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IB Biology D3.3: Homeostasis worked solutions

Worked solutions to every practice question in D3.3: Homeostasis of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which statement describes negative feedback? (A) amplifies the change (B) reverses the change (C) acts only above the set point (D) stops outside normal range
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Negative feedback reverses the change that started it, so the variable returns towards its set point (B). A and D describe responses that move the variable away from, or ignore, the set point. C is wrong because the loop also acts when the variable falls below the set point.

Final answer

B

Question A2

[1 mark]
Which row links the cell type, its hormone and the blood glucose that triggers secretion? (A) alpha, glucagon, high (B) beta, insulin, high (C) beta, glucagon, low (D) alpha, insulin, low
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Beta cells secrete insulin when blood glucose is high (B). Alpha cells secrete glucagon when it is low, so A (alpha, glucagon, high) and D (alpha, insulin) have a wrong pairing, and C pairs beta cells with glucagon.

Final answer

B

Question A3

[1 mark]
A person walks from a cold building into hot sunshine. Which pair of responses increases heat loss? (A) vasodilation, sweating (B) vasoconstriction, hairs raised (C) shivering, sweating (D) thyroxin, vasodilation
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Vasodilation of skin arterioles brings more blood to the surface and sweating lets evaporation remove heat (A). Vasoconstriction, hair erection, shivering and more thyroxin all conserve or generate heat.

Final answer

A

Question A4

[1 mark]
Which sequence shows the hormone route that raises metabolic rate in the cold? (A) hypothalamus, pituitary, thyroid (B) pituitary, hypothalamus, thyroid (C) thyroid, pituitary, hypothalamus (D) hypothalamus, thyroid, pituitary
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The hypothalamus signals the pituitary gland, which releases TSH; TSH makes the thyroid gland secrete thyroxin, which raises the metabolic rate (A). The other orders reverse the direction of control.

Final answer

A

Question A5

[1 mark]
Which statement about type 2 diabetes is correct? (A) Immune attack destroys beta cells. (B) Cells respond poorly to insulin. (C) Insulin is needed at diagnosis. (D) Alpha cells over-secrete glucagon.
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In type 2 diabetes target cells respond poorly to insulin, and obesity is a risk factor (B). Autoimmune destruction of beta cells describes type 1 (A). Type 2 is often first managed by diet and exercise, not insulin (C). Glucagon is not secreted in excess (D).

Final answer

B

Question A6

[1 mark]
Why does uncoupled respiration in brown adipose tissue release heat? (A) Protons speed through ATP synthase. (B) Protons bypass ATP synthase by a channel. (C) Glucose is stored as glycogen. (D) Muscle fibres hydrolyse ATP.
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Protons return across the inner mitochondrial membrane through a channel that bypasses ATP synthase, so the energy of the gradient is released as heat (B). ATP synthase is bypassed, not speeded up (A).

Final answer

B

Question A7

HL[1 mark]
Which part of a nephron lies mainly in the medulla of the kidney? (A) glomerulus (B) Bowman's capsule (C) proximal convoluted tubule (D) loop of Henle
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The loop of Henle dips into the medulla (D). The glomerulus, Bowman's capsule and proximal convoluted tubule are in the cortex.

Final answer

D

Question A8

HL[1 mark]
By which process is glucose reabsorbed from the proximal convoluted tubule? (A) osmosis (B) active transport (C) simple diffusion (D) endocytosis
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Glucose is reabsorbed by active transport, linked to the sodium gradient made by pumps that use ATP (B). The filtrate has the same glucose concentration as plasma, so diffusion and osmosis cannot return it.

Final answer

B

Question A9

HL[1 mark]
What does the ascending limb of the loop of Henle do to make concentrated urine possible? (A) filters plasma (B) reabsorbs glucose (C) pumps sodium ions out (D) secretes ADH
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It pumps sodium ions out into the medulla by active transport, and its wall is impermeable to water. This builds the hypertonic medulla (C). Filtration (A) happens in the glomerulus, and ADH is made in the hypothalamus (D).

Final answer

C

Question A10

HL[1 mark]
A person drinks a large volume of water. What happens at the collecting ducts? (A) More ADH; aquaporins enter the membrane. (B) Less ADH; aquaporins withdrawn into vesicles. (C) More ADH; ducts become less permeable. (D) Less ADH; ducts become more permeable.
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The blood is diluted, less ADH is secreted, and aquaporins are withdrawn into vesicles, so little water is reabsorbed (B).

