1.3

Transport into and out of cells

Diffusion, osmosis and active transport — what moves, which way, and whether energy is needed — with the osmosis practical and the percentage change in mass calculation that goes with it.

About 20 minutes to read

1.15Diffusion

Substances move into and out of cells across the cell membranes by diffusion.

Diffusion is the spreading out of the particles of any substance in solution, or particles of a gas, resulting in a net movement from an area of higher concentration to an area of lower concentration. Learn that sentence — the words "net movement" and "higher to lower" are where the marks are.

Substances transported this way include oxygen and carbon dioxide in gas exchange, and the waste product urea, which diffuses from cells into the blood plasma for excretion by the kidney.

Diffusion: particles spread out until they are evenly distributed, giving a net movement from higher to lower concentration.

Three factors affect the rate of diffusion:

• the difference in concentrations — the concentration gradient. A steeper gradient means a faster rate. • the temperature. Warmer particles have more energy and move faster. • the surface area of the membrane. More surface means more places to cross at once.

Practise drawing this one, not just reading it. You are expected to recognise, draw and interpret diagrams that model diffusion, and the sketch is quick: a container split by a membrane, particles crowded on one side and sparse on the other, and an arrow showing the net movement down the gradient. Add a second, smaller arrow the other way if you want to show that particles cross both ways.

Reading someone else's diagram works the same way in reverse: find the membrane, compare how many particles are on each side, then state which way the net movement goes and why.

Common mistake

Diffusion is a net movement, not a one-way one. Particles cross the membrane in both directions the whole time; there are simply more crossings from the crowded side. And diffusion needs no energy from respiration — the particles are already moving.

Size decides whether diffusion alone is enough. A single-celled organism has a relatively large surface area to volume ratio, which allows sufficient transport of molecules into and out of the cell to meet the needs of the organism.

As an organism gets bigger, its volume grows faster than its surface area, so the ratio falls. Every cell inside still needs supplying, but there is proportionally less surface to supply it through.

Surface area to volume ratio for cubes of side 1, 2 and 3 cm. The bigger the cube, the smaller the ratio.

Worked example

Calculate the surface area to volume ratio of a cube of side 2 cm, and compare it with a cube of side 1 cm.

  1. Surface area = 6 × (side)² = 6 × 2² = 24 cm².
  2. Volume = (side)³ = 2³ = 8 cm³.
  3. Ratio = 24 : 8 = 3 : 1.
  4. For the 1 cm cube: surface area 6 cm², volume 1 cm³, ratio 6 : 1.

Answer: 3 : 1, half the ratio of the smaller cube — the larger cube has less surface for each unit of volume.

That is why multicellular organisms need exchange surfaces and a transport system: their surface area to volume ratio is too small for diffusion across the outside to supply everything inside. Surfaces and organ systems are specialised for exchanging materials, so that enough molecules can be transported into and out of cells for the organism's needs.

The effectiveness of an exchange surface is increased by:

• having a large surface area • a membrane that is thin, to provide a short diffusion path • in animals, having an efficient blood supply • in animals, for gaseous exchange, being ventilated.

Figure: How exchange surfaces are adapted
Diffusion
the spreading out of the particles of any substance in solution, or particles of a gas, resulting in a net movement from an area of higher concentration to an area of lower concentration
Concentration gradient
the difference in concentration between two areas; the steeper the gradient, the faster the rate of diffusion
Surface area to volume ratio
the surface area of an organism divided by its volume; it falls as an organism gets larger, which is why large organisms need exchange surfaces and a transport system

1.15–1.17Osmosis

Water moves across cell membranes by osmosis.

Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane. "Partially permeable" means the membrane lets water through but not the dissolved solute, so only the water can move to even things out.

Osmosis: water diffuses through a partially permeable membrane from the dilute side to the concentrated side.

Draw this one too. A diagram that models osmosis needs three things to earn its marks: the partially permeable membrane down the middle, water molecules on both sides with more of them on the dilute side, and solute particles drawn large enough to show that they cannot get through. The arrow goes from dilute to concentrated.

Common mistake

Dilute and concentrated describe the solute, not the water. A dilute solution has little solute and therefore more water — so water moving "from dilute to concentrated" is still water moving from where there is more of it to where there is less. Say water, not "solution", when you describe what moves.

The maths here is percentages. You are expected to use simple compound measures such as rate of water uptake, to use percentages, and to calculate percentage gain and loss of mass of plant tissue — then plot, draw and interpret the graph.

