Before you start
Do the unit diagnostic first. Try each question before opening the answer.
Big ideas of the unit
- Distance decides the route; receptors decide who listens. Cells signal by direct contact, by local regulators or by hormones in the blood, and only cells with a matching receptor respond (Topic 4.1).
- The ligand decides where the receptor is. Large or polar ligands bind surface receptors; small nonpolar ones cross the membrane and bind receptors inside (Topic 4.2).
- Reception, transduction, response. A shape change in the receptor starts a relay of G proteins, second messengers such as cAMP and kinases. Enzyme steps amplify the signal (Topic 4.2).
- Fast responses change existing proteins; slow ones change genes. Responses include secretion, a new phenotype and apoptosis (Topic 4.3).
- Follow the change downstream. A mutation or chemical makes one step more or less active, and every later step follows (Topic 4.3).
- Feedback closes the loop. In negative feedback the response shrinks its own stimulus; in positive feedback it strengthens it until an outside event ends the loop (Topic 4.4).
- Copy, then share. DNA doubles in S phase; mitosis gives each daughter one complete set; cells can leave the cycle into G0 and come back (Topic 4.5).
- Division is a signalled decision. Checkpoints, cyclins and CDKs decide whether a cell moves on. Too little control leads towards cancer; apoptosis removes cells that should not divide (Topic 4.6).
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Receptor location | polar or large ligand → surface; small nonpolar ligand → cytoplasm or nucleus | 4.2 |
| Amplification | multiply the factor at each enzyme step; divide where several molecules activate one | 4.2 |
| Effect of a change | more or less active at that step → same direction at every later step; earlier steps unchanged | 4.3, 4.4, 4.6 |
| Feedback type | response reduces the change → negative; response increases it → positive | 4.4 |
| Mitotic index | cells in mitosis ÷ total cells × 100 | 4.5 |
| Stage length | fraction of cells in a stage × cycle length | 4.5 |
| Chromosome count | count centromeres; DNA doubles in S, chromosome number does not | 4.5 |
| CDK activity | follows cyclin concentration, not CDK amount | 4.6 |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All data, hormones and organisms are fictional.
Question 1 (multiple choice · mixed)
Two fictional hormones act on the same cells. Hormone P raises the activity of an enzyme within 30 seconds, even when transcription is blocked. Hormone S raises the amount of the same enzyme after 3 hours, and has no effect when transcription is blocked. Which pairing is most consistent with the data?
- (A) P: an intracellular receptor that acts as a transcription factor. S: a surface receptor and phosphorylation of existing enzyme.
- (B) Both: surface receptors; S is slower only because it travels further in the blood.
- (C) P: a surface receptor and phosphorylation of existing enzyme. S: an intracellular receptor that acts as a transcription factor.
- (D) Both: intracellular receptors; P is faster because it is a smaller molecule.
Answer and explanation
Answer: (C). P works in seconds without transcription, so it changes existing enzyme, as a kinase cascade from a surface receptor does. S needs transcription and hours, so it changes gene expression, as an intracellular receptor that binds DNA does.
- (A) swaps the two mechanisms.
- (B) ignores the result that S needs transcription.
- (D) cannot explain why P works with transcription blocked.
Question 2 (multiple choice · mixed)
Glucagon raises cAMP in liver cells. A drug slows the enzyme that breaks down cAMP. A person takes the drug, then fasts. Compared with fasting without the drug, which set of changes is most likely?
- (A) cAMP falls, the liver releases less glucose, and the α cells release more glucagon.
- (B) cAMP stays high for longer, but glucose release does not change, because cAMP acts only on the receptor.
- (C) cAMP stays high, the liver releases more glucose, and the α cells release more glucagon, because the loop is positive feedback.
- (D) cAMP stays high for longer, the liver releases more glucose, and the α cells then release less glucagon.
Answer and explanation
Answer: (D). Slower removal keeps the second messenger high, so glycogen breakdown stays on longer and blood glucose rises. Through negative feedback, higher glucose reduces the stimulus to the α cells, so they release less glucagon.
- (A) reverses the effect of slowing cAMP breakdown.
- (B) misplaces cAMP: it acts on protein kinase A, not the receptor.
- (C) gets the cell response right but the loop wrong: a response that removes its own stimulus is negative feedback.
Question 3 (multiple choice · mixed)
In a fictional cell, a growth factor binds a receptor. A relay protein activates a kinase cascade that ends with a transcription factor switching on a G1 cyclin gene. A mutation leaves the relay protein stuck in its active state. Which prediction is best supported?
- (A) The CDK gene is transcribed more, so more CDK is made and the cell divides faster only when the growth factor is present.
- (B) The cyclin is made without the growth factor, its CDK becomes active, and the cell passes the G1 checkpoint and divides without a signal.
- (C) The cell cannot divide, because the receptor no longer binds the growth factor.
