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Diffusion and osmosis: movement that costs nothing

Particles move about anyway, and given a gradient that random movement produces a net flow in one direction. Nothing in this lesson requires the cell to spend anything — which is precisely why water potential has to be defined carefully, or the whole argument runs backwards.

Before this The fluid mosaic model and the hydrophobic core · Surface area to volume ratio

Before you start

Osmosis is the movement of water from a region of high concentration to a region of low concentration. This sentence is in a great many notebooks and it fails the moment you ask a concentration of what. Of water? Concentrations describe solutes dissolved in water, not the water itself. Of solute? Then the sentence says water moves towards low solute concentration, which is the wrong way round. At A level the quantity you need is water potential, and it exists precisely because 'concentration of water' does not survive contact with a question about pressure.

What you should be able to do

Define simple and facilitated diffusion and say what distinguishes them.
State how rate of diffusion depends on surface area, concentration difference and thickness, and apply that to a named exchange surface.
Define osmosis using water potential, and use the relationship between water potential, solute potential and pressure potential.
Explain what happens to a plant cell and to an animal cell placed in a solution of lower water potential than their contents.
Describe how a serial dilution of sucrose is used to find the water potential of potato tissue.

Diffusion, and the three things that change its rate

Molecules and ions in a liquid or gas are in constant random motion. They do not know where the gradient is and they are not trying to go anywhere. But if there are more of them on one side than the other, more will happen to wander from the crowded side to the empty side than the reverse, and the net effect is movement down the concentration gradient until it disappears. That is diffusion, and it needs no energy from the cell: the kinetic energy the particles already have does the work.

Diffusion: The net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient, as a result of their random motion.
Simple diffusion: Diffusion directly through the phospholipid bilayer, available to small non-polar molecules such as oxygen and carbon dioxide.
Passive: Requiring no energy from ATP; driven by the kinetic energy of the particles themselves.

Three quantities set the rate, and they combine in a single relationship you should be able to quote and use.

rate of diffusion ∝ (surface area × difference in concentration) ÷ thickness of the exchange surfaceOften called Fick's law. The first two multiply the rate; the third divides it.

Two of these help by being large and one helps by being small. Questions that describe an adaptation are almost always describing one of the three.

Every gas exchange surface you will meet is an argument about that relationship. Human lungs hold roughly 70 m² of alveolar surface folded into a chest cavity, which serves the first term. The barrier between alveolar air and blood is around 0.3 µm thick — one flattened epithelial cell and one capillary endothelial cell — which serves the third. Ventilation and a continuous blood flow keep replacing the air and the blood on either side, which is what maintains the second, because a gradient that is allowed to even out stops driving anything.

Using the relationship rather than reciting it

A section of gut has a surface area of 4 arbitrary units and an epithelium 2 units thick. In a second species the same section has 12 units of surface area and an epithelium 3 units thick. Assuming the same concentration difference, how do the rates compare?

Take the rate as proportional to area divided by thickness. For the first species that is 4 ÷ 2 = 2. For the second, 12 ÷ 3 = 4.

The second species absorbs at twice the rate, even though its epithelium is thicker. Three times the surface area more than compensates for one and a half times the thickness.

The lesson is to work with the whole expression. Candidates who spot the thicker epithelium and answer 'slower' have used one term out of three.

Facilitated diffusion is still diffusion

Ions and larger polar molecules cannot cross the hydrophobic core on their own. They cross through proteins instead: through a channel protein, which is a hydrophilic pore, or on a carrier protein, which binds the molecule, changes shape and releases it on the other side. Either way the substance is still moving down its concentration gradient, and the cell still spends nothing.

Three routes, one gradient, no ATP anywhere. The protein changes which substances can cross, never which direction they cross in.

Because the routes go through a fixed number of proteins, facilitated diffusion behaves differently from simple diffusion when you push it hard. Raise the concentration gradient and simple diffusion speeds up roughly in proportion. Facilitated diffusion speeds up too, until every channel and carrier is occupied — after that the rate levels off, however steep the gradient becomes. A graph that plateaus is telling you that a protein is involved.

