Adaptations of gas exchange surfaces - AQA A-Level Biology (Paper 1) Exam Questions

Practise 19 exam-style questions on Adaptations of gas exchange surfaces for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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1–8 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Mosquito larvae live in still water. A larva hangs just below the water surface and takes in air through a short tube called a siphon, which has spiracles at its tip.

Spreading a thin layer of oil on the surface of the water kills mosquito larvae.

Suggest how the oil kills the larvae.

Model answer

The oil covers the spiracles at the tip of the siphon (and may enter the tracheae), so air cannot get into the tracheal system. Oxygen can no longer diffuse along the tracheae and tracheoles to the cells, so the cells cannot respire aerobically and not enough ATP is made for the larva to survive.

Explanation

An insect larva relies on its tracheal system for oxygen, and the only way into that system here is through the spiracles at the water surface. Blocking them stops air entering, so oxygen cannot diffuse to the tissues. Good answers link the blocked spiracles to the tracheal system and then to less aerobic respiration; 'it suffocates' on its own does not explain anything.

Mark scheme

1 mark
Oil blocks/covers the spiracles/siphon OR oil enters (and blocks) the tracheae
Max 2. Accept 'the larva cannot hang from the surface / the siphon cannot break through the oily surface, so it cannot reach the air' for this point. Ignore 'the oil poisons the larvae'.
1 mark
(So) air/oxygen cannot enter the tracheal system / less oxygen diffuses to the tissues/cells
Ignore 'the larva suffocates' unless linked to air or oxygen not entering the tracheal system.
1 mark
(So) little/no aerobic respiration, so not enough ATP / energy is released (for the larva to survive)
Reject 'energy is produced'.

Question 2

3 marks

A student dissected the gills of a fish obtained from a fishmonger. She cut off one gill filament, mounted it on a slide and viewed it with an optical microscope. The eyepiece graticule had been calibrated: 1 eyepiece division = 25 μm.

The student counted 18 lamellae along a length of 30 eyepiece divisions on one side of the filament.

Calculate the number of lamellae per mm of filament.

Suggest how the student could make her estimate of the number of lamellae per mm more representative of the fish's gills.

Model answer

Length counted = 30 × 25 = 750 μm = 0.75 mm
Lamellae per mm = 18 ÷ 0.75 = 24

She could count the lamellae on several filaments from different gills (and from more than one fish) and calculate a mean.

Explanation

Convert eyepiece divisions to a real length first (30 × 25 μm = 750 μm), then convert to mm (1 mm = 1000 μm) before dividing the number of lamellae by the length. One filament may not be typical, so sampling filaments from several gills, or several fish, and calculating a mean gives a more representative value.

Mark scheme

1 mark
Correct method: 18 ÷ (30 × 25 ÷ 1000) / 18 ÷ 0.75; or a correct final answer of 24 with no working
A correct answer of 24 lamellae per mm with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 750 μm or 0.75 mm (the length counted) OR 0.024 (lamellae per μm, not converted to mm) OR 0.042 (mm per lamella, inverted).
1 mark
24 (lamellae per mm)
1 mark
Count lamellae on several/different filaments (from different gills / different parts of the gill / different fish) and calculate a mean
Accept 'count at several places along the filament and calculate a mean'. Ignore 'repeat the count' unqualified.

Question 3

4 marks

Scientists measured the rate of carbon dioxide release by a resting desert beetle over 60 minutes. The beetle opened its spiracles only during the periods when carbon dioxide was released rapidly. For the rest of the time its spiracles were closed. The graph shows the results.

Use the graph to calculate the percentage of the 60 minutes for which the beetle's spiracles were open.

Suggest and explain why this pattern of spiracle opening is an advantage to the beetle.

Line graph of rate of carbon dioxide release in arbitrary units (y-axis, 0 to 2) against time in minutes (x-axis, 0 to 60). The line joins the points (0, 0.1), (10, 0.1), (11, 1.8), (12, 1.2), (13, 0.1), (28, 0.1), (29, 1.9), (30, 1.4), (31, 0.8), (32, 0.1), (47, 0.1), (48, 1.8), (49, 1.5), (50, 1), (51, 0.5), (52, 0.1), (60, 0.1).

Model answer

Spiracles open from 10 to 13, 28 to 32 and 47 to 52 minutes = 3 + 4 + 5 = 12 minutes
Percentage of time open = 12 ÷ 60 × 100 = 20%

Water vapour diffuses out of the tracheal system whenever the spiracles are open. Keeping them closed for 80% of the time reduces water loss, which matters for an insect living in a hot, dry desert. Because the beetle is resting, its rate of respiration is low, so opening the spiracles in short bursts is enough to let out the carbon dioxide that has built up and let in enough oxygen.

