Topic 11: Gas Exchange in Humans -- Challenge Exam 1
1 hour 15 minutes
80
7
75:00
0610
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 — One Wide Tube, Then a Million Narrow Ones
Total: 12 marks
Fig. 1.1 shows the human breathing system seen from the front, with an enlarged view of the muscles lying between two of the ribs. Nothing has been named for you. Work from the position and the shape of each structure.
(a)[4]
Name the structures labelled A, E, G and H on Fig. 1.1.
Model Answer — 1(a)
A — the larynx (accept voice box) [1]
E — a bronchiole [1]
G — the external intercostal muscle — it is the sheet on the outside of the body [1]
H — the internal intercostal muscle — the sheet lying deeper, on the lung side [1]
⚠ If you missed marks here: Two traps in one part. First, “windpipe” is not the larynx — A sits above the trachea and it is the voice box. Second, if you wrote just “intercostal muscle” for both G and H you scored one mark at most: Cambridge wants external and internal, and the whole of ventilation depends on telling them apart. The word external means nearer the outside of the body, which is why G is the sheet drawn on the left of the enlarged view.
(b)[2]
The structures labelled C are made of cartilage. State the function of these structures.
Model Answer — 1(b)
the C-shaped rings of cartilage hold the trachea open / keep it a wide tube [1]
so that it does not collapse when the pressure inside it falls below atmospheric pressure during inhalation, and air can always reach the lungs [1]
(extra, not needed for the marks: the rings are incomplete at the back, so that the oesophagus behind can bulge outwards when a mouthful of food is swallowed)
⚠ If you missed marks here: “To protect the trachea” is the answer nearly everyone gives and it earns nothing — cartilage here is not armour, it is scaffolding. The second mark is the one people never see: air is pushed in from outside because the pressure inside the thorax has dropped, and a soft-walled tube with a low pressure inside it would be squashed flat by the higher pressure outside. Also, cartilage, never bone: bone rings could not flex as you turn your neck.
(c)[3]
There is no cartilage in the wall of a bronchiole. Suggest three reasons why cartilage is needed in the structure labelled B but not in the structure labelled E.
Model Answer — 1(c)
the trachea is a single tube carrying all the air for both lungs, so if it collapsed the whole breathing system would be blocked; a bronchiole is one of millions, so the consequence of one closing is small [1]
the trachea lies in the neck with little around it to support it, whereas each bronchiole is buried inside the lung tissue, which holds it open from all sides [1]
rings of cartilage are rigid, so they would stop a bronchiole changing its diameter; a flexible wall also takes up less space, leaving more room for alveoli [1]
accept also: the pressure difference across the wall is greatest in the large airways nearest the outside air
⚠ If you missed marks here: A “suggest” question is not asking you to recall — it is asking you to take a fact you do know (cartilage holds a tube open against a pressure difference) and apply it to a tube you have never been taught about. The commonest empty answer is “because a bronchiole is small”. Small is not a reason on its own; say what being small changes — there are millions of them, they are supported by the lung tissue packed around them, and they need to stay flexible.
(d)[3]
Describe what the structures labelled G and J do when a breath is taken in, and state the effect this has on the space inside the thorax.
Model Answer — 1(d)
G, the external intercostal muscles, contract and pull the ribs upwards and outwards [1]
J, the diaphragm, contracts and flattens (so it moves downwards, losing its domed shape) [1]
the volume of the thorax therefore increases, so the pressure inside falls below atmospheric pressure and air is pushed in [1]
⚠ If you missed marks here: “The diaphragm moves down” on its own is not enough — the mark scheme wants contracts and flattens, because that is the bit that shows you know it is a muscle doing work rather than a floor sagging. And note that the diaphragm going down is what happens on breathing in; reversing those two is one of the most frequent errors in this whole topic.
Question 2 — The Air That Comes Back Out
Total: 12 marks
Table 2.1 compares inspired air with expired air. Three of the cells have been left empty.
component of the air
inspired air
expired air
oxygen
21%
.................
carbon dioxide
.................
4%
nitrogen
78%
.................
water vapour
variable, usually low
saturated
temperature
the temperature of the surroundings
about 37 °C
(a)[3]
Complete Table 2.1 by giving the three missing values.
