Free worked solutions · 2026 Edition

IB Biology C3.1: Integration of body systems worked solutions

Worked solutions to every practice question in C3.1: Integration of body systems of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

SL & HL
27 questions
Full worked solutions
Free, no sign-up

Start with Basic practice All topics

Basic practice

Question A1

[1 mark]

Which statement describes an emergent property of an organism? (A) A property of one cell, magnified by numbers (B) A property arising from the parts interacting (C) A property due to genes that single cells lack (D) A property that appears when every gene is expressed

Show solutionHide solution

B. An emergent property is one that no single part has and that appears because the parts interact. A treats it as a single-cell property that numbers amplify, C credits extra genes and D credits expression of every gene.

Final answer

Question A2

[1 mark]

A hand touches a hot surface and the arm muscle contracts. Which neuron carries the impulse to the muscle? (A) Sensory neuron (B) Interneuron (C) Motor neuron (D) Pain receptor

Show solutionHide solution

C. Motor neurons carry impulses from the CNS to effectors such as muscles. A sensory neuron carries impulses towards the CNS, an interneuron stays inside the CNS and a pain receptor is the start of the sensory pathway.

Final answer

Question A3

[1 mark]

Which statement about the movement of food along the small intestine is correct? (A) Voluntary; cerebral hemispheres coordinate it (B) Involuntary; the enteric nervous system coordinates it (C) Involuntary; hormones coordinate it directly (D) Voluntary; the spinal cord coordinates it

Show solutionHide solution

B. Peristalsis between swallowing and egestion is involuntary and is coordinated by the enteric nervous system in the gut wall. The CNS can modulate it but does not coordinate each wave, and hormones are not the direct controller.

Final answer

Question A4

[1 mark]

Which row correctly pairs a gland with its hormone and the action of that hormone? (A) Adrenal; epinephrine; prepares for activity (B) Pineal; epinephrine; sets the sleep-wake cycle (C) Thyroid; melatonin; sets metabolic rate (D) Adrenal; thyroxine; raises the heart rate

Show solutionHide solution

A. The adrenal glands release epinephrine, which prepares the body for vigorous activity. In B the hormone belongs to the adrenal glands, in C melatonin is from the pineal gland and thyroxine sets metabolic rate, and in D thyroxine is from the thyroid.

Final answer

Question A5

[1 mark]

A patient understands speech but sways and misses when reaching for a cup. Which region is damaged? (A) Cerebral hemispheres (B) Cerebellum (C) Hypothalamus (D) Medulla oblongata

Show solutionHide solution

B. The cerebellum coordinates skeletal muscle contraction and balance. The cerebral hemispheres give understanding and memory, which are intact. The hypothalamus controls hormones and the medulla controls breathing and heart rate.

Final answer

Question A6

[1 mark]

Which action is unconscious and integrated mainly in the spinal cord? (A) Deciding to wave to a friend (B) Pulling a foot off a sharp stone (C) Breathing faster during a run (D) Learning a new word

Show solutionHide solution

B. Pulling the foot from a sharp stone is a reflex integrated in the spinal cord. Waving and learning a word are conscious and involve the cerebral hemispheres. Breathing faster during a run is unconscious, but the medulla integrates it.

Final answer

Question A7

[1 mark]

Blood pressure in the carotid artery rises. Which sequence describes the response? (A) Chemoreceptors, hypothalamus, hormone, faster heart (B) Baroreceptors, medulla, pacemaker, slower heart (C) Baroreceptors, pituitary, epinephrine, slower heart (D) Chemoreceptors, medulla, pacemaker, slower heart

Show solutionHide solution

B. Baroreceptors detect the stretch from higher pressure and signal the medulla, which sends impulses to the sinoatrial node to slow the heart. Chemoreceptors detect chemical change, not pressure, and the medulla does not use a hormone to reach the heart.

