Cambridge International AS & A Level Biology 9700: Control and coordination (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 work through Topic 15, Control and coordination, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, one subtopic at a time: section 15.1 (Control and coordination in mammals, outcomes 1–12) and section 15.2 (Control and coordination in plants, outcomes 1–3). This is A Level content, examined in Paper 4, where AS knowledge is also needed. Calculators are allowed.

They differ from the topic-level practice set; related pages are under Next steps.

Questions

1. (15.1, outcomes 1 and 2) After a long run without drinking, a person’s blood becomes more concentrated and ADH is released.

(a) Using ADH, insulin and glucagon as examples, describe three features of the endocrine system. [3] (b) Compare the nervous system and the endocrine system. [3]

2. (15.1, outcomes 3–5) Taste-bud chemoreceptor cells detect sodium ions in salty food.

(a) State two differences in structure and one difference in function between the sensory neurone that carries this information and a motor neurone. [3] (b) State the role of intermediate neurones. [1] (c) Outline the role of a sensory receptor cell. [2] (d) Describe the events that lead from the arrival of sodium ions to an action potential in the sensory neurone. [5]

3. (15.1, outcome 6) An axon’s resting potential is −65 mV. During an action potential it rises to +30 mV, then falls to −80 mV before returning to −65 mV.

(a) Calculate the change in membrane potential from rest to the peak. [1] (b) Explain how the resting potential is maintained. [3] (c) Explain the changes in membrane potential from the start of the action potential until it reaches −80 mV. [4]

4. (15.1, outcomes 6 and 8) The refractory period of a neurone lasts 1.6 ms.

(a) Calculate the maximum frequency of impulses this neurone can carry. [2] (b) Explain how the resting potential is restored during the refractory period, and why this period limits impulse frequency. [3]

5. (15.1, outcome 7) In a myelinated axon the nodes of Ranvier are 1.2 mm apart, and impulses travel at 60 m s⁻¹.

(a) Calculate the time taken for an impulse to pass from one node to the next, in ms. [2] (b) In some diseases the myelin sheath is damaged. Suggest and explain the effect on impulse transmission. [3]

6. (15.1, outcome 9) (a) Describe the structure of a cholinergic synapse. [4] (b) A toxin blocks the voltage-gated calcium ion channels in the presynaptic membrane. Explain why no action potentials are produced in the postsynaptic neurone. [3]

7. (15.1, outcome 10) Describe how an action potential in a motor neurone leads to the release of calcium ions inside a striated muscle fibre. [5]

8. (15.1, outcome 11) An electron micrograph of relaxed striated muscle shows sarcomeres 2.6 µm long, with an A band of 1.6 µm and an H zone of 0.6 µm.

(a) Describe how you would identify one sarcomere, the A band and the I band on the micrograph. [3] (b) The sarcomere contracts to 2.2 µm. Calculate the new length of the H zone. [2] (c) Calculate the length of one thin (actin) filament, from the Z line to its free end. [2]

9. (15.1, outcome 12) (a) Explain the sliding filament model of muscle contraction, including the roles of calcium ions, troponin, tropomyosin and ATP. [6] (b) Suggest why muscles stay contracted if ATP runs out. [2]

10. (15.2, outcome 1) An insect bends the sensitive hairs on one lobe of an open Venus fly trap.

(a) Describe how stimulating the hairs leads to the trap closing. [3] (b) Explain how closure of the trap is achieved. [3]

11. (15.2, outcomes 2 and 3) (a) Explain how auxin causes elongation growth in cells behind a shoot tip. [4] (b) Describe the role of gibberellin in the germination of a barley seed. [5]

Answers

1. (a) Hormones are secreted into the blood by ductless glands: ADH by the posterior pituitary, insulin and glucagon by the islets of Langerhans [1]; they are carried in the blood around the body [1]; each acts only on target cells with specific receptors: ADH on collecting duct cells, insulin and glucagon on liver cells [1]. (b) Nervous: electrical impulses along neurones; endocrine: hormones carried in the blood [1]. Nervous responses are faster and usually short-lived; hormonal responses are slower and often longer-lasting [1]. Nervous signals reach specific effectors; hormones can affect widespread target cells [1]. Examiner insight: “Compare” needs both systems in each point; “the nervous system is fast” alone is not credited.

