IGCSE Biology Paper 4 (Theory / Extended)

Topic 9: Transport in Animals -- Challenge Exam 3

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 9. Like a real Cambridge paper it ranges across every sub-topic — 9.1 circulatory systems, 9.2 the heart, 9.3 blood vessels and 9.4 blood — and it mixes them inside single questions. All three Topic 9 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Four Vessels, Five Columns of Evidence

Total: 12 marks

Table 1.1 gives measurements made on four blood vessels taken from the same adult human. The vessels are labelled W, X, Y and Z. Oxygen content is given in cm³ of oxygen per 100 cm³ of blood.

Table 1.1	W	X	Y	Z
wall thickness / mm	2.00	0.80	0.001	0.60
lumen diameter / mm	25.0	24.0	0.008	30.0
mean blood pressure / kPa	13.0	3.0	3.0	0.5
valves along the length of the vessel	absent	absent	absent	present
oxygen content / cm³ per 100 cm³	20	14	falls from 20 to 14 along the vessel	14

(a) [4]

Identify each of W, X, Y and Z. Choose from: aorta, capillary, pulmonary artery, pulmonary vein, vena cava.

Model Answer — 1(a)

W — aorta: thickest wall, highest pressure by far, and blood is oxygenated [1]

X — pulmonary artery: thick wall and a pressure well above a vein, no valves, but the blood is deoxygenated [1]

Y — capillary: wall and lumen measured in thousandths of a millimetre, and the only vessel whose oxygen content changes along its length [1]

Z — vena cava: thin wall, the widest lumen, the lowest pressure, and the only vessel with valves [1]

⚠ If you missed marks here: The give-away for Y is not its size — it is the phrase “falls from 20 to 14 along the vessel”. A capillary is the only vessel where anything is unloaded, so it is the only one whose blood is different at the two ends. Every other vessel in the body is a pipe, not an exchange surface.

1 mark — W is the aorta 1 mark — X is the pulmonary artery 1 mark — Y is a capillary 1 mark — Z is the vena cava

(b) [3]

State two pieces of evidence in Table 1.1 that you used to identify X, and explain why a student who looked only at the oxygen content row would give the wrong answer.

Model Answer — 1(b)

evidence 1: the wall is 0.80 mm thick, far thicker than a vein of similar lumen, and the mean pressure is 3.0 kPa, six times that in Z [1]

evidence 2: there are no valves along its length — veins have them, arteries do not [1]

the oxygen content of X is only 14, so a student using that row alone would call it a vein; but a vessel is an artery because it carries blood away from the heart, not because of the oxygen in it. The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs [1]

⚠ If you missed marks here: “Arteries carry oxygenated blood” is the single most expensive sentence in Topic 9. It is a description that happens to be true of most arteries, not a definition. Learn the definition — direction of travel, not cargo — and the pulmonary artery and pulmonary vein stop being exceptions.

1 mark — thick wall and/or the relatively high pressure quoted from the table 1 mark — absence of valves 1 mark — an artery is defined by carrying blood away from the heart, so the pulmonary artery carries deoxygenated blood

Express the wall thickness of W as a percentage of its lumen diameter, and do the same for Z. Show your working. Then explain what the difference between your two answers tells you.

Model Answer — 1(c)

W: 2.00 ÷ 25.0 × 100 = 8.0 %

W = 8.0 % [1]

Z: 0.60 ÷ 30.0 × 100 = 2.0 %

Z = 2.0 % [1]

W has a wall four times thicker in proportion to its lumen. It carries blood at 13.0 kPa in surges straight from the left ventricle, so its thick muscular and elastic wall stops it bursting and its recoil keeps the blood moving between beats. Z carries blood at only 0.5 kPa, so it needs almost no wall — and a thin wall with a wide lumen offers little resistance to blood returning to the heart [1]

⚠ If you missed marks here: Notice that Z has the wider lumen of the two. A common answer says veins have thin walls because the blood has less far to travel; distance has nothing to do with it. Wall thickness tracks pressure, and only pressure.

1 mark — W = 8.0 % with working shown 1 mark — Z = 2.0 % with working shown 1 mark — thicker wall relates to the much higher pressure carried, not to distance

(d) [2]

A red blood cell is a disc about 0.007 mm across. Use this and Table 1.1 to suggest how red blood cells must travel through Y, and give one advantage of this.