Final answer

Question A8

HL[1 mark]

Ripening fruit makes ethylene, which speeds ripening and more ethylene. What is this? (A) Negative feedback; ripening reverses a change (B) Positive feedback; the response amplifies the change (C) Negative feedback; ethylene is a gas (D) No feedback; a gas is not a nerve signal

Show solutionHide solution

B. The response, ripening, increases ethylene production, so the change amplifies itself: positive feedback. Negative feedback would reverse the change, and a gas signal can still be part of feedback.

Final answer

Question A9

HL[1 mark]

A seedling on a window sill bends towards the glass. Which statement explains this? (A) Auxin on the lit side increases cell division (B) Auxin on the shaded side increases cell elongation (C) Cytokinin on the shaded side increases elongation (D) Auxin on the shaded side increases photosynthesis

Show solutionHide solution

B. Auxin accumulates on the shaded side, where it promotes cell elongation, so the shaded side grows longer and the shoot bends towards the light. A puts the auxin on the lit side and calls it division, C names the wrong hormone and D gives auxin a role in photosynthesis.

Final answer

Tutorial practice

Question B1

[5 marks]

During a 400 m race the breathing rate of an athlete rises. Explain how the rate is adjusted.

Show solutionHide solution

Any five of: faster respiration in the muscles produces more CO_2 (1); CO_2 forms carbonic acid, so blood pH falls (1); chemoreceptors in the carotid and aortic walls and the brainstem detect the change (1); impulses go to the breathing centre in the medulla (1); the medulla sends impulses to the diaphragm and external intercostal muscles (1); they contract more often and more strongly, so ventilation rate and depth rise (1); CO_2 is removed in the alveoli and pH returns to normal, which is negative feedback (1).

Final answer

Muscle respiration releases more CO_2, so blood CO_2 rises and pH falls. Chemoreceptors detect this and signal the breathing centre in the medulla. Impulses to the diaphragm and external intercostal muscles make them contract faster and harder. More CO_2 is removed and pH returns to normal: negative feedback.

Question B2

[4 marks]

A student measured the heart rate and stroke volume of a volunteer at three levels of activity. Rest 72 beats/min, 70 mL; walking 96, 80; running 150, 110. (a) Calculate the percentage increase in heart rate from rest to running. [2] (b) Cardiac output is heart rate multiplied by stroke volume. Calculate the cardiac output at rest and when running. [2]

Show solutionHide solution

(a) 150-72=78 beats per minute (1); 78/72\times100\approx108\% (1). (b) Rest: 72\times70=5040\,\mathrm{mL\,min^{-1}} (1); running: 150\times110=16500\,\mathrm{mL\,min^{-1}} (1).

Final answer

(a) 108%. (b) Rest 5040 mL per minute; running 16500 mL per minute.

Question B3

[4 marks]

The table shows illustrative resting energy use by four body parts of an adult. Brain 1.4 kg, 340 kcal/day; heart 0.3 kg, 130; kidneys 0.3 kg, 130; skeletal muscle 28 kg, 370. (a) Calculate the energy use of the brain per kilogram per day. [1] (b) Compare the rate for skeletal muscle with that for the brain. [1] (c) Suggest why the brain uses so much energy per kilogram even when the person is resting. [2]

Show solutionHide solution

(a) 340/1.4\approx243\,\mathrm{kcal\,kg^{-1}\,day^{-1}} (1). (b) Muscle 370/28\approx13.2; 243/13.2\approx18 times (1). (c) Neurons keep their ion gradients with pumps and fire impulses continuously (1); the brain integrates information and controls unconscious processes all the time, even at rest (1).

Final answer

(a) 243 kcal per kg per day. (b) Muscle uses 13.2 kcal per kg per day, so the brain's rate is about 18 times as great. (c) Neurons are active all the time: they keep ion gradients with pumps, send impulses and run synapses, and the brain integrates information continuously.