Model Answer — 2(a)
oxygen in expired air = 16% [1]
carbon dioxide in inspired air = 0.04% [1]
nitrogen in expired air = 78% (unchanged) [1]
⚠ If you missed marks here: If you wrote 0% for oxygen in expired air, look at what that would mean: nobody could ever be revived by mouth-to-mouth resuscitation, and yet they are. Expired air still holds 16% oxygen — more than three quarters of what you breathed in comes straight back out. The other classic slip is writing carbon dioxide in inspired air as 0.4% or 4%; it is 0.04%, a hundred times smaller than the expired figure, and that hundredfold jump is exactly what makes the limewater test so clear-cut.
(b)[3]
Explain the difference in the oxygen content and the carbon dioxide content of the two samples of air, and explain why the nitrogen content is the same in both.
Model Answer — 2(b)
oxygen falls from 21% to 16% because oxygen diffuses out of the air in the alveoli into the blood, down its concentration gradient, while the air is in the lungs [1]
carbon dioxide rises from 0.04% to 4% because carbon dioxide diffuses out of the blood into the air in the alveoli, down its concentration gradient [1]
nitrogen is unchanged because it is not used by the body and takes no part in gas exchange — it goes in and comes back out again [1]
⚠ If you missed marks here: Two vocabulary traps. Do not write that the blood “takes the oxygen” or that carbon dioxide is “pushed out” — nothing is doing anything on purpose. Both gases move by diffusion, down a concentration gradient, and that phrase is the mark. And do not say nitrogen is “used up more slowly”; it is not used at all. If a table you find elsewhere shows nitrogen as 79% in expired air, that is only because the other percentages changed and the total must still come to 100.
(c)[2]
Explain why expired air is warmer than inspired air and why it is saturated with water vapour.
Model Answer — 2(c)
while it is inside the body the air is in contact with the warm surfaces of the airways and alveoli, so heat is transferred to it and it leaves at body temperature, about 37 °C [1]
those surfaces are moist, and water evaporates from the lining into the air, so the expired air is saturated with water vapour [1]
⚠ If you missed marks here: The wrong answer here is that the water in expired air is a waste product being got rid of — it is not; it has evaporated from the moist lining of the alveoli, which is why breathing out on a cold day fogs a window. Notice too that this moist lining is not an accident: gases can only diffuse across a gas exchange surface once they have dissolved in it, so losing water in expired air is the unavoidable price of being able to breathe at all.
(d)[4]
Describe an experiment using limewater that would show that expired air contains more carbon dioxide than inspired air. Give the result you would expect, and name two variables that must be kept the same.
Model Answer — 2(d)
two boiling tubes 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]
result: the limewater in tube B (expired air) turns cloudy or milky after far fewer breaths than the limewater in tube A, which stays clear or takes very much longer to change [1]
variable kept the same, any one: the same volume of limewater in each tube [1]
variable kept the same, any one other: the same concentration of limewater / the same number of breaths / the same person breathing at the same rate [1]
⚠ If you missed marks here: Two things throw marks away here. First the result: limewater goes cloudy or milky. It does not go “white”, it does not “change colour”, and it certainly does not “go off”. Second, the point of the two-tube design: it is not two separate experiments, it is a controlled comparison in which the same person, breathing through one mouthpiece, tests both samples at once. Saying “blow into limewater and it goes cloudy” shows carbon dioxide is present but proves nothing about a difference, and the question asked for a difference.
Question 3 — Nothing Is Ever Sucked In
Total: 12 marks
A student is measured while breathing quietly. Table 3.1 gives the volume of her thorax and the pressure of the air inside it at two moments in one breath. Atmospheric pressure in the room was 101.3 kPa throughout.
moment in the breathing cycle
volume of the thorax / dm3
pressure inside the thorax / kPa
at the end of a quiet breath out
2.50
101.3
at the end of a quiet breath in
3.00
101.0
(a)[5]
Describe how the muscles of the body bring about the change shown in Table 3.1. In your answer refer to both sets of intercostal muscles and to the diaphragm.
Model Answer — 3(a)
the external intercostal muscles contract [1]
at the same time the internal intercostal muscles relax — the two sets are an antagonistic pair [1]
so the ribs are pulled upwards and outwards [1]
the diaphragm muscle contracts and flattens, losing its domed shape and moving downwards [1]
the volume of the thorax increases (2.50 to 3.00 dm3), so the pressure inside falls [1]
⚠ If you missed marks here: The mark that goes missing most often is the internal intercostals relaxing. If a question says “refer to both sets” it is telling you that one of the marks is for the set that is doing nothing, and “the intercostal muscles contract” on its own cannot score it because half of them do the opposite. The other regular loss is writing that the diaphragm “moves down” without saying it contracts and flattens. Check your answer says ribs up and out, not down and in — reversing them is the single commonest error in this topic.
