Questions on gas exchange in mammals: lungs and ventilation

Question 1

A spirometer trace shows 21 complete breaths in 84 seconds, each moving 0.55 dm³ of air. Calculate the pulmonary ventilation rate in dm³ min⁻¹. About 0.15 dm³ of every breath stays in the dead space; calculate the volume of air reaching the alveoli each minute.

Accepted Answer

M1 ventilation rate = 21 × (60 ÷ 84), converting the count to a number of breaths per minute A1 15 breaths per minute M1 pulmonary ventilation rate = tidal volume × ventilation rate = 0.55 × 15 A1 8.3 dm³ min⁻¹ (8.25 unrounded), with the unit given A1 alveolar volume = (0.55 − 0.15) × 15 = 6.0 dm³ min⁻¹, because the dead space is paid for on every breath

Question 2

The table shows the tissues found in the wall at three levels of the human airway. Compare the wall of the trachea with the walls of a small bronchiole and of an alveolus, and explain what each difference allows that part of the airway to do.

Accepted Answer

B1 the trachea has C-shaped rings of cartilage that hold it permanently open against the fall in pressure during inspiration, whereas neither the bronchiole nor the alveolus has any cartilage B1 the trachea is lined with ciliated epithelium and goblet cells, so mucus traps particles and the cilia sweep it up towards the throat, whereas the goblet cells and then the cilia are lost in the smallest bronchioles and the alveolus has neither B1 the bronchiole has a thick layer of smooth muscle and no cartilage, so contracting that muscle narrows the lumen and raises the resistance to air flow, while the alveolus has no smooth muscle B1 the alveolar wall is a single layer of squamous cells a fraction of a micrometre thick against a capillary, which gives the short diffusion path gas exchange needs, whereas the trachea's thick wall exchanges nothing, so the air in it is dead space B1 all three contain elastic fibres, which stretch during inspiration and recoil afterwards, driving quiet expiration and widening an airway again when its smooth muscle relaxes

Question 3

A spirometer measures the volume of air a subject moves in and out of the lungs, with soda lime in the circuit absorbing the carbon dioxide they breathe out. Evaluate this apparatus as a way of measuring the total volume of air the lungs can hold and the subject's rate of oxygen consumption.

Accepted Answer

B1 every volume that actually moves is recorded directly and continuously, so tidal volume, vital capacity and ventilation rate all come from one record of the same session B1 for oxygen consumption it is well suited: soda lime removes the carbon dioxide breathed out, so the volume of gas in the circuit falls by exactly the volume of oxygen absorbed, and the gradient of that fall is the rate B1 against it: the residual volume never leaves the lungs however hard the subject breathes out, so a spirometer cannot measure it and cannot give total lung capacity, which needs a different method such as dilution of an inert gas B1 against it: the readings depend on the subject's effort and on there being no leak at the mouthpiece, and the apparatus must be sterile with fresh soda lime and the subject free of respiratory infection, or the value is both unreliable and unsafe to obtain A1 a judgement: it is the right instrument for the volumes that move and for oxygen uptake, but it answers the question only in part, because the one quantity it cannot reach is the total volume the lungs hold

Question 4

A person with asthma was tested with a spirometer before and ten minutes after using a bronchodilator inhaler, which relaxes smooth muscle. The values are constructed for practice. Calculate the percentage increase in FEV₁ after the inhaler, and explain why FEV₁ changed much more than FVC.

Accepted Answer

M1 percentage increase = (3.2 − 1.8) ÷ 1.8 × 100 A1 78%, accepting 77.8% B1 the bronchodilator relaxes the smooth muscle in the walls of the bronchioles, so their lumen widens B1 the resistance to air flow falls, so air can be breathed out faster and more of it leaves in the first second B1 FVC depends on how far the lungs can fill and empty, which asthma barely changes, so given time almost the same volume leaves and FVC rises only slightly

Question 5

Smoking in enclosed public places and workplaces was banned across the UK in 2006 and 2007. Evaluate the evidence used to justify this ban, using what you know about studies of smokers and of non-smokers exposed to other people's smoke.

Accepted Answer

B1 for: large studies, such as the British Doctors Study, consistently found more lung cancer and chronic obstructive lung disease in smokers, with risk rising with the number smoked and falling after stopping B1 for: studies of non-smokers living or working with smokers found a raised risk of lung cancer (about 20 to 30 per cent; the figure is not required), and the ban protects people such as bar staff who had not chosen that risk B1 for: there is a plausible mechanism, since tar contains carcinogens and smoke damages cilia and the elastic fibres of the alveoli B1 against: the data are correlational rather than from randomised experiments, so confounding variables such as occupation, diet or air pollution cannot be ruled out entirely, and exposure to other people's smoke is self-reported and hard to measure B1 a judgement: no single study proves cause, but the consistency of many studies, the dose-response relationship and the mechanism together make a causal link very likely, so the evidence justified the ban

Question 6

Describe how air is drawn into the lungs during inspiration, giving the events in the order in which they happen.

