Cambridge International AS & A Level Biology 9700: Homeostasis (by subtopic) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, topic 14 Homeostasis, outcome by outcome: section 14.1 Homeostasis in mammals (learning outcomes 1 to 11) and section 14.2 Homeostasis in plants (outcomes 1 to 4). It is A Level content, examined on Paper 4. Calculators are allowed. Each question is tagged with the outcomes it tests.

This set differs from the topic-level homeostasis practice questions. Learn it first: homeostasis study guide · Recall: homeostasis revision notes · Course hub: Cambridge A Level Biology · Checklist: 9700 checklist

Questions

1. (14.1.1, 14.1.2) A woman walks from a warm house into air at 2 °C.

(a) Define homeostasis and explain why it matters in mammals. [3] (b) For her response to the cold, name the type of stimulus (internal or external), one type of receptor, the coordination system and one effector. [4]

2. (14.1.3) A man eats a meal very rich in protein. Explain why the concentration of urea in his blood rises over the next few hours. [3]

3. (14.1.4, 14.1.5) Identify:

(a) the tough outer layer of the kidney [1] (b) the kidney region, in pyramids, containing loops of Henle [1] (c) the space where urine collects before leaving the kidney [1] (d) in an electron micrograph, capillaries wrapped by cells with foot-like extensions [1] (e) in a photomicrograph, a cortex tubule of cuboidal cells with a dense brush border [1] (f) a medulla tube receiving filtrate from many nephrons. [1]

4. (14.1.5, 14.1.7) An electron micrograph of a glomerulus is printed at a magnification of ×40 000.

(a) The basement membrane measures 12 mm thick on the micrograph. Calculate its actual thickness in nm. [2] (b) Relate three features of the glomerulus and Bowman’s capsule to the formation of glomerular filtrate. [3]

5. (14.1.6, 14.1.7) A substance blocks the sodium-potassium pumps in the basal membranes of proximal convoluted tubule (PCT) cells. Explain why glucose then appears in the urine. [4]

6. (14.1.6, 14.1.10, 14.1.11) In one person, filtrate forms at 0.12 dm³ min⁻¹. The co-transporter proteins in the PCTs of both kidneys can reabsorb glucose at a maximum rate of 1.2 mmol min⁻¹. Glucose concentration in the filtrate equals that in the blood plasma.

(a) Calculate the blood glucose concentration above which glucose will appear in the urine. [2] (b) After a meal, the person’s blood glucose reaches 16 mmol dm⁻³. Calculate the rate at which glucose is lost in the urine, in mmol min⁻¹. [2] (c) Explain why reabsorption cannot rise above 1.2 mmol min⁻¹. [2]

7. (14.1.8) In one person, the collecting duct cells have no working ADH receptors. The hypothalamus and posterior pituitary gland work normally.

(a) Predict and explain the volume and water potential of this person’s urine. [3] (b) Suggest why the concentration of ADH in this person’s blood is higher than normal. [3]

8. (14.1.9) Glucagon binds to receptors on liver cells.

(a) Describe the events from the binding of glucagon to the activation of protein kinase A. [4] (b) In a model cascade, one active protein kinase A activates 20 molecules of the next enzyme, and each of those activates 30 glycogen phosphorylase molecules. Each glycogen phosphorylase releases 1000 glucose molecules per second. Calculate the glucose molecules released per second. [2]

9. (14.1.2, 14.1.10) A person drinks a glucose solution at time 0.

Time / min	0	30	120
Blood glucose / mmol dm⁻³	4.8	7.6	5.0

(a) Calculate the mean rate of fall in blood glucose between 30 and 120 minutes. Give your answer to 2 significant figures. [2] (b) Explain how insulin brings about this fall, with reference to muscle cells and liver cells. [5]

10. (14.1.11) A biosensor uses glucose oxidase immobilised on an electrode.