- (D) The extra cyclin triggers apoptosis, so the mutation protects against cancer.
Answer and explanation
Answer: (B). A relay protein stuck on switches on everything downstream, ending with the cyclin gene. The cyclin activates its CDK, which drives the cell past the G1 checkpoint with no growth factor.
- (A) confuses cyclin with CDK. CDK amount stays roughly constant; the pathway controls the cyclin.
- (C) treats a change downstream as if it affected the receptor, which is upstream.
- (D) is not supported: nothing described leads to apoptosis, and uncontrolled division points towards cancer.
Question 4 (constructed response · mixed)
In a fictional mammal, gland cells sense blood solute Z (set point 2.50 mmol L⁻¹). When Z falls, they release peptide hormone H. H binds a G protein-coupled receptor on kidney cells, cAMP rises, and the kidney returns more Z to the blood. After a sudden fall in Z:
| Time / min | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| Z / mmol L⁻¹ | 2.00 | 2.15 | 2.32 | 2.44 | 2.49 |
| H / pmol L⁻¹ | 60 | 48 | 30 | 15 | 10 |
(a) H is a peptide. Explain where its receptor is, and why liver cells exposed to H ignore it. (b) Classify each of these as reception, transduction or response: H binding the receptor; cAMP rising; the kidney returning Z to the blood. (c) Calculate the percentage of the fall in Z that is reversed by 60 min, and use the data to show that this loop is negative feedback. (d) A mutant’s kidney receptor binds H normally, but its intracellular domain cannot activate the G protein. Predict Z and H in the mutant after a fall in Z, and explain.
Worked solution and suggested Marlbridge rubric
(a) H is large and polar, so it cannot cross the membrane: its receptor is on the cell surface. Liver cells lack the receptor, so they ignore H.
(b) Binding: reception. cAMP rising: transduction (cAMP is a second messenger). Returning Z: response.
(c) Fall = 2.50 − 2.00 = 0.50. Reversed by 60 min = 2.44 − 2.00 = 0.44. 0.44 ÷ 0.50 × 100 = 88%. As Z returns towards the set point, H falls from 60 to 15 pmol L⁻¹ by 60 min (and 10 by 80 min). The response (more Z) removes the stimulus (low Z), so less H is released: negative feedback.
(d) The signal stops after binding: no G protein, no cAMP rise, no extra return of Z. Z stays low, near 2.00 mmol L⁻¹. The stimulus never goes away, so the gland keeps releasing H: H stays high or rises.
| Point | What earns it |
|---|---|
| 1 | Surface receptor (polar H); liver cells lack the receptor |
| 1 | All three stages classified correctly |
| 1 | 88% |
| 1 | H falls as Z rises, linked to the response removing the stimulus |
| 1 | Z stays low because transduction fails after binding |
| 1 | H stays high, because the stimulus is never removed |
Total: 6 points.
Question 5 (constructed response · mixed)
Onion root tips are treated for 4 hours with a fictional chemical that stops spindle fibres attaching to centromeres; untreated roots stay in water. 500 cells are counted in each.
| Roots | Interphase | Prophase | Metaphase | Anaphase | Telophase |
|---|---|---|---|---|---|
| Untreated | 440 | 30 | 15 | 10 | 5 |
| Treated | 380 | 28 | 88 | 2 | 2 |
(a) State the purpose of the untreated roots. (b) Calculate the mitotic index of each, and the percentage of dividing cells that are in metaphase. (c) Explain why metaphase cells build up in the treated roots. Name the checkpoint involved. (d) Explain why the number of interphase cells falls. (e) A student claims the chemical blocks DNA replication. Evaluate the claim using the data.
Worked solution and suggested Marlbridge rubric
(a) They are the control: they show the normal spread of stages, so any difference can be linked to the chemical.
(b) Untreated: 30 + 15 + 10 + 5 = 60; 60 ÷ 500 × 100 = 12.0%. Treated: 28 + 88 + 2 + 2 = 120; 24.0%. In metaphase: untreated 15 ÷ 60 = 25.0%; treated 88 ÷ 120 = 73.3%.
(c) The M (spindle) checkpoint delays anaphase until every chromosome is attached to fibres from both poles. With attachment blocked, the check is never passed, so cells that reach metaphase stay there.
(d) Cells in G2 still enter mitosis, but none can finish it, so interphase falls (440 to 380) while mitotic cells double.
(e) Not supported. Twice as many cells as in the control are in mitosis. To get there a cell must finish S phase and pass the G2 checkpoint, so replication happened; the chemical acts later, on the spindle.
| Point | What earns it |
|---|---|
| 1 | Control identified with its purpose |
| 1 | 12.0% and 24.0% |
| 1 | 25.0% and 73.3% |
| 1 | M checkpoint holds cells until all chromosomes attach to both poles |
| 1 | Cells keep entering mitosis but cannot leave it |
| 1 | Claim rejected: cells reached mitosis, so S phase was complete |
Total: 6 points.
