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AS 4

Cell membranes and transport

AS Level · Topic 4 · Papers 1, 2 and 3

🎯What you need to be able to do

  • Describe the fluid mosaic model, including the hydrophobic and hydrophilic interactions that form the bilayer.
  • Describe the arrangement and roles of phospholipids, cholesterol, glycolipids, proteins and glycoproteins.
  • Outline the main stages of cell signalling: ligand secretion, transport to target cells, and binding to cell surface receptors.
  • Describe and explain simple diffusion, facilitated diffusion, osmosis, active transport, endocytosis and exocytosis.
  • Calculate surface areas and volumes of simple shapes and explain why surface area to volume ratio falls as size rises.
  • Investigate the effect of solutions of different water potentials on plant tissue and use the results to estimate the water potential of the tissue.
  • Explain water movement in terms of water potential, and the different effects on plant and animal cells.

📚The biology

The fluid mosaic model

Phospholipids have a hydrophilic phosphate head and two hydrophobic fatty acid tails. In water they arrange themselves so the heads face the aqueous solutions on either side and the tails point inward, away from water: a bilayer forms spontaneously because that arrangement is the most stable. Proteins are scattered through it — some spanning the whole width (intrinsic or transmembrane), some on one surface (extrinsic).

Fluid, because the phospholipids move laterally within their layer and the membrane is flexible; mosaic, because the proteins are scattered irregularly like tiles.

  • Phospholipids — form the bilayer; the hydrophobic core is the barrier that makes the membrane partially permeable, letting small non-polar molecules through and blocking large or charged ones.
  • Cholesterol — sits between the phospholipids. It regulates fluidity: it stops the membrane becoming too fluid at high temperature and too rigid at low temperature, and it reduces permeability to water-soluble substances and ions. Also gives mechanical stability.
  • Glycolipids and glycoproteins — carbohydrate chains on the outer surface only. They act as cell surface antigens for recognition (see Topic 11), as receptors, and in cell adhesion.
  • Proteins — channel proteins (water-filled pores for specific ions), carrier proteins (change shape to move a specific solute), enzymes, and receptors for cell signalling.

Cell signalling

Three stages, and the syllabus wants them in order:

  1. A cell secretes a specific chemical (a ligand), such as a hormone.
  2. The ligand is transported to target cells, for example in the blood.
  3. The ligand binds to a complementary cell surface receptor on the target cell, triggering a response inside it.

Specificity comes entirely from the receptor: only cells carrying a receptor complementary to that ligand can respond, which is why a hormone circulating everywhere affects only certain tissues. What happens next — G proteins, second messengers, enzyme cascades — is Topic 14.

Movement across membranes

  • Simple diffusion — net movement of molecules from high to low concentration, down the gradient, passive. Small non-polar molecules (O₂, CO₂) pass straight through the bilayer.
  • Facilitated diffusion — still down the gradient and still passive, but through channel or carrier proteins. Needed by polar molecules and ions, which cannot cross the hydrophobic core. Because it depends on proteins, it can be saturated.
  • Osmosis — the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane. Much of it goes through aquaporin channel proteins.
  • Active transport — movement against the concentration gradient, through a carrier protein, using ATP. The carrier changes shape as ATP is hydrolysed. Rate depends on respiration, so it stops if a respiratory inhibitor is added — the standard way to distinguish it experimentally from facilitated diffusion.
  • Endocytosis — the membrane invaginates around bulk material and pinches off a vesicle inside the cell. Requires ATP.
  • Exocytosis — a vesicle fuses with the cell surface membrane and releases its contents outside. Requires ATP.

Surface area to volume ratio

Diffusion alone supplies a cell only if the surface is large enough relative to the volume it must serve. Consider a cube of side \( l \):

surface area \( = 6l^{2} \)
volume \( = l^{3} \)
ratio \( = \dfrac{6}{l} \)

Surface area rises with the square of the length while volume rises with the cube, so as an organism gets bigger the ratio falls. A 1 cm cube has a ratio of 6:1; a 2 cm cube, 3:1; a 3 cm cube, 2:1. That is the whole reason large organisms need specialised exchange surfaces and transport systems — the reason for Topics 7, 8 and 9.

For a cylinder of radius \( r \) and height \( h \): surface area \( = 2\pi r^{2} + 2\pi r h \), volume \( = \pi r^{2} h \). Agar blocks of different sizes soaked in a dye or an indicator show the effect directly — the small block is coloured through, the large one has an unstained core.