Accepted Answer

B1 the external intercostal muscles contract, pulling the ribcage upwards and outwards B1 the diaphragm contracts, flattens and moves downwards B1 the volume of the thoracic cavity increases, and the lungs are stretched with it A1 the pressure inside the lungs falls below atmospheric pressure, so air moves in down the pressure gradient

Question 7

Explain why inspiration always costs the body energy while quiet expiration costs almost none.

Accepted Answer

B1 inspiration depends on the diaphragm and external intercostal muscles contracting, and muscle contraction requires ATP B1 elastic fibres in the alveolar walls and airways are stretched as the lungs inflate, storing energy as they do so B1 in quiet expiration those muscles simply relax, and the stretched elastic tissue recoils, which needs no contraction of any muscle A1 the recoil reduces the volume of the thorax, so pressure in the lungs rises above atmospheric and air moves out

Question 8

During vigorous exercise, expiration stops being largely passive. Explain how the muscles of the chest and abdomen drive air out of the lungs faster than elastic recoil alone can, giving the events in causal order.

Accepted Answer

B1 the internal intercostal muscles contract while the external intercostals relax, the two being an antagonistic pair, and the ribcage is pulled downwards and inwards B1 the abdominal muscles contract at the same time and push the abdominal contents upwards against the relaxed diaphragm, forcing its dome further into the thorax B1 the volume of the thoracic cavity is therefore reduced further, and faster, than the recoil of the stretched elastic fibres alone would reduce it A1 pressure in the lungs rises further above atmospheric pressure, so the gradient is steeper, air leaves more quickly and more of the air in the lungs is expelled in each breath

Question 9

Each lung is covered by a pleural membrane, and a second pleural membrane lines the inside of the thoracic cavity. Between them is a thin film of fluid that lets the two membranes slide over each other but stops them being pulled apart. Explain how this arrangement allows the lungs to inflate during inspiration, and suggest why a wound that lets air into the space between the two membranes stops the lung on that side from inflating.

Accepted Answer

B1 the external intercostal muscles and the diaphragm contract, raising the ribs and flattening the diaphragm, so the volume of the thoracic cavity increases B1 because the two membranes cannot be pulled apart, the surface of the lung is carried outwards with the chest wall, so the lung expands with the cavity B1 the larger lung volume lowers the air pressure inside it below atmospheric pressure, so air flows in down the pressure gradient B1 air in the space lets the membranes separate, so the chest wall moves out without pulling the lung with it; the elastic lung recoils and collapses instead of inflating

Question 10

A person has a vital capacity of 4.6 dm³ and a residual volume of 1.2 dm³. Calculate their total lung capacity, and calculate the percentage of that total lung capacity which vital capacity represents.

Accepted Answer

M1 total lung capacity = vital capacity + residual volume, so 4.6 + 1.2 A1 5.8 dm³, the total lung capacity M1 percentage = (vital capacity ÷ total lung capacity) × 100, so (4.6 ÷ 5.8) × 100 A1 79%, accepting 79.3%, of total lung capacity

Question 11

The table shows the results of a forced breath out for three adults of similar age and height. The values are constructed for practice. Calculate FEV₁ ÷ FVC for each person, and use your answers to suggest which person has asthma and which has pulmonary fibrosis.

Accepted Answer

M1 ratio = FEV₁ ÷ FVC for each person, for example 1.4 ÷ 3.5 for Q A1 P 0.80, Q 0.40 and R 0.86; accept 0.9 for R B1 Q has asthma: the ratio is far below 0.7, so the airways are narrowed and air cannot be forced out quickly, an obstructive pattern B1 R has pulmonary fibrosis: the ratio is normal but the FVC is much smaller than P's, so the lungs cannot expand fully, a restrictive pattern

Question 12

In pulmonary fibrosis, scar tissue forms in the walls of the alveoli. Explain why a person with pulmonary fibrosis has a low concentration of oxygen in the blood leaving the lungs, and why their breathing is rapid and shallow.

