IGCSE Biology Paper 4 (Theory / Extended)

Topic 11: Gas Exchange in Humans -- 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 11. Like a real Cambridge paper it ranges across every sub-topic — 11.1 gas exchange surfaces and the breathing system, 11.2 inspired and expired air, 11.3 ventilation, and 11.4 exercise, breathing control and clean airways — and it mixes them inside single questions. All three Topic 11 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — A Jar, Two Balloons and a Sheet of Rubber

Total: 12 marks

Fig. 1.1 shows a piece of apparatus used in school laboratories to model the human thorax. The rubber sheet at the base has been pulled downwards and is being held there. Nothing on the diagram has been named.

(a) [4]

State the part of the human breathing system that each of A, B, C and D is intended to represent.

Model Answer — 1(a)

A — the ribs / ribcage (accept the wall of the thorax) [1]

B — the trachea, branching into the two bronchi [1]

C — a lung [1]

D — the diaphragm [1]

⚠ If you missed marks here: The commonest error is writing “alveoli” for C. A balloon is a single hollow bag, and a lung is not — a lung contains something like 300 to 500 million alveoli. That difference is the whole point of part (c). The second error is calling D “the muscles” without naming the diaphragm; Cambridge wants the named structure.

1 mark — A: the ribs / ribcage / wall of the thorax 1 mark — B: the trachea (branching into the bronchi) 1 mark — C: a lung 1 mark — D: the diaphragm

(b) [3]

Explain, in terms of volume and pressure, why the balloons inflate when D is pulled downwards.

Model Answer — 1(b)

pulling D down increases the volume of the sealed space inside the jar [1]

the same amount of air now occupies a larger space, so the pressure inside the jar decreases — it falls below atmospheric pressure [1]

the air outside is now at a higher pressure, so it is pushed down the tube into the balloons, which inflate; the balloons are stretched open from the inside, they do not pull themselves open [1]

⚠ If you missed marks here: The word to strike out of your vocabulary is sucked. Air is never sucked or pulled into the lungs; there is nothing in the chest capable of gripping it. Lowering the pressure inside simply removes the thing that was holding the outside air back, and the atmosphere pushes air in. The same logic explains why the lungs contain no muscle at all — they are elastic bags that are stretched by the thorax around them.

1 mark — volume of the sealed space inside the jar increases 1 mark — so the pressure inside decreases, falling below atmospheric pressure 1 mark — higher pressure outside pushes air down the tube into the balloons (pushed, not sucked)

Give three ways in which this apparatus is a poor representation of a real human thorax.

Model Answer — 1(c)

any three, one mark each:

there are no ribs and no intercostal muscles — the model changes volume only at the base, whereas a real thorax also raises the ribs up and out

the jar is rigid and cannot change shape, but the real ribcage moves at every breath

a balloon wall is thick rubber and is one single bag, whereas an alveolus wall is only one cell thick and there are hundreds of millions of alveoli — the model has a tiny surface area, the lungs have about 70 m²

there is no gas exchange at all in the model: no blood supply, no capillaries, so nothing diffuses in or out and the balloons are not moist

the sheet is pulled down by hand from outside, whereas the diaphragm is a muscle that contracts and flattens itself, and a muscle can only pull, never push

⚠ If you missed marks here: “It is made of plastic, not flesh” earns nothing — every model is made of the wrong material, so that is not a criticism of this model. A limitation only scores if it changes what the model can teach you. Notice how many of the good answers come back to surface area: one big balloon instead of hundreds of millions of tiny alveoli is the single most misleading thing about the apparatus.

1 mark — first valid limitation (e.g. no ribs or intercostal muscles) 1 mark — second valid limitation (e.g. balloon wall thick and a single bag, so tiny surface area) 1 mark — third valid limitation (e.g. no blood supply, so no gas exchange)

(d) [2]

A student says the apparatus shows “how we breathe”. Using precise biological terms, state which process the apparatus does model and which closely related process it does not, and give the definition of each.

Model Answer — 1(d)

it models ventilation — the movement of air into and out of the lungs [1]

it does not model gas exchange — the diffusion of oxygen and carbon dioxide across the alveolus wall, between the air in the alveolus and the blood in the capillary; there is no blood anywhere in the model [1]

⚠ If you missed marks here: Three words in this topic sound interchangeable and are not. Ventilation is air moving in and out. Gas exchange is diffusion across the alveolus wall. Respiration is the chemical reactions inside cells that release energy. A mark scheme will refuse the wrong one every time, so if you catch yourself writing “the lungs respire”, stop — the lungs ventilate, the alveoli exchange gases, and every cell in the body respires.