2. (a) The sensory neurone’s cell body is outside the CNS, part-way along; the motor neurone’s cell body is inside the CNS, at one end [1]. The sensory neurone has a long dendron to its cell body; the motor neurone has many short dendrites [1]. The sensory neurone carries impulses from a receptor to the CNS; the motor neurone carries them from the CNS to an effector [1]. (b) They connect sensory neurones to motor neurones in the CNS [1]. (c) It detects a stimulus and converts it into a receptor potential (a change in membrane potential) [1]; if this reaches threshold it stimulates an action potential in the sensory neurone [1]. (d) Na⁺ diffuse into the chemoreceptor cell through sodium ion channels [1]; the cell depolarises [1]; voltage-gated calcium ion channels open and Ca²⁺ diffuse in [1]; Ca²⁺ cause vesicles to release neurotransmitter onto the sensory neurone [1]; the neurone depolarises and, if threshold is reached, an action potential passes to the brain [1]. Examiner insight: Each step in (d) is a mark; skipping the Ca²⁺ step loses it.

3. (a) 30 − (−65) = 95 mV [1]. (b) Na⁺–K⁺ pumps use ATP to move 3 Na⁺ out per 2 K⁺ in [1]; the membrane is more permeable to K⁺ than Na⁺, so K⁺ diffuse out faster than Na⁺ leak in [1]; large negatively charged proteins stay inside [1]. (c) Voltage-gated Na⁺ channels open and Na⁺ diffuse in: depolarisation [1]; past threshold, more Na⁺ channels open and the inside becomes positive, up to +30 mV [1]; Na⁺ channels close and voltage-gated K⁺ channels open, so K⁺ diffuse out: repolarisation [1]; K⁺ channels close slowly, so too many K⁺ leave and the membrane is hyperpolarised to −80 mV [1]. Examiner insight: Name the channel type and the ion’s direction; in (a), 30 − 65 = −35 scores nothing.

4. (a) 1.6 ms = 0.0016 s; frequency = 1 ÷ 0.0016 [1] = 625 impulses per second (625 Hz) [1]. (b) K⁺ channels close and the Na⁺–K⁺ pump restores the resting distribution of ions [1]; meanwhile the Na⁺ channels cannot open, so a new action potential cannot start [1]; this sets a minimum time between impulses, which limits the maximum frequency and keeps action potentials separate [1]. Examiner insight: 1 ÷ 1.6 = 0.625 (ms not converted) loses the accuracy mark; the method mark can stand.

5. (a) 1.2 mm = 0.0012 m; time = 0.0012 ÷ 60 [1] = 2.0 × 10⁻⁵ s = 0.020 ms [1]. (b) The membrane is no longer insulated between nodes, so ions leak across it [1]; local circuits cannot reach the next node with enough current to bring it to threshold [1]; saltatory conduction is lost, so impulses are slower or fail altogether [1]. Examiner insight: “Impulses go slower” alone earns one mark at most.

6. (a) A presynaptic knob with many mitochondria and vesicles of acetylcholine [1]; voltage-gated calcium ion channels in the presynaptic membrane [1]; a synaptic cleft between the membranes [1]; acetylcholine receptors (ligand-gated Na⁺ channels) on the postsynaptic membrane [1]. Acetylcholinesterase in the cleft can replace any one point. (b) Ca²⁺ cannot enter the presynaptic knob when an action potential arrives [1]; vesicles do not fuse with the presynaptic membrane, so no acetylcholine is released [1]; no ACh binds to postsynaptic receptors, Na⁺ channels stay closed and threshold is not reached [1]. Examiner insight: Always say presynaptic or postsynaptic; “channels in the membrane” is too vague for credit.