Question B4

[3 marks]

The diagram shows a transverse section through a nerve. (a) Identify structures P and R. [2] (b) The nerve carries impulses to and from the CNS. State the two types of neuron whose fibres run in it. [1]

Show solutionHide solution

(a) P is the protective sheath around the whole nerve (1); R is an unmyelinated fibre, thinner and without a thick ring of myelin (1). (b) A nerve is a bundle of sensory fibres and motor fibres (1).

Final answer

(a) P is the protective sheath; R is an unmyelinated nerve fibre. (b) Sensory and motor.

Question B5

[4 marks]

The diagram shows the pathway of a pain withdrawal reflex. (a) Identify structures A, C and E. [3] (b) Explain why the hand is pulled away before the person feels pain. [1]

Show solutionHide solution

(a) A: pain receptor (1); C: interneuron (1); E: skeletal muscle, the effector (1). (b) The reflex is integrated in the spinal cord, which saves time, so the withdrawal occurs before the cerebral hemispheres register the pain (1).

Final answer

(a) A is the pain receptor, C the interneuron and E the skeletal muscle, the effector. (b) The impulses pass through the spinal cord only, with one synapse in the cord, so the reflex is complete before impulses reach and are processed by the brain.

Question B6

[3 marks]

A volunteer's blood melatonin was measured through the night on two occasions: after a normal evening, and after two hours of bright screen light that ended at midnight. Normal: 8, 25, 52, 70, 45, 12 pg/mL at 20:00, 22:00, 00:00, 02:00, 04:00, 06:00. Screen light: 7, 12, 30, 62, 50, 15. (a) State the time of the peak under normal conditions. [1] (b) Describe the effect of the screen light on the level at 00:00, with figures. [1] (c) Suggest a consequence of the screen light for the volunteer's sleep. [1]

Show solutionHide solution

(a) 02:00 (1). (b) 52-30=22\,\mathrm{pg\,mL^{-1}} lower, which is 22/52\times100\approx42\% (1). (c) Melatonin helps set the sleep-wake cycle, so the later and lower rise would delay sleep (1).

Final answer

(a) 02:00. (b) With screen light the level at midnight was 30, which is 22 pg per mL or 42% lower than the normal 52. (c) The lower level at midnight would delay the signal for sleep, so the volunteer would fall asleep later.

Question B7

HL[3 marks]

Undifferentiated plant cells were grown on gel with different amounts of two hormones. Medium 1 high auxin low cytokinin gives roots; 2 low auxin high cytokinin gives shoots; 3 equal gives unorganised cells. (a) Name the plant hormone made in root tips that promotes cell division. [1] (b) Deduce what the results show about the two hormones. [2]

Show solutionHide solution

(a) Cytokinin (1). (b) More auxin than cytokinin gave roots only (1); more cytokinin than auxin gave shoots only, and equal amounts gave unorganised cells, so the balance of the two hormones controls whether roots or shoots form (1).

Final answer

(a) Cytokinin. (b) The ratio of the hormones decides the result: more auxin than cytokinin gives roots, more cytokinin than auxin gives shoots, and equal amounts give an unorganised mass of cells.

Question B8

HL[4 marks]

The diagram shows three cells of a shoot. Auxin enters each cell by diffusion. Efflux carriers, shown in colour, occur only in the right-hand membrane of each cell. (a) State the direction of net auxin flow. [1] (b) Explain how the carriers produce a concentration gradient across the tissue. [2] (c) Outline how the gradient changes the growth of the shoot. [1]

Show solutionHide solution

(a) Net flow is to the right (1). (b) Auxin cannot diffuse out of a cell, so it leaves only through the carriers on the right (1); because every cell has carriers on the same side, auxin is carried across the tissue and the concentration on the right becomes higher than on the left (1). (c) Cells on the side with more auxin acidify and loosen their walls and elongate more, so the shoot bends (1).

Final answer

(a) To the right. (b) Auxin cannot diffuse out, so it leaves only where there are carriers; every cell passes it to its right neighbour, so auxin builds up on the right and falls on the left. (c) Auxin loosens walls and cells on the high-auxin side elongate more.

