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. A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data. All data sets are fictional; blood glucose is given in mg dL⁻¹.
Question 1 (multiple choice · foundation)
Which example is positive feedback?
- (A) When blood glucose falls, the pancreas releases glucagon, and the liver releases glucose.
- (B) When body temperature rises, sweating increases, and body temperature falls.
- (C) At a wound, activated platelets release chemicals that activate more platelets, which release more of those chemicals.
- (D) When the product of an enzyme pathway builds up, it inhibits the first enzyme in the pathway.
Answer and explanation
Answer: (C). The response (more activated platelets) increases the stimulus (the chemicals that activate platelets). The change grows with each round until the clot seals the wound.
- (A) The response raises glucose, opposing the fall: negative feedback.
- (B) Sweating lowers temperature, opposing the rise: negative feedback.
- (D) The build-up of product slows its own production: negative feedback at the molecular level.
Question 2 (multiple choice · core)
In a bacterium, the amino acid D is made by the pathway A → B → C → D. D normally binds an allosteric site on enzyme 1, which converts A to B. A mutation changes the allosteric site so that D can no longer bind, but the active site is unchanged. Which outcome is most likely?
- (A) The cell makes D continuously, even when D is plentiful, wasting A and energy.
- (B) The cell can no longer make any D.
- (C) The cell makes D only when D is plentiful.
- (D) Enzyme 1 stops working, so A builds up.
Answer and explanation
Answer: (A). The negative feedback link is broken. Enzyme 1 still works (its active site is unchanged), but it can no longer be slowed by D. So D is made at full rate even when the cell already has plenty.
- (B) would need the active site, or another enzyme, to be damaged. The question says the active site is unchanged.
- (C) reverses the logic: normally plenty of D switches off production.
- (D) confuses the allosteric site with the active site. A would be used up, not build up.
Question 3 (multiple choice · core)
The thyroid gland makes the hormone thyroxine, which contains iodine. Thyroxine inhibits the release of thyroid-stimulating hormone (TSH) from the pituitary gland, and TSH stimulates the thyroid to grow and make thyroxine. A person’s diet contains too little iodine. Which set of changes is expected?
- (A) Thyroxine low, TSH high, thyroid gland enlarges.
- (B) Thyroxine low, TSH low, thyroid gland shrinks.
- (C) Thyroxine high, TSH low, thyroid gland enlarges.
- (D) Thyroxine normal, TSH normal, thyroid gland unchanged.
Answer and explanation
Answer: (A). Without enough iodine, the thyroid cannot make enough thyroxine. Less thyroxine means less inhibition of the pituitary, so TSH rises. TSH keeps stimulating the thyroid, which grows larger, but it still cannot make enough hormone. The loop cannot close, so the signal stays high.
- (B) forgets that thyroxine inhibits TSH release: less thyroxine removes the brake.
- (C) has thyroxine rising, which cannot happen without enough iodine.
- (D) assumes the loop corrects the problem, but the broken link (no iodine) is in the effector, so no amount of TSH can fix it.
Question 4 (data analysis · core)
A volunteer fasted for 16 hours. Blood glucose and the hormone glucagon were measured every 4 hours.
| Hours since last meal | 0 | 4 | 8 | 12 | 16 |
|---|---|---|---|---|---|
| Glucose / mg dL⁻¹ | 95 | 90 | 86 | 84 | 83 |
| Glucagon / pg mL⁻¹ | 60 | 75 | 95 | 110 | 118 |
(a) Calculate the percentage change in glucose and in glucagon over the 16 hours. (b) Describe the relationship between the two variables. (c) Explain how the data show negative feedback. (d) A second person has a mutation that removes the glucagon receptor from liver cells. Predict how this person’s glucose and glucagon would compare with the volunteer’s by 16 hours, and justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) Glucose: (83 − 95) ÷ 95 × 100 = −12.6%, a fall of about 13%. Glucagon: (118 − 60) ÷ 60 × 100 = +97% (almost doubled). (Using the final value as the base gives 14.5% for glucose, which is the wrong base.)
(b) As glucose falls, glucagon rises: they are inversely related. The fall in glucose slows (95 → 90 in the first 4 hours, 84 → 83 in the last 4) while glucagon is high.
(c) The falling glucose is the stimulus. α cells respond by releasing more glucagon. Glucagon makes the liver break down glycogen and release glucose, which opposes the fall. That is why glucose levels off instead of continuing to drop.
(d) Glucose would be lower than 83 mg dL⁻¹, because liver cells cannot detect glucagon, so glycogen is not broken down and the fall is not opposed. Glucagon would be higher than 118 pg mL⁻¹, because glucose stays low, so the α cells keep being stimulated: the stimulus never goes away.
| Point | What earns it |
|---|---|
| 1 | Both percentage changes correct, with sign or direction |
| 1 | Inverse relationship described with data |
| 1 | Glucagon → glycogen breakdown → glucose release opposes the fall |
| 1 | Prediction of lower glucose, linked to no response by liver cells |
| 1 | Prediction of higher glucagon, linked to a stimulus that does not go away |
Question 5 (constructed response · core)
When a baby suckles, nerve signals from the nipple reach the mother’s brain. Her pituitary gland releases oxytocin, which makes milk flow, and prolactin, which makes the breast produce more milk.
(a) Explain why this is described as positive feedback. (b) Explain what ends the loop. (c) A mother regularly expresses milk with a pump between feeds. Predict the effect on her milk supply, and explain. (d) Explain why positive feedback, rather than negative feedback, suits this process.
Model answer and suggested Marlbridge rubric
(a) The response (milk flow and more milk production) leads to more feeding and more suckling, which is the stimulus. The response strengthens the stimulus instead of reducing it, so the process grows.
