IB DP Biology: Homeostasis: feedback, glucose, thermoregulation and the kidney (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover section D3.3 (Homeostasis) of the IB Biology guide, first assessment 2025. Questions 8 to 12 test D3.3.7 to D3.3.11 and are HL only; the rest are SL and HL. Read the study guide first and keep the revision notes beside you. Hub: IB DP Biology. Checklist: printable checklist. To find weak areas first, take the free 10-minute diagnostics.

Questions

1. State two of the homeostatic variables in humans named in the guide, other than body temperature. [2]

2. Explain why negative feedback rather than positive feedback is used to control body temperature. [3]

3. Describe how the body responds to a rise in blood glucose concentration after a meal. [5]

4. A glucose drink was given to a person without diabetes (control) and to a patient. Blood glucose (mmol L⁻¹) was measured:

Time / min	0	30	60	120
Control	4.6	7.1	5.4	4.7
Patient	7.2	10.8	11.5	9.9

(a) Calculate the percentage increase in the patient’s blood glucose from 0 to 60 minutes. [2] (b) Compare the responses of the two people. [2] (c) Suggest one reason why the patient’s glucose stays high. [1]

5. Compare the physiological changes, risk factors and treatment of type 1 and type 2 diabetes. [5]

6. Explain how the body responds to a fall in environmental temperature, including the role of thyroxin. [7]

7. Outline the responses that help a person lose heat on a hot day. [4]

8. (HL only) A kidney forms filtrate at 120 mL min⁻¹ and produces urine at 1.2 mL min⁻¹. (a) Distinguish between excretion and osmoregulation. [2] (b) Calculate the volume of filtrate formed per day in litres, and the percentage of filtrate that is reabsorbed. [3]

9. (HL only) Concentrations (g L⁻¹) in three fluids were:

Substance	Plasma	Filtrate	Urine
Glucose	1.0	1.0	0.0
Protein	70.0	0.0	0.0
Urea	0.30	0.30	21.0

(a) Explain why protein is absent from the filtrate. [2] (b) Explain why glucose is absent from the urine. [2] (c) Calculate the factor by which the urea concentration rises from plasma to urine. [1] (d) Urea and glucose have the same concentration in plasma and filtrate. Explain what this shows about ultrafiltration. [2]

10. (HL only) Explain the role of the loop of Henle in allowing the production of concentrated urine. [4]

11. (HL only) A person drinks a large volume of water after a long period without drinking. Explain the changes that follow in the control of blood osmotic concentration, and the effect on urine. [8]

12. (HL only) At rest, cardiac output is 5.5 L min⁻¹ and in vigorous exercise it is 22 L min⁻¹. Illustrative data on the share of cardiac output received by organs:

Organ	Rest / %	Exercise / %
Skeletal muscle	21	80
Gut	24	4
Brain	13	3.5
Kidneys	20	4

(a) Calculate the blood flow to skeletal muscle in exercise, in L min⁻¹. [1] (b) Calculate the percentage change in blood flow to the gut between rest and exercise. [2] (c) Explain how the change in flow to the muscle is brought about and why it is useful. [3]

Answers

1. Blood pH [1]; blood glucose concentration [1] (blood osmotic concentration is also accepted). [2] Examiner insight: Naming the variable is enough; adding a value is not needed, and “water” or “sugar” alone is too vague to be safe.

2. Negative feedback reverses a change, so a rise in temperature causes cooling responses [1] and a fall causes warming responses, returning the temperature to the set point from either side [1]. Positive feedback would amplify a change and drive the temperature further from the set point [1]. [3] Examiner insight: The strongest answers contrast the two directly. Describing only what negative feedback does leaves the “rather than” part of the question unanswered.

3. The beta cells of the pancreas detect the high glucose [1] and secrete insulin into the blood [1]. Insulin is carried in the plasma to target cells in the liver, muscle and fat [1], which take up glucose [1]. The liver and muscle convert glucose to glycogen; blood glucose falls, so insulin secretion falls (negative feedback) [1]. [5] Examiner insight: Name the cells and the target tissues. “The pancreas makes a hormone” is a description of the organ, not of the mechanism.

4. (a) Increase = 11.5 − 7.2 = 4.3 [1]. 4.3 / 7.2 × 100 = 59.7% [1]. [2] (b) The patient has a higher starting glucose (7.2 against 4.6) and higher values throughout [1]. The control returns almost to its starting value by 120 min (4.7), but the patient is still 2.7 above start (9.9) [1]. [2] (c) Target cells take up little glucose because they respond poorly to insulin, or too little insulin is secreted [1]. [1] Examiner insight: A percentage change needs the working shown; divide by the starting value, not the final one. A comparison needs figures from both rows.

5. Type 1: beta cells are destroyed, so little or no insulin is made [1]. Type 2: target cells respond poorly to insulin [1]. Risk factors: type 1 has a genetic predisposition and is not linked to lifestyle; type 2 is linked to excess body fat, inactivity, diet, age or family history [1]. Treatment: type 1 needs insulin by injection or pump [1]; type 2 may be managed with diet, exercise or medicines, with insulin in some cases [1]. [5] Examiner insight: A “compare” answer must be two-sided in each point. Giving type 1 facts in one paragraph and type 2 facts in another, without stating the contrast, loses the comparison.