Question B9

HL[6 marks]

Describe how auxin causes a shoot to grow towards light.

Show solutionHide solution

Any six of: light is detected at the tip (1); efflux carriers move to the shaded side of the cells (1); auxin is transported to the shaded side, giving a concentration gradient (1); auxin changes gene expression (1); hydrogen ions are pumped into the cell wall (1); the wall becomes acid and the cross-links between cellulose fibres loosen (1); the cells take up water and elongate (1); the shaded side grows faster, so the shoot bends towards the light (1). Write that the shoot grows or bends, not that it moves.

Final answer

Light from one side is detected at the tip. Efflux carriers move auxin to the shaded side. There auxin causes hydrogen ions to be pumped into the cell walls, which acidify and loosen. The cells elongate more than on the lit side, so the shoot bends towards the light.

Question B10

HL[5 marks]

A student tested whether the shoot tip is needed for phototropism. She used three groups of ten seedlings, all lit from the right: A intact, B tip removed, C tip covered by an opaque cap. After 6 hours the mean angles of curvature were A 38^{\circ}, B 2^{\circ} and C 3^{\circ}. (a) Identify the independent variable. [1] (b) State two variables that should be kept constant. [1] (c) Describe and explain the results. [3]

Show solutionHide solution

(a) The state of the tip: intact, removed or covered (1). (b) Any two of species, age, light intensity, temperature, duration (1). (c) Seedlings with intact tips bent about 38^{\circ}, while those with the tip removed or covered bent almost not at all (1); light is detected at the tip (1); auxin made in or redistributed from the tip to the shaded side causes the growth difference, so without a working tip there is no gradient (1).

Final answer

(a) The state of the tip (intact, removed or covered). (b) Any two: species, age, light intensity, temperature, time. (c) Only the seedlings with intact, uncovered tips bent. The tip detects the light and makes or redistributes auxin; without it no gradient forms.

Advanced practice

Question C1

HL[8 marks]

Maize seedlings were lit from the right for 6 hours. Six seedlings had normal auxin transport; six were treated with an inhibitor of auxin efflux carriers. The angle of curvature of each, in degrees, was measured with a protractor. Angles in degrees, seedlings 1 to 6. Normal: 36, 41, 39, 44, 38, 42. Inhibitor: 12, 9, 14, 10, 8, 13. (a) Calculate the mean angle for each treatment. [2] (b) Calculate the percentage reduction in mean angle caused by the inhibitor. [2] (c) Explain the effect of the inhibitor on the angle. [2] (d) Suggest two ways to improve the reliability of the investigation. [2]

Show solutionHide solution

(a) Control: (36+41+39+44+38+42)/6=240/6=40.0^{\circ} (1); inhibitor: (12+9+14+10+8+13)/6=66/6=11.0^{\circ} (1). (b) Difference =40.0-11.0=29.0^{\circ} (1); reduction =29/40\times100=72.5\% (1). (c) Efflux carriers move auxin to the shaded side (1); the inhibitor blocks them, so little gradient forms and the cells on the two sides elongate by similar amounts (1). (d) Any two of: more seedlings, repeats, measure at several times, a more precise method (1 each).

Final answer

(a) Control mean 40.0^{\circ}; inhibitor mean 11.0^{\circ}. (b) Difference 29^{\circ}, a reduction of 72.5\% (73%). (c) The inhibitor stops carriers moving auxin to the shaded side, so the gradient is smaller and the growth difference is smaller. (d) Any two: more seedlings, repeat the experiment, measure at several times, use a more precise instrument such as measuring from photographs.