(b) An outside event: the baby stops feeding (for example, when it is weaned). Without suckling, the nerve signals stop, oxytocin and prolactin release fall and milk production decreases.
(c) Supply should increase. Pumping acts as extra stimulation like suckling, so more prolactin is released and more milk is made.
(d) The aim is not to hold milk at a fixed level but to match supply to the baby’s demand. Positive feedback raises production as feeding increases. Negative feedback would cut production whenever milk was released, the opposite of what is needed.
| Point | What earns it |
|---|---|
| 1 | Response increases the stimulus, so it is positive feedback |
| 1 | Loop ends when suckling stops, with hormone release falling |
| 1 | Prediction of more milk, linked to extra stimulation |
| 1 | Reason positive feedback suits matching supply to demand |
Question 6 (constructed response · stretch)
Researchers measured the ethylene released by tomato fruits picked unripe (green) and kept at 20 °C. One group was treated with a chemical that blocks ethylene receptors. A third group was exposed to ethylene gas for 1 day, then moved to clean air before measurements began.
| Day | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Untreated / nL g⁻¹ h⁻¹ | 0.1 | 0.2 | 1.5 | 8.0 | 9.0 |
| Receptor blocked / nL g⁻¹ h⁻¹ | 0.1 | 0.1 | 0.2 | 0.2 | 0.3 |
| Pre-exposed to ethylene / nL g⁻¹ h⁻¹ | 0.1 | 3.0 | 9.5 | 10.0 | 9.0 |
(a) Describe the pattern for untreated fruit. (b) Use all three groups to justify the claim that fruit ripening involves positive feedback. (c) Predict what happens to green tomatoes stored in a closed box with one ripe fruit, and explain. (d) Suggest why the ethylene release levels off by day 8.
Model answer and suggested Marlbridge rubric
(a) Ethylene release stays low for 2 days (0.1 to 0.2), then rises steeply to 8.0 by day 6, about 40 times the day 2 value (8.0 ÷ 0.2 = 40), then levels off (9.0 by day 8).
(b) Claim: ethylene causes fruit to make more ethylene, so the change amplifies itself. Evidence: fruit briefly exposed to ethylene began releasing their own ethylene sooner (3.0 at day 2, against 0.2) even after the gas was removed. When fruit cannot detect ethylene, they release almost none (0.3 at day 8). Reasoning: the response (more ethylene) depends on detecting ethylene, and detecting ethylene increases the response. A response that strengthens its own stimulus is positive feedback.
(c) The green tomatoes should ripen faster. Ethylene from the ripe fruit builds up in the closed box, binds receptors in the green fruit, and triggers their own ethylene release, which speeds ripening further for all the fruit.
(d) Any reasonable suggestion, for example: ripening is complete, so the cells stop increasing production; the enzymes that make ethylene are working at their maximum rate; or the supply of the substrate for ethylene is running low. Positive feedback loops do not go on forever.
| Point | What earns it |
|---|---|
| 1 | Pattern: slow start, steep rise, levelling off, with numbers |
| 1 | Evidence from pre-exposed group: ethylene triggers more ethylene |
| 1 | Evidence from receptor-blocked group: detection needed for the rise |
| 1 | Reasoning linking the evidence to positive feedback |
| 1 | Prediction of faster ripening in the box, with mechanism |
Question 7 (calculation · stretch)
In a simple model of negative feedback, blood glucose is 162 mg dL⁻¹ just after a meal, and the set point is 90 mg dL⁻¹. In every 15-minute interval, person R’s feedback removes 40% of the remaining difference from the set point. Person S’s feedback is weaker: it removes only 20% each interval.
(a) Calculate each person’s glucose after 60 minutes. (b) Calculate how long each person takes to get within 10 mg dL⁻¹ of the set point. (c) State one way the model is unrealistic.
Worked solution
(a) Starting difference = 162 − 90 = 72 mg dL⁻¹. 60 min = 4 intervals. R keeps 60% each interval: 72 × 0.6⁴ = 72 × 0.1296 = 9.33, so glucose = 90 + 9.33 = 99 mg dL⁻¹. S keeps 80% each interval: 72 × 0.8⁴ = 72 × 0.4096 = 29.49, so glucose = 90 + 29.49 = 119 mg dL⁻¹.
(b) Need 72 × fⁿ < 10, where f is the fraction kept. R: after 3 intervals the difference is 15.6; after 4 it is 9.33. So 4 intervals = 60 min. S: n > log(10 ÷ 72) ÷ log 0.8 = 8.85, so 9 intervals = 135 min (difference 9.66; after 8 it is still 12.1).
(c) Any one: real glucose keeps rising for a while as food is digested, so it does not start falling at once; the fraction removed changes as insulin levels change; glucose can overshoot below the set point, and glucagon then acts.
Suggested mark points (3): 1 for both glucose values; 1 for 60 min and 135 min with working; 1 for a valid limitation.
Common error: subtracting 40% of the starting difference each time (72 → 43.2 → 14.4 → −14.4) treats the correction as constant. In negative feedback the correction shrinks as the stimulus shrinks.
How did you do?
- Q1 or Q2 wrong: re-read the comparison table and “Negative feedback inside cells” in the study guide.
- Q3 or Q4 incomplete: re-read “When feedback fails” and Worked example 1; always look for the knock-on effect of a loop that cannot close.
- Q5 or Q6 incomplete: re-read “Positive feedback: amplifying a change”, including what ends each loop.
- Q7 wrong: rework Worked example 2; each round multiplies the remaining difference by the same fraction.
Then tick off the topic checklist.