Movement into and out of cells

1. Overview

Water crosses cell membranes by 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. The direction and size of this movement are set by the difference in water potential between a cell and the solution around it. Because plant cells have a strong cellulose cell wall and animal cells do not, the same water movement produces very different effects in the two cell types. This topic also covers how the water potential of a plant tissue can be estimated experimentally by immersing it in a range of solutions of known water potential.

Osmosis is one of several ways substances cross membranes: simple and facilitated diffusion are also passive, while active transport, endocytosis and exocytosis use ATP. The topic ends with surface area to volume ratio.

Key Definitions

• 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.
• Water potential: a measure of the tendency of water molecules to move out of a system by osmosis, measured in kilopascals (kPa).
• Partially permeable membrane: a membrane that lets small molecules such as water pass through but restricts the movement of larger solute molecules.
• Turgid: the state of a plant cell that has taken in water until its protoplast pushes firmly against the cell wall and no more water can enter.
• Flaccid: the state of a plant cell that has lost water so it is soft and no longer pushes firmly against the wall, but the membrane has not yet pulled away.
• Plasmolysis: the pulling away of the cell surface membrane and cytoplasm from the cell wall as a plant cell loses water in a solution of lower water potential.
• Incipient plasmolysis: the point at which the cell membrane is just beginning to pull away from the wall, so the protoplast no longer presses against the wall.
• Haemolysis: the bursting of a red blood cell when it takes in so much water by osmosis that the cell surface membrane ruptures.
• Crenation: the shrinking and wrinkling of an animal cell as it loses water by osmosis to a solution of lower water potential.
• Simple diffusion: the net movement of molecules or ions from a region of higher concentration to a region of lower concentration, caused by their random movement.
• Facilitated diffusion: the diffusion of ions or polar molecules through specific channel or carrier proteins, down their concentration gradient and without using ATP.
• Active transport: the movement of molecules or ions against their concentration gradient through specific carrier proteins, using ATP.
• Endocytosis: the bulk uptake of material into a cell inside a vesicle formed from the cell surface membrane.
• Exocytosis: the bulk release of material from a cell when a vesicle fuses with the cell surface membrane.
• Surface area to volume ratio (SA:V): the surface area of an object divided by its volume.

Content

Simple and facilitated diffusion

Particles move randomly because of their kinetic energy, giving a net movement down a concentration gradient. Both kinds of diffusion are passive: they use no ATP.

• In simple diffusion, small non-polar molecules such as oxygen and carbon dioxide cross a membrane straight through the phospholipid bilayer.
• In facilitated diffusion, ions and polar molecules such as glucose, which cannot cross the hydrophobic core, pass through specific channel or carrier proteins (see topic 4.1). The rate stops rising once all these proteins are in use.

Diffusion is faster with a steeper concentration gradient, a higher temperature, a larger surface area and a shorter diffusion distance.

Water potential and the direction of osmosis

Water potential (symbol $\Psi$, the Greek letter psi) measures how likely water is to move out of a system by osmosis. It is measured in kilopascals (kPa). By definition, pure water at standard temperature and pressure has the highest possible water potential, which is set at 0 kPa. Adding any solute lowers the water potential, so every real solution, and almost every living cell, has a water potential that is a negative number. The more solute dissolved, the lower (more negative) the water potential.

Water always moves by osmosis from a higher water potential to a lower water potential - that is, from a less negative value towards a more negative value (for example, from -200 kPa to -500 kPa). It keeps moving until the two water potentials are equal, when there is no longer a net movement of water. Note that osmosis is the movement of water, while solutes (such as sugar) stay behind because they cannot cross the partially permeable membrane.

Effect of water movement on animal cells

Animal cells, such as red blood cells, have only a cell surface membrane and no wall to resist swelling, so they are very sensitive to the water potential of their surroundings:

• In a solution of higher water potential than the cell (a dilute or "hypotonic" solution), water enters by osmosis, the cell swells, and because there is no wall to stop it the membrane stretches and bursts. In a red blood cell this bursting is called haemolysis.
• In a solution of lower water potential than the cell (a concentrated solution), water leaves by osmosis, the cell shrinks and the membrane wrinkles - this is crenation.
• In a solution of the same water potential as the cell, there is no net movement and the cell keeps its normal shape.

