These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assume each origin of replication produces two forks that move apart, and that new DNA is made 5′ → 3′. A calculator is assumed. All organisms and data sets are fictional.
Question 1 (multiple choice · foundation)
As a replication fork moves, the double helix ahead of it becomes more and more tightly twisted. Which enzyme relieves this twisting?
- (A) Helicase
- (B) Topoisomerase
- (C) DNA polymerase
- (D) Ligase
Answer and explanation
Answer: (B). Topoisomerase acts on the unreplicated DNA ahead of the fork. It cuts the backbone, lets the DNA untwist, and rejoins it, relieving the supercoiling caused by unwinding.
- (A) Helicase causes the problem: it separates the strands at the fork, which overwinds the DNA in front. It does not relieve the twisting.
- (C) DNA polymerase builds new strands behind the fork; it does not change the twisting ahead.
- (D) Ligase joins lagging-strand fragments behind the fork.
Question 2 (multiple choice · core)
A cell’s DNA has radioactive label in both strands. The cell then divides twice in a medium with no label, so its DNA replicates twice. What percentage of the DNA molecules now contain any radioactive label?
- (A) 25%
- (B) 50%
- (C) 75%
- (D) 100%
Answer and explanation
Answer: (B). After round 1: 2 molecules, each with one labelled and one unlabelled strand. After round 2: 4 molecules. Each of the 2 labelled strands is still paired with a new unlabelled strand, so 2 of the 4 molecules (50%) contain label.
- (A) 25% is the fraction of strands that are labelled (2 of 8), not the fraction of molecules. It is also what the conservative model would predict.
- (C) 75% comes from subtracting the 25% of labelled strands from 100%, which mixes up strands and molecules.
- (D) 100% would fit the dispersive model, where every strand carries some old pieces, or the result after only one round.
Question 3 (multiple choice · core)
A template strand reads 5′-CAGTTACG-3′. Which new DNA strand is made on it, written 5′ to 3′?
- (A) 5′-GTCAATGC-3′
- (B) 5′-CGTAACTG-3′
- (C) 5′-GCATTGAC-3′
- (D) 5′-CGUAACUG-3′
Answer and explanation
Answer: (B). Pair each base (C–G, A–T, G–C, T–A), giving 3′-GTCAATGC-5′ under the template. Reading it from its 5′ end gives 5′-CGTAACTG-3′.
- (A) is the complement, but written in the wrong direction: its left end is really the 3′ end.
- (C) is the template simply reversed, with no pairing.
- (D) uses U instead of T. Replication makes DNA, which contains T. (U would appear only in an RNA primer or in transcription.)
Question 4 (visual representation · core)
The diagram shows a replication fork. The fork is moving to the right.
(a) Name the enzyme that acts at W, at X and at Y, and state what each one does. (b) Is the end of the new strand marked Z its 3′ end or its 5′ end? Explain. (c) Explain why the new DNA along the lower template is made in fragments. (d) Predict what would happen to the fork if the enzyme at W stopped working.
Model answer and suggested Marlbridge rubric
(a) W: topoisomerase, relieves the supercoiling (overwinding) of the parental DNA ahead of the fork. X: helicase, separates the two strands by breaking the hydrogen bonds between bases. Y: ligase, joins neighbouring fragments by forming a covalent bond in the backbone.
(b) 3′ end. The upper template has its 3′ end at the left, so the new strand runs 5′ at the left to 3′ at the right. DNA polymerase adds nucleotides only to the 3′ end, and this strand is growing towards the fork at Z.
(c) The lower template has its 5′ end at the left, so its new strand must run 5′ at the right to 3′ at the left. Building 5′ → 3′ therefore means moving away from the fork. The polymerase can only copy the stretch already exposed, then must start again near the fork with a new primer as more template opens. This is the lagging strand.
(d) Overwinding would build up ahead of the fork. Helicase would find it harder and harder to separate the strands, so the fork would slow and stop, and replication would be incomplete.
| Point | What earns it |
|---|---|
| 1 | All three enzymes named correctly |
| 1 | A correct job for each of the three |
| 1 | Z is the 3′ end, with reasoning from antiparallel strands or 3′ addition |
| 1 | Fragments explained by 5′ → 3′ synthesis pointing away from the fork |
| 1 | Prediction: twisting builds up, fork stalls |
Question 5 (data analysis · core)
Researchers study three fictional bacterial mutants. Each has one replication enzyme that stops working at 40 °C. They move cells to 40 °C and measure new DNA made in 10 minutes, as a percentage of normal (wild-type) cells.
| Mutant | New DNA (% of wild type) | Other observations |
|---|---|---|
| M1 | 98 | Many new DNA pieces about 1 000 nucleotides long that are never joined, alongside very long new strands |
| M2 | 2 | Parental strands stay paired; forks do not open |
| M3 | 11 | Short stretches of new DNA near each origin, then synthesis stops; the parental DNA just ahead of the forks is tightly overwound |
(a) Identify the defective enzyme in each mutant. Give one piece of evidence for each. (b) Explain why M1 makes almost as much new DNA as normal cells, yet cannot complete replication.