Model Answer — 1(d)

the lumen of Y is 0.008 mm and a red blood cell is 0.007 mm across, so the cells can only pass in single file, squeezing through one behind another [1]

every cell is therefore pressed right against the capillary wall, so the distance oxygen has to diffuse from the haemoglobin to the tissue is as short as possible; the tight fit also slows the blood, giving more time for exchange [1]

⚠ If you missed marks here: The word to reach for is diffusion distance. Writing that the capillary is narrow “so the cells go slowly” describes what happens without saying why it helps. Every capillary adaptation in the syllabus — wall one cell thick, huge total surface area, single-file flow — is a different way of making diffusion faster.

1 mark — cells pass in single file because 0.007 mm barely fits a 0.008 mm lumen 1 mark — short diffusion distance to the wall / more time for exchange

Question 2 — Reading a Pressure Profile

Total: 12 marks

Fig. 2.1 shows the blood pressure measured at points along one systemic circuit in a resting adult. Table 2.1 gives, for the same person, the total cross-sectional area of each class of vessel (the areas of all the vessels of that class added together) and the mean velocity of the blood through it.

Table 2.1 — vessel class	total cross-sectional area / cm²	mean velocity of blood / cm per second
aorta	4.5	20
arteries	18	5
arterioles	450	0.2
capillaries	4500	0.02
venae cavae	9	10

(a) [3]

Describe the shape of the curve in Fig. 2.1, quoting values from the graph.

Model Answer — 2(a)

in the arteries the pressure is high and fluctuating, swinging between roughly 10 and 15.5 kPa; each peak is one contraction of the left ventricle and is felt as the pulse [1]

the largest single fall happens across the arterioles, from about 14 kPa down to about 4.5 kPa [1]

through the capillaries and veins the pressure is low and steady — the fluctuations have gone — and it falls below 1 kPa by the time the blood reaches the vena cava [1]

⚠ If you missed marks here: “Describe” on a graph means shape plus numbers, and the disappearance of the pulsing is worth as much as the fall itself. The commonest wrong explanation is that the heart “used up” the pressure. It did not — the pressure was lost as friction against the walls of very narrow vessels.

1 mark — high and fluctuating in the arteries, with values quoted 1 mark — largest fall occurs across the arterioles, with values quoted 1 mark — low, steady and non-pulsating in capillaries and veins, below 1 kPa at the vena cava

(b) [2]

The volume of blood passing a point each second is the total cross-sectional area multiplied by the mean velocity. Use the aorta row of Table 2.1 to calculate the volume of blood leaving the heart in one minute. Give your answer in dm³ and show your working.

Model Answer — 2(b)

4.5 cm² × 20 cm/s = 90 cm³ per second

90 × 60 = 5400 cm³ per minute [1]

5400 ÷ 1000 = 5.4 dm³

5.4 dm³ per minute [1]

check: every other row gives the same figure — 18 × 5, 450 × 0.2, 4500 × 0.02 and 9 × 10 all come to 90 cm³ per second, as they must, since all the blood that leaves the heart has to come back

⚠ If you missed marks here: 1 dm³ = 1000 cm³, so the last step divides. Multiplying instead gives 5 400 000 dm³ a minute, which should stop you dead: a whole adult holds only about 5 dm³ of blood, so 5.4 dm³ per minute means the entire blood volume goes round once a minute at rest. Always ask whether the number is a sensible size for a body.

1 mark — 90 cm³ per second × 60 = 5400 cm³ per minute 1 mark — converted correctly to 5.4 dm³ per minute

Use Table 2.1 to explain why the blood moves so much more slowly in the capillaries than in the aorta. Support your answer with a calculation.

Model Answer — 2(c)

the circulation is a closed loop, so the same volume of blood must pass every class of vessel each second — 90 cm³ per second in this person [1]

4500 ÷ 4.5 = 1000

the capillaries have a total cross-sectional area 1000 times greater than the aorta, because there are so many of them [1]

20 ÷ 1000 = 0.02 cm/s

so the velocity must be 1000 times smaller, which is exactly the figure in the table [1]

⚠ If you missed marks here: The trap is to argue that each capillary is narrow, so the blood must be slow. A narrow pipe on its own makes flow faster, not slower — think of a thumb over a hose. What slows the blood is that the narrow pipes are in parallel, and there are billions of them, so their areas add up to something enormous.

1 mark — the same volume must pass every second, 90 cm³ per second 1 mark — total capillary cross-sectional area is 1000 times the aorta (4500 ÷ 4.5) 1 mark — so velocity is 1000 times lower, 20 ÷ 1000 = 0.02 cm per second

(d) [2]

Explain why slow flow in the capillaries is important.