	Simple diffusion	Facilitated diffusion
Route	Between the phospholipids	Through a channel or on a carrier protein
Substances	Small, non-polar: O₂, CO₂, steroids	Charged or large polar: Na⁺, K⁺, glucose, amino acids
Direction	Down the gradient	Down the gradient
ATP	None	None
At high gradient	Rate keeps rising	Rate plateaus once all proteins are occupied

One error is worth naming now, because it costs marks in the next lesson as well: a protein being involved does not mean ATP is involved. Facilitated diffusion is passive. The protein provides a route, not a push.

Osmosis, defined the way it needs to be

Water crosses membranes constantly, some of it through the bilayer and much of it through channel proteins called aquaporins. The direction it moves is set by water potential, given the symbol ψ (the Greek letter psi) and measured in kilopascals.

Water potential (ψ): The tendency of water molecules to move out of a solution, measured in kPa. Pure water at atmospheric pressure has a water potential of 0 kPa; adding any solute makes it negative.
Osmosis: The net movement of water molecules from a solution of higher water potential to a solution of lower water potential, through a partially permeable membrane.

Two consequences follow immediately, and they are what the definition is for. Pure water is the maximum, so every solution has a negative water potential — there is no such thing as a solution at +200 kPa. And 'higher' means less negative: water moves from −400 kPa to −1200 kPa, not the other way, because −400 is the higher of the two. Half the mistakes in this topic are arithmetic with negative numbers rather than biology.

Adding solute lowers water potential because the water molecules cluster around the dissolved particles and are less free to leave. A 1.0 mol dm⁻³ sucrose solution sits at about −3500 kPa; 0.2 mol dm⁻³ is around −540 kPa. Human blood plasma is roughly −800 kPa, which is the value your cells are built to sit in.

In a cell, two separate things affect water potential and they are added together.

ψ = ψ_s + ψ_pWater potential = solute potential + pressure potential. Solute potential is always negative; pressure potential in a living plant cell is zero or positive.

Solute potential (ψ_s, sometimes called osmotic potential) is the contribution from dissolved solutes, and it is negative in any real cell. Pressure potential (ψ_p) is the contribution from physical pressure. In a plant cell, water entering pushes the contents against the cellulose wall, the wall pushes back, and that push is a positive pressure potential that raises the cell's water potential back towards zero. An animal cell has no wall, so it has no meaningful pressure potential to call on — which is why the two behave so differently.

Plant cells, animal cells, and what a wall is worth

Three cells, three baths, one equation. Read the bottom row of each column against the bath printed above it: at equilibrium they match.

Put a plant cell in pure water and water enters by osmosis. The contents swell against the wall, pressure potential climbs, and water keeps entering until the cell's water potential has risen to equal the water outside. The cell is now turgid. The wall is strong enough to stop it bursting, and the pressure it generates is what holds a non-woody plant upright — a wilted plant is one whose cells have lost it.

Move that cell into a concentrated sucrose solution and the flow reverses. Water leaves, the contents shrink, and pressure potential falls. At the moment it reaches exactly zero the contents are just touching the wall but pushing on it with nothing: this is incipient plasmolysis, and at that point ψ = ψ_s, which is the whole reason the term is examined. Keep going and the membrane pulls away from the wall altogether, leaving a gap the external solution fills. The cell is plasmolysed.

An animal cell has no wall and gets neither outcome. In a solution of higher water potential than its contents — distilled water, for example — water enters, the membrane cannot resist the pressure, and the cell bursts. In a red blood cell this is haemolysis. In a solution of lower water potential, water leaves and the cell shrinks and puckers into a spiky shape: crenation. Keeping blood plasma close to −800 kPa is not a detail of physiology but a requirement for red cells to survive the trip.

