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. Useful relationship: percentage change = (new value − original value) ÷ original value × 100. A calculator is assumed. Keep full calculator values until the last step. All data sets and cell lines are fictional.
Question 1 (multiple choice · foundation)
What is checked at the G2 checkpoint?
- (A) Whether all of the DNA has been replicated and is undamaged, before the cell enters mitosis
- (B) Whether every chromosome is attached to spindle fibres from both poles, before anaphase begins
- (C) Whether the cell is large enough and has received signals to divide, before DNA replication begins
- (D) Whether a nuclear envelope has formed around each set of chromosomes, before cytokinesis begins
Answer and explanation
Answer: (A). The G2 checkpoint sits at the end of G2. It stops a cell from starting mitosis with incomplete or damaged DNA, which would give its daughter cells broken or missing genes.
- (B) describes the M (spindle) checkpoint, which acts inside mitosis.
- (C) describes the G1 checkpoint, which comes before S phase.
- (D) is not one of the main checkpoints; nuclear envelopes re-form in telophase as part of mitosis.
Question 2 (multiple choice · core)
Which statement correctly describes cyclins and cyclin-dependent kinases (CDKs)?
- (A) The amount of CDK rises and falls during the cycle, while the cyclin concentration stays constant.
- (B) Cyclins are kinases that add phosphate groups to CDKs, switching the CDKs off.
- (C) CDKs are active only when bound to a cyclin, so CDK activity rises and falls with the cyclin concentration.
- (D) An active cyclin-CDK complex removes phosphate groups from DNA, allowing it to replicate.
Answer and explanation
Answer: (C). CDK molecules are present all the time, but each needs a cyclin partner to work. Because cyclin is made and then destroyed in each cycle, CDK activity follows the cyclin concentration.
- (A) reverses the two proteins: it is cyclin that rises and falls, while the amount of CDK stays about constant.
- (B) gets the roles wrong. The CDK is the kinase; cyclin is its activating partner, and binding switches the CDK on.
- (D) A kinase adds phosphate groups (from ATP) to proteins, not removes them from DNA.
Question 3 (multiple choice · core)
A skin cell with working checkpoints is exposed to radiation, and its DNA is damaged so badly that it cannot be repaired. Which outcome is most likely?
- (A) The cell undergoes apoptosis, so the damaged DNA is not passed on to daughter cells.
- (B) The cell skips S phase so that the damaged DNA is not copied, then divides normally.
- (C) The cell divides faster, so that the damage is diluted among many daughter cells.
- (D) The cell enters mitosis and repairs the DNA during anaphase.
Answer and explanation
Answer: (A). A checkpoint that detects DNA damage first stops the cycle to allow repair. If the damage cannot be repaired, the cell is removed by apoptosis, which protects the tissue from cells carrying harmful mutations.
- (B) Skipping S phase would leave the cell unable to give each daughter a full set of chromosomes; checkpoints stop the cycle, they do not skip parts of it.
- (C) Damage is not diluted by division: each daughter cell inherits a copy of the damaged DNA. Faster division is what failed control looks like.
- (D) The G2 checkpoint stops cells with damaged DNA from entering mitosis; anaphase separates chromatids, it does not repair them.
Question 4 (data · core)
Researchers treat a culture of dividing cells with a drug that blocks the activity of the CDK needed to pass the G1 checkpoint. After 24 h they measure the percentage of cells in each stage.
| Culture | G1 / % | S / % | G2 and mitosis / % |
|---|---|---|---|
| Control (no drug) | 50 | 30 | 20 |
| Drug | 88 | 7 | 5 |
(a) Calculate the percentage change in the proportion of cells in G1, and in S. (b) Describe the effect of the drug, using the data. (c) Explain the result in terms of cyclins and CDKs. (d) A student suggests that such a drug could slow the growth of a tumour. Give one reason this is reasonable and one limitation.
Worked solution and suggested Marlbridge rubric
(a) G1: (88 − 50) ÷ 50 × 100 = +76%. S: (7 − 30) ÷ 30 × 100 = −77% (−76.7%).
(b) Cells pile up in G1 (from 50% to 88%), and far fewer are in S (30% to 7%) or in G2 and mitosis (20% to 5%). The drug stops most cells from moving past G1 into DNA replication.
(c) To pass the G1 checkpoint, a cyclin must bind to its CDK, and the active complex must phosphorylate the proteins that start S phase. If the CDK’s activity is blocked, even when cyclin is present, those targets are not phosphorylated. The cells cannot pass the checkpoint, so they stay in G1. The few cells still in S, G2 and mitosis had probably passed the checkpoint before the drug took effect.
(d) Reasonable: tumour cells need to pass through the cycle to multiply, so blocking G1 progression would slow the increase in cell number. Limitation (any one): normal dividing cells, for example in the skin, gut lining or bone marrow, would also be blocked; some tumour cells may have mutations that let them bypass this CDK; cells in culture may respond differently from a tumour in the body.
| Point | What earns it |
|---|---|
| 1 | Both percentage changes correct, with signs |
| 1 | Description quoting data: cells accumulate in G1 and fewer reach S, G2 and mitosis |
| 1 | Explanation: blocked CDK cannot phosphorylate targets, so the G1 checkpoint is not passed |
| 1 | One valid reason and one valid limitation |
Question 5 (constructed response · core)
During metaphase in a dividing cell, one chromosome is attached to spindle fibres from only one pole.
(a) Identify the checkpoint that responds to this, and describe what it does. (b) Explain why this response is important for the daughter cells. (c) In a mutant cell, this checkpoint does not work. Predict the effect on the chromosome numbers of the daughter cells of a cell with 46 chromosomes, and give one possible fate of a daughter cell with the wrong number.