Model Answer — 2(d)

the blood spends longer inside each capillary — at 0.02 cm per second it takes about a second to cross one [1]

so there is more time for diffusion: oxygen and glucose can pass out to the tissues and carbon dioxide and other waste can pass in, and the exchange can go close to completion before the blood leaves [1]

⚠ If you missed marks here: Diffusion is not instantaneous, and that single fact is why the capillary bed exists. It is also worth noticing that the capillary wall is one cell thick precisely because the pressure there is low — the two adaptations depend on one another.

1 mark — blood spends longer in the capillary 1 mark — more time for diffusion of oxygen and nutrients out and waste in

(e) [2]

Explain why the greatest fall in pressure in Fig. 2.1 happens across the arterioles.

Model Answer — 2(e)

arterioles have very narrow lumens and muscular walls that can narrow them further, so blood meets a large resistance and rubs against a great deal of vessel wall [1]

energy is lost as friction as the blood is forced through, and it is this loss — not the heart running out of push — that lowers the pressure; it also protects the capillaries beyond, whose walls are one cell thick and would be damaged by arterial pressure [1]

⚠ If you missed marks here: If pressure simply ran out with distance, it would fall smoothly all the way round the circuit, and Fig. 2.1 would be a straight slope. It is not — the loss is concentrated in one short stretch, which tells you the cause is local, and the only thing that is special about that stretch is how narrow the vessels are.

1 mark — arterioles have very narrow lumens, giving a large resistance 1 mark — energy lost as friction (and this protects the thin-walled capillaries)

Question 3 — Three Traces, One Cycle

Total: 12 marks

Fig. 3.1 shows the pressure recorded in the left atrium, the left ventricle and the aorta of a resting adult during one complete heartbeat. The lower panel repeats the atrium and ventricle traces on an expanded scale so that the low pressures can be read.

(a) [2]

Use Fig. 3.1 to find the time taken for one complete heartbeat, and calculate the heart rate in beats per minute. Show your working.

Model Answer — 3(a)

one complete cycle takes 0.8 s — the traces at 0.8 s have returned to the values they had at 0 s [1]

60 ÷ 0.8 = 75

75 beats per minute [1]

⚠ If you missed marks here: A cycle is measured between two points that are the same stage, not between the start and the peak. Multiplying by 0.8 instead of dividing gives 48 bpm, which is a trained athlete asleep — another case where checking whether the number is biologically sensible catches the slip.

1 mark — one cycle = 0.8 s read from the graph 1 mark — 60 ÷ 0.8 = 75 beats per minute

(b) [4]

A valve is pushed shut the moment the pressure behind it becomes lower than the pressure in front of it, and pushed open the moment it becomes higher. Use Fig. 3.1 to state the time at which each of the following happens: (i) the atrioventricular valve on the left side shuts; (ii) the semilunar valve at the start of the aorta opens; (iii) that semilunar valve shuts; (iv) the atrioventricular valve opens again.

Model Answer — 3(b)

(i) 0.15 s — in the lower panel the ventricle trace rises through the atrium trace, both at about 0.85 kPa [1]

(ii) 0.25 s — the ventricle trace meets the aorta trace at about 10.5 kPa and carries on rising [1]

(iii) 0.50 s — the falling ventricle trace crosses back below the aorta trace at about 11 kPa [1]

(iv) 0.60 s — in the lower panel the falling ventricle trace crosses back below the atrium trace at about 1.2 kPa [1]

⚠ If you missed marks here: Every answer is a crossing point, never a peak. A very common mistake is to say the semilunar valve opens when the ventricle pressure is at its maximum, at 0.35 s — but blood has been pouring into the aorta for a tenth of a second by then. The valve opened as soon as the ventricle beat the aorta, not when it won by the most.

1 mark — (i) 0.15 s (accept 0.14–0.16) 1 mark — (ii) 0.25 s (accept 0.24–0.26) 1 mark — (iii) 0.50 s (accept 0.49–0.51) 1 mark — (iv) 0.60 s (accept 0.59–0.61)

A doctor listening to this person hears two sounds in every beat. Using your answers to (b), state what makes each sound and at which time in Fig. 3.1 each would be heard.