Water potential

Water potential (Ψ) is the tendency of water to move out of a system, measured in kilopascals (kPa). Pure water has the highest possible value, defined as zero. Adding solute lowers water potential, so all solutions have negative values.

\[ \text{water moves from higher (less negative) } \Psi \text{ to lower (more negative) } \Psi \]

“More concentrated” therefore means “more negative”: −800 kPa is a lower water potential than −200 kPa, and water flows from the −200 to the −800. The syllabus does not require solute potential or pressure potential separately — only Ψ.

Plant cells have a cell wall, so they behave differently from animal cells:

  • In a solution of higher Ψ, water enters, the vacuole expands and pushes the membrane against the wall. The wall resists, so pressure builds and the cell becomes turgid. It does not burst.
  • In a solution of lower Ψ, water leaves, the cell becomes flaccid, and eventually the membrane pulls away from the wall: plasmolysis. The point at which plasmolysis just begins is incipient plasmolysis.

Animal cells have no wall. In a solution of higher Ψ water enters, the cell swells and bursts — haemolysis in a red blood cell. In a solution of lower Ψ water leaves and the cell shrinks and crenates. This is why animals must osmoregulate (Topic 14) and plants need not.

Never write “water moves down its concentration gradient” in a water potential question. The mark scheme wants water potential, and it wants the sign right: water moves to the more negative value. A common error is to reason that a “higher” number like −200 sounds like less water — it is the opposite.

✏️Worked example

Cylinders of potato tissue were blotted, weighed, and left for 24 hours in sucrose solutions of known water potential. The percentage change in mass was: 0.0 kPa: +18.0  ·  −200 kPa: +10.5  ·  −400 kPa: +3.0  ·  −600 kPa: −4.5  ·  −800 kPa: −12.0 (a) Estimate the water potential of the potato tissue. (b) Explain the sign of the change at −200 kPa and at −800 kPa. (c) Explain why percentage change in mass is used rather than change in mass.

(a) Plot percentage change in mass (y) against water potential of the solution (x) and draw a straight line of best fit. Where the line crosses the x-axis, the mass change is zero, so there was no net movement of water in either direction. That happens only when the water potential of the tissue equals that of the solution.

Between −400 kPa (+3.0%) and −600 kPa (−4.5%) the line crosses zero. Interpolating, the change of 7.5 percentage points is spread over 200 kPa, so zero lies \( \tfrac{3.0}{7.5} \times 200 = 80 \) kPa beyond −400. The water potential of the tissue is therefore about −480 kPa.

(b) At −200 kPa the solution has a higher (less negative) water potential than the tissue at −480 kPa, so water moves into the cells by osmosis down the water potential gradient, through the partially permeable cell surface membrane. The cells gain water and become turgid, so the mass increases. At −800 kPa the solution is lower (more negative) than the tissue, so water moves out of the cells; they become flaccid and may plasmolyse, so the mass decreases.

(c) The cylinders cannot be cut to identical starting masses. A cylinder of 4.0 g and one of 2.0 g absorbing water at the same rate per gram will show different absolute gains, so raw mass change cannot be compared between tubes. Expressing the change as a percentage of the initial mass standardises for that variation and makes the five results comparable.