Model answer and suggested Marlbridge rubric
(a) The M (spindle) checkpoint. It delays the start of anaphase until every chromosome is attached to spindle fibres from both poles.
(b) If anaphase started now, both sister chromatids of the unattached chromosome could be pulled to the same pole. Waiting allows the attachment to be corrected, so each daughter cell receives exactly one copy of every chromosome and a complete, identical genome.
(c) Anaphase may start anyway. If both chromatids of the faulty chromosome go to one pole, one daughter cell has 47 chromosomes and the other 45. Possible fates (any one): the abnormal cell is detected and removed by apoptosis; it stops dividing; or it survives and divides, and the wrong gene dosage, along with other mutations, may contribute to cancer.
| Point | What earns it |
|---|---|
| 1 | M (spindle) checkpoint named and its action (anaphase delayed until all attached) |
| 1 | Importance: each daughter gets one copy of every chromosome |
| 1 | 47 and 45 chromosomes |
| 1 | A valid fate linked to the abnormal chromosome number (apoptosis, or survival contributing to cancer) |
Question 6 (constructed response · stretch)
Researchers studied four groups of 50 fictional laboratory mice, each with a different genotype, and counted how many developed tumours by a set age. Mutation X makes a protein that drives the cell cycle permanently active. Mutation Y destroys a protein that triggers apoptosis in cells with DNA damage.
| Group | Mutations | Mice with tumours (out of 50) |
|---|---|---|
| 1 | none | 2 |
| 2 | X only | 9 |
| 3 | Y only | 11 |
| 4 | X and Y | 34 |
(a) Calculate the percentage of mice with tumours in each group. (b) A student claims: “Mutation Y alone causes cancer.” Evaluate this claim using the data. (c) If the effects of X and Y simply added together, the expected tumour rate for group 4 would be about 36%. Compare this with the data and explain the difference using your knowledge of cell cycle control.
Model answer and suggested Marlbridge rubric
(a) Group 1: 2 ÷ 50 × 100 = 4%. Group 2: 18%. Group 3: 22%. Group 4: 68%.
(b) The claim is only partly supported. Mutation Y raises the tumour rate from 4% to 22%, so it increases the risk (5.5 times the baseline). But 78% of the group 3 mice did not develop tumours, so Y alone is not enough to cause cancer in every case. Y raises the risk; other changes are usually also needed. (The baseline of 4% also shows tumours can arise without either mutation.)
(c) Expected if additive: 4% + (18 − 4)% + (22 − 4)% = 36%. Observed: 68%, almost twice as high. The two mutations work together. X makes cells divide when they should not; Y stops damaged cells being removed by apoptosis. With both, cells keep dividing and survive with damaged DNA, so mutations build up much faster in the dividing cells. Losing two different controls (the drive to divide and the safety net of apoptosis) has a larger effect than either alone, which fits the idea that cancer usually needs several mutations.
| Point | What earns it |
|---|---|
| 1 | All four percentages correct |
| 1 | Claim evaluated: Y raises the risk (with numbers) but is not sufficient alone |
| 1 | Observed 68% compared with the expected 36% |
| 1 | Explanation: overactive division plus loss of apoptosis lets damaged cells multiply, so mutations accumulate |
Question 7 (calculation · stretch)
A simple model of a tissue: in each cell cycle every cell divides into two, and then a fraction a of the new cells is removed by apoptosis. The tissue starts with 1000 cells.
(a) Show that with a = 0.50 the number of cells stays at 1000. (b) A mutation lowers apoptosis to a = 0.20. Calculate the number of cells after 5 cycles. (c) A different disruption raises apoptosis to a = 0.60. Calculate the number of cells after 5 cycles, and describe the likely effect on the tissue. (d) Use the model to explain why apoptosis is as important as cell division in keeping a tissue a steady size.
Worked solution and suggested Marlbridge rubric
(a) Each cycle multiplies the number of cells by 2 × (1 − a). With a = 0.50: 2 × 0.50 = 1.0, so 1000 × 1.0 = 1000 after every cycle.
(b) Factor = 2 × (1 − 0.20) = 1.6 per cycle. After 5 cycles: 1000 × 1.6⁵ = 1000 × 10.49 = 10 486 cells, about 10 500 (a rise of about 950%).
(c) Factor = 2 × (1 − 0.60) = 0.8 per cycle. After 5 cycles: 1000 × 0.8⁵ = 1000 × 0.328 = 328 cells (327.68). The tissue loses about two-thirds of its cells, so it would shrink and might not be able to do its job.
(d) A steady size needs cell production to be balanced by cell removal. In the model, division doubles the number each cycle, and apoptosis must remove exactly half of the new cells to hold the number constant. A small fall in apoptosis (0.50 to 0.20) produces a tenfold rise in only five cycles, like the growth of a tumour; a small rise (0.50 to 0.60) causes tissue loss.
| Point | What earns it |
|---|---|
| 1 | Factor 2 × (1 − a) shown to be 1.0 for a = 0.50 |
| 1 | About 10 500 cells for a = 0.20 |
| 1 | About 328 cells for a = 0.60, with the effect on the tissue |
| 1 | Balance of division and apoptosis explained, with reference to the model results |
Common error: using 1 − a without the factor of 2 gives 1000 × 0.8⁵ = 328 for a = 0.20, which ignores division.
How did you do?
- Q1 or Q5 wrong: re-read “Checkpoints: stop or go” in the study guide, with Figure 1.
- Q2 or Q4 wrong: re-read “Cyclins and CDKs: the engine of the cycle” and rework Worked example 1.
- Q3, Q6 or Q7 wrong: re-read “Apoptosis: programmed cell death” and “When control fails: cancer”, then rework Worked example 2.
Then tick off the topic checklist.