What you need to know (based on the AQA specification)
What you need to know (based on the AQA specification)
- The relationship between the size of an organism or structure and its surface area to volume ratio.
- Changes to body shape and the development of systems in larger organisms as adaptations that facilitate exchange as this ratio reduces.
- Students should be able to appreciate the relationship between surface area to volume ratio and metabolic rate.
Surface Area
Organisms exchange substances (oxygen, carbon dioxide, nutrients, waste) with their environment across a surface. This surface depends on the organism, for example:
- In single-celled organisms (e.g. an amoeba), the surface is the cell membrane
- In human lungs, the surfaces are the alveoli
- In fish, the surfaces are the gills
- In plants, the surfaces are the leaves
Volume
The amount of substance organisms need to exchange depends on their volume. If organisms have a larger volume, they have more cells, so a greater demand for nutrients.
The ratio between these two — surface area to volume ratio (SA:V) — determines whether the organisms can use simple diffusion, or whether specialised exchange surfaces and transport systems are needed.
Calculating SA:V Ratio
As an organism (or structure) gets larger, its volume increases faster than its surface area, so its SA:V ratio decreases.

The image above shows three worked examples.
- Doubling the side length of a cube (1 cm → 2 cm) halves the SA:V ratio (6:1 → 3:1).
- The 2 cm cube and the 4 × 2 × 1 cuboid have the same volume (8 cm³), but the cuboid has a higher SA:V (3.5:1 vs 3:1).
- Flatter / more elongated shapes give a higher SA:V ratio, even at the same volume.
You need to be able to calculate SA:V for any given shape (cube, cuboid, sphere) — exam questions may give the dimensions and ask you to work it out.
Single-celled vs Multicellular Organisms
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Single-celled organisms have a large SA:V ratio — they can absorb and release gases (and other substances) by diffusion directly across their outer surface.
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Multicellular organisms have a smaller SA:V ratio — diffusion across the body surface alone is too slow to meet the demands of all their cells (long diffusion distances, slow rate of diffusion)
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They compensate with adaptations to their body shape and development of specialised systems:
- Flattened or elongated body shapes (e.g. flatworms) — increase surface area relative to volume
- Folded / branched exchange surfaces — alveoli in lungs, villi in the small intestine, root hairs in plants — all increase surface area
- Mass transport systems — circulatory system in animals, xylem/phloem in plants — move substances quickly between exchange surfaces and tissues
SA:V and Metabolic Rate
Smaller organisms (or organisms in cold environments) have a higher SA:V ratio, so they lose heat to the environment more rapidly.
To maintain a stable body temperature, they need a higher metabolic rate. A higher metabolic rate means more respiration, so more heat is produced.
Examples include:
- Small mammals (e.g. shrews, mice) — high SA:V, so they lose heat rapidly. To compensate, they have a high metabolic rate and must eat frequently to fuel respiration.
- Large mammals (e.g. elephants, whales) — low SA:V, so they lose heat slowly. Because of this, they don’t need as high a metabolic rate per unit mass.
Animals in cold climates often have compact body shapes (e.g. seals are rounder with shorter limbs) to reduce SA:V and minimise heat loss. The reverse is true for animals in hot climates.
Exam Question Practice
Describe and explain the relationship between surface area to volume ratio of the human body and metabolic rate.
(3 marks)Hint
Focus on the human body as specified. What is the relationship between SA:V ratio and heat loss? How does this affect metabolic rate?
Mark Scheme
- As surface area to volume ratio increases, metabolic rate increases OR (humans with) a large surface area to volume ratio have a high/fast metabolic rate (1 mark)
- (A larger surface area to volume ratio will) lose more heat (1 mark)
- (A higher rate of metabolism/respiration) releases/provides/replaces heat OR maintains body temperature (1 mark)
Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.
(2 marks)Hint
Start with the size–SA:V relationship, then think about what a specialised exchange surface does to the diffusion pathway.
Mark Scheme
- Larger organisms have a smaller surface area : volume ratio OR smaller organisms have a larger SA:V ratio (1 mark)
- Overcomes long diffusion pathway OR faster diffusion (1 mark)
Comments from mark scheme
- Point 2: accept “short diffusion pathway”
- Point 2: accept “more” for “faster”
- Explain why larger animals have a smaller SA:V ratio — not just state it
Mammals such as a mouse and a horse are able to maintain a constant body temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse.
(3 marks)Mark Scheme
- Mouse (smaller, so) has a larger surface area to volume ratio (1 mark)
- More / faster heat loss (per gram / in relation to body size) (1 mark)
- (Faster rate of) respiration / metabolism releases heat (1 mark)
Comments from mark scheme
- Accept converse answers in relation to the horse
- Points 1 and 2 must be comparative
- Point 3: accept “respiration / metabolism replaces heat”
- Point 3: reject “produce / generate heat / energy” — say “heat is released in respiration”
- Ignore “heat lost more easily / readily”
Scientists calculated the surface area to volume ratios for each gill filament in two fish — one with healthy gills and one with damaged gills. Some of their results are shown in Table 4.
Complete Table 4. State your calculated volume and surface area:volume ratio to 2 significant figures.

Mark Scheme
1 mark for each correct row.
If no marks awarded, accept for 1 mark:
- 0.3217 (correct ratio calculation, but not given to 2 significant figures), OR
- A number that can be rounded to 85 000 (e.g. 84615 — correct calculation, but not given to 2 significant figures) (1 mark)
Comments from mark scheme
- Surface area to volume ratio = surface area ÷ volume (not volume ÷ surface area)
- Always give answers to the number of significant figures specified in the question
- Round correctly: 84 615 to 2 s.f. is 85 000, not 84 000
Comments from mark scheme