Accepted Answer

B1 scar tissue thickens the alveolar walls, so the diffusion pathway is longer and oxygen diffuses into the blood more slowly B1 scarring replaces some alveoli, so the surface area for gas exchange is smaller and less oxygen crosses each minute B1 the scarred lungs are stiff and cannot expand fully, so the tidal volume is reduced and the person cannot breathe deeply B1 to take in enough air the ventilation rate rises, so breathing becomes rapid as well as shallow

Question 13

A student collected data from ten towns and found a positive correlation between the mean concentration of fine particles in the air and the number of hospital admissions for asthma per 100 000 people. Explain why these data do not show that fine particles cause asthma attacks, and suggest two further pieces of evidence that would support a causal link.

Accepted Answer

B1 a correlation shows that two variables change together, not that a change in one causes the change in the other B1 a confounding variable could explain both, for example towns with more traffic may also have more smoking, more damp housing or a different age structure B1 any one of: admissions in the same town rise in weeks when particle levels are higher, or fall when pollution is reduced; a dose-response relationship, with admissions rising steadily with particle concentration; the link persisting after allowing for smoking and other confounding variables; a mechanism, such as experiments showing that fine particles inflame the airway lining or cause bronchiole smooth muscle to contract B1 a second, different piece of evidence from the list in the previous point

Question 14

A subject breathes through a spirometer with soda lime in the circuit. A line drawn through the midpoints of the breaths falls from 3.40 dm³ to 2.80 dm³ over 150 seconds. Calculate the rate of oxygen uptake in cm³ min⁻¹, and explain why the volume in the spirometer falls only because soda lime is present.

Accepted Answer

M1 fall in volume = 3.40 − 2.80 = 0.60 dm³, which is 600 cm³, and 150 s is 2.5 minutes A1 600 ÷ 2.5 = 240 cm³ min⁻¹ B1 soda lime absorbs the carbon dioxide the subject breathes out, so it is not returned to the circuit B1 the oxygen absorbed into the blood is therefore not replaced, so the total volume of gas in the spirometer falls by the volume of oxygen used

Question 15

A student breathes through a spirometer with soda lime in the circuit for 30 seconds immediately after exercise, taking 11 complete breaths with a mean tidal volume of 1.10 dm³, while a line through the troughs falls by 0.45 dm³ over the 30 seconds. These readings are invented for this question. Calculate the respiratory minute ventilation and the oxygen consumption per minute.

Accepted Answer

M1 breathing rate = 11 × 2 = 22 breaths per minute A1 minute ventilation = 1.10 × 22 = 24.2 dm³ min⁻¹ M1 0.45 dm³ is taken up in 0.5 minutes, so 0.45 ÷ 0.5 A1 oxygen consumption = 0.90 dm³ min⁻¹

Question 16

In emphysema the elastic fibres of the alveolar walls are destroyed and the walls themselves break down, so that many small alveoli are replaced by fewer, larger air spaces. Suggest why a person with emphysema becomes breathless on mild exertion.

Accepted Answer

B1 the total surface area available for gas exchange falls, and rate of diffusion is proportional to surface area B1 loss of elastic fibres weakens the recoil, so expiration is no longer largely passive and less air is expelled per breath A1 less oxygen therefore reaches the blood each minute, so ventilation has to rise steeply to meet the extra demand of exercise and the person feels breathless

Question 17

Describe the structure that holds the trachea permanently open, and describe why the rings are incomplete rather than complete circles.

Accepted Answer

B1 the trachea is held open by rings of cartilage in its wall, a firm but flexible supporting tissue B1 the rings are C-shaped rather than complete circles, with a gap at the back of the trachea B1 the oesophagus lies against that gap, so a swallowed mouthful of food can bulge into it rather than being crushed against a rigid, fully closed ring

Question 18

The alveoli are lined with a thin film of moisture, and surface tension in this film tends to pull the walls of each alveolus inwards, most strongly in the smallest alveoli. Specialised cells in the alveolar wall secrete surfactant, which lowers this surface tension. Premature babies are often short of surfactant. Explain why such a baby has a low concentration of oxygen in its blood and has to work harder to breathe.

Accepted Answer

B1 without enough surfactant, the surface tension pulls many of the smallest alveoli shut between breaths B1 a collapsed alveolus has no surface open to the air, so the total surface area for gas exchange falls and less oxygen diffuses into the blood B1 each collapsed alveolus has to be forced open again against the surface tension at every breath, so the breathing muscles must do more work

Question 19

During an asthma attack the smooth muscle in the walls of the bronchioles contracts strongly and narrows the airway. Suggest why bronchioles are affected in this way while the trachea and bronchi are not narrowed to nearly the same extent.