1 mark — models ventilation: the movement of air into and out of the lungs 1 mark — does not model gas exchange: diffusion of oxygen and carbon dioxide across the alveolus wall

Question 2 — Three Surfaces, One Argument

Total: 12 marks

You have already met two exchange surfaces that have nothing to do with breathing. A villus is a finger-like projection of the lining of the small intestine, about 1 mm long; its surface is a single layer of cells and inside it lies a dense network of blood capillaries. A root hair cell is a cell of the outer layer of a root that is drawn out into a long, narrow extension pushing between the soil particles; its wall and membrane are thin and the water entering it passes straight on into the root. An alveolus is a microscopic air sac at the end of a bronchiole, wrapped in capillaries. This question asks you to see what the three have in common.

(a) [4]

For each of the alveolus, the villus and the root hair cell, state one structural feature that gives it a large surface area. Then state the one further feature that all three share which keeps the distance a molecule must diffuse very short.

Model Answer — 2(a)

alveolus — there are hundreds of millions of them, each a separate tiny sac, so their total surface area is enormous (about 70 m²) [1]

villus — the lining of the small intestine is thrown into millions of finger-like projections, so far more surface is packed into the same length of gut [1]

root hair cell — the cell is drawn out into a long, narrow extension that reaches out between the soil particles [1]

all three — the exchange surface is only one cell thick (thin), so the diffusion distance is very short and diffusion is fast [1]

⚠ If you missed marks here: Watch the phrase “a large surface area” being used as though it were magic. A large surface area does not make oxygen or attract it — it simply means more molecules can cross per second, because diffusion happens everywhere on the surface at once. Also note that no single alveolus is large; it is the total of all of them that matters, and an answer saying “the alveolus is big” has the argument backwards.

1 mark — alveolus: hundreds of millions of tiny sacs give a huge total surface area 1 mark — villus: lining folded into millions of finger-like projections 1 mark — root hair cell: long narrow extension into the soil 1 mark — all three: surface only one cell thick, so a very short diffusion distance

(b) [3]

Two model organisms are made of solid living tissue. Model P is a cube of side 2 cm. Model Q is a cube of side 6 cm. Calculate the surface area to volume ratio of each, showing your working, and state what happens to this ratio as an organism becomes larger.

Model Answer — 2(b)

P: surface area = 6 × 2 × 2 = 24 cm²; volume = 2 × 2 × 2 = 8 cm³; ratio = 24 : 8 = 3 : 1 [1]

Q: surface area = 6 × 6 × 6 = 216 cm²; volume = 6 × 6 × 6 = 216 cm³; ratio = 216 : 216 = 1 : 1 [1]

as an organism becomes larger its surface area to volume ratio decreases — the volume grows faster than the surface area [1]

⚠ If you missed marks here: The trap here is a numerical coincidence: for a cube of side 6 the surface area and the volume are both 216, which tempts people to write “they are the same, so it does not matter”. They are the same number but not the same quantity — one is in cm² and the other in cm³. Always write the units, and always give the ratio in the form something : 1 so the two cubes can be compared at a glance.

1 mark — P: 24 cm² and 8 cm³, ratio 3 : 1 1 mark — Q: 216 cm² and 216 cm³, ratio 1 : 1 1 mark — the ratio decreases as the organism gets larger

Explain why a large, active animal cannot rely on diffusion across its outer body surface to obtain all the oxygen it needs.

Model Answer — 2(c)

a large animal has a small surface area compared with its volume, so there is not enough outer surface for the huge volume of respiring tissue inside it [1]

the diffusion distance from the outer surface to the cells deep inside is far too great, and diffusion over more than a fraction of a millimetre is far too slow [1]

so it needs a specialised gas exchange surface with a very large surface area folded up inside the body, together with a transport system (blood) to carry oxygen the rest of the way to the cells [1]

⚠ If you missed marks here: An answer that only says “it is too big” scores nothing. There are two separate problems and Cambridge wants both: not enough surface for the volume, and too far to travel once a molecule is across. Skin also fails a third test — it is thick and waterproof, and diffusion needs a thin, moist surface.