Model Answer — 3(c)

the first sound (“lub”) is the atrioventricular valves shutting, at 0.15 s [1]

the second sound (“dup”) is the semilunar valves shutting, at 0.50 s [1]

⚠ If you missed marks here: Both sounds are valves shutting; opening a valve is silent. Notice the gap in Fig. 3.1: 0.35 s from lub to dup, then 0.45 s of silence to the next lub. That uneven spacing is exactly what you hear, and it is why the sounds are written “lub dup … lub dup”.

1 mark — first sound is the atrioventricular valves shutting, at 0.15 s 1 mark — second sound is the semilunar valves shutting, at 0.50 s

(d) [2]

A student says the heart sends a signal to each valve telling it when to move. Use Fig. 3.1 to explain why this cannot be right.

Model Answer — 3(d)

every valve movement in Fig. 3.1 happens at the exact instant two pressure traces cross — the timing is set by the pressures themselves, not by anything arriving from outside [1]

a valve is a flap of tissue with no muscle and no nerve supply: blood flowing forwards pushes it open, and blood beginning to flow backwards fills it and pushes it shut [1]

⚠ If you missed marks here: This idea is worth marks well beyond the heart. Valves in veins work in exactly the same way, pushed shut by blood that starts to fall back down the leg — which is why a vein needs no nerve supply to keep blood moving one way.

1 mark — each valve moves exactly where two pressure traces cross, so pressure sets the timing 1 mark — blood pushes a valve open and pushes it shut; valves have no muscle or nerve supply

(e) [2]

At the moment the left ventricle reaches 16.0 kPa in Fig. 3.1, the right ventricle of the same person reaches only 3.2 kPa. Explain this difference, and state how the volume of blood leaving each ventricle compares.

Model Answer — 3(e)

the left ventricle drives blood into the aorta and round the whole body, a long circuit containing a huge number of narrow vessels, so a high pressure is needed to keep the blood moving all the way back to the heart; the right ventricle pumps only to the lungs, which are close by, and lung capillaries are so delicate that a high pressure would damage them and force plasma into the air spaces [1]

the volumes are the same. Whatever the right ventricle sends to the lungs must return and be pumped on by the left, so the thicker left wall gives a higher pressure, not a larger volume [1]

⚠ If you missed marks here: “The left ventricle pumps more blood” is the classic error, and it is impossible: the two sides are in series, so if one pumped more than the other, blood would pile up in the lungs within a minute. Thicker muscle buys pressure, and only pressure.

1 mark — left pumps to the whole body against a much greater resistance; right only to the nearby, delicate lungs 1 mark — the volumes are the same; the thicker wall gives higher pressure, not more blood

Question 4 — Four Blood Counts

Total: 12 marks

Table 4.1 shows the results of blood tests on four adults. Counts are given as the number of cells in one cubic millimetre of blood. Anya is healthy and her figures are typical of a healthy adult.

Table 4.1	Anya	Ben	Cara	Dev
red blood cells / per mm³	5 000 000	5 200 000	3 000 000	5 100 000
white blood cells / per mm³	5000	25 000	6500	7200
platelets / per mm³	250 000	260 000	240 000	40 000
haemoglobin / g per dm³	150	152	84	148

(a) [3]

Ben, Cara and Dev each have results that differ markedly from Anya. For each of the three, state which measurement is abnormal and suggest what it indicates.

Model Answer — 4(a)

Ben: white blood cell count is 25 000, five times Anya’s; his body is making extra phagocytes and lymphocytes, which suggests he is fighting an infection [1]

Cara: red blood cell count and haemoglobin are both far below the others (3 000 000 and 84 g per dm³); she is anaemic and her blood carries less oxygen [1]

Dev: platelet count is 40 000, about one sixth of Anya’s; his blood will clot slowly, so he bruises and bleeds easily [1]

⚠ If you missed marks here: White blood cells and platelets are different things doing different jobs, and mixing them up wrecks both Ben’s answer and Dev’s. Platelets are fragments of cells with no nucleus and they clot; white blood cells are whole cells with a nucleus and they defend. Note too that Cara’s two abnormal rows are really one problem — haemoglobin lives inside red cells, so fewer cells means less haemoglobin.

1 mark — Ben: raised white cell count, suggesting infection 1 mark — Cara: low red cell count and haemoglobin, i.e. anaemia 1 mark — Dev: low platelet count, so slow clotting

(b) [2]

Calculate the percentage by which Cara’s haemoglobin concentration is lower than Anya’s. Show your working.