Final answer

B

Tutorial practice

Question B1

[5 marks]
A healthy student ate a carbohydrate meal at time 0. Blood glucose / \mathrm{mmol\,dm^{-3}} at 0, 30, 60, 90, 120 and 180 min: 5.0, 7.8, 8.4, 6.6, 5.4, 4.9. (a) Identify the 30 minute interval of fastest fall and calculate that rate. [2] (b) Explain the fall in blood glucose over this interval. [2] (c) Outline how glucose is kept from falling too low if the student then fasts. [1]
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(a) The falls are 0.060 (60 to 90 min: 1.8/30), 0.040 (1.2/30) and 0.008\,\mathrm{mmol\,dm^{-3}\,min^{-1}} (0.5/60), so the fastest fall is from 60 to 90 min at 0.060 (1 for the interval, 1 for the rate). (b) Beta cells of the islets secrete insulin in response to the high glucose (1); insulin makes liver and muscle cells take up glucose and store it as glycogen (1). (c) If the person then fasts, low glucose makes alpha cells secrete glucagon, and the liver converts glycogen to glucose and releases it (1).

Final answer

(a) 60 to 90 min; 0.060\,\mathrm{mmol\,dm^{-3}\,min^{-1}}. (b) Beta cells secrete insulin; liver and muscle cells take up glucose and convert it to glycogen. (c) Alpha cells secrete glucagon; the liver converts glycogen to glucose.

Question B2

[6 marks]
A runner's blood temperature rises during a race on a hot day. (a) State the part of the brain that detects the rise. [1] (b) Explain how two responses of the skin lower body temperature. [4] (c) State the type of feedback involved. [1]
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(a) The hypothalamus detects the temperature of the blood (1). (b) Vasodilation: skin arterioles widen (1), so more blood flows near the surface and loses heat by radiation (1). Sweating: sweat glands secrete sweat (1) and its evaporation takes heat from the skin (1). (c) Negative feedback: the response reverses the rise in temperature (1).

Final answer

(a) Hypothalamus. (b) Vasodilation of skin arterioles lets more blood flow near the surface and lose heat; sweat glands secrete sweat, and evaporation removes heat. (c) Negative feedback.

Question B3

[4 marks]
Core temperature / ^\circ\mathrm{C} of a volunteer, in a cold room and then a warm room, every 10 min from 0 to 70: 37.0, 36.5, 36.2, 36.7, 37.0, 37.4, 37.7, 37.0. (a) State the set point and give the evidence for it. [1] (b) Describe the two disturbances, using values. [2] (c) State the type of feedback and justify it. [1]
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(a) The set point is 37.0\,{}^\circC, the value the temperature returns to at 0, 40 and 70 min (1). (b) Below the set point: a fall of 37.0-36.2=0.8\,{}^\circC by 20 min (1); above the set point: a rise to 37.7, which is 0.7\,{}^\circC above it, at 60 min (1). (c) It is negative feedback because each disturbance was followed by a response that reversed it, from below and from above the set point; name one response for each direction, for example vasoconstriction or shivering after the fall and sweating after the rise (1).

Final answer

(a) 37.0\,{}^\circC, the value regained. (b) Fell by 0.8\,{}^\circC to 36.2\,{}^\circC at 20 min, then rose by 0.7\,{}^\circC to 37.7\,{}^\circC at 60 min. (c) Negative feedback; shivering (or vasoconstriction) after the fall and sweating (or vasodilation) after the rise.

Question B4

[5 marks]
Type 1 and type 2 diabetes both raise the blood glucose concentration. (a) Distinguish the cause of type 1 from the cause of type 2 diabetes. [2] (b) State two changes in lifestyle that reduce the risk of type 2 diabetes. [2] (c) Predict the effect on glucose if a person with type 1 diabetes omits insulin at a meal. [1]
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(a) Type 1 is an autoimmune attack that destroys the beta cells, so little or no insulin is made (1); in type 2 the target cells are less responsive to insulin (insulin resistance) (1). (b) Two from: keep a healthy body mass, exercise regularly, eat less sugar and saturated fat (1 each). (c) Without insulin the liver and muscle cells do not take up glucose and store it, so the concentration stays high after the meal (1).

Final answer

(a) Type 1: autoimmune destruction of beta cells, so little insulin. Type 2: target cells respond poorly to insulin. (b) Any two of: a healthy body mass, regular exercise, less sugar and fat in the diet. (c) Glucose stays high.

