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Every cell has to take in the substances it needs — oxygen, glucose, water — and get rid of its waste, such as carbon dioxide. The simplest way substances move in and out of cells is by diffusion. But diffusion only works well over short distances, and that creates a problem as organisms get bigger: a large body cannot supply all of its cells by diffusion alone. This lesson, right at the heart of Topic B2 of your OCR Gateway Combined Science course, explains diffusion, the factors that affect its rate, and the all-important surface area to volume ratio that shows why large organisms need special exchange surfaces and transport systems. The cube calculation here is one of the most reliable sources of marks in the whole topic.
By the end of this lesson you should be able to define diffusion, list the factors affecting its rate, calculate and compare surface area to volume ratios, and explain why larger organisms need exchange surfaces and transport systems.
This lesson builds AO1 (understanding of diffusion and the factors affecting its rate) and AO2 (applying it in the surface-area-to-volume calculation to explain why large organisms need exchange surfaces).
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration — that is, down a concentration gradient. It happens because particles are always moving about randomly; over time this random movement spreads them out until they are evenly distributed.
A few key points:
Substances that move in and out of cells by diffusion include oxygen and carbon dioxide (in gas exchange) and small soluble molecules. They diffuse across the cell membrane, which is partially permeable.
Exam Tip: Always use the word net and the phrase down a concentration gradient (high to low). State that diffusion needs no energy — this is what tells it apart from active transport later in the topic.
OCR expects you to know what makes diffusion faster or slower. There are four main factors:
| Factor | Effect on rate of diffusion |
|---|---|
| Concentration gradient (difference) | The steeper the gradient (bigger difference), the faster the diffusion |
| Temperature | Higher temperature → particles have more energy and move faster → faster diffusion |
| Surface area | A larger surface area allows more diffusion at once → faster overall |
| Diffusion distance (membrane thickness) | A shorter distance / thinner membrane → faster diffusion |
Higher tier only: these factors can be combined into a single relationship. The rate of diffusion is proportional to the surface area multiplied by the concentration difference, divided by the diffusion distance: rate of diffusion∝diffusion distancesurface area×concentration difference This one expression captures all three key factors at once: increasing the surface area or the concentration difference speeds diffusion up, while increasing the diffusion distance slows it down.
Exam Tip: When a question asks how to increase the rate of diffusion across a surface, the strong answers are: increase the surface area, increase the concentration gradient, decrease the diffusion distance (make it thinner), or raise the temperature.
Here is the central idea of "scaling up". A cell or organism exchanges substances across its surface, but it uses those substances throughout its volume. So what really matters is the surface area to volume ratio (SA:V) — how much surface is available for each unit of volume.
The crucial fact is this: as an object gets bigger, its volume increases faster than its surface area, so its surface area to volume ratio falls. A small object has a large SA:V (plenty of surface for its size); a large object has a small SA:V (relatively little surface for its size).
Cubes make a simple model of cells because the formulae are easy:
Let us work out the SA:V for cubes of side 1, 2 and 3 units.
Cube of side 1 unit:
surface area=6×12=6volume=13=1 SA:V=16=6:1
Cube of side 2 units:
surface area=6×22=6×4=24volume=23=8 SA:V=824=3:1
Cube of side 3 units:
surface area=6×32=6×9=54volume=33=27 SA:V=2754=2:1
Collecting the results:
| Cube side | Surface area | Volume | SA:V ratio |
|---|---|---|---|
| 1 unit | 6 | 1 | 6 : 1 |
| 2 units | 24 | 8 | 3 : 1 |
| 3 units | 54 | 27 | 2 : 1 |
The pattern is clear: as the cube gets bigger, the SA:V ratio falls (from 6:1 down to 2:1). The volume (which needs supplying) grows much faster than the surface area (which does the supplying). This is exactly why a single very large cell, or a large solid animal, could not survive on diffusion across its outer surface alone — there simply would not be enough surface for its volume.
Common error: calculating surface area as side2 (one face) instead of 6×side2 (six faces), or writing the ratio the wrong way round. SA:V puts surface area first, and for a cube the surface area uses the factor of 6.
Exam Tip: Learn the two cube formulae cold: surface area =6×side2 and volume =side3. Then SA:V =volumesurface area. Show every step — examiners award method marks for the working even if the final division slips.
It is not only size that affects the surface area to volume ratio — shape matters as well. Compare two objects with the same volume of 8 cubic units:
Even though both have the same volume, the flatter, thinner slab has a larger surface area to volume ratio (4.25:1 against 3:1). This is exactly why many exchange surfaces in living things are flat or thin rather than compact — a flat shape squeezes more surface around the same volume. It is one reason a leaf is broad and thin, and why the lining of the gut is folded into flat villi.
Common error: forgetting that a non-cube box has different-sized faces. For a slab you must add up all six faces, which come in three matching pairs, not multiply one face by six.
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