Solution	Plant cell	Animal cell
Higher ψ than the cell (e.g. pure water)	Water enters, cell becomes turgid, wall prevents bursting	Water enters, cell swells and bursts (haemolysis)
Same ψ as the cell	Flaccid; no net movement	No net movement; normal shape
Lower ψ than the cell (e.g. 1.0 mol dm⁻³ sucrose)	Water leaves, membrane pulls from wall (plasmolysis)	Water leaves, cell shrinks and crenates

Turgid: A plant cell whose contents press against the wall, giving a positive pressure potential.
Incipient plasmolysis: The point at which the pressure potential of a plant cell has just fallen to zero, so its water potential equals its solute potential.
Plasmolysis: The pulling away of the cell surface membrane from the cell wall as water leaves a plant cell.
Haemolysis: The bursting of a red blood cell when water enters it by osmosis.
Crenation: The shrinking and puckering of an animal cell when water leaves it by osmosis.

Finding the water potential of potato tissue

The investigation is a serial dilution, and the point of it is that you cannot measure a cell's water potential directly. What you can do is find the external solution that produces no net movement, because that solution has the same water potential as the cells.

Make sucrose solutions from a 1.0 mol dm⁻³ stock — 0.8, 0.6, 0.4, 0.2 and 0.0 mol dm⁻³ is a standard set, each made by mixing a measured volume of stock with a measured volume of distilled water. Cut cylinders of potato with a cork borer, trim them to the same length, blot them dry and record the mass of each. Leave one cylinder in each solution for a fixed time, then blot and reweigh.

Work in percentage change in mass, not raw change. The cylinders start at slightly different masses, and a percentage removes that difference: percentage change = (change in mass ÷ initial mass) × 100. Plot it against sucrose concentration and draw a line of best fit. Where the line crosses zero, the cylinder neither gained nor lost water, so that solution's water potential equals the potato's. Read the concentration off, then convert it with the standard table.

TRY IT — Two cylinders, two directions

A potato cylinder in 0.2 mol dm⁻³ sucrose gains 6% of its mass. An identical cylinder in 0.8 mol dm⁻³ sucrose loses 11%. Using water potential, explain both results, and state what you can conclude about the water potential of the potato cells.

Check your answer

The 0.2 mol dm⁻³ solution has a water potential of about −540 kPa, which is higher — less negative — than the water potential of the potato cells. Water therefore moved by osmosis from the solution into the cells, through their partially permeable membranes, and the cylinder gained mass.

The 0.8 mol dm⁻³ solution is about −2580 kPa, which is lower than the water potential of the cells. Water moved out of the cells into the solution, the vacuoles and cytoplasm lost volume, and the cylinder lost mass. The larger percentage change suggests a steeper water potential difference in this direction.

So the water potential of the potato cells lies between −540 kPa and −2580 kPa. To narrow it down you need the concentration at which the percentage change is zero, which is what plotting all six points and reading off the intercept gives you.

Notice that neither answer mentions the concentration of water. Both compare two water potentials and let the sign of the difference give the direction.

In the exam

Water moves from higher to lower water potential, and higher means less negative. Sketch a number line if you have to: −400 is above −1200.
Define osmosis with water potential and a partially permeable membrane. 'From high to low concentration of water' is not accepted at A level.
Every solution has a negative water potential and pure water is 0 kPa. A positive value for a solution is an error, not an unusual case.
If a question gives you two of ψ, ψs and ψp, it wants the third. The arithmetic is trivial and the signs are where marks are lost.
Facilitated diffusion needs no ATP. Writing that it does is one of the most reliable ways to lose a mark in this unit.
For the potato practical, use percentage change in mass and say why: the cylinders do not start at identical masses.
When a rate-against-concentration graph plateaus, name the reason — all the carrier or channel proteins are occupied.

Check yourself

A plant cell has a solute potential of −1400 kPa and a pressure potential of +500 kPa. It is placed in a sucrose solution of water potential −1100 kPa. Calculate the cell's water potential, state the direction of net water movement, and explain what will happen to the cell's pressure potential as a result.

Answer

Water potential is solute potential plus pressure potential: ψ = −1400 + 500 = −900 kPa.

The cell is at −900 kPa and the solution is at −1100 kPa. The cell has the higher (less negative) water potential, so water moves by osmosis out of the cell into the solution, through the partially permeable cell surface membrane.

