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How do substances move into and out of cells and across exchange surfaces?

The fluid mosaic model of the cell membrane, diffusion, osmosis and active transport, the role of water as a solvent and its properties, and the features of efficient exchange surfaces.

An Edexcel A-Level Biology B (Salters-Nuffield) answer on cell transport, covering the fluid mosaic membrane model, diffusion, facilitated diffusion, osmosis and active transport, the properties of water, and the features of efficient gas-exchange surfaces.

Generated by Claude Opus 4.810 min answer

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

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  1. What this dot point is asking
  2. The fluid mosaic model
  3. Movement across membranes
  4. Properties of water
  5. Efficient exchange surfaces
  6. Examples in context
  7. Try this

What this dot point is asking

Edexcel wants you to describe the fluid mosaic model of the cell membrane, explain diffusion, osmosis and active transport, describe the properties of water that make it a good solvent and transport medium, and explain the features that make a gas-exchange surface efficient. Osmosis calculations on potato or plant tissue are a recurring practical-based question.

The fluid mosaic model

Cholesterol fits between phospholipids and regulates fluidity; proteins act as channels, carriers, receptors and enzymes.

Movement across membranes

  • Diffusion: the net movement of particles from a region of high to low concentration, down a gradient, passively. Small, non-polar molecules such as oxygen and carbon dioxide cross the bilayer directly.
  • Facilitated diffusion: diffusion of polar or charged molecules through channel or carrier proteins, still down the gradient and without ATP.
  • Osmosis: the net movement of water across a partially permeable membrane from a higher (less negative) to a lower (more negative) water potential.
  • Active transport: movement against the concentration gradient using carrier proteins and ATP.

Properties of water

Water is a polar molecule that forms hydrogen bonds. This makes it a good solvent for ions and polar molecules (so it transports substances in blood and cytoplasm), gives it a high specific heat capacity (stabilising temperature) and high cohesion (helping transport in xylem).

Efficient exchange surfaces

Surfaces such as the alveoli are efficient because they have a large surface area, a thin barrier (short diffusion distance), and a steep concentration gradient maintained by ventilation and a good blood supply. Fick's law summarises this: rate of diffusionsurface area×concentration differencediffusion distance\text{rate of diffusion} \propto \frac{\text{surface area} \times \text{concentration difference}}{\text{diffusion distance}}. So a bigger area, a steeper gradient and a thinner barrier all speed diffusion.

Small organisms exchange gases over their whole body surface because their surface area to volume ratio is large. As organisms get bigger this ratio falls, so they need specialised exchange surfaces (lungs, gills) and a transport system to keep gradients steep.

Examples in context

Example 1. The alveoli. Each lung has hundreds of millions of alveoli, giving a total surface area of around 70 m270 \text{ m}^2. The alveolar wall and capillary wall are each one cell thick, so the diffusion distance is under a micrometre. Breathing constantly replaces alveolar air and blood flow constantly removes oxygenated blood, keeping a steep oxygen gradient. Every term in Fick's law is maximised, which is why this is the standard exam example of an efficient surface.

Example 2. Active transport in the gut. Glucose is absorbed from the small intestine even when its concentration in the gut is lower than in the blood. Sodium ions are pumped out of the epithelial cells by active transport (using ATP), creating a sodium gradient that drags glucose in by co-transport. This shows active transport moving a substance against its gradient at the expense of ATP, a favourite contrast with passive diffusion.

Try this

Q1. Explain why oxygen can diffuse directly through the phospholipid bilayer but glucose cannot. [2 marks]

  • Cue. Oxygen is small and non-polar so it crosses the hydrophobic tails; glucose is large and polar, so it needs a transport protein.

Q2. State two ways an exchange surface is adapted for rapid diffusion. [2 marks]

  • Cue. Large surface area and thin barrier (short diffusion distance); a steep gradient maintained by blood flow.

Exam-style practice questions

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

Edexcel 20184 marksExplain how the structure of the cell surface membrane allows it to control which substances enter and leave the cell.
Show worked answer →

Markers want structure linked to selective transport.

The phospholipid bilayer has hydrophobic fatty acid tails in the centre, so small non-polar molecules (oxygen, carbon dioxide) dissolve and diffuse straight through while charged or polar molecules cannot. Channel proteins allow specific ions through by facilitated diffusion, and carrier proteins change shape to move specific molecules, including by active transport using ATP against the gradient. Because each protein is specific, the membrane is partially permeable and controls what crosses.

Award marks for: hydrophobic core blocks polar or charged solutes; non-polar molecules diffuse through; channel and carrier proteins are specific; active transport against the gradient needs ATP.

Edexcel 20225 marksPieces of potato were placed in sucrose solutions of different concentration. A piece in a 0.3 mol dm30.3 \text{ mol dm}^{-3} solution had a mass of 4.20 g4.20 \text{ g} before and 3.78 g3.78 \text{ g} after. Calculate the percentage change in mass and explain, in terms of water potential, why the mass changed.
Show worked answer →

A worked osmosis calculation with explanation.

Percentage change =3.784.204.20×100=0.424.20×100=10%= \frac{3.78 - 4.20}{4.20} \times 100 = \frac{-0.42}{4.20} \times 100 = -10\%. The mass decreased, so water left the potato cells by osmosis. This means the sucrose solution had a lower (more negative) water potential than the cell contents, so water moved from the higher (less negative) water potential inside the cells to the lower water potential outside, down the water potential gradient.

Markers reward: correct 10%-10\% (with negative sign), water left by osmosis, and the explanation that the external solution had a lower or more negative water potential.

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