Question 6 (constructed response · mixed)
In a fictional fish, immune cells carry a death ligand L fixed in their membrane. When one touches a gut cell with damaged DNA, L binds receptor DR. DR starts an enzyme cascade: one DR activates 4 initiator enzymes, each initiator activates 60 executioner enzymes, and each executioner cuts 150 target proteins per minute. The gut cell dies by apoptosis once 1.8 × 10⁷ proteins have been cut.
(a) Name this type of cell communication, and explain why it suits removing only damaged cells. (b) Calculate the proteins cut per minute by one bound DR, and the time to apoptosis if 25 DR molecules are bound. (c) Explain why apoptosis, rather than the cell bursting, protects the gut lining. (d) A mutant fish has DR with a changed intracellular domain that cannot start the cascade. Predict the long-term effect on the gut, and explain why more L would not help.
Worked solution and suggested Marlbridge rubric
(a) Direct contact through a surface molecule. L is fixed to the immune cell, so only the cell it touches is signalled; healthy neighbours are left alone.
(b) One DR: 4 × 60 × 150 = 36 000 proteins per minute. 25 DR: 25 × 36 000 = 900 000 per minute. Time = 1.8 × 10⁷ ÷ 9.0 × 10⁵ = 20 minutes.
(c) In apoptosis the cell is packaged into membrane-wrapped fragments that other cells engulf. Its contents do not leak, so nearby tissue is not damaged and no inflammation follows.
(d) L still binds, but the signal stops at the receptor: no cascade, no apoptosis. Damaged cells survive and divide, passing on DNA damage. Further mutations build up, raising the chance of a tumour. More L cannot help, because binding is not the faulty step.
| Point | What earns it |
|---|---|
| 1 | Direct contact, with only the touched cell receiving the signal |
| 1 | 36 000 per minute |
| 1 | 20 minutes, with working |
| 1 | Apoptosis packages the cell so contents do not leak |
| 1 | Damaged cells survive and divide, raising tumour risk |
| 1 | More ligand cannot help: the fault is after binding |
Total: 6 points.
Question 7 (constructed response · mixed)
60% of the liver is removed from fictional laboratory animals. The percentage of liver cells in S phase and the liver mass (as a percentage of the original) are measured.
| Day | 0 | 1 | 2 | 3 | 5 | 7 | 10 |
|---|---|---|---|---|---|---|---|
| Cells in S phase / % | 1 | 2 | 25 | 18 | 8 | 2 | 1 |
| Liver mass / % of original | 40 | 42 | 55 | 72 | 90 | 99 | 100 |
(a) Before surgery, most liver cells are in G0. Explain how the data show that G0 cells can re-enter the cycle. (b) Calculate the mean rate of increase in liver mass between day 2 and day 5. (c) A student claims that growth is controlled by negative feedback. Use the data to support the claim. (d) In a mutant, the receptor for the liver’s growth signal is stuck in its active shape. Predict the result, and explain.
Worked solution and suggested Marlbridge rubric
(a) At day 0 only 1% of cells are replicating DNA; by day 2, 25% are, 25 times as many. These cells were not dividing before, so G0 cells have returned to the cycle in response to a cue.
(b) (90 − 55) ÷ (5 − 2) = 35 ÷ 3 = 11.7 percentage points per day.
(c) S-phase cells peak (25%) early, then fall (18, 8, 2, 1%) as mass returns towards 100%. Growth slows as the gap closes and stops at the original mass. The response (more liver) reduces the stimulus (missing liver): negative feedback.
(d) A receptor stuck “on” runs the pathway without the growth signal. Cells keep entering S phase after the liver reaches 100%, so it grows past its original size: a gain-of-function change, like those in cancers.
| Point | What earns it |
|---|---|
| 1 | 1% → 25% in S phase linked to G0 cells re-entering the cycle |
| 1 | 11.7 percentage points per day |
| 1 | S-phase fall matched to mass approaching 100% |
| 1 | Response reduces the stimulus, so negative feedback |
| 1 | Mutant: division continues without the signal, liver overgrows |
Total: 5 points.
How did you do?
Add up your points: 3 for multiple choice and 23 for the constructed responses. Compare it only with your next attempt; it does not predict an exam score.
- Lost points on receptors and pathway steps (Questions 1, 4(a)–(b), 6(a)): reread Topic 4.1 and Topic 4.2.
- Lost points on predictions (Questions 2, 3, 4(d), 6(d), 7(d)): practise “more or less active at this step, so every later step…” from Topic 4.3.
- Lost points on feedback (Questions 4(c), 7(c)): reread Topic 4.4.
- Lost points on the cycle (Questions 5, 7(a)): rework Topic 4.5 and Topic 4.6.
- Quick recheck: retake the unit diagnostic.
Then tick off each topic checklist: