IGCSE Biology Paper 4 (Theory / Extended)

Topic 14: Coordination and Response -- Challenge Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 14. Like a real Cambridge paper it ranges across every sub-topic — 14.1 neurones, the reflex arc and synapses, 14.2 sense organs and the eye, 14.3 hormones, 14.4 homeostasis, and 14.5 tropic responses and auxin — and it mixes them inside single questions. All three Topic 14 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Timing a Nervous System

Total: 12 marks

A class measures the time between a stimulus and a response for four different situations. The mean results are shown in Table 1.1.

situation	response	mean time / s
P	knee jerk when the tendon below the kneecap is tapped	0.05
Q	hand withdrawn from a sharp point	0.08
R	button pressed as soon as a light comes on	0.22
S	button pressed only if the light is red and not if it is green	0.41

(a) [2]

Using Table 1.1, identify which of P, Q, R and S are reflex actions, and state two features of a reflex action.

Model Answer — 1(a)

P and Q are reflex actions [1]

a reflex action is automatic / involuntary and rapid [1]

⚠ If you missed marks here: Most candidates write only “fast” for the second half, and lose the mark for automatic — which is the feature that actually separates a reflex from a very quick voluntary action. Note that R and S both involve a decision, and S involves a choice between two decisions, which is why it is slowest of all; deciding takes time and that time happens in the brain.

1 mark — P and Q are the reflex actions 1 mark — automatic / involuntary and rapid

(b) [4]

Explain the difference in time between situation Q and situation R.

Model Answer — 1(b)

Q is a reflex whose pathway lies within the spinal cord, so the impulse passes from sensory neurone to relay neurone to motor neurone without going to the brain [1]

this is a short pathway crossing only two synapses [1]

in R the impulse must travel to the brain, be processed and travel back out to the effector, which is a much longer pathway with many more synapses [1]

each synapse takes time because neurotransmitter has to be released and diffuse across the gap, so more synapses means a longer response time [1]

⚠ If you missed marks here: “A reflex is faster” restates the data instead of explaining it, and earns nothing. There are four marks here and the last one is the one nearly everybody omits: why a synapse costs time. Note also what the answer must not claim — the impulses themselves do not travel at different speeds in the two situations; it is the route that differs.

1 mark — Q is a spinal reflex, sensory to relay to motor, not involving the brain 1 mark — short pathway with only two synapses 1 mark — R requires impulses to travel to and from the brain, a longer pathway with more synapses 1 mark — each synapse takes time because neurotransmitter must be released and diffuse

A student calculates that an impulse takes 0.0002 s to travel 1 cm along a neurone, and that crossing one synapse takes 0.0005 s. A pathway is 55 cm long and contains 8 synapses. Calculate the total time, showing your working.

Model Answer — 1(c)

travel time = 55 × 0.0002 = 0.0110 s [1]

synapse time = 8 × 0.0005 = 0.0040 s [1]

total = 0.0110 + 0.0040 = 0.0150 s [1]

⚠ If you missed marks here: Write the two components on separate lines. The first two marks are for method, so even an arithmetic slip at the end still scores two if the working is visible — and an answer consisting of a single number with no working scores nothing at all if the number is wrong. Notice what the figures show: eight synapses cost about a third as much time as 55 cm of neurone, which is precisely why a reflex arc keeps the count of synapses low.

1 mark — travel time = 55 × 0.0002 = 0.0110 s 1 mark — synapse time = 8 × 0.0005 = 0.0040 s 1 mark — total = 0.0150 s

(d) [3]

Some invertebrates possess a single giant nerve fibre running the length of the body, with no synapses along it, used only for escape responses. Suggest one advantage and two disadvantages of such an arrangement.

Model Answer — 1(d)

advantage: no time is lost crossing synapses, so the escape response is as fast as possible [1]

disadvantage: without synapses there is nothing to make transmission one-way, so an impulse could travel in either direction [1]

disadvantage: the impulse always produces the same single response, since it cannot be directed to different effectors or combined with information from other neurones [1]

⚠ If you missed marks here: A “suggest” question expects you to apply what you know to something you have never met, and here the reasoning runs backwards from a fact you do know: synapses ensure one-way transmission, so removing them must remove that property. The disadvantage most people miss is the loss of flexibility — junctions are where a nervous system does its choosing, and a wire with no junctions can only ever do one thing.

1 mark — advantage: no time lost crossing synapses, so the response is faster 1 mark — disadvantage: nothing ensures one-way transmission 1 mark — disadvantage: only one fixed response is possible

Question 2 — A Retina Mapped Out

Total: 12 marks

Fig. 2.1 shows the number of rods and cones at different distances from the fovea, measured along the retina of one eye.

(a) [3]

Identify line J and line K, giving a reason for your answer, and state what region M must be.

Model Answer — 2(a)

K is the cones, because the number is highest at the fovea (distance 0) and falls sharply away from it [1]

J is the rods, because there are almost none at the fovea and large numbers further out [1]

M is the blind spot, because both lines fall to zero there — there are no light receptor cells at all [1]

⚠ If you missed marks here: The question said “giving a reason”, so naming the lines without saying what in the graph told you scores half. Region M is the part most candidates skip: a place on the retina where both cell types are absent can only be where the optic nerve leaves. Do not confuse it with the fovea, which is at the other end of the axis and is where the cones are most crowded.

1 mark — K is cones, highest at the fovea and falling away 1 mark — J is rods, almost absent at the fovea and numerous further out 1 mark — M is the blind spot, where there are no receptor cells

(b) [3]

Use Fig. 2.1 to explain why a very faint star can be seen more easily when a person looks slightly to one side of it than when they look directly at it.

Model Answer — 2(b)

looking directly at an object places its image on the fovea, where the graph shows cones are packed and rods are almost absent [1]

cones need bright light and are not stimulated by a very faint star [1]

looking to one side places the image where rods are numerous, and rods have the greater sensitivity, so they are stimulated and the star is detected [1]

⚠ If you missed marks here: The command was “use Fig. 2.1”, so refer to what the lines actually do — an answer that recites rods and cones without touching the graph loses marks even if every sentence is true. And be careful with the sensitivity mark: rods are more sensitive to light, not “better”. They give no colour and less detail, which is the price of that sensitivity.

1 mark — looking directly places the image on the fovea, which is cone-rich and rod-poor 1 mark — cones need bright light and are not stimulated by faint light 1 mark — looking to one side uses rods, which have greater sensitivity

A person walks from a dark cinema into bright sunlight and immediately reads a sign on the far side of the road. Describe the changes in the iris and in the lens, naming the muscles involved in each.

Model Answer — 2(c)

iris: the circular muscles contract and the radial muscles relax [1]

so the pupil diameter decreases, reducing the light reaching the retina and protecting the receptor cells [1]

lens: the ciliary muscle relaxes, so the ring widens and the suspensory ligaments become taut [1]

the lens is pulled thinner and refracts the light less, which is what is needed for a distant object [1]

⚠ If you missed marks here: This question puts the two eye mechanisms side by side on purpose, because candidates who have learned them as one blur produce ciliary muscles in the iris half. Keep the stimuli separate: brightness drives the iris, distance drives the lens. The step that goes wrong most often is that a relaxed ciliary ring is a wide ring, which pulls the ligaments tight — the opposite of what feels mechanically obvious.

1 mark — circular muscles contract, radial muscles relax 1 mark — pupil diameter decreases, protecting the retina 1 mark — ciliary muscle relaxes and suspensory ligaments become taut 1 mark — lens is pulled thinner and refracts light less

(d) [2]

The pupil reflex is a reflex action, yet it takes about 0.30 s — longer than the knee jerk. Suggest why.

Model Answer — 2(d)

the pupil reflex is coordinated in the brain rather than the spinal cord, so its pathway is longer than that of a spinal reflex [1]

the effector is a ring of muscle in the iris, which takes time to contract and to change the pupil diameter measurably [1]

⚠ If you missed marks here: The trap is assuming that “reflex” means “spinal”. It does not: reflex describes how a response is controlled — automatically and rapidly — not where. Plenty of reflexes are coordinated in the brain, including this one and the salivation reflex, and they are still completely involuntary.

1 mark — coordinated in the brain, so a longer pathway than a spinal reflex 1 mark — the iris muscles take time to contract and change the pupil diameter

Question 3 — The Liver Does the Work

Total: 12 marks

Fig. 3.1 shows the blood glucose concentration of a healthy person over one day. Meals were eaten at the times marked with arrows.

(a) [2]

Name the hormone that is secreted in the largest amount at point A and the hormone secreted in the largest amount at point B, and name the gland that secretes both.

Model Answer — 3(a)

A — insulin; B — glucagon [1]

both are secreted by the pancreas [1]

⚠ If you missed marks here: Read the graph rather than the labels: A is a peak, so the concentration is above the set point and must be brought down, which calls for insulin; B is a trough, so glucagon. And check your spelling — glucagon is the hormone from the pancreas, glycogen is the store in the liver, and a mark scheme will not treat one as the other.

1 mark — A: insulin; B: glucagon 1 mark — both secreted by the pancreas

(b) [4]

Explain what happens in the body between point A and point B.

Model Answer — 3(b)

the pancreas detects the high blood glucose concentration and secretes insulin into the blood [1]

insulin is carried to the liver, which it stimulates to convert glucose into glycogen for storage [1]

muscle cells also take up more glucose from the blood [1]

so the concentration falls, and continues to fall past the set point because glucose is also being used in respiration and none is being absorbed from the gut [1]

⚠ If you missed marks here: Two organs, two jobs, and mixing them up is the most expensive error in this sub-topic: the pancreas detects and secretes, the liver converts and stores. Do not write that insulin “converts” glucose — it is a hormone, not an enzyme. And notice the last mark: the line keeps falling below the set point, which needs a reason, and that reason is that respiration goes on using glucose all the time.

1 mark — pancreas detects the rise and secretes insulin into the blood 1 mark — insulin stimulates the liver to convert glucose into glycogen 1 mark — muscle cells take up more glucose 1 mark — the concentration falls, and keeps falling as glucose is used in respiration

Explain what is meant by negative feedback and a set point, using Fig. 3.1 as your example.

Model Answer — 3(c)

the set point is the level the system is controlled around — the dashed line on the graph [1]

a change away from the set point is detected, and triggers a response that reverses the change: a rise above it leads to insulin, a fall below it to glucagon [1]

the level is therefore never perfectly constant but fluctuates around the set point, because a correction cannot begin until a change has already happened [1]

⚠ If you missed marks here: The word that earns the second mark is reverses — “keeps it constant” and “stops the change” are both refused, because neither describes what actually happens. The third mark is the one most candidates never think to write, and it is the one that shows real understanding: a wavy line is what success looks like, and demanding a flat one asks for something impossible.

1 mark — the set point is the level the system is controlled around 1 mark — a change is detected and triggers a response that reverses it 1 mark — the level fluctuates around the set point because correction follows the change

(d) [3]

Outline the treatment of Type 1 diabetes.

Model Answer — 3(d)

insulin is injected, in an amount matched to the meal being eaten [1]

the diet is managed, particularly the amount and timing of carbohydrate [1]

blood glucose concentration is monitored regularly, and exercise is taken into account when deciding the dose [1]

⚠ If you missed marks here: “Outline” means give the components without a long explanation of each, so three short sentences is exactly right. Two claims that lose marks: insulin cannot be given as a tablet, because it is a protein and would be digested into amino acids; and the diet is managed, not carbohydrate-free — the aim is to keep the concentration near its set point, not permanently low.

1 mark — insulin injected in amounts matched to meals 1 mark — diet managed, especially carbohydrate amount and timing 1 mark — blood glucose monitored regularly and exercise allowed for

Question 4 — Two Runners and a Wet Afternoon

Total: 12 marks

Two runners of similar fitness ran the same 10 km route on two different days. Both days had an air temperature of 30 °C. Table 4.1 shows the conditions and the results.

	runner G (dry air)	runner H (humid air)
mass of sweat produced / g	1150	1420
mass of sweat evaporated / g	1090	510
core temperature at finish / °C	38.2	39.6
skin appearance at finish	flushed, slightly damp	flushed, dripping wet

(a) [3]

Describe the differences between the two runners shown in Table 4.1, using figures.

Model Answer — 4(a)

H produced more sweat than G (1420 g against 1150 g) [1]

but H evaporated far less of it (510 g against 1090 g, less than half as much) [1]

H finished with a core temperature 1.4 °C higher than G (39.6 against 38.2 °C) [1]

⚠ If you missed marks here: The result that overturns the expected answer is in row two: H sweated more, not less, which rules out “H did not sweat enough” before you even start part (b). Attach a figure to every statement — “H got hotter” is worth one mark of three, and the difference of 1.4 °C is the number the rest of the question turns on.

1 mark — H produced more sweat (1420 g against 1150 g) 1 mark — H evaporated far less sweat (510 g against 1090 g) 1 mark — H finished 1.4 °C hotter (39.6 against 38.2 °C)

(b) [4]

Explain why runner H became so much hotter than runner G, even though the air temperature was the same on both days.

Model Answer — 4(b)

cooling comes from the evaporation of sweat, not from the sweat itself [1]

as water evaporates it takes heat energy from the skin and from the blood flowing beneath it [1]

humid air already contains a great deal of water vapour, so sweat evaporates much more slowly and runs off the skin instead [1]

H therefore lost far less heat by evaporation despite producing more sweat, so more of the heat released by respiration in the muscles was retained and the core temperature rose further [1]

⚠ If you missed marks here: “Sweat cools you down” is not an explanation, and it is exactly the sentence this question is designed to defeat — both runners sweated, and H sweated more. The mark is evaporation, and the reason it fails in humid air is that the air can take up very little more water vapour. Do not attribute H’s temperature to a failure of vasodilation: both runners finished flushed, which tells you that mechanism was working in both.

1 mark — cooling comes from evaporation, not from sweat itself 1 mark — evaporation takes heat energy from the skin 1 mark — in humid air sweat evaporates more slowly and runs off 1 mark — H lost less heat by evaporation, so the core temperature rose further

Both runners had flushed skin at the finish. Explain what causes this and how it helps.

Model Answer — 4(c)

vasodilation: the arterioles supplying the surface capillaries widen [1]

so more blood flows through the capillaries near the skin surface, which is what makes the skin look red [1]

more heat is therefore lost by radiation from the skin to the surroundings [1]

⚠ If you missed marks here: Two precise points and both are frequently lost. It is the arterioles that dilate, because a capillary wall is one cell thick and contains no muscle; and blood vessels change diameter, never position, so “the vessels move to the surface” is refused. The redness is worth pointing out for what it is: direct visual evidence of the mechanism.

1 mark — vasodilation: arterioles supplying the surface capillaries widen 1 mark — more blood flows through capillaries near the surface, making the skin red 1 mark — more heat is lost by radiation from the skin

(d) [2]

Explain why a core temperature of 41 °C would be dangerous.

Model Answer — 4(d)

enzymes would begin to be denatured — the shape of the active site changes permanently so the substrate no longer fits [1]

every reaction in the body is catalysed by an enzyme, so the reactions of metabolism would slow down or stop [1]

⚠ If you missed marks here: Use denatured, never “killed” — an enzyme is not alive. And a bare statement that enzymes are denatured is only half an answer; the second mark is for the consequence. Note also why heat is worse than cold: being chilly slows enzymes down reversibly, whereas denaturing does not undo itself.

1 mark — enzymes would be denatured, the active site changing shape permanently 1 mark — the reactions of metabolism would slow down or stop

Question 5 — A Chemical Sent to Everybody

Total: 10 marks

Table 5.1 gives some information about four hormones named in the syllabus.

hormone	gland	one effect
adrenaline	…………	…………
…………	pancreas	decreases blood glucose concentration
testosterone	testes	…………
oestrogen	…………	development of female secondary sexual characteristics

(a) [4]

Complete Table 5.1 by giving the four missing entries.

Model Answer — 5(a)

adrenaline: gland = adrenal glands [1]

adrenaline: effect = increased heart rate, or increased breathing rate, or increased pupil diameter, or increased blood glucose concentration [1]

pancreas: hormone = insulin [1]

testosterone: effect = development of male secondary sexual characteristics; oestrogen: gland = ovaries [1]

⚠ If you missed marks here: Four glands and four hormones is a small enough list that there is no excuse for losing marks on it, so write it out from memory once a week until it is automatic. The trap in this table is the pancreas row: it also secretes glucagon, but the effect given — decreases blood glucose — can only be insulin, so read the effect before you fill in the name.

1 mark — adrenaline is secreted by the adrenal glands 1 mark — a correct effect of adrenaline 1 mark — the pancreatic hormone that decreases blood glucose is insulin 1 mark — testosterone: male secondary sexual characteristics; oestrogen: ovaries

(b) [3]

Insulin is carried in the blood to every organ of the body, yet only certain organs respond to it. Explain this, and explain why hormonal control is slower to act than nervous control.

Model Answer — 5(b)

the blood reaches every organ, but only the specific target organs are able to respond to that hormone [1]

hormonal control is slower because the hormone must be carried in the blood from the gland to the target organ, which takes seconds or minutes [1]

whereas a nervous impulse travels electrically along a neurone that already reaches the effector, taking a fraction of a second [1]

⚠ If you missed marks here: The selectivity happens at the receiving end, so any answer suggesting the blood only goes to certain organs misunderstands what a circulatory system does. For the second half, a comparison needs both systems in the same sentence — “hormones are slow” states a fact about one of them and compares nothing.

1 mark — only the specific target organs are able to respond 1 mark — hormones must be carried in the blood, which takes time 1 mark — impulses travel electrically along neurones, taking a fraction of a second

A sprinter waiting on the starting line has a blood glucose concentration above their normal resting value, although they have not eaten for four hours. Suggest an explanation, and state why this is useful.

Model Answer — 5(c)

adrenaline has been secreted by the adrenal glands in anticipation of hard physical effort [1]

one of its effects is to increase the blood glucose concentration [1]

this makes more glucose available to the muscle cells for respiration, releasing energy for contraction [1]

⚠ If you missed marks here: Notice what rules out the obvious answers: the sprinter has not eaten, so glucose is not arriving from the gut, and the concentration is up, which is the opposite of what insulin would do. That leaves adrenaline. This also shows that adrenaline is secreted before the effort, not in response to it — preparation is the whole point of “fight or flight”.

1 mark — adrenaline has been secreted by the adrenal glands 1 mark — adrenaline increases the blood glucose concentration 1 mark — more glucose is available for respiration in the muscles

Question 6 — The Tip That Was Cut Off and Put Back

Total: 12 marks

Fig. 6.1 shows four seedlings, each lit from the left for three days.

seedling	increase in height / mm	angle of bending / degrees
P	30	33
Q	4	0
R	27	29
S	3	0

(a) [4]

State the four points about the role of auxin required by the syllabus.

Model Answer — 6(a)

auxin is made in the shoot tip [1]

it diffuses through the plant from the shoot tip [1]

it is unequally distributed in response to light and gravity [1]

it stimulates cell elongation [1]

⚠ If you missed marks here: Write the four as four short clauses on separate lines so the examiner can see all of them. The one candidates leave out is diffuses, and it is the one this very experiment tests. Do not say auxin is “carried in the phloem” or “in the plant’s blood” — plants have no blood, and borrowing a transport tissue from Topic 8 is an invented mechanism.

1 mark — made in the shoot tip 1 mark — diffuses through the plant from the shoot tip 1 mark — unequally distributed in response to light and gravity 1 mark — stimulates cell elongation

(b) [4]

Explain the results for seedlings Q and R.

Model Answer — 6(b)

in R, agar is permeable, so auxin made in the tip can diffuse through it into the shoot below [1]

more auxin passes down the shaded side, those cells elongate more and the shoot bends towards the light, growing 27 mm and bending 29° — almost the same as P [1]

in Q the plastic is impermeable, so no auxin can reach the cells below [1]

without auxin there is almost no cell elongation, so Q grew only 4 mm and could not bend, since bending requires one side to grow more than the other [1]

⚠ If you missed marks here: Quote the figures — 27 mm against 4 mm is the evidence, and an answer without numbers reads as theory rather than interpretation. The point worth grasping is what the pair of results proves: the tip was physically separated in both, so living contact is not needed, and the only difference is whether a substance could diffuse through. That is the classic evidence for a chemical plant growth substance.

1 mark — agar is permeable, so auxin diffuses through into the shoot 1 mark — more auxin on the shaded side, so R grew 27 mm and bent 29° 1 mark — plastic is impermeable, so no auxin reaches the cells below 1 mark — without auxin there is no elongation, so Q grew 4 mm and could not bend

Explain why seedling S was included in the investigation.

Model Answer — 6(c)

S is a control showing what happens when the tip is absent altogether and no auxin is produced [1]

since Q behaves almost identically to S (4 mm against 3 mm), it shows that the plastic has effectively removed the tip’s influence, rather than the cutting itself being responsible for the result in R [1]

⚠ If you missed marks here: “It is a control” on its own is worth nothing; the mark is for saying what it controls for. Here it does two jobs at once: it shows what a seedling with no tip does, and comparing it with Q shows that the plastic achieved exactly that. Comparing controls with each other, rather than only with the untreated plant, is where the harder marks in experimental questions live.

1 mark — S shows the result when no tip and therefore no auxin is present 1 mark — Q resembles S, showing the plastic removed the tip’s influence

(d) [2]

A student concludes from these results that auxin is destroyed by light on the lit side of the shoot. Evaluate this conclusion.

Model Answer — 6(d)

the conclusion is consistent with the results, since destroying auxin on the lit side would leave more on the shaded side and produce this bending [1]

but it is not shown by them: no auxin was measured, and unequal redistribution to the shaded side would give exactly the same result, so the two explanations cannot be distinguished by this experiment [1]

⚠ If you missed marks here: An evaluation must say what the data do support before it attacks — an answer that only criticises is incomplete. The skill being tested is spotting that a named chemical has been introduced into a conclusion when nothing chemical was measured, and examiners reward that kind of scepticism heavily. If you want to separate the two ideas, compare the total amount of auxin below a lit tip with that below an evenly lit one.

1 mark — the conclusion is consistent with the results 1 mark — but no auxin was measured, and redistribution would give the same result

Question 7 — The Substance That Blocks a Junction

Total: 10 marks

Read all of this information before you begin. Everything that is new is given here.

Where a motor neurone meets a muscle there is a junction that works in the same way as a synapse: an impulse causes vesicles to release a neurotransmitter into a gap, the molecules diffuse across, and they bind with receptor proteins on the muscle, which then contracts.

A plant produces a substance, curare, which binds to those receptor proteins on the muscle without causing contraction, and cannot be removed. An animal given curare becomes completely unable to move, but experiments show that it can still detect touch, light and sound normally.

(a) [4]

Explain why an animal given curare cannot move.

Model Answer — 7(a)

impulses still travel normally along the motor neurone to the junction [1]

neurotransmitter is still released from the vesicles and still diffuses across the gap [1]

but the receptor proteins are occupied by curare, so the neurotransmitter cannot bind to them [1]

the muscle, which is the effector, is therefore never stimulated to contract, so no response can occur [1]

⚠ If you missed marks here: Write the normal chain first and then delete one link — two of the four marks here are for saying what still works. An answer that begins “the impulse cannot get through” skips three stages and shows nothing about which of them failed. Note also that curare does not damage the neurone or the muscle; the failure is entirely at the binding step.

1 mark — impulses still travel along the motor neurone to the junction 1 mark — neurotransmitter is still released and still diffuses across the gap 1 mark — the receptor proteins are blocked, so it cannot bind 1 mark — the muscle, the effector, is not stimulated, so there is no response

(b) [3]

Explain why the animal can still detect touch, light and sound, and state which stages of a reflex arc are still working.

Model Answer — 7(b)

curare acts only at the junction between the motor neurone and the muscle, which is at the end of the arc [1]

the receptors and the sensory neurones are unaffected, so stimuli are still detected and impulses are still produced [1]

the relay and motor neurones also conduct normally; only the effector stage fails, so the animal detects everything and can respond to nothing [1]

⚠ If you missed marks here: Detection and response are separate stages of the same arc, and this question tests whether you can hold them apart. Name the stages rather than describing them vaguely: receptor, sensory neurone, relay neurone, motor neurone, effector — and say which one is broken. It is worth pausing on how unpleasant this combination would be, and it is a fair reminder that the arc is a chain, not a single event.

1 mark — curare acts only at the junction with the muscle, at the end of the arc 1 mark — receptors and sensory neurones are unaffected, so stimuli are still detected 1 mark — relay and motor neurones conduct normally; only the effector stage fails

Suggest why an animal given curare would also be unable to maintain its body temperature in a cold room. Refer to two different responses in your answer.

Model Answer — 7(c)

shivering is rapid contraction of muscles, and curare prevents any muscle from being stimulated to contract, so the animal cannot shiver [1]

so the increase in respiration that would release extra heat energy does not occur [1]

the hair erector muscles also cannot contract, so the hairs cannot be raised and no thicker insulating layer of air is trapped [1]

(vasoconstriction depends on muscle in the walls of the arterioles, so heat conservation by that route may also be impaired)

⚠ If you missed marks here: The question is asking you to notice that most temperature responses are carried out by muscles — shivering, raising the hairs, and narrowing the arterioles are all muscular. The one that is not is sweating, which is a gland, and that is worth saying because it shows you have thought about the exception. Keep the shivering chain complete: contraction needs energy from respiration, and it is respiration that releases the heat, not friction.

1 mark — the animal cannot shiver, because muscles cannot be stimulated to contract 1 mark — so respiration does not increase and no extra heat energy is released 1 mark — the hair erector muscles cannot contract, so no insulating layer of air is trapped

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