Model Answer — 4(b)

150 − 84 = 66 g per dm³

difference calculated correctly [1]

66 ÷ 150 × 100 = 44 %

44 % lower [1]

⚠ If you missed marks here: A percentage change is always divided by the value you started from — here Anya’s 150, not Cara’s 84. Dividing by 84 gives 78.6 %, which is a different (and unasked) statement: how much higher Anya is than Cara. Write down which number is the “before” before you divide.

1 mark — difference of 66 g per dm³ found 1 mark — 66 ÷ 150 × 100 = 44 %

Cara becomes breathless and tired after climbing a single flight of stairs. Explain this using her figures in Table 4.1. Respiration is the set of chemical reactions in cells that break down nutrient molecules and release energy.

Model Answer — 4(c)

haemoglobin, carried inside red blood cells, combines with oxygen in the lungs to form oxyhaemoglobin, and releases it again in the tissues [1]

Cara has 44 % less haemoglobin and only 3 000 000 red cells per mm³, so each cm³ of her blood carries much less oxygen [1]

less oxygen reaches her muscle cells, so less energy is released by respiration; she tires quickly and breathes faster to try to make up the shortfall [1]

⚠ If you missed marks here: Cambridge wants the word oxyhaemoglobin — it is the name of the compound, and “the haemoglobin holds oxygen” will not do. Note also that Cara’s problem is not that she cannot breathe in enough oxygen; the air reaching her lungs is perfectly normal. Her problem is the carrying, not the collecting.

1 mark — haemoglobin combines with oxygen to form oxyhaemoglobin 1 mark — with 44 % less haemoglobin her blood carries much less oxygen 1 mark — less oxygen to the cells means less energy released by respiration

(d) [2]

Describe what normally happens at a cut in the skin, naming the two proteins involved, and state how Dev’s response would differ.

Model Answer — 4(d)

platelets gather at the wound and set off the conversion of the soluble plasma protein fibrinogen into insoluble fibrin, which forms a mesh of fibres across the cut; red blood cells are trapped in the mesh and the clot dries to a scab [1]

Dev has only 40 000 platelets per mm³, so the mesh forms slowly or incompletely: he loses more blood and the wound stays open to pathogens for longer [1]

⚠ If you missed marks here: Fibrinogen and fibrin sound almost the same and that is the whole difficulty. Fibrinogen is the one dissolved in the plasma, waiting; fibrin is the one that comes out of solution as threads. Also, clotting has two roles — stopping blood escaping and stopping pathogens entering — and the second is the one that gets forgotten.

1 mark — fibrinogen converted to fibrin, forming a mesh that traps red cells 1 mark — Dev clots slowly, so more blood is lost and pathogens can enter

(e) [2]

State two things about a person’s blood that the measurements in Table 4.1 cannot show.

Model Answer — 4(e)

any two of the following, 1 mark each:

nothing about the plasma, which is more than half the volume of blood and transports ions, nutrients, urea from the liver to the kidneys, hormones and carbon dioxide [1]

whether Ben’s extra white cells are lymphocytes or phagocytes, and which pathogen (if any) is responsible — a raised count has many possible causes [1]

whether the haemoglobin present is actually carrying oxygen; the appearance or shape of the cells; or how any of these values changes over time, since each is a single snapshot [1]

⚠ If you missed marks here: Questions asking what data cannot show are asking you to notice what was never measured. The biggest omission here is the largest component of blood by volume: counting cells tells you nothing at all about the liquid they are floating in.

1 mark — first valid limitation, e.g. nothing about the plasma and what it transports 1 mark — second valid limitation, e.g. cannot tell lymphocytes from phagocytes or identify a cause

Question 5 — A Trout and a Rabbit

Total: 10 marks

Table 5.1 compares measurements made on a trout, a fish, and a rabbit, a mammal. The trout exchanges gases at its gills and the rabbit at its lungs.

Table 5.1	trout	rabbit
body temperature / °C	12 (the same as the water)	39
oxygen used / cm³ per kg per hour	70	700
blood pressure entering the gas exchange organ / kPa	4.0	3.0
blood pressure leaving the gas exchange organ / kPa	1.2	1.0
blood pressure entering the organs of the body / kPa	1.2	16.0
times the blood passes through the heart in one complete circuit	1	2

(a) [2]

Calculate the percentage of its blood pressure that the trout loses in passing through the gills, and state how many times greater the rabbit’s oxygen use is than the trout’s. Show your working.

