1.3.3Cell Biology

Diffusion in Multicellular Organisms

5 min read 9 parts Lesson 13 of 17 1 quiz

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Section 1 of 9The problem of size

A single-celled organism has a large surface area compared with its volume, so substances reach all parts quickly by diffusion.

A multicellular organism has a much lower surface area to volume ratio, and many of its cells are deep inside the body, too far from the surface for diffusion alone to supply them.

2Specialised exchange surfaces

These organisms need specialised exchange surfaces and a transport system (such as the blood circulatory system) to move substances to and from the cells.

Section 2 of 9Features of an efficient exchange surface

Good exchange surfaces share several features:

Features of an efficient exchange surface

Large surface area

a large surface area, often through folding or branching;

Thin membrane

a thin membrane or wall, giving a short diffusion distance;

Blood supply

a good blood supply in animals, to carry away the substance and maintain a steep gradient;

Ventilation

in the gas exchange system, ventilation (breathing) to keep a steep gradient of oxygen and carbon dioxide.

Section 3 of 9Examples in animals

Small Intestine, Villus & Blood Capillary

Examples in animals

Alveoli in the lungs

Alveoli in the lungs the lungs contain millions of tiny air sacs.

Their walls are one cell thick and covered in capillaries.
They give a huge surface area, a short diffusion distance, a rich blood supply,
- and ventilation brings in fresh air,
- so oxygen diffuses into the blood and carbon dioxide diffuses out.

Villi in the small intestine

Villi in the small intestine the lining is folded into finger-like villi.

Each villus has a thin wall and a network of capillaries; digested food molecules like glucose and amino acids pass into the blood,
- and the large surface area and blood flow keep the gradient steep.
The surface of the villi cells has microvilli, tiny projections that increase the area further.

Gills in fish

Gills in fish each gill has many thin filaments with a large surface area, and water flows over them in the opposite direction to blood in the capillaries so that a gradient is kept along the whole length (counter-current).

Section 4 of 9Examples in plants

Root hair cells

Root hair cells the long extension increases the surface area for the absorption of water (by osmosis) and mineral ions (by active transport).

Leaves

Leaves broad and flat, with thin blades, so that carbon dioxide only needs to diffuse a short distance to the cells.

On the underside, small pores called stomata let carbon dioxide in and oxygen and water vapour out.
The air spaces inside the leaf give a large surface area of cell surface exposed to gases.
Guard cells open and close the stomata.

Exchange surface	Substance	Adaptations
Alveoli	O₂, CO₂	Many sacs, thin walls, capillaries, ventilation
Villi	Glucose, amino acids	Folds, thin wall, capillaries
Gills	O₂	Filaments, thin, counter-current flow
Root hairs	Water, ions	Long extension, thin wall
Leaf	CO₂	Flat, thin, stomata, air spaces

Section 5 of 9Writing good answers

In an examination, take any surface and explain each adaptation by linking it to the factor it affects: 'The walls of the alveoli are one cell thick, so the diffusion distance is short, so diffusion is faster.' Many students list adaptations but do not say what they do. Also be careful to say 'blood supply maintains a steep concentration gradient' rather than that the blood 'speeds up' the diffusion directly.

You should also be able to use the idea of surface area to volume ratio to explain why a flatworm can rely on diffusion across its flat body whereas a human cannot, and why elephants lose less heat per kilogram than mice.

Section 6 of 9Surface area to volume ratio in whole organisms

Surface Area : Volume — 3D Step-by-Step Simulation

Examples of Diffusion in Living Organisms

Gas exchange by diffusion in a single-celled protist

Body shape and size also reflect the need for exchange.

Flatworms are flat, so every cell is close to the surface, and they need no lungs.

Earthworms use their moist skin for gas exchange and have a blood system to carry oxygen.

Insects have tubes (tracheae) that carry air directly to the tissues.

Larger animals evolved internal organs such as lungs and kidneys that give a very large internal surface area packed into a small volume.

Estimating the area of the lungs is a good example.

Surface area to volume ratio in whole organisms

A pair of human lungs has an alveolar surface area of about 70 square metres, around the size of half a tennis court, yet it is packed into the chest because of the millions of tiny alveoli.

The same idea applies to the villi, where the inner surface of the small intestine, about 9 metres long, has an area much larger than a smooth tube would have.

In plants, a leaf is thin and flat to maximise light absorption and to keep the diffusion distance for carbon dioxide short.

Section 7 of 9Exam-style questions

Describe three ways in which the alveoli are adapted for gas exchange. [3]3 marks
Explain why a large animal needs a transport system but a single-celled organism does not. [3]3 marks
Explain how the structure of a villus helps absorption. [3]3 marks

Section 8 of 9Key terms

Term	Meaning
Exchange surface	A part of an organism where substances move in or out.
Alveolus	Air sac in the lungs where gas exchange occurs.
Villus	Finger-like projection of the small intestine wall that absorbs food.
Stomata	Pores in leaves for gas exchange.
Ventilation	Movement of air in and out of the lungs.
Capillary	A very thin blood vessel where exchange occurs.
Counter-current	Flow of two fluids in opposite directions to maintain a gradient.

Section 9 of 9Exam technique

Listing adaptations without explaining their effect on the rate of diffusion.

Saying the alveolus is one cell thick; the wall is one cell thick.

Confusing villi (small intestine) with alveoli (lungs).

Forgetting that the blood supply maintains the gradient.

Cambridge 9700 AS Biology Core Formula

Magnification Triangle & Unit Conversion

Formula: M = I / A ⇔ A = I / M ⇔ I = M × A

Units: 1 mm = 1,000 μm = 1,000,000 nm.
Rule: Convert Image size (ruler mm) to μm by ×1,000 before dividing by Actual size.