6. Peripheral thermoreceptors in the skin detect the fall [1]. They send signals to the hypothalamus, which also monitors blood temperature [1]. The hypothalamus signals the pituitary gland, which stimulates the thyroid to release thyroxin [1]. Thyroxin raises metabolic rate, so more heat is released [1]. Skeletal muscle shivers, releasing heat from respiration [1]. Brown adipose tissue releases heat by uncoupled respiration, and skin arterioles constrict to reduce heat loss [1]. As temperature rises towards the set point, the responses weaken: negative feedback [1]. [7] Examiner insight: Cover receptor, control centre and effectors in sequence. Naming thyroxin without saying what it does to metabolic rate leaves the answer incomplete.

7. Skin arterioles dilate (vasodilation) [1], so extra blood passes close to the skin surface and heat is radiated to the surroundings [1]. Sweat glands secrete sweat [1], and evaporation of sweat removes heat from the skin [1]. [4] Examiner insight: Each response needs its effect stated. “Sweating” alone is a name, not an explanation.

8. (a) Excretion is the removal of metabolic waste from the body [1]. Osmoregulation is the regulation of osmotic concentration [1]. [2] (b) Filtrate = 120 × 1440 = 172 800 mL = 172.8 L per day [1]. Reabsorbed = (120 − 1.2) / 120 × 100 = 99.0% [1]. Both answers carry correct units or a percentage sign [1]. [3] Examiner insight: Convert mL per minute to a daily volume before comparing. State the units with each answer, and take the percentage of the filtrate volume, not of the urine volume.

9. (a) Protein molecules are too large to pass through the basement membrane [1], so they stay in the blood [1]. [2] (b) Glucose passes into the filtrate but is reabsorbed [1] by transport proteins in the cells of the proximal convoluted tubule [1]. [2] (c) 21.0 / 0.30 = 70 [1]. [1] (d) Both are small molecules [1] that pass freely through the capillary wall, basement membrane and podocyte slits, so filtration removes solutes by size, not by whether they are wanted [1]. [2] Examiner insight: For “explain absent from”, name the process and where it happens. “It is reabsorbed” is incomplete without the location. For part (d), link the equal concentrations to molecular size.

10. The ascending limb actively transports sodium ions out into the medulla [1]. This keeps the medulla at a high osmotic concentration [1]. Water leaves the collecting duct by osmosis [1] down this gradient, so the urine becomes concentrated [1]. [4] Examiner insight: The key idea is sodium ions being pumped out of the ascending limb. Writing that the loop “filters” or “reabsorbs glucose” is off target.

11. The water lowers blood osmotic concentration [1]. Osmoreceptors in the hypothalamus detect the fall [1]. The pituitary gland secretes less ADH [1]. Less ADH reaches the collecting duct cells [1]. Aquaporins are moved from the cell membrane into vesicles [1]. The collecting duct becomes less permeable to water [1]. Less water is reabsorbed, so a large volume of dilute urine is produced [1]. Blood osmotic concentration returns to the set point: negative feedback [1]. [8] Examiner insight: The location switch of aquaporins is the guide’s key mechanism. Answers that say “ADH is made” or “aquaporins are destroyed” miss it.

12. (a) 22 × 0.80 = 17.6 L min⁻¹ [1]. [1] (b) Rest: 5.5 × 0.24 = 1.32. Exercise: 22 × 0.04 = 0.88 [1]. Change = (0.88 − 1.32) / 1.32 × 100 = −33.3% (a fall of 33.3%) [1]. [2] (c) Smooth muscle in the muscle arterioles relaxes (vasodilation) [1], while arterioles to organs such as the gut constrict [1]. Muscle receives more oxygen and glucose for respiration and carbon dioxide is removed faster [1]. [3] Examiner insight: Calculate the actual flow from the percentage and the cardiac output. The gut’s share fell, but because cardiac output rose, the fall in flow is smaller than the share suggests.

Where marks are usually lost

• Describing the pancreas as “producing a hormone” without naming beta cells or alpha cells.
• Stating that negative feedback “keeps the variable constant” without the return to the set point.
• Giving type 1 and type 2 diabetes the same cause.
• Listing responses to heat without explaining how they lose heat.
• Writing “brown fat burns fat” instead of uncoupled respiration.
• Calculating percentage change from the final value.
• Giving no units or the wrong units for osmotic concentration and flow rates.
• Confusing filtration, reabsorption and secretion.
• Saying ADH creates or destroys aquaporins instead of moving them.

Next steps

• Revise with the revision notes and re-read the study guide.
• Course hub: IB DP Biology. Checklist: printable checklist.
• Try all free 10-minute diagnostics to find your weakest areas.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Section D3.3, Homeostasis.