1 mark — small surface area compared with volume, so too little surface for the respiring tissue 1 mark — diffusion distance to the inner cells is too great and diffusion is too slow 1 mark — so a specialised gas exchange surface and a transport system are needed

(d) [2]

Both the alveolus and the villus are described as having a good blood supply. A student writes that this matters “because the blood brings oxygen to the alveoli”. Explain what a good blood supply actually does at an exchange surface, and explain why the student has the argument the wrong way round.

Model Answer — 2(d)

the flowing blood constantly carries away the oxygen that has just diffused in and constantly brings carbon dioxide to be removed, so the concentration of oxygen in the blood is always kept lower than in the alveolus and the concentration of carbon dioxide always higher — that is, the concentration gradients are maintained, and diffusion continues rapidly [1]

the blood arriving at an alveolus is deoxygenated — it does not deliver oxygen to the alveolus, it collects oxygen there; the oxygen comes from the air. The villus works the same way: the blood carries the absorbed nutrients away, it does not bring them [1]

⚠ If you missed marks here: Remember the direction of the pulmonary circulation: the pulmonary artery carries deoxygenated blood to the lungs and the pulmonary vein carries oxygenated blood away. Blood arriving at an alveolus is therefore the low-oxygen side of the gradient. And notice that good ventilation does exactly the same job from the other side — it keeps replacing the alveolar air, so neither a good blood supply nor good ventilation “gives more oxygen”; both simply maintain the gradient.

1 mark — blood flow carries substances away and so maintains the concentration gradient 1 mark — blood arriving is deoxygenated; it collects oxygen from the air rather than delivering it

Question 3 — Marking Somebody Else’s Table

Total: 12 marks

A candidate was asked to complete a table comparing inspired air with expired air. Their answers are shown in Table 3.1. Four of the five rows contain a mistake. One row is completely correct.

component or property	inspired air — candidate’s answer	expired air — candidate’s answer
oxygen	21%	0%
carbon dioxide	0.04%	4%
nitrogen	78%	60%
water vapour	saturated	low
temperature	37 °C	the temperature of the room

Table 3.1

(a) [4]

Identify the four rows that are wrong and, for each one, give the correct entry.

Model Answer — 3(a)

oxygen — expired air contains about 16% oxygen, not 0% [1]

nitrogen — expired air contains 78% nitrogen, unchanged, not 60% [1]

water vapour — the two entries are the wrong way round: inspired air is variable and usually low, expired air is saturated [1]

temperature — also reversed: inspired air is at the temperature of the surroundings, expired air is at about 37 °C, body temperature [1]

(the carbon dioxide row, 0.04% rising to 4%, is correct)

⚠ If you missed marks here: Two of these four errors are simply the columns swapped over, and that is worth noticing as a habit: before you write anything, ask which column is the air going in. Air going in has whatever temperature and humidity the room has; only air that has been inside you can be at 37 °C and saturated. If a table ever tells you that inspired air arrives at body temperature, it is describing a room at 37 °C, not a person.

1 mark — oxygen in expired air is about 16%, not 0% 1 mark — nitrogen in expired air is 78%, unchanged 1 mark — water vapour entries reversed: inspired low/variable, expired saturated 1 mark — temperature entries reversed: inspired at room temperature, expired about 37 °C

(b) [3]

Explain why expired air still contains about 16% oxygen, and explain why the percentage of nitrogen does not change.

Model Answer — 3(b)

only some of the oxygen in each breath diffuses into the blood — the air stays in the alveoli for a very short time and diffusion never gets anywhere near evening out the concentrations [1]

so more than three quarters of the oxygen breathed in is breathed straight back out again; this is exactly why mouth-to-mouth resuscitation works — the air the rescuer blows in still has plenty of oxygen in it [1]

nitrogen is not used by the body and none is produced by the body, so none is taken into the blood and the same percentage comes back out [1]

⚠ If you missed marks here: “Expired air has no oxygen left” is the single most common wrong idea in this topic, and it is worth being annoyed about, because if it were true nobody could ever be resuscitated by another person’s breath. The nitrogen mistake is subtler: some tables print 79% and 79%, others print figures that differ by a fraction, and that tiny difference is only because the other percentages changed — the actual amount of nitrogen is untouched.

1 mark — only some of the oxygen diffuses into the blood in the short time the air is in the alveoli 1 mark — so more than three quarters is breathed straight out (the basis of mouth-to-mouth resuscitation) 1 mark — nitrogen is not used or produced by the body, so its percentage is unchanged

Explain why expired air is warmer than inspired air and why it contains more water vapour.

