Exam questions · Biology · Transport In and Out of Cells
Surface Area to Volume Ratio and Exchange Surfaces
- 8 exam questions
- 24 marks
- 8 quick checks
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1 State [1 mark]
State what is meant by surface area to volume ratio.
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Model answer
The surface area of an object divided by its volume.
Mark scheme
- Surface area divided by volume — 1 mark
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2 Calculate [2 marks]
A cube has sides of 2 cm. Calculate its surface area to volume ratio.
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Model answer
Surface area = 6 × 2 × 2 = 24 cm². Volume = 2 × 2 × 2 = 8 cm³. Ratio = 24 ÷ 8 = 3, so 3:1.
Mark scheme
- Surface area 24 cm² and volume 8 cm³ — 1 mark
- Ratio of 3:1 — 1 mark
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3 Calculate [3 marks]
Cube A has sides of 1 cm. Cube B has sides of 3 cm. Calculate the surface area to volume ratio of each cube, and state which has the larger ratio.
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Model answer
Cube A: 6 ÷ 1 = 6, so 6:1. Cube B: 54 ÷ 27 = 2, so 2:1. Cube A, the smaller cube, has the larger ratio.
Mark scheme
- Ratio of 6:1 for cube A — 1 mark
- Ratio of 2:1 for cube B — 1 mark
- Cube A has the larger ratio — 1 mark
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4 Explain [2 marks]
Explain why a single-celled organism does not need a transport system.
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Model answer
It has a large surface area to volume ratio. So diffusion across its cell membrane is fast enough to supply the whole cell.
Mark scheme
- Large surface area to volume ratio — 1 mark
- Diffusion across the surface is fast enough / short distances — 1 mark
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5 Explain [3 marks]
Explain why a large multicellular organism needs exchange surfaces and a transport system.
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Model answer
A large organism has a small surface area to volume ratio, so its body surface is too small to supply all its cells. Cells deep inside are far from the outside, so diffusion over that distance would be too slow. Exchange surfaces swap materials quickly and a transport system carries them to every cell.
Mark scheme
- Small surface area to volume ratio — 1 mark
- Diffusion distances too long / too slow — 1 mark
- Exchange surface takes in materials and the transport system carries them to the cells — 1 mark
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6 Name [3 marks]
Figure 1 shows a gas exchange surface in the lungs. Name the parts labelled A, B and C.
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Model answer
A is an alveolus (air sac). B is the thin wall of the alveolus (one cell thick). C is a capillary.
Mark scheme
- A: alveolus — 1 mark
- B: wall (of the alveolus) — 1 mark
- C: capillary — 1 mark
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7 Explain [4 marks]
Explain how the small intestine is adapted for the absorption of digested food.
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Model answer
The inner wall is covered with villi, which give a large surface area. Each villus has a thin wall, so the diffusion path is short. Each villus has a rich blood supply, which carries the absorbed food away and keeps the concentration gradient steep. This allows food to be absorbed quickly.
Mark scheme
- Villi give a large surface area — 1 mark
- Thin wall, so a short diffusion path — 1 mark
- Rich blood supply — 1 mark
- Carries absorbed food away / keeps a steep concentration gradient — 1 mark
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8 Explain [6 marks]
Explain how the structure of the lungs and the structure of a fish's gills are adapted for gas exchange. In your answer, refer to surface area to volume ratio.
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Model answer
Large organisms have a small surface area to volume ratio, so they need specialised gas exchange surfaces. The lungs contain millions of alveoli, which give a very large surface area. Each alveolus has a wall one cell thick, so the diffusion path is short, and is surrounded by capillaries, so the blood supply is rich. Breathing ventilates the lungs, so oxygen is always at a higher concentration in the alveoli than in the blood. In a fish, the gills are made of many thin filaments, which give a large surface area. The filaments have thin surfaces and a rich blood supply, and water flowing over them maintains the concentration gradient.
Mark scheme
- Small surface area to volume ratio means a specialised surface is needed — 1 mark
- Alveoli / gill filaments give a large surface area — 1 mark
- Thin walls, short diffusion path — 1 mark
- Rich blood supply in both — 1 mark
- Ventilation of the lungs / water flowing over the gills — 1 mark
- Keeps a steep concentration gradient — 1 mark
Quick check
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1
What is the surface area to volume ratio of a cube with sides of 2 cm?
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C: 3:1
Surface area is 6 × 4 = 24 cm² and volume is 8 cm³, so the ratio is 24 ÷ 8 = 3:1.
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2
What happens to the surface area to volume ratio of an organism as it gets bigger?
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B: It gets smaller
Volume grows faster than surface area, so the ratio gets smaller.
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3
Why does a large animal need a transport system?
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D: Diffusion over long distances is too slow
Its surface area to volume ratio is small, so diffusion alone is too slow to supply cells deep inside.
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4
Which feature of an exchange surface keeps the concentration gradient steep in an animal?
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A: An efficient blood supply
The blood carries substances away from the exchange surface, so the concentration there stays low and the gradient stays steep.
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5
What are the tiny air sacs in the lungs called?
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C: Alveoli
Alveoli are the tiny air sacs where gas exchange happens.
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6
How do villi help the small intestine absorb food?
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B: They increase the surface area
Villi are finger-like folds, so they greatly increase the surface area for absorption.
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7
Which of these is an exchange surface in a plant?
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D: A root hair cell
Root hair cells have long thin extensions that give a large surface area for taking up water and ions.
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8
A cube is made 3 times wider. What happens to its surface area to volume ratio?
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B: It falls to a third
A 1 cm cube has a ratio of 6:1 and a 3 cm cube has 2:1, so the ratio falls to a third.