization activity remains unaffected. What is the most likely consequence of this mutation for the bacteria?

  1. The bacteria will have a significantly higher rate of spontaneous point mutations. (correct answer)
  2. DNA replication will be unable to initiate, as the origin will not be recognized.
  3. Okazaki fragments will be synthesized but will not be joined into a continuous strand.
  4. The replication forks will stall because the polymerase cannot move along the template.
Explanation: The 3' to 5' exonuclease activity is the 'proofreading' function of DNA polymerase. It removes incorrectly incorporated nucleotides. If this function is lost, errors made during polymerization will not be corrected, leading to a much higher mutation rate.

Question 8

At a single, advancing replication fork, which statement correctly describes the synthesis of the two new DNA strands?

  1. Both new strands are synthesized discontinuously away from the replication fork.
  2. Both new strands are synthesized continuously towards the replication fork.
  3. One strand is synthesized towards the fork, while the other is synthesized away from the fork. (correct answer)
  4. Both strands are synthesized in the same physical direction, but one is faster than the other.
Explanation: Due to the antiparallel nature of DNA and the 5' to 3' directionality of DNA polymerase, synthesis occurs differently on each template. The leading strand is synthesized continuously in the same direction as the fork's movement (towards the fork). The lagging strand is synthesized discontinuously in fragments in the overall opposite direction (away from the fork).

Question 9

[HL] During DNA synthesis, a small amount of dideoxycytidine triphosphate (ddCTP), which lacks a 3'-OH group, is added to the reaction. What is the most likely outcome when a ddCTP molecule is incorporated into a growing DNA strand?

  1. Elongation of that specific strand is terminated immediately. (correct answer)
  2. The DNA polymerase stalls and reverses direction to remove the ddCTP.
  3. Synthesis continues, but a mutation is created that will be repaired later.
  4. The ddCTP base pairs with guanine but is then immediately replaced with a normal dCTP.
Explanation: DNA polymerase requires a free 3'-OH group on the preceding nucleotide to form a phosphodiester bond with the next incoming nucleotide. Because dideoxynucleotides lack this 3'-OH group, no further nucleotides can be added once one is incorporated. This results in the termination of strand elongation.

Question 10

Which of these enzymes involved in DNA replication works ahead of the replication fork to relieve the torsional strain caused by unwinding?

  1. Helicase
  2. Primase
  3. DNA polymerase
  4. Topoisomerase (correct answer)
Explanation: As helicase unwinds the DNA at the replication fork, the parental DNA ahead of the fork becomes overwound and builds up torsional stress (supercoiling). Topoisomerase works ahead of the fork to cut, swivel, and rejoin the DNA strands, relieving this strain and allowing replication to proceed.

Question 11

Bacteria were cultured in a medium containing a heavy isotope of nitrogen, ¹⁵N, and then transferred to a new medium containing the normal light isotope, ¹⁴N. If the bacteria were allowed to complete three full generations of DNA replication in the ¹⁴N medium, what would be the expected ratio of hybrid (¹⁵N-¹⁴N) DNA molecules to light (¹⁴N-¹⁴N) DNA molecules?

  1. 1:1
  2. 1:3 (correct answer)
  3. 1:4
  4. 1:7
Explanation: After the first generation, all DNA molecules are hybrid (100%). After the second generation, there are 2 hybrid and 2 light molecules, a 1:1 ratio. After the third generation, the 2 hybrid molecules replicate to form 2 hybrid and 2 light molecules, and the 2 light molecules from the previous generation replicate to form 4 more light molecules. This results in a total of 2 hybrid and 6 light molecules, which simplifies to a 1:3 ratio.

Question 12

[HL] A 4,000 base pair segment of DNA is replicated as a lagging strand. In this organism, an Okazaki fragment is on average 200 base pairs long and requires one primer. Approximately how many RNA primers and DNA ligase reactions are needed to synthesize this segment?

  1. 20 primers and 20 ligase reactions. (correct answer)
  2. 1 primer and 20 ligase reactions.
  3. 20 primers and 19 ligase reactions.
  4. 1 primer and 19 ligase reactions.
Explanation: First, calculate the number of Okazaki fragments: 4,000 bp / 200 bp/fragment = 20 fragments. Each fragment requires its own RNA primer for initiation, so 20 primers are needed. DNA ligase must seal the nicks between these 20 fragments and connect the final fragment to the adjacent DNA, requiring 20 ligation events. A common mistake is to calculate N-1 ligations, but for a segment within a larger strand, the number of ligations equals the number of fragments.

Question 13

The addition of a nucleotide to a growing DNA strand is an endergonic reaction. What is the immediate source of the energy that allows DNA polymerase to catalyze this phosphodiester bond formation?

  1. The hydrolysis of separate ATP molecules which provide energy to the polymerase enzyme.
  2. The formation of hydrogen bonds between the complementary base pairs on opposite strands.
  3. The hydrolysis of the two terminal phosphate groups from the incoming deoxynucleoside triphosphate. (correct answer)
  4. The release of potential energy as helicase unwinds the DNA double helix.
Explanation: Each incoming nucleotide is a deoxynucleoside triphosphate (dNTP). DNA polymerase cleaves off the two terminal phosphate groups (pyrophosphate), and the energy released from breaking these high-energy bonds is used to form the phosphodiester bond that attaches the nucleotide to the growing strand.