Model Answer — 3(c)

the air has been in contact with the warm surfaces of the airways and the lungs, so heat is transferred to it and it leaves at about body temperature, 37 °C [1]

the lining of the alveoli and airways is moist, and water evaporates from that moist lining into the air, so the air leaves saturated with water vapour [1]

⚠ If you missed marks here: A popular wrong answer is that the water vapour is “produced by respiration” and blown out. Some of it is, but that is not what this question is testing and it is not what makes expired air saturated — the marking point is evaporation from the moist lining. That moist lining is not an accident either: gases must dissolve before they can diffuse across the alveolus wall, so the surface has to stay wet, and losing water vapour is the price the body pays for it.

1 mark — warmed by contact with the warm surfaces of the airways and lungs, to about 37 °C 1 mark — water evaporates from the moist lining into the air, so it leaves saturated

(d) [3]

Describe how limewater could be used to show that expired air contains more carbon dioxide than inspired air. State the result you would expect and give two variables that must be kept the same.

Model Answer — 3(d)

two boiling tubes, A and B, each containing limewater, joined by a T-piece with one-way valves to a single mouthpiece, so that breathing in draws room air through tube A and breathing out pushes expired air through tube B [1]

the limewater in tube B turns cloudy (milky) after only a few breaths, while the limewater in tube A stays clear for far longer [1]

any two kept the same: the same volume of limewater in each tube; the same concentration of limewater; the same number of breaths through each; the same person breathing [1]

⚠ If you missed marks here: Limewater goes cloudy or milky. Not white, not “it changes”, not “it goes off” — those score nothing. The second thing examiners look for is that tube A is a genuine control: the point of the experiment is not that expired air turns limewater cloudy, it is that it does so in far fewer breaths than ordinary room air, which is why the number of breaths has to be the same in both tubes.

1 mark — two tubes of limewater with one-way valves so one is breathed in through and one out through 1 mark — the expired-air tube turns cloudy (milky) in far fewer breaths 1 mark — two suitable controlled variables (volume, concentration, number of breaths, same person)

Question 4 — Reading the Breathing Trace

Total: 12 marks

A student breathes in and out through a machine that records the volume of air it holds. Every breath she takes out puts air into the machine and every breath she takes in removes air from it, so the trace rises and falls once for every breath. Fig. 4.1 shows one minute of recording. For the first thirty seconds she sat still; she then started to exercise.

(a) [2]

Use Fig. 4.1 to determine, for the period when she was sitting still, the volume of air moved in one breath and the number of breaths taken per minute.

Model Answer — 4(a)

the trace swings between 2.0 and 2.5 dm³, so the volume of one breath (the tidal volume) is 0.5 dm³, which is 500 cm³ [1]

6 complete breaths in 30 s, so 6 × 2 = 12 breaths per minute [1]

⚠ If you missed marks here: Two reading traps. First, the tidal volume is the difference between the top and the bottom of a swing, not the height of the peak above zero — 2.5 dm³ is not the answer. Second, the graph only lasts 30 s at rest, so you must double the count to get breaths per minute; six is the number of breaths, not the rate.

1 mark — tidal volume 0.5 dm³ (500 cm³), read as 2.5 − 2.0 1 mark — 6 breaths in 30 s, so 12 breaths per minute

(b) [3]

The volume of air moved in and out of the lungs in one minute is called the ventilation rate. Calculate the ventilation rate while she was sitting still and while she was exercising, showing your working, and state how many times greater the second value is.

Model Answer — 4(b)

sitting still: 12 breaths per minute × 0.5 dm³ = 6 dm³ per minute [1]

exercising: the trace now swings between 2.0 and 4.5 dm³, so the tidal volume is 2.5 dm³, and there are 12 breaths in 30 s, so 24 breaths per minute; 24 × 2.5 = 60 dm³ per minute [1]

60 ÷ 6 = 10 times greater [1]

⚠ If you missed marks here: Ventilation rate is a product of two things, rate and depth, and the commonest mistake is to change only one of them. If you had kept the tidal volume at 0.5 dm³ you would have got 12 dm³ per minute — a doubling instead of a tenfold rise — and you would have missed the entire point of the graph.