(b)(i)[2]
Calculate the percentage increase in the volume of the thorax during this breath. Show your working.
Model Answer — 3(b)(i)
increase in volume = 3.00 − 2.50 = 0.50 dm3, and percentage increase = (0.50 ÷ 2.50) × 100 [1]
= 20.0% [1]
⚠ If you missed marks here: A percentage change is always divided by the starting value, not the final one. Dividing by 3.00 gives 16.7%, which is the answer to a different question, and it is the wrong answer a mark scheme deliberately watches for. Always write the subtraction down as a separate line: the first mark here is for the method, so even a slip in the arithmetic still scores if the working is visible.
(b)(ii)[2]
Use the pressure values in Table 3.1 to explain why air enters the lungs during this breath.
Model Answer — 3(b)(ii)
as the thorax gets bigger the pressure inside it falls from 101.3 to 101.0 kPa, which is 0.3 kPa below the atmospheric pressure outside [1]
air therefore moves down this pressure gradient, from the higher pressure outside to the lower pressure inside — it is pushed in from outside [1]
⚠ If you missed marks here: Look at how small the difference is — 0.3 kPa out of 101.3, less than a third of one per cent. That tiny gap is all it takes, and it is why quiet breathing costs so little effort. The wording mark is the second one: air is pushed in by the higher pressure outside. “The lungs suck the air in” is marked wrong every time, because nothing in the body can pull on a gas.
(c)[3]
A student writes: “The lungs expand and this pulls the air into them, and the rings of cartilage in the trachea are squeezed shut as the air passes.” Explain why both parts of this statement are wrong.
Model Answer — 3(c)
the lungs contain no muscle, so they cannot expand themselves — they are elastic bags [1]
they are stretched by the movement of the thorax around them, and air is never pulled or sucked — it is pushed in by the higher atmospheric pressure outside once the pressure inside has fallen [1]
the cartilage rings do the opposite of what the student says: they hold the trachea open and stop it collapsing — which is exactly what is needed during inhalation, because the pressure inside the airway has fallen below atmospheric and a floppy tube would be squashed shut by the higher pressure outside [1]
⚠ If you missed marks here: This is worth getting straight once, because the same idea is examined again and again. The chain runs muscles → thorax → lungs → air, in that order, and never backwards. If the lungs really did pull, a hole in the chest wall would make no difference — but it does, because it lets outside air into the space around the lung and the lung collapses at once. Beware the word “suck” entirely; it does not appear in any mark scheme in this topic. The second half of the statement is the same misunderstanding wearing a different hat: cartilage is not muscle and cannot squeeze anything, and its whole purpose is to resist the very pressure drop that the first half of the question is about.
Question 4 — Six Minutes on a Bicycle
Total: 12 marks
A girl sat still for two minutes, then pedalled hard on an exercise bicycle for six minutes, then sat still again. Her breathing rate and the depth of each breath were recorded throughout. The results are shown in Fig. 4.1.
(a)[4]
Describe the changes shown in Fig. 4.1. Use figures from the graph in your answer.
Model Answer — 4(a)
at rest both are steady: breathing rate 14 breaths per minute and depth 500 cm³ for the first 2 minutes [1]
when exercise starts both rate and depth rise sharply — rate to about 40 breaths per minute and depth to about 2500 cm³ [1]
both then level off from about 6 to 8 minutes, and both fall once exercise stops at 8 minutes [1]
they do not return to the resting values immediately — at 10 minutes the rate is still about 22 breaths per minute, and resting values are only reached at about 13 minutes, roughly five minutes after exercise ended [1]
⚠ If you missed marks here: “Describe” means say what the lines do, with numbers and times attached — not why. Two marks are regularly thrown away: forgetting that there are two lines, so that depth never gets mentioned at all; and missing the slow recovery, which is the most interesting feature on the whole graph. Notice also that breathing has already begun to rise within the first minute of exercise, long before the muscles could have run short of anything.
(b)[3]
The volume of air moved in and out of the lungs in one minute is found by multiplying the breathing rate by the depth of each breath. Calculate this volume at 1 minute and at 6 minutes, and state how many times greater the larger value is.