Model Answer — 5(a)

(4.0 − 1.2) ÷ 4.0 × 100 = 2.8 ÷ 4.0 × 100 = 70 %

70 % of the pressure is lost at the gills [1]

700 ÷ 70 = 10

the rabbit uses 10 times as much oxygen per kilogram per hour [1]

⚠ If you missed marks here: Read the units in the oxygen row: they are per kilogram, so the comparison already allows for the two animals being different sizes. Without that phrase the figures would tell you nothing, because a bigger animal uses more oxygen simply for being bigger.

1 mark — 70 % pressure lost at the gills, with working 1 mark — rabbit uses 10 times as much oxygen per kg per hour

(b) [3]

Use the last row of Table 5.1 to describe the circulation of the trout and the circulation of the rabbit, naming both types.

Model Answer — 5(b)

the trout has a single circulation: the blood passes through the heart once for each complete circuit of the body [1]

its route is heart → gills → body → heart, so blood leaving the gills goes straight on to the body without returning to the heart [1]

the rabbit has a double circulation: the blood passes through the heart twice per circuit, once round the pulmonary circuit (heart → lungs → heart) and once round the systemic circuit (heart → body → heart) [1]

⚠ If you missed marks here: Double circulation does not mean two hearts — it means one heart entered twice, which is exactly what the last row of the table is telling you. The mammalian heart is better thought of as two pumps built side by side and kept apart by the septum.

1 mark — trout: single circulation, blood through the heart once per circuit 1 mark — trout route heart → gills → body → heart 1 mark — rabbit: double circulation, through the heart twice, pulmonary and systemic circuits

Use the pressure rows of Table 5.1 to explain the advantage of the rabbit’s type of circulation.

Model Answer — 5(c)

in the trout the blood has already lost 70 % of its pressure in the narrow gill capillaries, so it must travel round the entire body on only 1.2 kPa [1]

in the rabbit the blood returns to the heart after the lungs and is pumped a second time, so it sets off round the body at 16.0 kPa instead of 1.0 kPa [1]

higher pressure means the blood flows faster, so oxygen and nutrients are delivered and carbon dioxide removed far more quickly — which is what a fast rate of respiration needs [1]

⚠ If you missed marks here: The advantage is pressure restored, and the table hands you the evidence: 1.2 against 16.0. It is not that mammals separate oxygenated from deoxygenated blood — a fish does that perfectly well too. What a fish cannot do is give the blood a second push after the gills have flattened it.

1 mark — trout blood reaches the body at only 1.2 kPa after the gills 1 mark — rabbit blood is returned to the heart and repressurised to 16.0 kPa 1 mark — faster flow delivers oxygen and nutrients and removes carbon dioxide more quickly

(d) [2]

Suggest, using two other rows of Table 5.1, why a single circulation is nevertheless enough for the trout.

Model Answer — 5(d)

the trout’s body temperature is 12 °C, the same as the water, so it uses no energy keeping itself warm, whereas the rabbit holds itself at 39 °C [1]

its oxygen use is therefore only one tenth of the rabbit’s per kilogram per hour, so slow delivery at low pressure supplies all the oxygen it needs (being supported by water also means it uses less energy moving about) [1]

⚠ If you missed marks here: “Suggest” means the mark is for reasoning from the data, so the two rows have to be quoted. It is also worth seeing that the fish is not badly designed — its circulation matches its demand exactly. A structure is only inadequate relative to what is being asked of it.

1 mark — trout is at the temperature of the water and uses no energy keeping warm 1 mark — its oxygen demand is only one tenth, so slow low-pressure delivery is enough

Question 6 — Nineteen Thousand Men, Ten Years

Total: 12 marks

Table 6.1 shows the results of a study in which 19 000 men aged 45–64 were placed in four groups according to how many cigarettes they smoked each day, and then followed for ten years. Incidence is the number of men in every 1000 who developed coronary heart disease during those ten years.

Table 6.1 — cigarettes smoked per day	number of men in the group	number developing coronary heart disease in 10 years	incidence per 1000 men
0	8000	240	30
1–9	4000	160	40
10–19	5000	300	to be calculated
20 or more	2000	180	90

(a) [2]

Calculate the missing value in Table 6.1, showing your working, and state how many times greater the incidence is in the heaviest-smoking group than in the group who did not smoke.

Model Answer — 6(a)

300 ÷ 5000 × 1000 = 60

incidence for the 10–19 group = 60 per 1000 men [1]

90 ÷ 30 = 3

the heaviest smokers have three times the incidence of the non-smokers [1]

⚠ If you missed marks here: Compare the raw counts and the story reverses: 240 non-smokers developed the disease against only 180 heavy smokers. The raw number is meaningless because the groups are different sizes — that is exactly why the incidence column exists, and why converting to a rate is the first thing to do with data like this.