This is why the water potential of blood plasma and tissue fluid must be tightly controlled in the body.

Effect of water movement on plant cells

Plant cells have a strong cellulose cell wall outside the membrane. The wall is fully permeable to water and solutes but is inelastic, so it stops the cell bursting.

When the outside solution has a higher water potential than the cell:

• Water enters the cell by osmosis.
• The protoplast swells and pushes outwards against the wall.
• The inelastic wall pushes back, building up an inward pressure.
• This pressure rises until no more water can enter, and the cell becomes firm, or turgid.

Turgor matters because it supports non-woody (herbaceous) plant tissues and keeps leaves and young stems upright.

When the outside solution has a lower water potential than the cell:

• Water leaves the cell by osmosis.
• The protoplast shrinks, so the cell first becomes flaccid (soft).
• If water loss continues, the membrane and cytoplasm pull away from the wall. This separation is plasmolysis, and the cell is described as plasmolysed.
• The exact point where the membrane just begins to pull away is incipient plasmolysis.

The key difference is that the wall lets a plant cell become turgid without bursting, whereas an animal cell with no wall would burst (lyse) in the same dilute solution. The table below sets the two cell types side by side.

Surrounding solution	Plant cell	Animal cell (e.g. red blood cell)
Higher water potential (dilute)	Water enters; cell becomes turgid (wall prevents bursting)	Water enters; cell swells and bursts (haemolysis)
Same water potential	No net movement; cell stays the same	No net movement; cell keeps normal shape
Lower water potential (concentrated)	Water leaves; cell becomes flaccid, then plasmolysed	Water leaves; cell shrinks and wrinkles (crenation)

It also helps to remember the order in which a plant cell changes as it loses water: turgid -> flaccid -> incipient plasmolysis -> fully plasmolysed.

Active transport

Active transport moves molecules or ions against their concentration gradient through specific carrier proteins (pumps). Hydrolysis of ATP from respiration makes the carrier change shape and release the particle on the other side; the sodium-potassium pump, for example, moves 3 Na⁺ out of a cell for every 2 K⁺ in (see topic 15.1). Cells doing a lot of active transport have many mitochondria, and a lack of oxygen slows active transport but not diffusion.

Endocytosis and exocytosis

These move large particles or large amounts of material in bulk, inside vesicles, using ATP.

• In endocytosis, the cell surface membrane surrounds the material and pinches off as a vesicle, as when a phagocyte engulfs a bacterium (see topic 11.1).
• In exocytosis, vesicles, often from the Golgi body, fuse with the cell surface membrane and release their contents, as in secreting enzymes or hormones.

Investigating diffusion and osmosis with non-living materials

Visking (dialysis) tubing is an artificial partially permeable membrane: water and small molecules such as glucose pass through it, but large molecules such as starch do not.

• Diffusion: stand tubing filled with starch and glucose solution in distilled water. The water later gives a positive Benedict's test but a negative iodine test: only glucose diffused out.
• Osmosis: weigh tubing filled with concentrated sucrose solution, stand it in distilled water and reweigh it at intervals. The mass rises as water enters by osmosis.

Agar made up with dilute sodium hydroxide and phenolphthalein is pink. Dilute hydrochloric acid diffusing in neutralises the alkali, so the pink fades from the outside inwards. Timing this tests a factor such as temperature or acid concentration, with the others kept constant.

Estimating the water potential of a plant tissue

Because there is no net water movement when the water potential of the tissue equals that of the surrounding solution, the tissue's water potential can be estimated by finding the external solution at which the tissue neither gains nor loses water.

A standard approach using a tissue such as potato:

1. Cut several pieces of tissue of equal size, blot off surface liquid and measure each one (record its mass, or its length).
2. Make a range of solutions of known water potential - for example, several concentrations of sucrose (or salt) solution - and add equal volumes to labelled tubes. Include a distilled-water control (water potential 0 kPa).
3. Leave a piece of tissue in each solution for a set time at constant temperature.
4. Remove, blot, and re-measure each piece. Calculate the percentage change in mass (or length): $\text{percentage change} = \dfrac{\text{final} - \text{initial}}{\text{initial}} \times 100$. Using percentage change rather than the raw figure removes the effect of small differences in starting size.
5. Plot percentage change (y-axis) against the water potential of the solution (x-axis). Where the best-fit line crosses zero (no change in mass), the external water potential equals the water potential of the tissue, so read this value off the x-axis.