1 mark — at rest: 12 × 0.5 = 6 dm³ per minute 1 mark — exercising: 24 × 2.5 = 60 dm³ per minute 1 mark — 10 times greater

Two separate things about her breathing changed after 30 s. Using figures from Fig. 4.1, describe each change, and explain what is meant by the depth of breathing.

Model Answer — 4(c)

the rate of breathing doubled, from 12 to 24 breaths per minute [1]

the depth of breathing increased five-fold, the tidal volume rising from 0.5 to 2.5 dm³ [1]

the depth of breathing means the volume of air moved in a single breath, whereas the rate means the number of breaths per minute; here the far bigger change was in depth, not rate [1]

⚠ If you missed marks here: “Her breathing got faster” is only half the answer and will only ever score half the marks. On a trace, rate is how many peaks and depth is how tall they are, and Cambridge asks about both because most candidates only notice the first. Say “rate” and “depth” explicitly, and quote a figure for each.

1 mark — rate doubled, 12 to 24 breaths per minute 1 mark — depth increased, tidal volume 0.5 to 2.5 dm³ 1 mark — depth = volume of air per breath; rate = number of breaths per minute

(d) [4]

About 150 cm³ of every breath never reaches the alveoli at all: it stays in the trachea, bronchi and bronchioles, where no gas exchange can happen. Using this figure, calculate the volume of fresh air actually reaching her alveoli each minute while sitting still and while exercising. Then state whether breathing more deeply or breathing more often is the more efficient way to increase that volume, and explain why.

Model Answer — 4(d)

sitting still: (500 − 150) × 12 = 350 × 12 = 4200 cm³ per minute, which is 4.2 dm³ per minute [1]

exercising: (2500 − 150) × 24 = 2350 × 24 = 56 400 cm³ per minute, which is 56.4 dm³ per minute [1]

breathing more deeply is the more efficient way [1]

because the 150 cm³ is wasted once for every breath, so taking more breaths wastes it more often; a deeper breath spreads that same fixed waste over a much larger volume, so a greater proportion of each breath reaches the alveoli — at rest only 70% of a breath is useful, but during exercise 94% of it is [1]

⚠ If you missed marks here: The commonest slip is subtracting the 150 cm³ once at the end instead of once per breath — the wasted air sits in the airways on every single breath, so the subtraction must happen inside the bracket, before you multiply. The second slip is mixing units: work in cm³ throughout and convert at the end, because 1 dm³ = 1000 cm³ and a stray factor of a thousand ruins an otherwise perfect answer. And notice what the numbers are telling you: rapid shallow panting can move a lot of air and still deliver very little of it where it is needed.

1 mark — at rest 350 × 12 = 4200 cm³ (4.2 dm³) per minute 1 mark — exercising 2350 × 24 = 56 400 cm³ (56.4 dm³) per minute 1 mark — breathing more deeply is the more efficient way 1 mark — because the fixed 150 cm³ is wasted once per breath, so deeper breaths waste a smaller proportion

Question 5 — Two Minutes in a Bag

Total: 10 marks

In a supervised investigation, a student sat still and breathed in and out of a sealed 5 dm³ plastic bag for two minutes, so that she breathed the same air over and over again. A teacher stood beside her throughout and the investigation was to be stopped at once if she felt at all unwell. Sensors sampled the air inside the bag every 30 seconds while an observer counted her breaths. The results are in Table 5.1.

time / s	oxygen in the bag / %	carbon dioxide in the bag / %	breathing rate / breaths per minute
0	21.0	0.04	13
30	18.6	2.1	15
60	16.4	4.0	22
90	14.5	5.8	31
120	12.9	7.4	40

Table 5.1

(a) [2]

Describe the changes in the composition of the air inside the bag during the two minutes, using figures from Table 5.1.

Model Answer — 5(a)

the oxygen falls steadily, from 21.0% to 12.9% — a fall of 8.1 percentage points [1]

the carbon dioxide rises steadily, from 0.04% to 7.4% — a rise of about 7.4 percentage points; the two changes are roughly equal and opposite [1]

⚠ If you missed marks here: “Oxygen goes down and carbon dioxide goes up” is a description of every rebreathing experiment ever done and will usually score one mark at most. The question says using figures, so quote the start value, the end value and the change. Noticing that the two changes are nearly equal is the extra step that turns a description into data handling.

1 mark — oxygen falls from 21.0% to 12.9% (a fall of 8.1) 1 mark — carbon dioxide rises from 0.04% to 7.4%, a roughly equal and opposite change

(b) [2]

Describe how her breathing changed during the two minutes.