(a) Explain why the biosensor gives a reading for glucose but not for fructose. [2] (b) Current is directly proportional to glucose concentration. A standard of 5.0 mmol dm⁻³ gives 0.80 µA. A blood sample gives 1.36 µA. Calculate its glucose concentration. [2] (c) Name the second enzyme on a glucose test strip and state what it produces. [2]

11. (14.2.1, 14.2.2) A plant had 12 hours of light and 12 of dark, then 24 hours of continuous light. In both periods, mean stomatal aperture rose from 2.4 µm at 04:00 to 9.6 µm at 10:00 and fell in the late afternoon.

(a) Calculate the percentage increase in aperture between 04:00 and 10:00. [1] (b) Explain what the continuous-light results suggest about the control of stomatal opening. [2] (c) Explain why opening stomata during the day and closing them at night is an advantage to the plant. [3]

12. (14.2.3, 14.2.4)

(a) Describe the mechanism by which guard cells open a stoma in the light. [4] (b) Explain how the structure of the guard cell wall causes the stoma to open when the cells become turgid. [2] (c) A substance that stops calcium ions entering the cytoplasm is applied to guard cells of open stomata, then abscisic acid (ABA) is added. Predict and explain the effect on the stomata. [3]

Answers

1. (a) Keeping the internal environment (such as core temperature) constant [1], within narrow limits, despite changes inside or outside the body [1]; so enzymes and cells work efficiently near their optimum [1]. [3] (b) External stimulus [1]; thermoreceptors in the skin [1]; nervous system (the hypothalamus) [1]; skeletal muscles (shivering), or arterioles in the skin [1]. [4]

Examiner insight: Each part of (b) is a separate mark; “the brain” or plain “muscles” may cost a mark each.

2. Excess amino acids cannot be stored [1]; in the liver they are deaminated (amino group removed) [1]; the amino groups are converted to urea, which enters the blood [1]. [3]

Examiner insight: “Proteins become urea” skips deamination and the liver, scoring one mark at most.

3. (a) Fibrous capsule [1] (b) Medulla [1] (c) Renal pelvis [1] (d) Glomerulus (the cells are podocytes) [1] (e) Proximal convoluted tubule [1] (f) Collecting duct [1] [6]

Examiner insight: “Ureter” for (c) is not credited; the ureter carries urine away from the kidney.

4. (a) 12 mm = 12 000 µm; 12 000 ÷ 40 000 = 0.3 µm [1] = 300 nm [1]. [2] (b) Afferent arteriole wider than efferent, giving high hydrostatic pressure in the glomerulus [1]; capillary endothelium has pores and the basement membrane acts as the filter, holding back proteins and cells [1]; podocytes leave filtration slits, so filtrate passes easily into the capsule [1]. [3]

Examiner insight: In (a), an answer left in mm or µm, or with no unit, loses the accuracy mark.

5. Na⁺ is no longer pumped out of the cell, so the Na⁺ concentration inside the cell rises [1]; there is no Na⁺ concentration gradient from filtrate into the cell [1]; so co-transporter proteins cannot carry glucose in with Na⁺ [1]; glucose stays in the filtrate and passes into the urine [1]. [4]

Examiner insight: Saying glucose is “actively pumped” loses the co-transport mark; only the Na⁺–K⁺ pump uses ATP directly.

6. (a) 1.2 ÷ 0.12 [1] = 10 mmol dm⁻³ [1]. [2] (b) Filtered = 0.12 × 16 = 1.92 mmol min⁻¹ [1]; lost = 1.92 − 1.2 = 0.72 mmol min⁻¹ [1]. [2] (c) The number of co-transporter proteins is limited [1]; at high glucose concentration they are all in use (saturated) [1]. [2]

Examiner insight: In (b), 16 − 1.2 mixes a concentration with a rate and earns nothing; find the filtered amount first.

