Question 1
Paper 1A style
Which change happens as an organism of the same shape increases in size?
- Surface area increases and the surface area-to-volume ratio increases
- Surface area increases and the surface area-to-volume ratio decreases
- Volume decreases relative to surface area
- The distance from the centre to the exterior decreases
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Answer: B [1]
Question 2
Paper 1A style
What is the role of surfactant in the alveoli of mammalian lungs?
- It dissolves oxygen so that it can diffuse into the blood
- It kills bacteria that enter the alveoli with inhaled air
- It reduces surface tension so that the alveoli do not collapse
- It increases the thickness of the alveolus wall to prevent damage
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Answer: C [1]
Question 3
Paper 2A style
Outline how concentration gradients are maintained at the gas-exchange surface of a mammal.
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- dense network of capillaries close to the alveoli [1]
- continuous blood flow carries oxygenated blood away / brings deoxygenated blood [1]
- ventilation replaces air in alveoli / brings fresh air with high O₂ concentration and removes CO₂ [1]
- keeps O₂ concentration higher in alveoli than in blood / CO₂ higher in blood than in alveoli [1]
- max 3
Question 4
Paper 1B style
A student measured her lung volumes with a spirometer at rest. The values are practice data. (a) Calculate her vital capacity. (b) Calculate her total lung volume. (c) State which value in the table cannot be measured with a spirometer and explain why. (d) Explain how contraction of the external intercostal muscles and diaphragm causes inspiration.
| Volume | Value / dm³ |
|---|---|
| Tidal volume | 0.5 |
| Inspiratory reserve volume | 2.6 |
| Expiratory reserve volume | 1.0 |
| Residual volume | 1.2 |
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- (a) 0.5 + 2.6 + 1.0 = 4.1 dm³ [1]
- (b) 4.1 + 1.2 = 5.3 dm³ [1] ECF from (a)
- (c) residual volume, because this air always remains in the lungs / cannot be breathed out into the spirometer [1]
- (d) ribs move up and out and diaphragm flattens/moves down, so thorax volume increases [1]
- (d) pressure in the thorax/lungs falls below atmospheric so air flows in [1]
Question 5
Paper 2A style
Explain why plants cannot avoid losing water by transpiration.
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- stomata must be open to allow CO₂ to enter for photosynthesis [1]
- water evaporates from (moist) mesophyll cell walls into air spaces [1]
- water vapour diffuses out through open stomata down a concentration gradient [1]
- gas exchange surface must be moist/permeable so water is lost from it [1]
- max 3
Question 6
Paper 1B style
Leaf casts were made of the lower epidermis of two species. Stomata were counted in five fields of view at ×400. The field of view diameter was 0.45 mm. (a) Calculate the mean stomatal density of species X. Give your answer in stomata per mm². (b) Suggest why five fields of view were counted. (c) Species Y lives in a hot, dry habitat. Suggest how its stomatal density helps it survive.
| Species | Count 1 | Count 2 | Count 3 | Count 4 | Count 5 |
|---|---|---|---|---|---|
| X | 31 | 35 | 29 | 33 | 32 |
| Y | 12 | 15 | 11 | 14 | 13 |
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- (a) mean count = 160 ÷ 5 = 32 [1]
- (a) area = π × 0.225² = 0.159 mm²; density = 32 ÷ 0.159 = 201 stomata per mm² (accept 200–202) [1]
- (b) stomatal number varies across the leaf/biological variability, so replicates give a reliable mean [1]
- (c) fewer stomata so less water vapour lost by transpiration [1]