Model Answer — 5(b)

her breathing rate increased, from 13 to 40 breaths per minute — roughly a threefold rise [1]

the increase was small at first (only 13 to 15 in the first 30 s) and then became much steeper (22, 31, 40) [1]

⚠ If you missed marks here: When a table gives you five readings rather than two, the examiner wants the shape of the change as well as its direction. Look at the differences between consecutive rows — here they run 2, 7, 9, 9 — and say that the rate of increase itself increased. Also note what the table does not record: the depth of her breathing, which would almost certainly have risen too.

1 mark — rate increased from 13 to 40 breaths per minute, about threefold 1 mark — the increase was slow at first and then much steeper

Explain, as fully as you can, why her breathing changed in the way you have described.

Model Answer — 5(c)

the carbon dioxide she breathed out was trapped in the bag and breathed straight back in, so the concentration of carbon dioxide in the alveoli rose and less could diffuse out of the blood — the concentration of carbon dioxide in her blood rose [1]

this rise was detected by the brain [1]

the brain sent nerve impulses to the diaphragm and the intercostal muscles [1]

so both the rate and the depth of breathing increased, which would normally remove more carbon dioxide and bring the concentration back down [1]

⚠ If you missed marks here: This is a four-mark chain and each link is a separate mark, so write it as a chain: carbon dioxide rises in the blood → detected by the brain → impulses to the diaphragm and intercostal muscles → rate and depth increase. The link candidates leave out most often is the middle one — something has to detect the change, and in this topic Cambridge is happy with “the brain”. Note also the cruelty of the bag: the corrective response cannot work, because the extra breathing only pushes the same carbon dioxide back into the same bag.

1 mark — rebreathed carbon dioxide builds up, so the concentration in the blood rises 1 mark — the rise is detected by the brain 1 mark — the brain sends nerve impulses to the diaphragm and intercostal muscles 1 mark — both the rate and the depth of breathing increase, to remove more carbon dioxide

(d) [2]

Another student concludes: “Her breathing sped up because the oxygen in the bag was running out.” Use the data in Table 5.1 to explain why this conclusion is wrong.

Model Answer — 5(d)

at 60 s the bag still contained 16.4% oxygen — as much as ordinary expired air, and plenty to breathe — yet her breathing rate had already risen from 13 to 22, so oxygen was clearly not short at the time the change began [1]

the trigger is the rise in the concentration of carbon dioxide, detected by the brain, not a fall in oxygen; the carbon dioxide had risen a hundredfold by then, from 0.04% to 4.0% [1]

⚠ If you missed marks here: “You breathe faster because you need more oxygen” feels obviously true and is the wrong answer in every Cambridge mark scheme. The body monitors carbon dioxide, and this data set is designed to prove it: at 16.4% the oxygen is nowhere near running out, and yet the breathing has already changed. When you are asked to disprove a conclusion, always do it the way this answer does — quote the number that contradicts it, then supply the real cause.

1 mark — quotes data: 16.4% oxygen still present at 60 s while the rate had already risen 1 mark — the trigger is the rising carbon dioxide detected by the brain, not falling oxygen

Question 6 — The Muscles Between the Ribs

Total: 12 marks

Fig. 6.1 shows the ribcage from the side at the two extremes of a breath, together with an enlarged view of a single gap between two ribs. The two sheets of muscle filling that gap are drawn with their fibres running in the directions they really take.

(a) [2]

State what happens to each of the two sheets of muscle shown in Fig. 6.1 during inhalation.

Model Answer — 6(a)

the external intercostal muscles contract [1]

the internal intercostal muscles relax [1]

⚠ If you missed marks here: The word that costs marks is relax. Candidates write “the internal intercostals expand” or “stretch” or “do nothing”. Muscles contract or they relax, and Cambridge marks the word. It is also worth memorising which is which by the picture rather than the name: the external pair are the outer ones and they are the ones that lift the ribs, so external goes with inhalation.

1 mark — external intercostal muscles contract 1 mark — internal intercostal muscles relax

(b) [4]

Describe fully how the volume of the thorax is increased during inhalation and explain how this causes air to enter the lungs.