Question C2

[8 marks]

A hiker's resting heart rate is 68 beats per minute at sea level. At a camp at 4000 m, where the air holds less oxygen, her resting heart rate is 84 and her breathing rate rises from 14 to 19 per minute. (a) Calculate the percentage increase in resting heart rate at the camp. [1] (b) Explain how the body detects the low oxygen and raises both heart rate and breathing rate. [4] (c) A falling rock frightens her and she runs. Outline the role of epinephrine. [3]

Show solutionHide solution

(a) (84-68)/68\times100=23.5\% (1). (b) Chemoreceptors in the carotid and aortic walls detect the low oxygen (1); impulses pass to the medulla (1); the medulla sends impulses to the diaphragm and external intercostal muscles, so ventilation rises (1) and to the sinoatrial node, so heart rate rises (1). (c) The adrenal glands release epinephrine into the blood (1); the heart beats faster and harder, and the airways widen (1); glycogen is broken down to glucose for the muscles, preparing her for vigorous activity (1).

Final answer

(a) 23.5%. (b) Chemoreceptors in the carotid and aortic walls detect low oxygen (and high CO_2) and send impulses to the medulla, which sends impulses to the diaphragm and intercostal muscles and to the sinoatrial node, raising both rates. (c) The adrenal glands release epinephrine; the blood carries it to the heart and other targets; heart rate and strength rise, airways widen, and glucose is released.

Question C3

[5 marks]

Three patients each have damage confined to one brain region. P cannot stand steadily and her limbs jerk when she reaches, yet she speaks and remembers normally. Q breathes normally and pulls back from a hot object, but cannot form new memories or plan. R cannot breathe unaided and his heart rate no longer rises in exercise. (a) Identify the damaged region in each patient. [3] (b) Explain why Patient Q still pulls his hand from a hot object. [2]

Show solutionHide solution

(a) P: cerebellum, which coordinates skeletal muscle contraction and balance (1); Q: cerebral hemispheres, which give learning, memory and higher-order function (1); R: medulla oblongata, which controls breathing and heart rate (1). (b) Withdrawal from the heat is a reflex, an unconscious response integrated in the spinal cord (1), so it does not need the cerebral hemispheres (1).

Final answer

(a) P: cerebellum. Q: cerebral hemispheres. R: medulla oblongata. (b) The withdrawal is a reflex integrated in the spinal cord, which does not need the cerebral hemispheres.

Question C4

HL[7 marks]

In an experiment on coleoptiles, agar blocks containing different concentrations of auxin were placed on the right-hand side of the cut top of decapitated stumps. The mean angle of curvature of eight stumps was measured after 90 minutes. 0 mg/L: 0 degrees; 0.5: 7; 1.0: 13; 2.0: 21. (a) A block with auxin is placed on the right. State the direction of curvature. [1] (b) Describe the trend in the data. [2] (c) Calculate the percentage increase in angle from 1.0 to 2.0 mg per litre. [1] (d) Evaluate the claim that these data show auxin alone controls phototropism. [3]

Show solutionHide solution

(a) The stump bends to the left, because the cells on the right, below the block, elongate more (1). (b) The angle rises steadily, from 0^{\circ} to 21^{\circ}, as the concentration increases (1) with figures quoted, such as 13^{\circ} at 1.0 and 21^{\circ} at 2.0 (1). (c) (21-13)/13\times100=61.5\%\approx62\% (1). (d) Any three: auxin applied to a cut stump, with no light, does not show how light causes the redistribution (1); only one hormone was tested (1); n=8 is small and the spread is not given (1); other hormones and the lag before the response were not examined (1); coleoptiles are not mature shoots (1).

Final answer

(a) To the left, away from the block. (b) The angle increases as the auxin concentration increases. (c) 62%. (d) The data support auxin as a growth regulator in this tissue but do not prove that auxin alone controls phototropism: the stumps had no light stimulus, only one hormone was tested, n=8, and other hormones and a lag in the response were not examined.