Question B5

HL[5 marks]
Albumin is a plasma protein with a large molecular size. (a) Define excretion. [1] (b) Explain why albumin is absent from the glomerular filtrate. [2] (c) Distinguish excretion from osmoregulation, with one kidney example of each. [2]
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(a) Excretion is the removal of waste products of metabolism from the body (1). (b) Plasma proteins are too large to pass through the pores of the capillary wall and the basement membrane of the glomerulus (1), so they stay in the blood although small solutes are forced into the capsule by high pressure (1). (c) Excretion removes metabolic waste such as urea, which is filtered and left in the urine; osmoregulation regulates the osmotic concentration of body fluids, such as ADH changing water reabsorption in the collecting duct (1 for each correct distinction with example, maximum 2).

Final answer

(a) The removal of waste products of metabolism from the body. (b) Albumin is too large to pass the capillary pores and basement membrane. (c) Excretion removes metabolic waste, for example urea filtered into the urine; osmoregulation controls the osmotic concentration of the blood, for example ADH adjusting water reabsorption.

Question B6

HL[4 marks]
Urine of a healthy person, patient P and patient Q: protein 0, 2.4, 0\,\mathrm{g\,dm^{-3}}; glucose 0, 0, 12 and urea 250, 245, 240\,\mathrm{mmol\,dm^{-3}}. (a) Identify the patient with a fault in filtration and justify your choice. [2] (b) Suggest why glucose appears in the urine of the other patient. [2]
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(a) Patient P (1): plasma protein, which should stay in the blood, is in the urine at 2.4\,\mathrm{g\,dm^{-3}} against 0, so the filtration barrier of the glomerulus is damaged (1). (b) In patient Q the filtrate carries glucose at the plasma concentration (1); glucose is in the urine because the amount exceeds what the tubule transporters can reabsorb, as when blood glucose is high in untreated diabetes (1).

Final answer

(a) Patient P: protein (2.4 compared with 0) has crossed the filter, so the glomerulus is leaking. (b) Patient Q: glucose reached the filtrate faster than the proximal tubule could reabsorb it, for example because blood glucose was very high.

Question B7

HL[4 marks]
A student eats very salty food and drinks nothing for six hours. (a) Explain the effect on ADH secretion. [2] (b) State the effect on the collecting ducts and on the urine. [1] (c) State what happens to ADH secretion once the blood is back to normal. [1]
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(a) The solute concentration of the blood rises (1); osmoreceptors in the hypothalamus detect it and the pituitary gland secretes more ADH (1). (b) The collecting ducts become more permeable to water, so little, concentrated urine is made (1). (c) Once the blood concentration is back to normal, less ADH is secreted: negative feedback (1).

Final answer

(a) Plasma solute concentration rises; osmoreceptors in the hypothalamus detect it, and the pituitary gland releases more ADH. (b) A small volume of concentrated urine; the collecting duct is more permeable to water. (c) ADH secretion falls.

Question B8

HL[4 marks]
Blood flow / \mathrm{L\,min^{-1}} at rest and in maximal exercise: muscle 1.20 and 15.50; kidneys 1.10 and 0.55; brain 0.75 and 0.75. (a) Calculate the percentage change in kidney blood flow. [1] (b) Explain the benefit of the change in muscle flow. [2] (c) State the change in brain blood flow. [1]
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(a) \dfrac{0.55-1.10}{1.10}\times100=-50\% (1). (b) Wider arterioles bring more blood, so more oxygen and glucose reach the muscle for respiration (1), and carbon dioxide and heat are carried away (1). (c) The brain flow is unchanged at 0.75\,\mathrm{L\,min^{-1}} (1).

Final answer

(a) -50\%. (b) More oxygen and glucose reach the muscle, and carbon dioxide and heat are removed. (c) No change: the brain keeps its supply.

Advanced practice

Question C1

[8 marks]
Adults with a healthy body mass (group A) and adults with obesity (group B) took a glucose tolerance test (n=20 in each group). Means with standard deviations are given. Data: , group A, group B; Fasting glucose / \mathrm{mmol\,dm^{-3}}, 4.9\pm0.4, 5.8\pm0.5; Fasting insulin / \mathrm{mU\,L^{-1}}, 6\pm2, 18\pm5; Glucose at 120 min / \mathrm{mmol\,dm^{-3}}, 5.2\pm0.6, 8.1\pm1.1. (a) Calculate the percentage difference in fasting insulin of group B relative to group A. [1] (b) Compare the glucose concentrations of the two groups 120 min after the glucose drink. [2] (c) Explain what the insulin and glucose data suggest about the response of the target cells in group B. [3] (d) Evaluate the claim that the data prove that obesity causes type 2 diabetes. [2]
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(a) \dfrac{18-6}{6}\times100=+200\% (1). (b) Group B has the higher value at 120 min, 8.1 against 5.2\,\mathrm{mmol\,dm^{-3}}, a difference of 2.9 (1); the standard deviation ranges do not overlap (4.6 to 5.8 and 7.0 to 9.2), so the difference is likely to be real (1). (c) Group B secretes three times as much insulin (1) but cannot lower its glucose as well (1), so its target cells respond poorly to insulin: insulin resistance, the pattern of type 2 diabetes (1). (d) The data show an association between obesity and a weaker response, not proof of cause (1); each group has only 20 people and age, diet and activity were not controlled, so a longer study with measured changes in body mass is needed (1).