Check it. Two checks. First, the sign of the answer: a living plant tissue must have a negative water potential, and a value of a few hundred negative kPa is the expected order — a positive answer, or one of −5 kPa, means an error. Second, the intercept must lie between the last positive and the first negative reading, so between −400 and −600 kPa. −480 does. Any answer outside that bracket has been read off the wrong axis or the line has been drawn through the origin out of habit.
Reading the line the wrong way round, and forgetting to blot. It is the x-intercept that matters, not the y-intercept: the y-intercept (+18.0% at 0 kPa) merely tells you the tissue gains most in pure water, which was never in doubt. In the practical itself, the cylinders must be blotted identically before both weighings — surface water left on after the second weighing appears as a mass gain and shifts the whole line up, producing an estimate that is too negative.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A cell is placed in a solution of water potential −350 kPa. The cell’s water potential is −700 kPa. State the direction of net water movement and explain.
Water moves into the cell. Water always moves by osmosis from a region of higher (less negative) water potential to lower (more negative) water potential, through a partially permeable membrane. The solution at −350 kPa has the higher water potential; the cell at −700 kPa has the lower. So there is a net movement of water molecules down the water potential gradient into the cell. If it were a plant cell it would become turgid; if an animal cell, it would swell and could burst, since there is no wall to generate an opposing pressure.
2. Explain how you would show experimentally that the uptake of an ion by root cells is by active transport rather than facilitated diffusion.
Set up two identical samples of root tissue in the same solution of the ion, at the same temperature and pH, and measure uptake over time. To one, add a respiratory inhibitor such as cyanide (or deprive it of oxygen). If uptake is by active transport, it requires ATP from respiration, so uptake will fall sharply or stop in the inhibited sample. If it were facilitated diffusion, which is passive, uptake would be unaffected because no ATP is used. Two supporting observations: active transport can move the ion against its concentration gradient, so uptake continuing when the internal concentration exceeds the external one is itself evidence; and active transport rate is strongly temperature-dependent through its effect on respiration.
3. Calculate the surface area to volume ratio of a cube of side 2 mm and one of side 5 mm, and explain the biological significance.
For a cube of side \( l \): surface area \( = 6l^{2} \), volume \( = l^{3} \), ratio \( = 6/l \). Side 2 mm: SA = 6 × 4 = 24 mm², V = 8 mm³, ratio = 3:1. Side 5 mm: SA = 6 × 25 = 150 mm², V = 125 mm³, ratio = 1.2:1. Surface area increases with the square of the linear dimension but volume with the cube, so the ratio falls as size rises. Biologically, a small organism has enough surface per unit of volume for diffusion alone to supply oxygen and remove waste over a short distance. A large organism does not, so it needs a specialised exchange surface with a large area and thin barrier (alveoli, gills, root hairs) and a mass transport system to carry substances the longer distances involved.
4. Describe the role of cholesterol in the cell surface membrane and predict the effect of removing it.
Cholesterol molecules sit between the phospholipids in the bilayer, interacting with the hydrophobic tails. They regulate fluidity: at higher temperatures they restrict the movement of phospholipids and prevent the membrane becoming too fluid; at lower temperatures they prevent the tails packing tightly together, so the membrane does not become too rigid. They also add mechanical stability and reduce permeability to water-soluble substances and ions. Removing cholesterol would leave the membrane more fluid and less stable at high temperature, more rigid at low temperature, and more permeable, so ions and polar molecules would leak across and the cell could not maintain concentration gradients.
5. Red blood cells burst when placed in distilled water, but plant cells in distilled water do not. Explain.
In distilled water (Ψ = 0 kPa) both cells have a lower water potential than the surroundings, so water enters both by osmosis. A red blood cell has only a cell surface membrane, which is flexible and cannot resist the increasing volume, so the cell swells until the membrane ruptures — haemolysis. A plant cell is enclosed by an inelastic cellulose cell wall. As water enters, the vacuole and cytoplasm press the membrane against the wall, and the wall pushes back, generating a pressure that rises until it prevents further net entry of water. The cell becomes turgid and stops there. The wall does not stop the osmosis; it stops the swelling, which is what prevents bursting.
6. Distinguish between facilitated diffusion and active transport in terms of gradient, protein type, energy and the effect of increasing solute concentration.
Gradient: facilitated diffusion moves solute down its concentration gradient; active transport moves it against the gradient. Proteins: facilitated diffusion uses channel or carrier proteins; active transport uses carrier proteins only, which change shape during the cycle. Energy: facilitated diffusion is passive and needs no ATP, driven by the kinetic energy of the molecules themselves; active transport hydrolyses ATP from respiration. Increasing solute concentration: in facilitated diffusion, raising the external concentration steepens the gradient and increases rate until all the proteins are occupied and the rate plateaus (saturates); in active transport, raising external concentration reduces the gradient the pump works against but rate is limited by the number of carriers and the supply of ATP, so it too plateaus. Both saturate — that is a shared feature, not a difference, and only the ATP requirement cleanly separates them experimentally.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Nuffield Foundation practical biology — the potato osmosis and agar block diffusion protocols, with the blotting and repeat-reading detail that examiners look for
  • “Learn Genetics” (University of Utah) — an animated membrane where you can watch each transport mechanism separately
  • BioMan Biology or similar interactive osmosis simulators — useful for building intuition about the sign convention on water potential before you meet it in exam data