Biology • Year 11 • Module 2 • Lesson 10
Gas Exchange in Animals
Apply diffusion principles and gas exchange adaptations to real oxygen-uptake data, a counter-current exchange graph, and a scenario comparing tracheal vs lung-based systems.
1. Interpret oxygen-uptake data across animal groups
The table below shows oxygen-uptake rates measured in three animals under resting conditions. A researcher is comparing the efficiency of each gas exchange system. 7 marks
| Animal | Gas exchange structure | Resting O₂ uptake (mL O₂ g⁻¹ h⁻¹) | Body mass (g) |
|---|---|---|---|
| Grasshopper (insect) | Tracheal system | 0.14 | 2 |
| Carp (fish) | Gills (counter-current) | 0.09 | 250 |
| Mouse (mammal) | Alveolar lungs | 2.50 | 25 |
1.1 Describe the difference in mass-specific oxygen uptake between the grasshopper and the mouse. What does this suggest about the metabolic demands of each animal? 2 marks
1.2 The carp has the lowest mass-specific O₂ uptake, yet it is a much larger animal than the grasshopper. Explain why gills (rather than tracheoles) are the appropriate exchange structure for the carp's body size. Use SA:V ratio in your answer. 2 marks
1.3 The researcher notes that the tracheal system limits insect body size. Using the data and lesson content, predict what would happen to oxygen supply if an insect had the same body mass as the carp (250 g). Justify your prediction. 3 marks
2. Interpret a counter-current exchange graph
The diagram below models O₂ concentration (%) along the length of a fish gill lamella, comparing concurrent flow (blood and water moving in the same direction) with counter-current flow (blood and water moving in opposite directions). 7 marks
Stylised model, illustrative of counter-current exchange efficiency.
2.1 In the concurrent model, what is the approximate maximum O₂ concentration the blood can reach? Where along the lamella does this occur? 2 marks
2.2 In the counter-current model, the blood exits with approximately 90% O₂. Explain why a concentration gradient is maintained along the full length of the lamella in this arrangement. 2 marks
2.3 Why would gills be less efficient if a fish's body movements were to stop and water no longer flowed over the lamellae? Use the concept of "maintained concentration gradient" in your answer. 3 marks
3. Apply to a new scenario, a terrestrial crustacean
Woodlice (slaters) are small crustaceans that live on land under rocks and bark. Unlike insects, they lack a tracheal system and instead exchange gases across moist, thin respiratory surfaces on their underside. They die rapidly if exposed to dry air. 6 marks
3.1 Identify which features of an efficient exchange surface the woodlouse respiratory surface already possesses. Explain how each feature supports gas exchange. 3 marks
3.2 Using the lesson's four features of efficient exchange surfaces, explain why woodlice die in dry air. 2 marks
3.3 A biologist argues that the woodlouse exchange surface is more "primitive" than alveoli but solves the same fundamental problem. Identify the one feature that alveoli have to a far greater degree than the woodlouse surface, and explain why this matters for mammals but is less critical for a small crustacean. 1 mark
Q1.1, O₂ uptake comparison, metabolic demand
The mouse (2.50 mL O₂ g⁻¹ h⁻¹) has a mass-specific O₂ uptake approximately 18 times greater than the grasshopper (0.14 mL O₂ g⁻¹ h⁻¹). This suggests the mouse has a much higher metabolic rate, it is endothermic (warm-blooded) and must generate body heat, requiring far more oxygen per gram of tissue.
Q1.2, Why gills suit a larger animal
The carp has a much larger body and hence a lower SA:V ratio than the grasshopper. A tracheal system can only deliver oxygen directly to cells within a few millimetres of a tracheole, workable for a 2 g insect, but completely inadequate for a 250 g fish. Gills, paired with a blood-based transport system, allow oxygen to be collected at one surface and carried to all tissues throughout the body, regardless of body size.
Q1.3, Prediction for a 250 g insect with tracheal system
A 250 g insect with a tracheal system would be unable to supply oxygen to its internal cells. The tracheal system has no mechanism to carry oxygen long distances: tracheoles can only reach cells that are a few millimetres away. A 250 g body would contain internal cells many centimetres from any spiracle [1]. At the grasshopper's O₂ uptake rate (0.14 mL g⁻¹ h⁻¹), a 250 g insect would need 35 mL O₂ h⁻¹, but diffusion along tracheoles would be far too slow to meet that demand [1]. The animal would be unable to survive, this is why tracheal systems constrain insect body size and why no truly large insects exist [1].
Q2.1, Concurrent model maximum O₂
In the concurrent model, blood and water both converge at approximately 50% O₂this occurs at the midpoint (equilibrium point) along the lamella. After equilibrium, the gradient disappears and no further diffusion occurs, so blood cannot acquire more oxygen for the remainder of the surface.
Q2.2, Why counter-current maintains a gradient
In counter-current flow, blood travels in the opposite direction to water. As blood picks up oxygen and its O₂ concentration rises along the lamella, it always encounters water that has not yet lost its oxygen, water flowing in from the inlet still carries a higher O₂ concentration than the blood it meets. A small gradient persists all the way along the gill surface, so diffusion never stops. This is why blood can exit with ~90% O₂ instead of being capped at ~50%.
Q2.3, Without water flow
If water stopped flowing over the lamellae, the water immediately adjacent to the gill surface would quickly lose its oxygen to the blood [1]. With no fresh oxygenated water arriving to replace it, the O₂ concentration on the water side would fall to match the blood side [1]. Once the concentration gradient disappears, net diffusion stops, and the fish cannot take up oxygen even though its gill surface is structurally intact, showing that maintaining the gradient by ventilation is just as essential as having the surface itself [1].
Q3.1, Exchange surface features of woodlice
Thin surface: The respiratory surfaces are thin, reducing diffusion distance so gases cross quickly. Moist surface: They are kept moist, allowing O₂ and CO₂ to dissolve before crossing membranes, essential for gas exchange. Maintained concentration gradient: Blood (haemolymph) circulating below the surface carries O₂ away from the exchange site and CO₂ towards it, maintaining gradients. (Award 1 mark per correctly identified feature with a correct explanation, max 3.)
Q3.2, Why woodlice die in dry air
Woodlice die in dry air because their respiratory surface loses its moisture [1]. Without a moist surface, O₂ and CO₂ cannot dissolve and cross cell membranes, so gas exchange stops even though the surface area and thin membrane remain intact, the "moist surface" feature has been removed [1].
Q3.3, Alveoli vs woodlouse: key difference
Alveoli provide a far greater surface area (hundreds of square metres in a human) compared to the woodlouse surface. This matters for a large mammal because its low SA:V ratio and high metabolic rate demand an enormous exchange area. A woodlouse is tiny, so its high SA:V ratio means a modest external surface is sufficient for its gas exchange needs, it does not require the vast internal folding that alveoli provide.