Model Answer — 4(b)
at 1 minute: 14 × 500 = 7000 cm³ per minute (7 dm³ per minute) [1]
at 6 minutes: 40 × 2500 = 100 000 cm³ per minute (100 dm³ per minute) [1]
100 000 ÷ 7000 = about 14 times greater [1]
⚠ If you missed marks here: The point of this calculation is worth more than the marks. The rate roughly trebles and the depth roughly quintuples, and because the two are multiplied together the volume of air handled goes up about fourteenfold. Anyone who says exercise makes you “breathe faster” has caught only the smaller half of what is happening. Always carry the unit through the working: an answer of “100 000” with no unit is a naked number, and per minute is part of the meaning.
(c)[5]
Explain what causes the changes in breathing shown between 2 and 8 minutes.
Model Answer — 4(c)
the muscles are contracting more, so the muscle cells respire faster and produce more carbon dioxide [1]
so the concentration of carbon dioxide in the blood rises [1]
this rise is detected by the brain — the trigger is high carbon dioxide, not a shortage of oxygen [1]
the brain sends nerve impulses to the diaphragm and to the intercostal muscles [1]
so both the rate and the depth of breathing increase; more carbon dioxide is removed and more oxygen taken in, and the carbon dioxide concentration falls back towards normal [1]
⚠ If you missed marks here: The answer nearly everybody gives is “the muscles need more oxygen, so you breathe faster”. That is the wrong trigger, and it is worth understanding why: oxygen is still present at 16% in every breath you push back out, so the body is never anywhere near running out — but carbon dioxide changes a hundredfold, which makes it a far more sensitive signal to monitor. Write out the whole chain in order, ending with the carbon dioxide concentration returning to normal, because that last step is what makes it a control system rather than a one-way effect.
Question 5 — Where the Air Meets the Blood
Total: 10 marks
Fig. 5.1 shows one alveolus and the capillary running over its surface. The enlarged panel on the right shows the layers that a gas molecule must cross, drawn to no particular scale but with the real thicknesses given.
(a)[4]
State the four features of an efficient gas exchange surface, and for each one give the feature of the alveolus shown in Fig. 5.1 that provides it.
Model Answer — 5(a)
large surface area — there are hundreds of millions of alveoli, giving a total area of roughly 70 m² [1]
thin surface — the wall of the alveolus is one cell thick and the capillary wall is one cell thick, so the diffusion distance is about one micrometre [1]
good blood supply — a dense network of capillaries carries oxygen away and brings carbon dioxide, so the concentration gradients are maintained [1]
good ventilation with air — breathing continually replaces the air in the alveolus, keeping the oxygen concentration high and the carbon dioxide concentration low inside it [1]
⚠ If you missed marks here: These are the same four words you already used for the villus and for the root hair cell, and they are worth learning as a set: large surface area, thin, good blood supply, good ventilation. The two that get muddled are the last pair: a good blood supply and good ventilation do exactly the same job — they maintain the concentration gradient. Neither of them “gives more oxygen”, and a large surface area does not “make” oxygen; it lets more diffusion happen at once.
(b)[2]
Name the gas moving in the direction of arrow P and the gas moving in the direction of arrow Q, and in each case explain why it moves in that direction.
Model Answer — 5(b)
P is oxygen, diffusing from the alveolar air into the blood because its concentration is higher in the alveolus than in the blood — it moves down its concentration gradient [1]
Q is carbon dioxide, diffusing from the blood into the alveolar air because its concentration is higher in the blood than in the alveolus [1]
⚠ If you missed marks here: The explanation mark is not for the word diffusion on its own — it is for saying where the concentration is higher. Watch out for the phrase “the blood carries oxygen to the alveoli”, which reverses the whole thing: blood arriving at the alveolus is the deoxygenated blood pumped there by the right ventricle, and it leaves oxygenated.
(c)[2]
Use Fig. 5.1 to calculate the total distance, in µm, that a molecule of oxygen must diffuse to travel from the air in the alveolus into the blood. Give this distance in millimetres as well.
Model Answer — 5(c)
0.40 + 0.25 + 0.35 = 1.00 µm [1]
1 µm is one thousandth of a millimetre, so 1.00 µm = 0.001 mm (1 × 10−3 mm) [1]
⚠ If you missed marks here: Two things to watch. All three layers count — leaving out the film of tissue fluid gives 0.75 µm, and the tissue fluid is real, because a gas has to dissolve before it can cross a membrane. And the conversion goes divide by 1000 to turn µm into mm: multiplying instead gives 1000 mm, a metre of lung wall, which should stop you at once. Getting a feel for the answer matters more than the mark: one micrometre is why a gas crosses in a fraction of a second.
