A Level · Biology 9700 · Energy & control

Control & coordination: A Level Biology questions with answers

Free Control & coordination questions for A Level Biology (9700), part of Energy & control. Each question has the answer and the marking points an examiner would look for.

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Control & coordination practice questions

Question 1Structured · 10 marksA* challenge
(a)

Explain how a resting potential is maintained in a neurone.

[3]
(b)

Describe the events at a cholinergic synapse that lead to an action potential in the post-synaptic neurone.

[5]
(c)

Explain why conduction is faster in myelinated neurones.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)Na+/K+ pump moves 3 Na+ out, 2 K+ in; membrane more permeable to K+ which diffuses out; inside negative, about −70 mV
sodium–potassium pump 3 out 2 in B1
K+ leak channels B1
inside negative (−70 mV) B1
3
(b)action potential opens voltage-gated Ca2+ channels; Ca2+ enters; vesicles fuse, ACh exocytosis; ACh diffuses and binds receptors; Na+ channels open, depolarisation; acetylcholinesterase breaks down ACh
Ca2+ influx B1
vesicles release acetylcholine by exocytosis B1
ACh binds receptors on post-synaptic membrane B1
Na+ channels open, depolarisation beyond threshold B1
acetylcholinesterase hydrolyses ACh B1
5
(c)saltatory conduction: action potentials only at nodes of Ranvier, jump between nodes
depolarisation only at nodes of Ranvier B1
saltatory conduction B1
2
Question 2Structured · 3 marksA* challenge

Plants respond to their environment using hormones.

Explain the role of abscisic acid in the closing of stomata.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
ABA binds guard cells; K⁺ leave; water leaves; flaccid
ABA binds receptors on guard cells B1
K⁺ ions leave, raising water potential B1
water leaves by osmosis and guard cells become flaccid, closing the stoma B1
3
Question 3Structured · 12 marks

This question is about control and coordination in plants.

(a)

Describe how abscisic acid causes stomata to close during drought.

[4]
(b)

Explain how auxin causes cell elongation in shoots.

[3]
(c)

Explain how gibberellin is involved in the germination of barley seeds.

[3]
(d)

State one similarity and one difference between plant and animal hormones.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)ABA binds receptors on guard cells; Ca²⁺ enter, act as second messenger; K⁺ leave; water potential rises, water leaves, guard cells flaccid
ABA binds to receptors on guard cell membranes B1
Ca²⁺ ions enter the cytoplasm (second messenger) B1
K⁺ channels open and K⁺ ions leave the guard cells B1
water potential rises so water leaves by osmosis, guard cells become flaccid and the stoma closes B1
4
(b)auxin stimulates proton pumps; H⁺ into cell wall, acidifies; expansins loosen cellulose; cell elongates as water enters
auxin binds receptors and stimulates proton pumping into the cell wall B1
low pH activates expansins which loosen bonds between cellulose microfibrils B1
the cell takes up water by osmosis and elongates B1
3
(c)gibberellin made by embryo; stimulates aleurone layer to make amylase; starch → maltose for respiration and growth
water absorbed, embryo produces gibberellin B1
gibberellin causes the aleurone layer to synthesise amylase B1
amylase breaks down starch in the endosperm to maltose / glucose for respiration of the embryo B1
3
(d)both chemical messengers effective in low concentration; plant hormones not made in glands / move by diffusion or phloem
both chemical messengers effective at low concentration B1
plant hormones are not made in specialised endocrine glands / travel by diffusion, active transport or phloem B1
2
Question 4Structured · 9 marks

This question is about synapses and drugs.

(a)

Explain why synapses ensure one-way transmission of impulses.

[2]
(b)

Explain how summation at synapses allows integration of information.

[3]
(c)

Nicotine has a similar shape to acetylcholine. Suggest its effect at cholinergic synapses.

[2]
(d)

Some insecticides inhibit acetylcholinesterase. Explain why this kills insects.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)neurotransmitter vesicles only presynaptic; receptors only postsynaptic
vesicles of neurotransmitter are only in the presynaptic knob B1
receptors are only on the postsynaptic membrane B1
2
(b)temporal: repeated impulses add; spatial: several neurones add; excitatory and inhibitory inputs combine to decide threshold
several impulses arriving in quick succession (temporal summation) or from several neurones (spatial summation) add up B1
to reach the threshold in the postsynaptic neurone B1
excitatory and inhibitory synapses combine so the neurone integrates inputs B1
3
(c)binds ACh receptors, causes depolarisation; not broken down by acetylcholinesterase
binds to acetylcholine receptors and opens Na⁺ channels, depolarising the postsynaptic membrane B1
it is not broken down by acetylcholinesterase so the effect is prolonged B1
2
(d)ACh not broken down; continuous stimulation, muscles cannot relax (paralysis)
acetylcholine is not hydrolysed so it stays bound to receptors B1
postsynaptic neurones / muscles are continuously stimulated, causing paralysis B1
2
Question 5Structured · 14 marks

In an investigation, a nerve impulse travels 80 cm along a motor neurone in 8 ms.

(a)

Calculate the speed of conduction of the impulse, in m s⁻¹.

[2]
(b)

Use your answer to (a) to suggest whether this neurone is myelinated. Explain your answer.

[2]
(c)

Explain how myelination increases the speed of conduction.

[3]

Fig. 5.1 shows an action potential recorded from this neurone.

12345−100−80−60−40−2002040600time / mspotential difference / mVABCDE
Fig. 5.1
(d)(i)

State the resting potential shown in Fig. 5.1.

[1]
(ii)

Explain the change between C and E.

[3]
(e)

Describe how the resting potential of a neurone is maintained.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)100 m s−1
0.80 m ÷ 0.008 s C1
100 m s−1 A1
2
(b)myelinated
myelinated B1
the speed is high: myelinated neurones conduct at tens of m s−1, unmyelinated at about 0.5–2 m s−1 B1
2
(c)myelin insulates; depolarisation only at nodes; impulse jumps (saltatory)
myelin is an electrical insulator / prevents ion movement B1
depolarisation (action potentials) occur only at the nodes of Ranvier B1
local circuits form between nodes, so the impulse jumps from node to node (saltatory conduction) B1
3
(d)(i)−70 mV
−70 mV B1
1
(ii)Na⁺ channels close; K⁺ channels open, K⁺ out; overshoot / hyperpolarisation
sodium ion channels close B1
voltage-gated potassium ion channels open and K⁺ diffuses out (repolarisation) B1
K⁺ channels are slow to close so the p.d. overshoots below −70 mV (hyperpolarisation) B1
3
(e)Na⁺/K⁺ pump 3 out, 2 in; membrane more permeable to K⁺; K⁺ diffuses out
sodium–potassium pump moves 3 Na⁺ out and 2 K⁺ in using ATP B1
membrane is more permeable to K⁺ (more K⁺ channels open) B1
K⁺ diffuses out, so the inside is negative (about −70 mV) B1
3
Question 6Structured · 14 marks

This question is about nervous coordination.

(a)

Explain how a resting potential is maintained across the membrane of a neurone.

[3]
(b)

Describe the events that produce an action potential.

[4]
(c)

Explain why myelinated neurones conduct impulses faster than unmyelinated neurones.

[2]
(d)

Describe how an impulse is transmitted across a cholinergic synapse.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)Na⁺/K⁺ pump: 3 Na⁺ out, 2 K⁺ in; membrane more permeable to K⁺, K⁺ diffuses out; inside negative −70 mV
sodium–potassium pump actively transports 3 Na⁺ out and 2 K⁺ in B1
the membrane is more permeable to K⁺ so K⁺ diffuses out B1
leaving the inside negative (about −70 mV) B1
3
(b)stimulus opens Na⁺ channels; Na⁺ in, depolarisation to +30 mV; Na⁺ channels close, K⁺ channels open; K⁺ out, repolarisation, hyperpolarisation
a stimulus opens voltage-gated Na⁺ channels B1
Na⁺ diffuse in, depolarising the membrane (to about +30 mV) B1
Na⁺ channels close and voltage-gated K⁺ channels open B1
K⁺ diffuse out (repolarisation), overshooting to hyperpolarisation B1
4
(c)myelin insulates; saltatory conduction between nodes of Ranvier
myelin insulates the axon so depolarisation occurs only at the nodes of Ranvier B1
the impulse jumps from node to node (saltatory conduction) B1
2
(d)impulse opens Ca²⁺ channels; Ca²⁺ in, vesicles fuse, ACh released by exocytosis; diffuses, binds receptors; Na⁺ channels open, depolarisation; acetylcholinesterase breaks down ACh
action potential arrives and voltage-gated Ca²⁺ channels open B1
Ca²⁺ diffuse in, causing vesicles to fuse with the presynaptic membrane and release acetylcholine (exocytosis) B1
ACh diffuses across the cleft and binds to receptors on the postsynaptic membrane B1
Na⁺ channels open and the postsynaptic membrane depolarises B1
acetylcholinesterase hydrolyses ACh so the signal stops B1
5

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in