Question 14

In an in vitro DNA replication experiment, a potent inhibitor of primase is added to the reaction mixture. Which statement best predicts the outcome?

  1. Replication will proceed, but the Okazaki fragments will fail to join together.
  2. Only the leading strand will be synthesized, since it requires only a single primer.
  3. The DNA helix will unwind, but synthesis of new DNA will not be initiated. (correct answer)
  4. Replication will be completed, but the RNA primers will remain in the new DNA.
Explanation: Primase synthesizes the RNA primers that provide the necessary 3'-OH group for DNA polymerase to begin synthesis. Without primase activity, no primers can be made. Since DNA polymerase cannot initiate synthesis de novo (from scratch), no new DNA will be made on either the leading or the lagging strand, even if the helix is unwound by helicase.

Question 15

[HL] Which sequence correctly outlines the process of joining Okazaki fragments during lagging strand synthesis?

  1. DNA ligase seals nicks → DNA polymerase removes RNA primer → DNA polymerase synthesizes DNA.
  2. DNA polymerase removes RNA primer → DNA polymerase synthesizes DNA → DNA ligase seals nicks. (correct answer)
  3. DNA polymerase removes RNA primer → DNA ligase seals nicks → DNA polymerase synthesizes DNA.
  4. DNA ligase seals nicks → DNA polymerase synthesizes DNA → DNA polymerase removes RNA primer.
Explanation: After an Okazaki fragment is synthesized, the RNA primer of the preceding fragment must be removed and replaced with DNA. This is done by a DNA polymerase (like DNA Pol I in prokaryotes). Once the RNA is replaced with DNA, a nick remains in the sugar-phosphate backbone, which is then sealed by DNA ligase.

Question 16

[HL] Most normal human somatic cells have a finite number of cell divisions they can undergo, known as the Hayflick limit. This cellular aging is primarily a consequence of which molecular process?

  1. The gradual depletion of the cell's supply of deoxynucleoside triphosphates over time.
  2. An increase in the error rate of DNA polymerase as the cell ages.
  3. The inability of DNA polymerase to replicate the final 5' end of the lagging strand. (correct answer)
  4. The progressive cross-linking of DNA strands, which prevents helicase from unwinding them.
Explanation: The 'end-replication problem' occurs because the lagging strand cannot be fully replicated to its very end. When the final RNA primer is removed from the 5' end, there is no existing 3' end for DNA polymerase to build from, leaving a gap. This causes the chromosome to shorten with each round of replication, eventually leading to senescence.

Question 17

[HL] A researcher compares the activity of telomerase in different human cells. Which observation would be most consistent with the known function of this enzyme?

  1. High telomerase activity in differentiated muscle cells to support their high metabolic rate.
  2. Telomerase activity is highest just before cell division in all cell types to prepare chromosomes.
  3. Low telomerase activity in most somatic cells, but high activity in germ-line stem cells. (correct answer)
  4. Identical levels of telomerase activity in both young and senescent somatic cells.
Explanation: Telomerase counteracts the end-replication problem by lengthening telomeres. This is crucial in cells that divide indefinitely, such as germ-line stem cells that produce gametes. Most differentiated somatic cells have very low or no telomerase activity, which is linked to cellular aging and the Hayflick limit.

Question 18

What is the role of single-strand binding (SSB) proteins in DNA replication?

  1. They prevent the separated template strands from re-annealing. (correct answer)
  2. They unwind the DNA double helix at the replication fork.
  3. They act as a primer for DNA polymerase to initiate synthesis.
  4. They join the Okazaki fragments on the lagging strand.
Explanation: After helicase unwinds the DNA, the separated single strands have a natural tendency to base-pair and re-form a double helix. Single-strand binding proteins coat the separated strands, keeping them apart and accessible as templates for DNA polymerase.

Question 19

The fundamental constraint that DNA polymerase can only add nucleotides to the 3' end of a growing strand is the direct cause of which feature of replication?

  1. The need for the enzyme helicase to unwind the parent DNA molecule.
  2. The semi-conservative nature of the overall replication process.
  3. The requirement for RNA primers to initiate any new DNA synthesis.
  4. The synthesis of one strand continuously and the other discontinuously. (correct answer)
Explanation: Because the two parent strands are antiparallel, and DNA polymerase can only synthesize in the 5' to 3' direction, only one of the new strands (the leading strand) can be synthesized continuously towards the replication fork. The other strand (the lagging strand) must be synthesized in the opposite direction, away from the fork, in short segments called Okazaki fragments.

Question 20

In the Meselson-Stahl experiment, control samples of DNA from bacteria grown exclusively in ¹⁵N and exclusively in ¹⁴N were used. What was the primary scientific purpose of these controls?

  1. To confirm that the bacteria could grow and replicate in both types of nitrogen media.
  2. To establish the reference banding positions for pure heavy and pure light DNA. (correct answer)
  3. To prove that the centrifugation process did not alter the structure of DNA.
  4. To determine the exact molecular weight of DNA containing each nitrogen isotope.
Explanation: These samples served as crucial controls or standards. By centrifuging them, Meselson and Stahl could determine the exact position in the density gradient where 100% heavy DNA and 100% light DNA would settle. This allowed them to unambiguously interpret the position of the DNA band from the experimental samples in subsequent generations.