(d)[2]
Oxygen keeps diffusing into the blood for as long as blood flows past the alveolus. Explain how the concentration gradient for oxygen is maintained.
Model Answer — 5(d)
oxygen entering the blood immediately combines with haemoglobin in the red blood cells to form oxyhaemoglobin, so the concentration of free dissolved oxygen in the plasma stays low; the flow of blood also carries the oxygen away [1]
meanwhile ventilation continually replaces the air in the alveolus with fresh air containing 21% oxygen, so the concentration on the air side stays high — both together keep the gradient steep [1]
⚠ If you missed marks here: This is the part of the topic that separates a grade A from a grade C. Diffusion does not need pumping; it needs a difference, and everything the body does here is about refusing to let that difference disappear — haemoglobin mops the oxygen up on one side, breathing tops it up on the other. If your answer said only “the blood takes the oxygen away” you had half of it; add haemoglobin and add ventilation.
Question 6 — The Staircase That Carries Dirt Out of the Lungs
Total: 12 marks
Fig. 6.1 shows part of the layer of cells lining the trachea, seen at high magnification. Two kinds of cell are present, labelled P and Q.
(a)[4]
Name the cells labelled P and Q, and state what each one does.
Model Answer — 6(a)
P is a goblet cell [1]
it secretes mucus, which traps dust particles and pathogens in the air that has been breathed in [1]
Q is a ciliated cell [1]
its cilia beat, wafting the mucus up the trachea, away from the lungs, to the back of the throat where it is swallowed [1]
⚠ If you missed marks here: The two jobs get swapped constantly: cilia do not make mucus, and goblet cells do not beat. Keep the shapes attached to the names — a goblet cell is the one shaped like a wine glass, packed with mucus and with no cilia at all. And do not say the cilia “push the dust out”: they move the mucus, and the dust is a passenger stuck in it.
Pieces of trachea lining were kept alive in a dish and exposed to tobacco smoke. Table 6.1 shows what happened to the cilia.
time of exposure to tobacco smoke / minutes
mean rate at which the cilia beat / beats per second
percentage of ciliated cells with cilia still beating / %
0
12.0
100
5
9.5
96
10
6.2
81
20
2.4
48
40
0.0
9
(b)(i)[2]
Describe the effect of tobacco smoke on the cilia, using figures from Table 6.1.
Model Answer — 6(b)(i)
the longer the exposure, the more slowly the cilia beat — from 12.0 beats per second at 0 minutes to 2.4 at 20 minutes, and they have stopped completely (0.0) by 40 minutes [1]
and fewer cells have any beating cilia at all — falling from 100% to 48% at 20 minutes and to only 9% at 40 minutes; the effect on the beat rate appears first and is proportionally larger at every time [1]
⚠ If you missed marks here: There are two columns of data, so there are two things to describe: how fast the cilia beat, and how many of them are beating at all. Quoting one figure and stopping loses the second mark almost every time. Notice at 5 minutes: the beat rate has already dropped by a fifth while 96% of cells are still going — the cilia are being paralysed before they are destroyed, which is why even a light smoker has a cough.
(b)(ii)[2]
Calculate the percentage decrease in the mean rate at which the cilia beat between 0 and 20 minutes. Show your working.
⚠ If you missed marks here: Divide by the starting value, 12.0, not by the final value 2.4 — dividing by 2.4 gives 400%, and a decrease can never be more than 100%. That impossibility is a free check: if a percentage decrease comes out above 100 you have divided the wrong way round.
(c)[4]
Explain the consequences for a person who smokes tobacco every day of the changes shown in Table 6.1.
Model Answer — 6(c)
the cilia are paralysed and destroyed, so the mucus is no longer swept up the trachea and it collects in the airways and lungs [1]
the only remaining way to shift it is to cough, which is why a regular smoker develops a persistent cough [1]
the trapped pathogens stay in the lungs instead of being swallowed, so they can multiply there and chest infections such as bronchitis become more frequent [1]
the collected mucus and the swollen lining narrow the airways, so less air reaches the alveoli each breath and the person becomes breathless on exertion [1]
⚠ If you missed marks here: Notice that a smoker does not produce a cough because smoke “irritates the throat” — it is because the escalator that normally removes the mucus has stopped, so coughing is the emergency substitute. Swallowing the mucus matters too: acid in the stomach destroys the pathogens caught in it, so once the mucus stays put in the lungs the bacteria have somewhere warm and moist to multiply. Build your answer as a chain of consequences rather than a list of bad things.