Model Answer — 6(b)

the contraction of the external intercostal muscles pulls the ribs up and out [1]

the diaphragm muscle contracts and flattens, moving downwards from its domed shape [1]

both together increase the volume of the thorax, so the pressure inside falls below atmospheric pressure [1]

the higher pressure of the air outside therefore pushes air in through the nose or mouth and down the trachea, along the pressure gradient — the air is pushed in, not sucked in [1]

⚠ If you missed marks here: Two precise phrases are being marked here. “The diaphragm moves down” on its own is not enough — write “contracts and flattens”, because that says it is a muscle doing the work. And a good many candidates write that the diaphragm moves up during inhalation, which is exactly backwards: it is domed at rest and pulls itself flat, and flat is lower than domed.

1 mark — ribs move up and out 1 mark — diaphragm contracts and flattens 1 mark — volume of the thorax increases, so pressure falls below atmospheric 1 mark — higher pressure outside pushes air in down the pressure gradient

The internal and external intercostal muscles are described as an antagonistic pair. Explain what this means. Then explain why a quiet, resting exhalation still happens even though the internal intercostal muscles are barely used during it.

Model Answer — 6(c)

antagonistic means the two muscles have opposite effects and work as a pair: when one contracts the other relaxes, and each pulls the ribs in the opposite direction to the other [1]

a muscle can only pull, never push, so a second muscle is needed to move the ribs back again — contracting the externals raises the ribs, contracting the internals lowers them [1]

a quiet exhalation is largely passive: the external intercostals and the diaphragm simply relax, and the elastic recoil of the stretched lungs together with the weight of the ribcage falling back reduces the volume of the thorax without any muscular effort [1]

⚠ If you missed marks here: Do not imagine the internal intercostals are useless. They earn their place in a forced exhalation — blowing out candles, shouting, or coughing — where the ribs have to be dragged down and in hard and fast. That is exactly the distinction part (d) turns on. Note too the last piece of vocabulary: the lungs recoil because they are elastic, not because they contain muscle. They contain none.

1 mark — antagonistic: opposite effects, one contracts while the other relaxes 1 mark — a muscle can only pull, so a second muscle is needed to reverse the movement 1 mark — quiet exhalation is passive: elastic recoil of the lungs and the ribs falling back

(d) [3]

A patient has an injury to the spinal cord which has completely paralysed the intercostal muscles on both sides of the chest. The nerve supplying the diaphragm leaves the spinal cord above the level of the injury and is undamaged, so the diaphragm still works normally. The patient breathes comfortably while sitting still, but becomes breathless at the slightest exertion and cannot cough forcefully enough to clear mucus. Explain all three of these observations.

Model Answer — 6(d)

quiet breathing is still possible because the diaphragm alone, contracting and flattening, increases the volume of the thorax enough to draw in a resting tidal volume of about 500 cm³ [1]

he becomes breathless on exertion because the extra volume normally gained by the ribs moving up and out is lost, so he cannot increase the depth of his breathing; his ventilation rate cannot rise anything like tenfold, and the extra carbon dioxide produced by his working muscles cannot be removed fast enough [1]

he cannot cough forcefully because a forced exhalation needs the internal intercostal muscles to contract and pull the ribs sharply down and in, raising the pressure in the thorax rapidly; without them the air leaves only by elastic recoil, which is far too gentle to shift mucus [1]

⚠ If you missed marks here: Questions like this reward you for having noticed that ventilation has two engines, the ribs and the diaphragm, working in parallel. Lose one and quiet breathing survives but the reserve does not. The cough mark is the one most often missed, because it is easy to forget that coughing is a forced exhalation — and a patient who cannot cough cannot clear the mucus that traps pathogens, which is why chest infections are the great danger after this kind of injury.

1 mark — the diaphragm alone can increase the thorax volume enough for quiet breathing 1 mark — without rib movement the depth of breathing cannot increase, so exertion causes breathlessness 1 mark — a forced exhalation (cough) needs the internal intercostals to pull the ribs down and in

Question 7 — The Escalator in Your Windpipe

Total: 10 marks

A pathogen is an organism that causes disease. You already know that white blood cells destroy pathogens once they have got inside the body; the airways deal with them before that stage is ever reached. Table 7.1 compares a large group of long-term smokers with a group of non-smokers of the same ages.

measurement	non-smokers	smokers
mean number of cilia per mm² of trachea lining / thousands	62	21
mean volume of mucus produced per day / cm³	95	240
mean number of chest infections per year	1.2	4.1
people reporting a cough every day / %	6	71

Table 7.1

(a) [3]

Describe how mucus and cilia work together to keep the lungs clean, naming the cells responsible for each.