Final answer

(a) +200\%. (b) At 120 min group B is 2.9\,\mathrm{mmol\,dm^{-3}} higher (8.1 against 5.2) and the ranges do not overlap (4.6 to 5.8 against 7.0 to 9.2). (c) Group B has three times the insulin yet higher glucose, so the target cells respond less to insulin: insulin resistance, as in type 2 diabetes. (d) Not proved: association only.

Question C2

[6 marks]
A volunteer rested for an hour in a chamber at each air temperature. Data: Air temperature / ^\circ\mathrm{C}, 10, 20, 30, 40; Core temperature / ^\circ\mathrm{C}, 36.9, 37.0, 37.0, 37.3; Skin temperature / ^\circ\mathrm{C}, 29, 32, 34, 36; Oxygen use / \mathrm{mL\,kg^{-1}\,min^{-1}}, 6.8, 3.5, 3.2, 3.4; Sweat rate / \mathrm{g\,h^{-1}}, 0, 5, 40, 380. (a) Compare the change in core temperature with the change in air temperature across the range. [2] (b) Explain the oxygen use at 10\,{}^\circC. [2] (c) Suggest why the sweat rate rises steeply between 30 and 40\,{}^\circC. [2]
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(a) The air temperature rose by 40-10=30\,{}^\circC (1) but the core temperature rose by only 37.3-36.9=0.4\,{}^\circC, so the core is held close to 37\,{}^\circC by negative feedback (1). (b) At 10\,{}^\circC oxygen use is 6.8, about twice the 3.2 at 30\,{}^\circC (1): shivering, more thyroxin and brown fat raise the rate of aerobic respiration, and the heat released replaces the heat lost (1). (c) At 40\,{}^\circC the skin (36\,{}^\circC) is only 4\,{}^\circC cooler than the air, so little heat leaves by radiation, conduction or convection (1); evaporation of sweat, which rises to 380\,\mathrm{g\,h^{-1}}, removes the heat (1).

Final answer

(a) Air rose by 30\,{}^\circC but core temperature by only 0.4\,{}^\circC, so core temperature is held nearly constant. (b) In the cold, shivering and more thyroxin (and brown fat) raise the metabolic rate: more aerobic respiration, so more oxygen is used, and the heat released keeps the body warm. (c) At 40 C skin and air differ by only 4 C, so little heat leaves by radiation or conduction; evaporation of sweat removes heat.

Question C3

HL[8 marks]
A hiker loses 1.5\,\mathrm{dm^3} of water as sweat on a hot day and does not drink. Explain how the kidneys help to restore water balance.
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One mark for each point, maximum 8, in this order: the solute concentration of the blood rises (1); osmoreceptors in the hypothalamus detect the change (1); the pituitary gland releases more ADH (1); ADH binds to receptors on the collecting duct cells and aquaporins move from vesicles into the plasma membrane (1), making the duct more permeable to water (1); the ascending limb of the loop of Henle pumps sodium ions into the medulla by active transport and makes it hypertonic (1); water leaves the collecting duct by osmosis and returns to the blood (1); a small volume of concentrated urine is produced (1); as the blood concentration returns to the set point, ADH secretion falls: negative feedback (1). Credit also: ultrafiltration of plasma in the glomerulus and reabsorption of glucose and some water in the proximal tubule as the first stages (1), and the distinction that this is osmoregulation while urea removal is excretion (1).

Final answer

Eight points from: blood solute concentration rises; osmoreceptors in the hypothalamus detect it; pituitary releases more ADH; ADH moves aquaporins into the collecting duct membrane; duct permeable to water; loop of Henle ascending limb pumps Na+ so the medulla is hypertonic; water leaves the duct by osmosis; small volume of concentrated urine; blood concentration returns to normal; ADH secretion falls (negative feedback); ultrafiltration and reabsorption in the proximal tubule earlier.