1 mark — 300 ÷ 5000 × 1000 = 60 per 1000 1 mark — 90 ÷ 30 = 3 times greater

(b) [2]

Describe the trend shown by Table 6.1.

Model Answer — 6(b)

as the number of cigarettes smoked per day increases, the incidence of coronary heart disease increases — from 30 to 40 to 60 to 90 per 1000 (a positive correlation) [1]

the rise is not proportional: the steps between successive groups get bigger, +10, then +20, then +30, so the incidence climbs more steeply the more a man smokes [1]

⚠ If you missed marks here: “It goes up” is one mark at most. The second mark is always for the shape of the going-up — steady, levelling off, or accelerating — and the way to find it is to work out the differences between consecutive rows rather than looking at the values themselves.

1 mark — incidence increases with cigarettes smoked, values quoted 1 mark — the increase accelerates / is not proportional (+10, +20, +30)

Explain why these data alone do not prove that smoking causes coronary heart disease.

Model Answer — 6(c)

the study shows a correlation, and a correlation on its own never proves cause — the two things rise together, which is not the same as one producing the other [1]

the four groups may differ in other ways that were not measured: heavy smokers might also eat more fatty food, take less exercise, be more stressed, be older, or have a genetic predisposition, and any of these is itself a risk factor [1]

the sample is limited — only men aged 45–64, so nothing can be said about women or other ages — and the heaviest-smoking group contained only 2000 men, so its figure is the least reliable of the four [1]

⚠ If you missed marks here: Saying “correlation is not causation” and stopping there is worth one mark. The other two come from naming a specific alternative explanation and a specific weakness in the sample. And note what the question is not asking: smoking really does cause coronary heart disease, but it took experimental evidence about how the chemicals damage artery walls to prove it, not a table like this one.

1 mark — a correlation does not prove cause 1 mark — a named confounding factor that was not measured, e.g. diet, exercise, stress, age 1 mark — a limitation of the sample, e.g. only men 45–64, or the small 2000-man group

(d) [3]

Describe how coronary heart disease develops, and its consequences for the heart muscle.

Model Answer — 6(d)

fatty material, containing cholesterol, builds up in the wall of a coronary artery, so the lumen narrows [1]

less blood can flow through, so the heart muscle receives less oxygen and glucose; during exercise, when it needs most, this causes chest pain [1]

if a coronary artery becomes completely blocked, often by a clot forming on the fatty deposit, the muscle beyond it cannot respire and that region dies — a heart attack [1]

⚠ If you missed marks here: The vessel that blocks must be named as a coronary artery. The heart is full of blood at every moment, but its own wall is far too thick to be supplied by diffusion from inside, so it has its own arteries branching from the aorta — and blocking one starves the muscle even though blood is pouring through the chambers a millimetre away.

1 mark — fatty material builds up in a coronary artery, narrowing the lumen 1 mark — less blood, so less oxygen and glucose reach the heart muscle 1 mark — complete blockage kills that region of muscle (heart attack)

(e) [2]

Smoking is one risk factor. Describe how changing diet and taking regular exercise each reduce the risk of coronary heart disease.

Model Answer — 6(e)

diet: eating less saturated fat and cholesterol, and less salt, slows the build-up of fatty deposits in the artery walls and helps keep blood pressure down [1]

exercise: regular exercise strengthens the heart muscle so it pumps more blood per beat, helps control body mass and lowers blood pressure, so the heart works less hard for the same output [1]

⚠ If you missed marks here: Each mark needs a mechanism, not an instruction. “Eat healthily and exercise more” is advice, not biology. Note also that some risk factors — age, sex and genetic predisposition — cannot be changed at all, which is why the two that can matter so much.

1 mark — diet: less saturated fat/cholesterol/salt slows fatty deposits and lowers blood pressure 1 mark — exercise: stronger heart muscle, lower body mass and blood pressure

Question 7 — Ravi Runs Up the Stairs

Total: 10 marks

Ravi wanted to find out how physical activity affects heart rate. This is exactly what he did.

1  He counted the beats at his wrist for 10 seconds and multiplied by 6.
2  He ran up and down a flight of stairs as fast as he could until he felt tired.
3  He immediately counted the beats for 10 seconds again and multiplied by 6.
4  He waited 3 minutes and counted once more.
5  He then asked his friend Meera, who plays hockey every day, to do exactly the same.
6  He recorded each reading once and wrote his results in Table 7.1.