Model Answer — 7(a)

goblet cells secrete mucus onto the lining of the trachea and bronchi [1]

the sticky mucus traps pathogens and dust particles in the air breathed in [1]

ciliated cells carry cilia which beat and sweep the mucus upwards, away from the lungs, to the back of the throat, where it is swallowed [1]

⚠ If you missed marks here: Two cell types, two jobs, and they are constantly swapped over in exam answers. Goblet cells make the mucus and cannot move it; ciliated cells move it and cannot make it. Also check the direction: the cilia sweep the mucus up, out of the lungs. An answer that has them sweeping it down into the lungs has described the disease rather than the defence.

1 mark — goblet cells secrete mucus 1 mark — mucus traps pathogens and dust particles 1 mark — ciliated cells have cilia that beat and sweep the mucus up to the throat, where it is swallowed

(b) [3]

Use the data in Table 7.1 to explain why the smokers suffer more chest infections and why almost three quarters of them cough every day.

Model Answer — 7(b)

tobacco smoke paralyses and destroys the cilia — the count has fallen from 62 to 21 thousand per mm², only about a third of the normal number [1]

yet the mucus produced has more than doubled, from 95 to 240 cm³ per day, so far more mucus must be shifted by far fewer cilia; the mucus is not swept up and instead collects in the lungs — and coughing becomes the only way left to move it, which is why 71% cough every day compared with 6% [1]

the pathogens trapped in that mucus are not removed; they stay in the warm, moist airways where they can reproduce and reach the bronchioles and alveoli, so chest infections rise from 1.2 to 4.1 per year [1]

⚠ If you missed marks here: The mistake to avoid is treating the cough as the illness. The cough is the replacement for the cilia — a crude, exhausting substitute for a mechanism that used to run silently day and night. Notice as well that the table contains a double blow: fewer cilia and more mucus. Quoting both figures is what turns a recall answer into a data-handling answer, and Cambridge gives credit for the comparison, not for the memory.

1 mark — smoke paralyses and destroys cilia; number falls from 62 to 21 thousand per mm² 1 mark — mucus more than doubles (95 to 240 cm³) and collects, so coughing is needed to shift it 1 mark — trapped pathogens are not removed and reach the lungs, so infections rise from 1.2 to 4.1 a year

Explain how a thick layer of mucus collecting in the airways reduces the efficiency of gas exchange. Refer to the features that make a good gas exchange surface.

Model Answer — 7(c)

a layer of mucus over the surface increases the diffusion distance — instead of about one micrometre, oxygen must also cross the mucus, so it diffuses more slowly [1]

mucus blocking a bronchiole stops air reaching the alveoli beyond it, so those alveoli are not ventilated: the concentration gradient is no longer maintained and the effective surface area available for gas exchange is reduced [1]

⚠ If you missed marks here: Cambridge gives you a checklist of four features for a good gas exchange surface — large surface area, thin surface, good blood supply, good ventilation — and questions like this are simply asking which of the four has been damaged. Run through all four whenever a lung disease appears in a question; here mucus damages two of them at once.

1 mark — mucus increases the diffusion distance, so diffusion is slower 1 mark — blocked bronchioles mean alveoli are not ventilated, so the gradient is not maintained and the surface area is reduced

(d) [2]

Some people are born with a condition in which their cilia cannot beat. They suffer repeated lung infections from early childhood, even though their white blood cells work perfectly normally. Suggest an explanation.

Model Answer — 7(d)

the mucus, with the trapped pathogens and dust still in it, is never swept up out of the airways, so it collects in the lungs and the pathogens reproduce there [1]

the mucus and cilia system is a barrier that stops pathogens entering the body in the first place, whereas white blood cells only act after pathogens have got in; healthy white blood cells therefore cannot prevent the constant arrival of new pathogens, they can only fight the infections once they have begun [1]

⚠ If you missed marks here: This question is really asking you to see the difference between keeping pathogens out and destroying pathogens that are already in. Skin, stomach acid, and this mucus-and-cilia escalator all belong to the first category; white blood cells belong to the second. A person can have a flawless second line of defence and still be ill constantly if the first one has failed — which is also, in slower motion, exactly what a lifetime of smoking does.

1 mark — mucus with trapped pathogens is not swept out, so it collects and pathogens reproduce in the lungs 1 mark — mucus and cilia are a barrier keeping pathogens out; white blood cells act only after entry

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