Accepted Answer

B1 the trachea and the bronchi are held open by cartilage, whether as C-shaped rings or irregular blocks, and cartilage does not contract, so it resists narrowing regardless of what the smooth muscle in the wall does B1 the wall of a bronchiole contains a conspicuous layer of smooth muscle and no cartilage at all, so there is nothing rigid to resist the muscle when it contracts strongly B1 the bronchiole's lumen therefore narrows sharply during an asthma attack, raising resistance to air flow and reducing the volume of air that can reach the alveoli each breath, whereas the trachea and bronchi stay open

Question 20

Describe how an alveolus can be identified in a section of lung tissue, distinguishing it from a bronchiole.

Accepted Answer

B1 an alveolus is lined by a single layer of very flattened, squamous epithelial cells, rather than by the cuboidal or columnar ciliated cells found further up the airway B1 an alveolus wall contains no smooth muscle and no cartilage, and is wrapped in a dense network of capillaries visible pressed against it B1 a bronchiole, by contrast, has a conspicuous layer of smooth muscle in its wall and a visible lumen wide enough to be a distinct airway, rather than being one of many tiny air spaces in a cluster

Question 21

Explain why the FEV₁ of a person with emphysema is lower than that of a healthy person of the same size, and why a bronchodilator inhaler improves it very little.

Accepted Answer

B1 the elastic fibres in the alveolar walls are destroyed, so the lungs lose the elastic recoil that helps drive air out B1 without the support of the elastic tissue, the small airways collapse when the person breathes out hard, trapping air, so less leaves in the first second B1 the narrowing comes from damaged lung tissue rather than from contracted smooth muscle, so relaxing smooth muscle with a bronchodilator changes little

Question 22

In active tuberculosis, areas of lung tissue are destroyed and later replaced by scar tissue. Suggest how this reduces the rate of gas exchange in the lungs.

Accepted Answer

B1 alveoli are destroyed, so the surface area for gas exchange is reduced B1 scar tissue is thicker than the alveolar wall it replaces, so the diffusion pathway is longer B1 damaged and scarred regions are poorly ventilated, so less fresh air reaches them and the concentration gradient for oxygen is smaller

Question 23

A person with emphysema has a pulmonary ventilation rate of 9.6 dm³ min⁻¹ at rest and a tidal volume of 0.40 dm³. Calculate their ventilation rate, and suggest why it is higher than the 12 to 15 breaths per minute of a healthy person at rest.

Accepted Answer

M1 ventilation rate = pulmonary ventilation rate ÷ tidal volume = 9.6 ÷ 0.40 A1 24 breaths per minute B1 the surface area of the alveoli is reduced, so less oxygen enters the blood with each breath, and more breaths are needed each minute to take up enough oxygen and remove carbon dioxide

Question 24

Explain how each of the following would affect the oxygen consumption calculated from a spirometer trace: a leak at the nose clip, and soda lime that has been used up. Explain why the used-up soda lime is also a hazard to the subject.

Accepted Answer

B1 a leak lets gas escape from the circuit, so the volume falls faster and oxygen consumption is overestimated B1 used-up soda lime does not absorb the carbon dioxide breathed out, which partly replaces the oxygen taken up, so the volume falls less and oxygen consumption is underestimated B1 carbon dioxide builds up in the air the subject rebreathes, which can cause breathlessness, headache or dizziness

Question 25

A section through an airway shows irregular blocks of cartilage in the wall, together with a ciliated epithelium containing goblet cells. Name this airway, and name the tissue that forms a conspicuous layer in the wall of a bronchiole in place of cartilage.

Accepted Answer

A1 the airway is a bronchus, identified from the blocks of cartilage in its wall A1 a bronchiole has a conspicuous layer of smooth muscle, with elastic fibres, and no cartilage at all

Question 26

Name the volume of air remaining in the lungs after the hardest possible breath out, and name the reason a spirometer cannot measure it directly.

Accepted Answer

A1 the residual volume A1 a spirometer can only measure air that actually moves in and out of the circuit, and the residual volume never leaves the lungs however hard the subject breathes out

Feature	Trachea	Small bronchiole	Alveolus
Cartilage	C-shaped rings	none	none
Lining	ciliated epithelium with goblet cells	cuboidal epithelium, goblet cells gone, cilia lost in the smallest	a single layer of squamous cells
Smooth muscle	present	a thick, conspicuous layer	none
Elastic fibres	present	present	present
Thickness of the wall	many layers of tissue	several layers of tissue	a fraction of a micrometre, against a capillary

When tested	FEV₁ / dm³	FVC / dm³
Before the inhaler	1.8	3.9
After the inhaler	3.2	4.1

Person	FEV₁ / dm³	FVC / dm³
P	3.2	4.0
Q	1.4	3.5
R	1.9	2.2