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How do substances move into and out of cells, and why do larger organisms need exchange surfaces?

Diffusion, osmosis and active transport as ways substances move across membranes, the factors affecting the rate of diffusion, surface area to volume ratio, and the adaptations of exchange surfaces such as alveoli, villi and root hair cells.

A focused answer to the OCR Gateway GCSE Combined Science A topic B2 on transport across membranes, covering diffusion, osmosis and active transport, the factors affecting diffusion rate, surface area to volume ratio, and adaptations of exchange surfaces.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Diffusion, osmosis and active transport
  3. Surface area to volume ratio
  4. Exchange surfaces

What this topic is asking

OCR wants you to define diffusion, osmosis and active transport, explain the factors that affect the rate of diffusion, calculate and use surface area to volume ratio, and describe how exchange surfaces are adapted for efficient transport.

Diffusion, osmosis and active transport

  • Diffusion is passive, meaning it needs no energy from the cell. Examples include oxygen and carbon dioxide moving across the membranes of the alveoli, and dissolved food and oxygen passing into cells. The rate of diffusion increases with a steeper concentration gradient, a higher temperature (particles move faster), and a larger surface area.
  • Osmosis moves water specifically: water moves from a dilute solution (high water concentration) to a more concentrated solution (low water concentration) across a partially permeable membrane. This is how water enters root hair cells and how plant cells become firm (turgid).
  • Active transport moves substances the "wrong" way, against the gradient from low to high concentration, so it needs energy from respiration. Examples are mineral ions moving from dilute soil water into root hair cells, and glucose being reabsorbed from the kidney tubules and the gut even when its concentration there is already low.

Surface area to volume ratio

You can calculate the ratio for a cube of side ss: surface area =6s2= 6s^2, volume =s3= s^3, so the ratio is 6s2s3=6s\dfrac{6s^2}{s^3} = \dfrac{6}{s}. As ss increases, 6s\dfrac{6}{s} decreases, which is the mathematical reason the ratio falls with size. OCR may give you the dimensions of a model organism and ask you to compare ratios.

Exchange surfaces

Exchange surfaces are adapted to make diffusion fast by having a large surface area, a short diffusion distance and a steep concentration gradient. Key examples are the alveoli in the lungs (millions of tiny air sacs give a huge surface area, with thin walls one cell thick and a rich blood supply), the villi lining the small intestine (folds and finger-like projections increase the surface area for absorbing digested food, with a thin lining and a good blood supply), and root hair cells (each long projection increases the surface area for absorbing water and mineral ions from the soil). Ventilation (in lungs) and a flowing blood supply maintain the concentration gradient so diffusion keeps going.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksDescribe how to investigate the effect of different salt solutions on the mass of potato cylinders by osmosis, and explain the expected results.
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A required-practical (osmosis) question. Reward a clear method: cut potato cylinders of equal size, measure and record the starting mass of each, place them in solutions of different salt concentration for a set time, then dry the surface and re-weigh. For the explanation: in a concentrated salt solution (lower water concentration, or lower water potential, than the cell) water leaves the cells by osmosis across the partially permeable membrane, so the cylinders lose mass; in pure water (higher water concentration) water enters and they gain mass; at the concentration where the mass does not change, the solution and the cell contents have equal water concentration. Markers credit drying before reweighing, controlling size and time, and linking the direction of water movement to the mass change.

OCR 20203 marksExplain why a single-celled organism does not need a specialised exchange surface but a large animal does, using the idea of surface area to volume ratio.
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A B2 question on surface area to volume ratio. Reward: a single-celled organism is small and has a large surface area compared with its volume, so substances can diffuse in and out fast enough across its surface to meet its needs. As an organism gets larger its volume increases faster than its surface area, so the surface area to volume ratio falls, and diffusion across the outer surface alone is too slow to supply the inner cells. Large animals therefore need specialised exchange surfaces (such as lungs) and transport systems (such as blood) to move substances quickly. Markers want the ratio comparison and the link to diffusion distance and rate.

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