Explanation

Read the start and end of each burst from the graph (where the line leaves and returns to the low baseline) and add the three periods before dividing by 60. The pattern is a compromise: gas exchange needs open spiracles, but open spiracles also let water vapour diffuse out, so a desert insect keeps them closed as much as its low resting respiration rate allows. Answers that say only 'it saves water' need to link this to the spiracles being closed.

Mark scheme

1 mark
Correct method: total time open ÷ 60 × 100, e.g. (3 + 4 + 5) ÷ 60 × 100; or a correct final answer of 20 (accept 18 to 22) with no working
Max 4. A correct answer of 20% (accept 18 to 22) with no working scores 2 marks (points 1 and 2). The spiracles were open from 10 to 13, 28 to 32 and 47 to 52 minutes, 12 minutes in total. Accept for 1 mark: evidence of a total time open of 11 to 13 minutes OR an answer of 80 (the percentage of time closed).
1 mark
20 (%) (accept 18 to 22)
1 mark
Water (vapour) is lost / diffuses out (of the tracheal system) through open spiracles
Points 3 to 6: max 2. Accept 'water evaporates through the spiracles'.
1 mark
Spiracles are closed for most of the time / for 80% of the time, so less water is lost, which is important in a hot/dry (desert) habitat
Ignore 'to conserve water' unless linked to the spiracles being closed.
1 mark
At rest the rate of respiration is low, so short periods of opening let in enough oxygen
1 mark
Carbon dioxide builds up (in the tracheae) while the spiracles are closed and is released in short bursts when they open

Question 4

6 marks

Describe and explain how the gas exchange system of a terrestrial insect allows efficient gas exchange while limiting water loss.

Model answer

Air enters through spiracles on the sides of the body and passes along tracheae, which are kept open by rings of chitin, into tracheoles. The tracheoles are very numerous and highly branched, giving a large surface area, and their walls are thin, giving a short diffusion distance. They reach right into the tissues, so oxygen diffuses directly into the respiring cells without any need for blood to carry it. Respiration uses oxygen and produces carbon dioxide, which maintains the diffusion gradients, and in active insects abdominal pumping ventilates the tracheae. During activity, fluid is withdrawn from the tracheole ends so air reaches the cells and diffusion is faster.
Water loss is limited because the spiracles can be closed by valves when the insect does not need much oxygen, which reduces the diffusion of water vapour out of the tracheal system. The body is covered by a waterproof, waxy cuticle that reduces evaporation, and hairs around the spiracles trap moist air, reducing the water vapour gradient.

Explanation

A terrestrial insect needs a large, thin exchange surface (the tracheoles) close to its cells, but any surface open to the air also loses water. The compromise is that the spiracles can be shut and the rest of the body is waterproofed by a waxy cuticle. Strong answers describe the structure (spiracles, tracheae, tracheoles) and explain each feature in terms of surface area, diffusion distance or diffusion gradient, then give the water-saving features. Writing about lungs or blood transport loses credit.

Mark scheme

1 mark
Air enters through spiracles and passes along tracheae (supported by chitin) to tracheoles
Max 4 if the answer does not refer to limiting water loss. Max 6. Reject lungs, alveoli or bronchi.
1 mark
Tracheoles are highly branched/numerous, so there is a large surface area (for diffusion)
Ignore 'surface area to volume ratio' for this point.
1 mark
Tracheoles have thin walls, so there is a short diffusion distance/pathway
1 mark
Tracheoles reach/enter the tissues/cells, so oxygen diffuses directly into the cells / no (blood) transport system is needed
Accept 'tracheoles are in contact with muscle fibres'.
1 mark
Use of oxygen and production of carbon dioxide in (aerobic) respiration maintains the diffusion/concentration gradients
1 mark
(Abdominal) pumping / ventilation movements move air in and out (maintaining the gradients)
1 mark
Fluid in the ends of the tracheoles is withdrawn into the cells (during activity), so air reaches further and diffusion is faster / surface area for gas diffusion increases
1 mark
Spiracles can be closed (by valves), reducing the loss of water vapour
Points 8, 9 and 10 are ways of limiting water loss.
1 mark
Waxy cuticle / exoskeleton is waterproof/impermeable, so it reduces evaporation from the body surface
Accept 'water cannot diffuse through the cuticle'.
1 mark
Hairs around the spiracles trap moist air, reducing the water vapour (concentration) gradient
Accept 'reduces the water potential gradient'.

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