Question C4

HL[7 marks]
The maximum urine concentrations of a human and of a desert rodent are compared. Data: , human, desert rodent; Plasma / \mathrm{osmol\,L^{-1}}, 0.30, 0.32; Maximum urine / \mathrm{mosmol\,L^{-1}}, 1200, 5400. (a) Calculate the ratio of maximum urine concentration to plasma concentration for each species. [2] (b) Explain how the loop of Henle makes it possible for urine to be more concentrated than plasma. [3] (c) Suggest the difference in the length of the loop of Henle between the two species and give the reason. [2]
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(a) Convert first: urine =1.20 and 5.40\,\mathrm{osmol\,L^{-1}} (1). Ratios: human \dfrac{1.20}{0.30}=4.0, rodent \dfrac{5.40}{0.32}=16.9 (1). (b) The ascending limb of the loop pumps sodium ions into the medulla by active transport (1) and is impermeable to water, so the medulla becomes hypertonic (1); with ADH the collecting duct is permeable, so water leaves by osmosis into the medulla and the urine left behind is more concentrated than plasma (1). (c) The rodent probably has longer loops of Henle (1), which build a steeper gradient in the medulla and allow more water to be reabsorbed, an advantage where water is scarce (1).

Final answer

(a) Human 4.0; rodent 16.9. (b) The ascending limb pumps sodium ions out and is impermeable to water, so the medulla becomes hypertonic; when the collecting duct is permeable (ADH, aquaporins) water leaves by osmosis, so urine becomes more concentrated than plasma. (c) Longer loop of Henle in the rodent: a steeper gradient, so more water is conserved.

Connected questions

Question X1

Connected questions · HL[6 marks]
D3.3 + B2.1. Water moves from the fluid in a collecting duct, 0.15\,\mathrm{osmol\,L^{-1}}, into the surrounding medulla, 1.20\,\mathrm{osmol\,L^{-1}}. (a) Explain why water crosses the plasma membrane of a collecting duct cell slowly when no ADH is present. [2] (b) Describe how ADH increases the water permeability of the membrane. [2] (c) State the direction of net water movement and name the process. [2]
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(a) Water is polar, so it crosses the hydrophobic fatty acid core of the phospholipid bilayer only slowly (1); without ADH there are few aquaporins in the membrane (1). (b) ADH binds to its receptor and vesicles carrying aquaporins fuse with the plasma membrane (1), adding channel proteins through which water passes rapidly (1). (c) Net movement is from the duct to the medulla (1) by osmosis, from the lower to the higher solute concentration (1).

Final answer

(a) The hydrophobic core of the bilayer is a barrier to polar water molecules, and there are few channels. (b) ADH binds a receptor; vesicles carrying aquaporins fuse with the membrane. (c) From the duct into the medulla, by osmosis, because the medulla has the higher solute concentration (lower water potential).

Question X2

Connected questions[6 marks]
D3.3 + C2.1 + B1.2. Insulin is a polypeptide hormone. (a) Explain why insulin binds to a receptor on the cell surface and does not enter the cell. [2] (b) Outline how fewer functional insulin receptors cause high blood glucose after a meal. [2] (c) Explain why insulin is injected and not taken by mouth. [2]
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(a) Insulin is a large, polar polypeptide, so it cannot pass through the hydrophobic core of the plasma membrane (1) and acts by binding to a receptor on the outer surface (1). (b) With fewer working receptors, the signal is weaker (1), so less glucose is taken up and stored as glycogen and the blood glucose stays high after a meal (1). (c) A polypeptide taken by mouth is digested by proteases in the gut (1) to amino acids, so no active hormone would reach the blood (1).

Final answer

(a) Insulin is large and polar, so it cannot cross the hydrophobic core of the membrane; it binds to a receptor on the surface. (b) Fewer working receptors means the cells take up less glucose and the liver stores less. (c) A polypeptide is digested to amino acids by proteases, so it would be inactive.

Question X3

Connected questions[6 marks]
D3.3 + A1.1 + C1.2. Sweating and brown adipose tissue both help to control body temperature. (a) Explain how evaporation of sweat cools the skin, with reference to hydrogen bonds. [2] (b) State the effect of the uncoupling channel on ATP production and on heat production. [2] (c) Explain why uncoupling raises the rate of oxygen consumption of these cells. [2]
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(a) Water molecules are held by hydrogen bonds (1); evaporation needs energy to break them, taken from the skin, so the skin cools (1). (b) Less ATP is made (1) because protons return through the uncoupling channel, and more of the energy is released as heat (1). (c) The proton gradient is dissipated, so the electron transport chain runs faster to rebuild it (1), and this uses oxygen as the final electron acceptor, so oxygen consumption rises (1).

Final answer

(a) Hydrogen bonds between water molecules must be broken for evaporation; the energy comes from the skin, which cools. (b) ATP synthesis falls; more of the energy is released as heat. (c) The gradient is dissipated, so the electron transport chain runs faster to rebuild it and more oxygen is used.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Homeostasis from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book