Connected questions

Question X1

Connected questions[6 marks]

C3.1 + B3.1. During hard exercise the breathing rate of a runner rises to three times its resting value. (a) Explain how the rise in breathing rate is brought about. [3] (b) Explain why the faster breathing helps remove CO_2 from the blood. [3]

Show solutionHide solution

(a) More CO_2 and lower pH are detected by chemoreceptors (1); the medulla sends impulses to the diaphragm and external intercostal muscles (1); ventilation rate and depth increase and the change is reversed (1). (b) Ventilation replaces alveolar air with air that has a low CO_2 concentration (1); the concentration gradient between blood and air stays steep (1); CO_2 diffuses faster from blood into the alveoli and is breathed out (1).

Final answer

(a) Muscle respiration raises blood CO_2 and lowers pH; chemoreceptors signal the medulla, which sends impulses to the diaphragm and external intercostal muscles. (b) Faster ventilation brings fresh air with low CO_2 into the alveoli, so CO_2 stays steeply down its gradient from blood to alveolar air; faster blood flow carries CO_2 to the lungs.

Question X2

Connected questions[6 marks]

C3.1 + C2.2. In a pain withdrawal reflex an impulse passes from a sensory neuron to an interneuron across a synapse in the spinal cord. (a) Outline how the impulse crosses the synapse. [3] (b) Explain why a reflex is quicker than a response decided by the cerebral hemispheres. [3]

Show solutionHide solution

(a) The impulse depolarizes the presynaptic membrane and neurotransmitter is released by exocytosis (1); it diffuses across the synaptic cleft and binds to receptors on the postsynaptic membrane (1); this opens ion channels and starts an impulse in the interneuron if the threshold is reached (1). (b) The pathway has only a few synapses, each with a delay (1); myelinated fibres conduct impulses faster by saltatory conduction (1); the cerebral hemispheres are not part of the pathway (1).

Final answer

(a) The impulse arrives at the presynaptic terminal and neurotransmitter is released by exocytosis; it diffuses across the cleft and binds to receptors on the interneuron, which depolarizes and may fire. (b) The reflex arc has few synapses, so synaptic delay is short, and the myelinated fibres speed conduction by saltatory conduction; no processing in the cerebral hemispheres is needed.

Question X3

Connected questions[6 marks]

C3.1 + C1.2. A sprinter's adrenal glands release epinephrine just before the start of a race. (a) Outline how epinephrine prepares the body for vigorous activity. [3] (b) Explain how these changes increase the rate of ATP production in the leg muscle. [3]

Show solutionHide solution

(a) Epinephrine is carried in the blood to the heart (1); the heart beats faster and harder (1); glycogen is broken down to glucose (1). (b) More blood delivers more oxygen and glucose to the muscle (1); aerobic cell respiration in the mitochondria makes ATP faster (1); the ATP supplies energy for muscle contraction (1).

Final answer

(a) Epinephrine from the adrenal glands travels in the blood to the heart and raises heart rate and stroke volume, and the liver breaks glycogen down to glucose. (b) Glucose and oxygen reach the muscle faster; aerobic respiration makes ATP at a higher rate, which powers contraction.

Question X4

Connected questions · HL[6 marks]

C3.1 + D3.3. Ripening fruit and the control of blood glucose both involve feedback. (a) Distinguish between positive and negative feedback, with one example of each. [4] (b) Explain why positive feedback would be unsuitable for controlling blood glucose. [2]

Show solutionHide solution

(a) Negative feedback: a change in a variable is detected and the response reverses it, returning the value to normal, for example blood glucose falls after insulin is released (1)(1); positive feedback: the response amplifies the original change, for example ripening raises ethylene production, which speeds ripening (1)(1). (b) Positive feedback would drive blood glucose further from the normal range (1); the value would never settle, whereas homeostasis needs a stable internal condition (1).

Final answer

(a) Negative feedback: a change triggers a response that reverses it, as when high blood glucose causes insulin release that lowers glucose. Positive feedback: a response amplifies the change, as ripening raises ethylene production. (b) With positive feedback blood glucose would keep rising or falling away from the normal range, with no stable value.

From the book

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

Integration of body systems from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book