Diffusion, osmosis and active transport

Cells must take in substances such as oxygen, glucose, water and mineral ions, and get rid of waste such as carbon dioxide and urea. These move across the cell membrane in three ways: diffusion, osmosis and active transport. The bigger an organism is, the smaller its surface area is compared with its volume, so large organisms need special exchange surfaces.

Key facts

  • Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration. It needs no energy. Oxygen and carbon dioxide diffuse in the lungs, and urea diffuses from cells into the blood.
  • Diffusion is faster with a steeper concentration gradient, a higher temperature and a larger surface area of membrane.
  • Osmosis is the diffusion of water from a dilute solution to a more concentrated solution through a partially permeable membrane.
  • Active transport moves substances from a more dilute solution to a more concentrated solution, against the concentration gradient. It needs energy from respiration. Root hair cells take in mineral ions this way, and the gut absorbs sugar into the blood.
  • For a cube of side L, surface area = 6L² and volume = L³. Small organisms have a large surface area to volume ratio, so diffusion across their surface is enough; large organisms need exchange surfaces and transport systems.
  • A good exchange surface has a large surface area, a thin membrane (a short diffusion path), a good blood supply and, in animals, ventilation. Examples are the alveoli, the villi, fish gills and leaves.
  • Percentage change in mass = (final mass − initial mass) ÷ initial mass × 100. Plant tissue gains mass in a dilute solution and loses mass in a concentrated one.

Worked examples

Example 1

A cube of agar jelly has sides of 2 cm. Find its surface area to volume ratio.

  1. Surface area: 6 faces, each 2 × 2 = 4 cm². 6 × 4 = 24 cm²
  2. Volume: 2 × 2 × 2. 8 cm³
  3. Divide the surface area by the volume. 24 ÷ 8 = 3
  4. So the ratio is 3 : 1.

Answer: 3 : 1

Example 2

A potato cylinder had a mass of 2.50 g. After 30 minutes in a sugar solution its mass was 2.20 g. Calculate the percentage change in mass, and explain it.

  1. Change in mass: 2.20 − 2.50 = −0.30 g
  2. Divide by the starting mass and multiply by 100. −0.30 ÷ 2.50 × 100 = −12%
  3. The mass went down, so water moved out of the cells by osmosis.
  4. The sugar solution was more concentrated than the cytoplasm of the potato cells.

Answer: −12%: water left the cells by osmosis, because the solution was more concentrated than the cytoplasm

Example 3

Potato cylinders were left in sugar solutions of different concentrations. Use the graph to estimate the concentration of the solution inside the potato cells.

  1. Above the axis the potato gained mass; below it, it lost mass.
  2. Where the change in mass is 0%, there is no net movement of water.
  3. So the solution outside has the same concentration as the cytoplasm.
  4. The line crosses 0% at about 0.3 mol/dm³.

Answer: about 0.3 mol/dm³

Example 4

Root hair cells take in mineral ions from the soil. Explain why they use active transport, and why they have many mitochondria.

  1. Soil water has a very low concentration of mineral ions.
  2. The cell already has a higher concentration, so the ions cannot diffuse in.
  3. Active transport moves them in, against the gradient, using energy.
  4. The energy comes from respiration in the mitochondria, so the cell has many of them.

Answer: the ions are more concentrated inside the cell than in the soil, so they must be moved against the gradient using energy from respiration, which happens in mitochondria