Table 7.1	at rest / beats per minute	immediately after stopping / beats per minute	3 minutes after stopping / beats per minute
Ravi	84	156	120
Meera	60	144	78

(a) [2]

State the independent variable and the dependent variable in Ravi’s investigation.

Model Answer — 7(a)

independent variable: the amount of physical activity — in practice, whether the person has just exercised and how long ago they stopped [1]

dependent variable: the heart rate, measured as pulse in beats per minute [1]

⚠ If you missed marks here: The independent variable is the one he changes on purpose; the dependent variable is the one he then measures. Writing “Ravi and Meera” as the independent variable confuses a second comparison he happens to be making with the variable his question is actually about.

1 mark — independent variable: amount of physical activity / time since exercise 1 mark — dependent variable: heart rate in beats per minute

(b) [4]

Identify two faults in Ravi’s method and describe an improvement for each.

Model Answer — 7(b)

any two faults, each with its improvement, 1 mark for the fault and 1 for the improvement:

fault: “until he felt tired” is not a fixed amount of exercise, so Ravi and Meera almost certainly did different amounts and the comparison is not fair — improvement: a set exercise for a set time, for example stepping on and off a bench at a rate fixed by a metronome for exactly 3 minutes

fault: each reading was taken only once, so an anomalous count cannot be spotted and no mean can be calculated — improvement: repeat on several days and take a mean, and test several people at each level of fitness

fault: counting for only 10 seconds multiplies any miscount by six — improvement: count for a full 60 seconds, or use a heart-rate monitor

fault: “immediately” is vague and the rate falls fast, so a few seconds’ delay changes the reading — improvement: take readings at fixed times, e.g. 0, 1, 2 and 3 minutes after stopping

fault: nothing was controlled — food, caffeine, time of day, and how long each person rested before the resting reading — improvement: state and fix these conditions for both people

⚠ If you missed marks here: An improvement has to be something you could actually write into the method. “Be more accurate” and “do it properly” earn nothing; “step on and off a 30 cm bench 30 times a minute for 3 minutes” earns the mark because another person could follow it and get comparable results.

1 mark — first fault correctly identified 1 mark — a workable improvement for the first fault 1 mark — second fault correctly identified 1 mark — a workable improvement for the second fault

Every value in Table 7.1 is a whole number of sixes. Explain why, and state what this tells you about the precision of Ravi’s results.

Model Answer — 7(c)

he counted whole beats in 10 seconds and multiplied by 6, so every result he can possibly get is a multiple of 6 (84 = 14 × 6, 156 = 26 × 6, 78 = 13 × 6, and so on) [1]

the readings are therefore precise only to the nearest 6 beats per minute, and a single miscounted beat shifts the answer by 6; counting for a full minute would be six times more precise [1]

⚠ If you missed marks here: This is the difference between precision and accuracy. Multiplying up does not make a measurement better; it just carries the same uncertainty into bigger units. Whenever a method says “count for n seconds and multiply”, the smallest difference the method can detect is the multiplier itself.

1 mark — counted whole beats in 10 s and multiplied by 6, so all values are multiples of 6 1 mark — precise only to the nearest 6 bpm; one miscount changes the result by 6

(d) [2]

Explain why heart rate rises during physical activity, and suggest why Meera’s rate had almost returned to her resting value after 3 minutes while Ravi’s had not.

Model Answer — 7(d)

during activity the muscles respire faster, so they need oxygen and glucose delivered more quickly and carbon dioxide removed more quickly; the heart beats faster so that blood circulates faster and carries these substances at the rate required [1]

Meera plays hockey every day, so she is fitter — her heart muscle is stronger and pushes out more blood per beat, so she needs fewer beats and recovers sooner (her rate fell from 144 to 78, only 18 above resting, while Ravi’s was still 36 above his). Neither returns instantly, because the muscles still need extra oxygen to clear the carbon dioxide built up during the exercise [1]

⚠ If you missed marks here: Compare the right numbers. Meera’s peak (144) is lower than Ravi’s (156), but the fitness evidence is the recovery: how far above her own resting rate she still is. Comparing two people’s raw pulse figures without comparing each to their own resting value is the commonest way to misread a table like this one.

1 mark — muscles need faster delivery of oxygen and glucose and faster removal of carbon dioxide 1 mark — Meera is fitter, so recovers sooner; rates stay raised for a while after stopping

Self-Assessment

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