The graph shown for this topic illustrates the shape of these results: pieces in solutions of higher water potential gain mass (water enters) and lie above the zero line, while those in solutions of lower water potential lose mass (water leaves) and lie below it; the cross-over point where the best-fit line meets zero gives the estimate. The axes are illustrative, so the figure shows the shape of the relationship rather than exact readings. Because cells vary, this is an estimate and the result is an average for the tissue.

Worked example

Exam-style question: Identical cubes of beetroot tissue were each blotted, weighed, and left for 60 minutes in one of several sucrose solutions ranging from a high (less negative) to a low (more negative) water potential. The cubes were then blotted and re-weighed. A graph of percentage change in mass against the water potential of each solution was plotted, and the line crossed zero at -720 kPa. Explain what these results show about the beetroot tissue, and explain why percentage change in mass, rather than the actual change in mass, was plotted. [4]

Model answer:

• The water potential of the beetroot tissue is approximately -720 kPa, because at this external water potential there was no net movement of water, so no change in mass.
• In solutions of higher water potential than the tissue, water entered the cells by osmosis and the cubes gained mass; in solutions of lower water potential, water left the cells and the cubes lost mass.
• At the cross-over point the water potential of the solution equals the water potential of the tissue, so this value is read from the graph.
• Percentage change is used so that all cubes are compared on the same scale even if their starting masses differed slightly, making the results a fair comparison.

Worked example

Exam-style question: A student measured the percentage change in mass of potato cylinders left in a range of sucrose solutions. Their results are shown below.

Water potential of solution / kPa	Percentage change in mass / %
-100	+6.0
-300	+3.0
-550	0.0
-800	-4.5
-1100	-8.0

(a) State the water potential of the potato tissue and explain how you obtained it. (b) Predict whether a potato cell placed in a solution of water potential -300 kPa would gain or lose water, and state whether it would become more turgid or more flaccid. [4]

Model answer:

• (a) The water potential of the tissue is -550 kPa, because this is the solution at which there was 0% change in mass, showing no net movement of water - so the solution and the tissue have the same water potential.
• (a) (At -550 kPa the percentage change crosses zero, so the external water potential here equals the water potential of the tissue.) Here the zero point falls exactly on a tabulated row; if it fell between two points - say between -300 kPa (+3.0%) and -550 kPa (0.0%) - you would read the value off the best-fit line where it crosses zero (interpolation), not simply quote the nearest row.
• (b) The solution at -300 kPa has a higher (less negative) water potential than the tissue (-550 kPa), so water moves into the cell by osmosis; the cell gains water.
• (b) Because water enters, the protoplast pushes harder against the wall, so the cell becomes more turgid.

Surface area to volume ratio

Substances cross a cell's surface but are used or made throughout its volume. As an object gets larger, its volume increases faster than its surface area, so its surface area to volume ratio (SA:V) falls:

Length of side of cube / cm	Surface area / cm²	Volume / cm³	SA:V
1	6	1	6 : 1
2	24	8	3 : 1
4	96	64	1.5 : 1

Doubling the side multiplies the surface area by 4 but the volume by 8, so SA:V halves, and the distance to the centre doubles. This is one reason cells are small and large organisms need specialised exchange surfaces and transport systems.

Investigating SA:V with agar blocks

Put cubes of the pink agar with sides of, say, 0.5 cm, 1 cm and 2 cm in equal volumes of the same acid at the same temperature, and time how long each takes to become completely colourless (repeat for a mean). The smallest clears first: the acid moves in at about the same speed in each, but a smaller cube has a larger SA:V and a shorter distance to its centre.