7. Acetylcholine is released at the neuromuscular junction and crosses the cleft [1]; it binds to receptors and depolarises the sarcolemma [1]; the depolarisation spreads along the sarcolemma [1] and down the T-tubules into the fibre [1]; the sarcoplasmic reticulum releases Ca²⁺ into the sarcoplasm [1]. Examiner insight: Write “sarcolemma” and “sarcoplasmic reticulum”, not “cell membrane” and “endoplasmic reticulum”.

8. (a) A sarcomere runs from one Z line to the next [1]; the A band is dark, the length of the thick (myosin) filaments [1]; the I band is light, either side of a Z line, with thin (actin) filaments only [1]. (b) The sarcomere shortens by 2.6 − 2.2 = 0.4 µm, and the H zone by the same amount [1]; 0.6 − 0.4 = 0.2 µm [1]. (c) Half the I band = (2.6 − 1.6) ÷ 2 = 0.5 µm; overlap each side = (1.6 − 0.6) ÷ 2 = 0.5 µm [1]; thin filament = 1.0 µm [1]. Examiner insight: Show each subtraction so a method mark survives an error; give units.

9. (a) At rest tropomyosin covers the myosin-binding sites on actin [1]; Ca²⁺ bind to troponin, which changes shape and moves tropomyosin aside [1]; myosin heads bind to actin, forming cross-bridges [1]; the heads tilt (power stroke), releasing ADP and Pi and pulling actin towards the M line [1]; ATP binds to the myosin head, which detaches [1]; the head hydrolyses ATP and returns upright, ready to bind again [1]. (b) Myosin heads cannot detach from actin [1], so cross-bridges stay attached [1]. Examiner insight: ATP binding and ATP hydrolysis earn separate marks; “ATP provides energy” earns neither.

10. (a) Bending a hair opens ion channels in its cells, giving a receptor potential [1]; a second stimulation within a short time is needed [1]; then an action potential spreads across the lobes [1]. (b) Cells in the outer layers of the lobes and midrib pump H⁺ into their cell walls, loosening them [1]; water enters by osmosis and these cells expand rapidly [1]; the lobes flip from convex to concave, shutting the trap [1]. Examiner insight: Repeating the action potential in (b) earns nothing.

11. (a) Auxin binds to receptors in cells of the elongation zone [1]; it stimulates proton pumps to pump H⁺ into the cell wall, lowering its pH [1]; at low pH expansins loosen the links between cellulose microfibrils [1]; water enters by osmosis and turgor stretches the loosened wall, so the cell elongates [1]. (b) The seed absorbs water and the embryo releases gibberellin [1]; gibberellin reaches the aleurone layer [1]; it causes breakdown of DELLA proteins, which normally inhibit a factor that promotes transcription [1]; the amylase gene is transcribed and amylase is secreted into the endosperm [1]; amylase hydrolyses starch to maltose, used by the embryo for respiration and growth [1]. Examiner insight: “Auxin makes cells divide” scores nothing; in (b), gibberellin does not “activate amylase” directly.

Where marks are usually lost

• Saying the inside of the axon becomes “negative” during an action potential, instead of positive at the peak.
• Leaving out voltage-gated calcium ion channels in the taste-bud cell or the presynaptic knob.
• Saying the A band or the filaments shorten.
• Giving ATP only an “energy” role instead of detachment and re-cocking.

Next steps

• Control and coordination revision notes
• Control and coordination study guide
• Topic-level control and coordination practice set
• Cambridge A Level Biology hub
• Printable Biology checklist
• A Level Biology 9700 diagnostic, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, published by Cambridge University Press & Assessment. Topic 15, Control and coordination: sections 15.1 and 15.2.