Question 7 — The Patient Whose Diaphragm Stopped Working
Total: 10 marks
Read all of this information before you begin. Everything you need that is new is given here.
The diaphragm is made to contract by impulses that travel to it along a nerve from the brain. A man was injured in an accident that damaged this nerve on both sides of his body. His diaphragm can no longer contract, so it stays in its domed shape all the time. His brain, his ribs and his intercostal muscles were not damaged and work normally. He is awake, and he is breathing.
Of every breath a healthy person takes, about 150 cm³ never reaches an alveolus. That air stays in the trachea, the bronchi and the bronchioles, where there is no gas exchange, and it is breathed straight back out unchanged. Table 7.1 gives this man’s figures.
before the accident
after the accident
volume of each breath / cm³
500
250
volume of each breath that never reaches an alveolus / cm³
150
150
(a)[3]
Explain why the man is still able to breathe, and explain why the volume of each breath has fallen to 250 cm³.
Model Answer — 7(a)
his external intercostal muscles still contract and can still pull the ribs upwards and outwards, so the volume of the thorax can still be increased [1]
the pressure inside the thorax therefore still falls below atmospheric pressure and air is still pushed in [1]
but the diaphragm no longer contracts or flattens, so one of the two ways of enlarging the thorax has been lost; the increase in volume is much smaller, so a much smaller volume of air enters — the depth of each breath is halved [1]
⚠ If you missed marks here: In an unfamiliar question like this one, go back to the mechanism rather than hunting for a remembered fact. Ventilation has exactly two engines — the ribs and the diaphragm — so losing one halves the job but does not stop it. Watch the word “domed” in the stem: it tells you the diaphragm is stuck in its relaxed position, which is a quiet reminder that a relaxed diaphragm is domed and a contracted one is flat.
(b)[3]
Using Table 7.1, predict how the composition of this man’s expired air now differs from that of a healthy person. Explain your answer.
Model Answer — 7(b)
before the accident 500 − 150 = 350 cm³ of each breath reached the alveoli; now only 250 − 150 = 100 cm³ does [1]
so a much larger proportion of each breath — 150 out of 250, which is 60% instead of 30% — never touches a gas exchange surface and is breathed out unchanged [1]
his expired air would therefore be closer in composition to inspired air: more than 16% oxygen and less than 4% carbon dioxide [1]
⚠ If you missed marks here: The intuitive answer — less oxygen in his expired air, because he is short of oxygen — is exactly backwards, and it is worth understanding why. His blood is worse off, but the air coming out of his mouth is a mixture of air that reached the alveoli and air that never left the tubes. Shrink the breath while the 150 cm³ of tubes stays the same size, and the unchanged portion dominates the mixture. Whenever a question gives you two numbers that behave differently, do the subtraction before you write a word.
(c)[2]
Predict the effect of the injury on the man’s breathing rate while he is sitting still, and explain your prediction.
Model Answer — 7(c)
his breathing rate would be higher than a healthy person’s at rest [1]
because carbon dioxide is now removed less effectively, so its concentration in the blood rises; this is detected by the brain, which sends more impulses to the intercostal muscles — and since he cannot increase the depth of his breathing, taking more breaths per minute is the only way left to move the same volume of air [1]
⚠ If you missed marks here: The control system in this topic monitors carbon dioxide, so any prediction about breathing rate has to be argued through carbon dioxide, never through “he needs more oxygen”. The satisfying part is the last step: rate and depth are the two things the brain can change, and when one of them is unavailable the other is pushed harder. That is precisely the relationship you calculated in Question 4.
(d)[2]
A newspaper reports the case with the headline: “Man survives despite losing the muscle that pulls his lungs open.” Evaluate this statement.
Model Answer — 7(d)
the diaphragm is correctly described as a muscle, and it is true that he has survived without it — but the biology in the headline is wrong [1]
the lungs are never pulled open: they contain no muscle and nothing is attached to them for the diaphragm to pull on. The diaphragm and the intercostal muscles enlarge the thorax, the pressure inside falls, and air is pushed into the lungs by the higher atmospheric pressure outside, which stretches them [1]
⚠ If you missed marks here: “Evaluate” does not mean “disagree”. Say what is right before you say what is wrong, then correct it precisely — that structure is what earns both marks. The specific error here is the one this whole topic keeps coming back to: nothing pulls, nothing sucks. Muscles move the thorax; the atmosphere does the rest.
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