Worked example

Exam-style question: Block P of the pink agar is a cuboid measuring 4.0 cm × 1.0 cm × 0.50 cm, and block Q is a cylinder of radius 0.50 cm and length 3.0 cm. (a) Calculate the SA:V of each block. (b) Both blocks were put in the same dilute hydrochloric acid. Predict which became colourless first, and explain your answer. [5]

Model answer:

• (a) Block P: surface area $= 2[(4.0 \times 1.0) + (4.0 \times 0.50) + (1.0 \times 0.50)] = 2(4.0 + 2.0 + 0.50) =$ 13 cm²; volume $= 4.0 \times 1.0 \times 0.50 =$ 2.0 cm³; SA:V $= 13 \div 2.0 =$ 6.5 : 1.
• (a) Block Q: surface area $= 2\pi r^2 + 2\pi rh = 2\pi(0.50)^2 + 2\pi(0.50)(3.0) = 1.571 + 9.425 =$ 11.0 cm²; volume $= \pi r^2 h = \pi(0.50)^2(3.0) =$ 2.356 cm³; SA:V $= 10.996 \div 2.356 = 4.67$, so 4.7 : 1.
• (b) Block P. Its larger SA:V gives more surface for acid to enter for each cm³ of agar to be neutralised.
• (b) Its centre is also only 0.25 cm from its largest faces, but the centre of Q is 0.50 cm from its curved surface, so the diffusion distance is shorter in P.

Key Equations

This topic is mainly qualitative, but two calculations are used. The first is used when estimating tissue water potential: $$\text{percentage change in mass} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100$$ Water potential is measured in kilopascals (kPa); pure water = 0 kPa, and all solutions (and almost all cells) have negative values.

The second is the surface area to volume ratio, given as a ratio to 1 (such as 6.5 : 1): $$\text{SA:V} = \frac{\text{surface area}}{\text{volume}}$$ For a cuboid $l \times w \times h$, surface area $= 2(lw + lh + wh)$ and volume $= lwh$. For a cylinder of radius $r$ and length $h$, surface area $= 2\pi r^2 + 2\pi rh$ and volume $= \pi r^2 h$.

Common Mistakes to Avoid

• Writing "concentration of water" instead of "water potential". Osmosis is driven by water potential, not by a "concentration of water molecules". Always describe water moving from a higher water potential to a lower water potential.
• Getting the sign of water potential the wrong way round. Pure water is 0 kPa (the highest), and adding solute makes the value more negative. So -200 kPa is a higher water potential than -500 kPa, and water moves from the -200 kPa region to the -500 kPa region.
• Saying a plant cell "bursts" in a dilute solution. The cellulose cell wall is inelastic and stops the cell bursting; the cell becomes turgid instead. Only cells without a wall, such as red blood cells, burst (haemolysis).
• Confusing flaccid with plasmolysed. A cell that has lost some water is flaccid; only when the membrane has pulled away from the wall is it plasmolysed.
• Saying the membrane "breaks" during plasmolysis. In plasmolysis the membrane stays intact but detaches from the wall as the protoplast shrinks.
• Forgetting to control variables in the experiment. Keep temperature, time and tissue size constant, and use a range of solutions; otherwise the cross-over point is unreliable.
• Recording units inside the table. Put units only in the column headings (for example "Mass / g" or "Time / min") and leave the cells of the table for numbers alone.
• Saying facilitated diffusion uses ATP. It is passive, down the concentration gradient. Only active transport, endocytosis and exocytosis use ATP.

Exam Tips

• State the direction of osmosis explicitly: from higher water potential to lower water potential, through a partially permeable membrane. Markers look for all three parts.
• When comparing plant and animal cells, build your answer around the presence or absence of the cell wall - this single difference explains turgidity versus bursting.
• Use the precise vocabulary for each state: turgid, flaccid, plasmolysed for plant cells; haemolysis and crenation for animal cells.
• Watch the command word for cell-effect questions: "describe" wants the observable change (the cell swells, becomes turgid, or bursts), but "explain" must give the water potential reason - water enters because the outside has a higher water potential, moving through the partially permeable membrane.
• In data questions, remember the tissue's water potential is the value where the percentage-change line crosses zero on the x-axis - describe it as an estimate and quote it with units (kPa).
• When asked to plan or improve the experiment, mention repeats, blotting before weighing, a constant temperature, and using percentage change so different starting sizes can be compared fairly.