Worked example

A potato cylinder has a mass of 4.2 g at the start and 4.8 g after an hour in distilled water. Calculate the percentage change in mass.

  1. Change in mass = 4.8 − 4.2 = 0.6 g.
  2. Percentage change = (change ÷ starting mass) × 100.
  3. (0.6 ÷ 4.2) × 100 = 14.28...

Answer: A gain of 14.3% (3 significant figures). Water has entered the cells by osmosis, so the solution outside was more dilute than the cell contents.

Worked example

A plant takes up 12 cm³ of water in 40 minutes. Calculate the rate of water uptake in cm³ per hour.

  1. Rate is a compound measure: amount ÷ time.
  2. 12 cm³ ÷ 40 minutes = 0.3 cm³ per minute.
  3. Scale up to an hour: 0.3 × 60.

Answer: 18 cm³ per hour.

Required practical 3

Investigate the effect of a range of concentrations of sugar solution on the mass of plant tissue

  1. Cut cylinders of potato to the same length, blot them dry and record the mass of each one.
  2. Put each cylinder into a different concentration of sugar solution, including distilled water as a 0 mol/dm³ point.
  3. Leave them all for the same length of time, at the same temperature.
  4. Remove each cylinder, blot it dry the same way as before, and record its new mass.
  5. Calculate the percentage change in mass for each concentration, and plot percentage change against concentration.

Watch out: Percentage change, not change in grams — the cylinders do not start at identical masses, and only a percentage makes them comparable. The concentration where the line crosses zero is the one that matches the cell contents, so there is no net movement of water.

Plotting and reading the graph. Concentration goes on the x-axis, because it is what you chose; percentage change in mass goes on the y-axis, because it is what you measured. Pick a scale that fills at least half the grid, mark each point with a small cross, and draw one smooth line of best fit rather than joining the dots.

Then interpret it. Points above the axis mean the tissue gained water, so that solution was more dilute than the cell contents; points below mean it lost water to a more concentrated solution. Where the line crosses zero there is no net movement, so the solution outside matches the concentration inside the cells — reading off that crossing point is a standard question, and it is an estimate from your line, not one of your measurements.

An application worth knowing: sports drinks. An isotonic drink is made at roughly the same concentration as body fluids, so there is no large net movement of water by osmosis either way — it rehydrates without upsetting the balance, which is what a runner wants during a race. A high energy drink is more concentrated, carrying more sugar for respiration; because it is more concentrated than body fluids it draws water into the gut by osmosis, so it fuels well but rehydrates more slowly. Which is the better choice depends on whether the athlete needs water or energy.

Osmosis
the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane
Partially permeable membrane
a membrane that lets water through but not the dissolved solute, so only the water can move
Isotonic
at the same concentration as body fluids, so there is no large net movement of water by osmosis in either direction

1.15Active transport

Active transport moves substances from a more dilute solution to a more concentrated solution — against a concentration gradient. This requires energy from respiration.

Two examples to know. Active transport allows mineral ions to be absorbed into plant root hairs from very dilute solutions in the soil, and plants require those ions for healthy growth. It also allows sugar molecules to be absorbed from lower concentrations in the gut into the blood, which has a higher sugar concentration; those sugar molecules are then used for cell respiration.

Active transport: particles are moved against the concentration gradient, which costs energy from respiration.

You need to be able to describe how substances are transported into and out of cells by diffusion, osmosis and active transport, and to explain the differences between the three processes. The differences come down to three questions: what moves, which way, and does it cost energy?

The three processes side by side: diffusion and osmosis run down a gradient for free, active transport runs up one and must be paid for.
Figure: Diffusion, osmosis and active transport compared

Common mistake

If a question says a substance moved from a low concentration to a high one, the answer is active transport and the reason is energy from respiration — which is why root hair cells are packed with mitochondria.

Active transport
the movement of substances from a more dilute solution to a more concentrated solution, against a concentration gradient, using energy from respiration

Before you practise

  • Diffusion is the net movement of particles from a higher to a lower concentration, and needs no energy.
  • Osmosis is the movement of water across a partially permeable membrane, from a dilute solution to a concentrated one.
  • Active transport moves substances against the concentration gradient and requires energy from respiration.
  • percentage change in mass = (change in mass ÷ starting mass) × 100 — and the sign matters.
  • Where the line crosses zero on a potato graph is the concentration inside the cells.
  • Blot the cylinders before weighing, or surface water is weighed as though it were absorbed.

Now practise it

Reading is the easy half. Answer questions on 1.3 Transport into and out of cells and find out which parts you actually know.