7. (a) Large volume of urine [1] with a high (less negative) water potential [1], because no aquaporins are inserted, so little water is reabsorbed by osmosis from the collecting duct [1]. [3] (b) The water potential of the blood stays low [1]; osmoreceptors in the hypothalamus stay stimulated [1]; so the posterior pituitary keeps releasing ADH; negative feedback never switches it off [1]. [3]

Examiner insight: In (b), “the body makes more ADH” earns nothing without the osmoreceptor and negative feedback links.

8. (a) Binding causes a conformational change in the receptor [1]; this activates a G-protein [1]; the G-protein stimulates adenylyl cyclase [1], which converts ATP into cyclic AMP, and cAMP activates protein kinase A [1]. [4] (b) 1 × 20 × 30 = 600 glycogen phosphorylase molecules [1]; 600 × 1000 = 600 000 glucose molecules per second [1]. [2]

Examiner insight: Each link in (a) is a separate mark, so skipping the G-protein drops a mark.

9. (a) (7.6 − 5.0) ÷ 90 [1] = 0.029 mmol dm⁻³ min⁻¹ [1]. [2] (b) β cells detect the rise and secrete insulin [1]; insulin binds to receptors on muscle and liver cells [1]; in muscle cells, vesicles with GLUT4 transporter proteins fuse with the membrane, so more glucose enters by facilitated diffusion [1]; in liver and muscle cells, glycogenesis is activated (glucose → glycogen) [1]; more glucose is respired, so blood glucose falls towards the set point [1]. [5]

Examiner insight: A rate needs a “per time” unit; 0.029 alone, or 2.6, loses the accuracy mark.

10. (a) Glucose oxidase is specific: its active site is complementary only to glucose [1]; fructose cannot bind, so no current is produced [1]. [2] (b) 0.80 ÷ 5.0 = 0.16 µA per mmol dm⁻³ [1]; 1.36 ÷ 0.16 = 8.5 mmol dm⁻³ [1]. [2] (c) Peroxidase [1]; it uses the hydrogen peroxide to oxidise a colourless chromogen to a coloured product [1]. [2]

Examiner insight: In (c), naming glucose oxidase again earns nothing; peroxidase is the second enzyme.

11. (a) (9.6 − 2.4) ÷ 2.4 × 100 = 300% [1]. [1] (b) The 24-hour cycle continued without darkness [1], so the rhythm is controlled inside the plant, not only by light [1]. [2] (c) By day, open stomata let CO₂ diffuse in for photosynthesis [1]; at night there is no photosynthesis, so CO₂ is not needed [1]; closing reduces water loss by transpiration [1]. [3]

Examiner insight: In (c), “to save water” alone earns one mark at most; state the CO₂ and water trade-off.

12. (a) Proton pumps use ATP to pump H⁺ out of the guard cells [1]; K⁺ channels open and K⁺ diffuses in [1]; the water potential of the guard cells falls [1]; water enters by osmosis and the cells become turgid [1]. [4] (b) The wall next to the pore is thicker (less stretchy) than the outer wall [1]; with radial cellulose microfibrils, the cells curve outwards, so the pore widens [1]. [2] (c) The stomata stay open [1]; ABA binds its receptors, but without Ca²⁺ there is no second messenger [1]; K⁺ does not leave, so the guard cells stay turgid [1]. [3]

Examiner insight: In (c), a bare prediction earns one mark; the explanation must name Ca²⁺ as the second messenger.

Where marks are usually lost

• Saying glucose is actively pumped into PCT cells; it enters by co-transport.
• Leaving the G-protein or adenylyl cyclase out of the glucagon chain.
• Saying glucagon acts on muscle cells.
• Calling ABA, not Ca²⁺, the second messenger.

Next steps

• Homeostasis revision notes
• Homeostasis study guide
• Topic-level homeostasis practice questions
• Cambridge A Level Biology hub
• 9700 printable checklist
• Free 9700 A Level diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment – Topic 14, Homeostasis (sections 14.1 Homeostasis in mammals and 14.2 Homeostasis in plants).