As water leaves, the volume of the cell contents falls, so they press less hard against the cellulose wall and the pressure potential drops below +500 kPa. The solute potential becomes slightly more negative too, since the same solutes are now dissolved in less water, but the fall in pressure potential is the larger effect.

Water continues to leave until the cell's water potential has fallen to −1100 kPa and matches the solution. If the pressure potential reaches zero before that happens, the cell is at incipient plasmolysis, and any further loss of water pulls the membrane away from the wall.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 14 marks

A potato cylinder of initial mass 4.20 g is left in a sucrose solution for thirty minutes, blotted dry and reweighed at 3.78 g. Calculate the percentage change in mass and state the direction of net water movement.

Mark scheme

M1 change in mass = 3.78 − 4.20 = −0.42 g
M1 percentage change = change in mass divided by initial mass, multiplied by one hundred: (−0.42 ÷ 4.20) × 100
A1 −10.0 per cent, that is a loss of a tenth of the initial mass
B1 water moved out of the cylinder into the solution, so the solution had the lower, more negative, water potential

Question 24 marks

Explain what happens to a red blood cell placed in distilled water, and what happens to an identical red blood cell placed in a concentrated sodium chloride solution.

Mark scheme

B1 distilled water has a water potential of 0 kPa, which is higher, that is less negative, than the roughly −800 kPa of the cell contents
B1 water therefore enters by osmosis through the partially permeable cell surface membrane, and with no cell wall to resist the pressure the cell bursts, which is haemolysis
B1 the concentrated sodium chloride solution has a lower, more negative, water potential than the cell contents, so water leaves the cell by osmosis
B1 the cell shrinks and puckers into a spiky shape, which is crenation

Question 34 marks

The uptake of glucose by red blood cells rises as the external glucose concentration rises and then levels off, while the uptake of oxygen by the same cells continues to rise across the whole range tested. Suggest an explanation for both results.

Mark scheme

B1 glucose is a large polar molecule and cannot cross the hydrophobic core unaided, so it enters by facilitated diffusion through carrier proteins
B1 the rate levels off once every carrier protein is occupied, so a steeper concentration gradient cannot raise it any further
B1 oxygen is small and non-polar, so it dissolves in the bilayer and crosses by simple diffusion between the phospholipids
B1 simple diffusion uses no proteins, so there is nothing to become saturated and the rate keeps rising with the concentration difference

Question 43 marks

A plant cell has a water potential of −1150 kPa and a pressure potential of +450 kPa. Calculate the solute potential of the cell, giving your answer in kPa.

Mark scheme

M1 rearranges the relationship to give solute potential = water potential − pressure potential
M1 substitutes correctly: −1150 − (+450)
A1 −1600 kPa, negative as a solute potential must always be

Question 53 marks

Explain how the alveoli of the lungs and the blood flowing past them maintain a high rate of diffusion of oxygen into the blood.

Mark scheme

B1 the alveoli give an enormous total surface area, about 70 m² in a human lung, and rate is proportional to surface area
B1 the barrier is only about 0.3 µm thick, one flattened epithelial cell and one capillary endothelial cell, and rate is inversely proportional to thickness
B1 ventilation and continuous blood flow keep replacing the air and the blood, which maintains a steep concentration difference that would otherwise even out

Question 62 marks

State the water potential of pure water at atmospheric pressure, and state what this tells you about the sign of the water potential of any solution.

Mark scheme

B1 pure water at atmospheric pressure has a water potential of 0 kPa
B1 any solution therefore has a negative water potential, because adding solute can only lower it

Worth remembering

Rate of diffusion rises with surface area and concentration difference, and falls as the exchange surface gets thicker.
Facilitated diffusion uses proteins but no ATP, and its rate plateaus when they are all occupied.
Pure water is 0 kPa; every solution is negative; water moves from the less negative to the more negative.
ψ = ψs + ψp, with ψs always negative and ψp zero or positive in a living plant cell.
At incipient plasmolysis ψp = 0, so ψ = ψs.

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