Model answer and suggested Marlbridge rubric
(a)
- M1: ligase. Short fragments (lagging-strand pieces) are made but never joined; long continuous strands (leading strands) are normal.
- M2: helicase. The strands are never separated, so there is no template for DNA polymerase: almost no new DNA (2%).
- M3: topoisomerase. Forks start, but overwinding builds up ahead of them and stops further unwinding after a short distance (11%).
(b) In M1, helicase, primers and DNA polymerase all work, so both strands are copied and the amount of new DNA is nearly normal (98%). But the lagging strand stays as separate fragments, because nothing seals the backbone between them. The daughter molecules have breaks in one strand, so replication is not complete.
| Point | What earns it |
|---|---|
| 1 | M1 = ligase, with evidence |
| 1 | M2 = helicase, with evidence |
| 1 | M3 = topoisomerase, with evidence |
| 1 | M1: polymerase still copies both strands, so amount is near normal |
| 1 | M1: fragments not joined, so the lagging strand has gaps |
Question 6 (calculation · stretch)
A fictional eukaryotic cell must copy a linear chromosome of 1.5 × 10⁸ base pairs during an S phase lasting 8.0 hours. Each fork copies 30 nucleotides per second.
(a) Calculate the minimum number of origins of replication needed. (b) If the chromosome actually has 120 evenly spaced origins, how long would replication take? (c) Give one reason why a real cell might need more origins than your answer to (a).
Worked solution and suggested Marlbridge rubric
(a) S phase = 8.0 × 3 600 = 28 800 s. One origin gives two forks, so it can copy 2 × 30 × 28 800 = 1 728 000 bp. Origins needed = 1.5 × 10⁸ ÷ 1 728 000 = 86.8, so at least 87 origins (round up: 86 would not be enough).
(b) Each origin covers 1.5 × 10⁸ ÷ 120 = 1.25 × 10⁶ bp, so each fork copies 6.25 × 10⁵ bp. Time = 6.25 × 10⁵ ÷ 30 = 20 833 s ≈ 5.8 hours.
(c) Any one: origins do not all start at the same moment; forks may pause or stall at damaged or tightly packed DNA; origins are not evenly spaced; histones must be removed and replaced around the fork.
| Point | What earns it |
|---|---|
| 1 | Correct capacity per origin, using two forks and seconds |
| 1 | 87 origins, rounded up with a reason |
| 1 | About 5.8 h for 120 origins |
| 1 | A sensible biological reason |
Common error: forgetting that each origin has two forks gives 174 origins in (a) and 11.6 h in (b).
Question 7 (constructed response · stretch)
Replication is under way in a fictional bacterium. Researchers add a drug that blocks the enzyme that makes RNA primers. Over the next 2 minutes they measure how much new DNA is added to leading and lagging strands, compared with untreated cells.
| Strand | New DNA added (fraction of untreated) |
|---|---|
| Leading strand | 0.95 |
| Lagging strand | 0.04 |
Make a claim about why the drug affects the two strands so differently. Support it with evidence from the table and reasoning about how each strand is made.
Model answer and suggested Marlbridge rubric
Claim: the drug blocks lagging-strand synthesis much more than leading-strand synthesis because the lagging strand needs a new primer for every fragment, while each leading strand needs only one primer, which was already in place.
Evidence: leading-strand synthesis continued at 0.95 of the untreated value, but lagging-strand synthesis fell to 0.04.
Reasoning: DNA polymerase cannot start a strand; it can only add to the 3′ end of an existing primer or strand. The leading strand grows continuously towards the fork from one primer laid down at the origin, so once started it no longer needs new primers. The lagging strand grows away from the fork in short fragments, and each new fragment must begin with a fresh RNA primer. With no new primers, no new fragments can start; the small amount (0.04) is probably fragments that were already primed when the drug was added.
| Point | What earns it |
|---|---|
| 1 | Claim links the difference to primer use on each strand |
| 1 | Evidence quotes both values |
| 1 | DNA polymerase needs a primer (cannot start a chain) |
| 1 | Leading: continuous from one primer; lagging: one primer per fragment |
How did you do?
- Q1, Q4 or Q5 wrong: re-read “At the replication fork” in the study guide, with Figure 2 and the enzyme table.
- Q2 wrong: rework Worked example 2 and look again at Figure 1; separate “molecules” from “strands”.
- Q3 or Q4(b) wrong: rework Worked example 1; always reverse the complement to write it 5′ to 3′.
- Q6 or Q7 wrong: rework Worked example 3 and re-read “Leading and lagging strands”.
Then tick off the topic checklist.