GCSE Biology Transport in cells revision

Diffusion is the spreading out of the particles of any substance in solution, or of a gas, giving a net movement from where they are more concentrated to where they are less concentrated. No energy from respiration is spent — the movement comes from the random motion of the particles themselves. The cell membrane is the gate this happens through: small molecules such as oxygen slip between its phospholipids. Classic examples: oxygen diffuses from the lungs into the blood and from blood into respiring cells; carbon dioxide diffuses the other way; and the waste product urea diffuses out of cells into the blood plasma for removal by the kidneys.

A single-celled organism has a relatively large surface area to volume ratio, so diffusion across its membrane supplies everything it needs — oxygen in, carbon dioxide out — fast enough to survive. As organisms get bigger, volume grows faster than surface area, and deep interior cells sit too far from the outside world. Multicellular organisms solve this twice over: specialised exchange surfaces absorb fast enough, and transport systems (blood, xylem, phloem) carry substances from the exchange surface to every cell.

Osmosis is the diffusion of water from a dilute solution (high water concentration) to a concentrated solution (low water concentration) through a partially permeable membrane — a barrier with tiny holes that let water molecules through but hold back bigger dissolved solute molecules. In animal cells osmosis is risky: in pure water a red blood cell swells and can burst, and in a salty solution it shrinks. Plant cells are safer — the inelastic cellulose wall stops bursting; a cell in dilute solution swells and becomes turgid, and a cell in a concentrated solution loses water, the cytoplasm pulling away from the wall (plasmolysed).

1. 1Cut and dry the chips

   Cut potato cylinders of identical size with a cork borer, blot them dry, and record each starting mass.

2. 2Set up the concentrations

   Place one chip in each of a range of sugar (or salt) solutions — for example 0, 0.25, 0.5, 0.75 and 1.0 mol/dm³ — plus distilled water.

3. 3Control the conditions

   Use the same volume of solution and the same temperature for every tube, and leave them for the same time (typically 30–60 minutes).

4. 4Reweigh

   Blot each chip dry again and record its final mass.

5. 5Calculate and plot

   Find the percentage change in mass for each chip and plot it against solution concentration; the concentration where the line crosses 0% change is the concentration matching the potato cell sap, where there is no net water movement.

Variables: the independent variable is the solution concentration; the dependent variable is the percentage change in mass; controls include chip size and starting mass, solution volume, temperature and time. Risks: glass tubes break; the cork borer and scalpel cut — cut onto a white tile, away from your hand. Typical results: chips gain mass in dilute solutions (water moves in) and lose mass in concentrated solutions (water moves out), with a straight-ish trend that crosses zero at the concentration matching the potato’s own cell sap.

Active transport moves substances from a more dilute solution to a more concentrated solution — against the concentration gradient — and it requires energy from respiration. Two exam-favourite examples: root hair cells absorb mineral ions from very dilute soil solutions, because plants need those ions for healthy growth; and cells lining the gut absorb sugar molecules from lower concentrations in the gut into the blood, where the sugar concentration is already higher — the sugar then fuels cell respiration. This is why both cell types are packed with mitochondria.