Cambridge IGCSE0610

Diffusion

Biology 0610 Chapter Notes

What this chapter covers

DiffusionOsmosisActive transport

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In this chapter

1What diffusion is and where its energy comes from
2Diffusion into and out of cells
3Concentration gradient and temperature
4Surface area and distance
5Investigating diffusion
6Water as a solvent
7Osmosis and partially permeable membranes
8Investigating osmosis
9Plant tissue in different solutions
10Water pressure supports plants
11Osmosis and water potential
12Turgid, flaccid and plasmolysed cells
13Water potential in organisms
14Moving against the concentration gradient
15Energy from respiration
16Why cells need it: ion uptake by root hairs
17Protein carriers

1. What diffusion is and where its energy comes from

Diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration (i.e. down a concentration gradient), as a result of their random movement.

The energy for diffusion comes from the kinetic energy of random movement of molecules and ions. It is not energy from respiration.

Examiner insight

“Very occasionally, candidates contradicted themselves by referring to particles moving against their concentration gradient.” “Incorrect responses included activation energy, passive energy and suggestions that energy was not required.”

Worked example 11 mark

Which row shows correct features of diffusion?

	direction of net movement of particles	always across a cell membrane	requires energy from respiration
A	high concentration to low concentration	no	no
B	high concentration to low concentration	yes	yes
C	low concentration to high concentration	no	yes
D	low concentration to high concentration	yes	no

1
Row A: the net movement is from high concentration to low concentration, it is not always across a cell membrane and it needs no energy from respiration. [1]

2. Diffusion into and out of cells

Some substances move into and out of cells by diffusion through the cell membrane. Gases and solutes both move by diffusion.

Carbon dioxide diffuses through the stomata into a leaf, where it is needed for photosynthesis. Oxygen passes out of a leaf by diffusion. In the lungs, oxygen moves by diffusion from a region of higher concentration in the alveoli to a region of lower concentration in the blood, then into the red blood cells. Glucose, a solute, enters cells by both active transport and diffusion.

Examiner insight

“A large proportion of candidates incorrectly chose transpiration as the process by which oxygen passes out of a leaf.” “Some candidates did not describe it fully, for example, stating that carbon dioxide is absorbed rather than diffuses through stomata.”

Worked example 12 marks

On a sunny day, carbon dioxide moves from the air into a leaf through the stomata. State the name of this process. Explain why the carbon dioxide moves into the leaf.

1
diffusion [1]
2
the concentration of carbon dioxide is higher in the air than in the leaf, so it moves down the concentration gradient [1]

3. Concentration gradient and temperature

The rate of diffusion increases as the concentration gradient gets steeper. A smaller difference in concentration means slower diffusion.

A higher temperature gives the particles more kinetic energy, so they move faster. Increased temperature increases the rate of diffusion.

Examiner insight

“Some incorrectly thought that decreasing the concentration gradient would increase the rate of diffusion.”

Worked example 13 marks

A student investigated the effect of surface area to volume ratio on the time taken for hydrochloric acid to move through cubes of agar jelly. The student recorded the time taken for each cube to become completely red.

The investigation was repeated at a higher temperature. Predict and explain the effect of this on the results.

1
the time taken for all cubes to turn red decreases [1]
2
increasing temperature increases the kinetic energy of the particles [1]
3
this increases the rate of diffusion [1]

4. Surface area and distance

The higher the surface area, the greater the rate of diffusion. A greater distance for the particles to move through makes diffusion slower.

A bigger cube has a smaller surface area to volume ratio. There is a greater distance for the particles to move through to reach its centre, so the time taken increases.

Examiner insight

“Some candidates chose decreasing the surface area or decreasing the concentration gradient.”

Worked example 13 marks

A student investigated the effect of surface area to volume ratio on the time taken for hydrochloric acid to move through cubes of agar jelly. The cubes contained universal indicator that changed colour from green to red at low pH values. The student recorded the time taken for each cube to become completely red.

Describe and explain the results in the table.

cube	surface area : volume ratio	time taken for the cube to become completely red / s
A	6 : 1	15.2
B	3 : 1	28.5
C	2 : 1	43.0
D	1.5 : 1	57.8

1
as the surface area to volume ratio decreases, the time taken for the cube to turn completely red increases [1]
2
correct reference to diffusion [1]
3
a bigger cube has a greater distance for the acid particles to move through to reach the centre [1]

5. Investigating diffusion

Two practicals time a colour change. Cylinders of beetroot are cut into 1, 2, 4 or 8 pieces and put in hot water. The time taken for the liquid to reach the same colour is recorded. Agar cubes with universal indicator are put in hydrochloric acid, and the time taken to turn red is recorded.

The independent variable is what you change, such as surface area. The dependent variable is what you measure, which is time. A conclusion must be about the rate of diffusion, not only about time.

Examiner insight

“To construct a valid conclusion, candidates should go back to the stem of the question and use the underlying science behind any dependent variable measurement.” “Candidates must state the variable that was measured, rather than a calculated value.”

Worked example 12 marks

A student investigated the effect of surface area on the diffusion of pigment out of cylinders of beetroot tissue. A cylinder was cut into 1, 2, 4 or 8 pieces and put in hot water. The student timed how long the liquid took to become the same colour as the liquid in test-tube 1.

State the independent and dependent variables in this investigation.

1
independent variable: surface area (of beetroot) [1]
2
dependent variable: time (for the liquid to become the same colour as test-tube 1) [1]

Recap

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient.
It happens because of the random movement of particles.
The energy comes from the kinetic energy of the molecules and ions. It does not come from respiration.
Gases and solutes move into and out of cells by diffusion through the cell membrane. Carbon dioxide diffuses into a leaf through the stomata, and oxygen diffuses from the alveoli into the blood.
Diffusion is faster with a steeper concentration gradient, a higher temperature, a larger surface area and a shorter distance.
In a practical, state the variable you measured (time), then link the conclusion to the rate of diffusion.

6. Water as a solvent

Water is a solvent: other substances dissolve in it.

In organisms, water is the solvent in digestion, excretion and transport. In digestion, water is the solvent for most of the molecules that enzymes have digested.

When a question asks for a use of water, give a specific one, such as support, transport or acting as a solvent.

Examiner insight

“Many candidates gave generic answers rather than stating specific uses of water, such as photosynthesis, support, transport or to act as a solvent.” “Very few candidates stated that water is the substance that act as a solvent for most molecules that have been digested by enzymes.”

Worked example 11 mark

State the name of the substance that acts as a solvent for most molecules that have been digested by enzymes.

1
water. Hydrochloric acid, gastric juice and bile are not credited [1]

7. Osmosis and partially permeable membranes

Water diffuses through partially permeable membranes by osmosis. Water moves into and out of cells by osmosis through the cell membrane.

Only water molecules move by osmosis. Osmosis does not use energy from respiration, and it does not need a cell wall.

Examiner insight

“Many candidates did not appreciate that although the process of osmosis requires a partially permeable membrane, diffusion does not.” “Only water moves across partially permeable membranes by osmosis.”

Worked example 11 mark

Which statements describe features of osmosis?

1 a partially permeable membrane · 2 the movement of any liquid · 3 using energy from respiration · 4 water molecules moving by diffusion

A 1 and 2 · B 1 and 4 · C 2 and 3 · D 3 and 4

1
B: a partially permeable membrane (1) and water molecules moving by diffusion (4) [1]

8. Investigating osmosis

Cut potato cylinders to the same length, for example 40 mm. Put one in each of several salt solutions and measure the final length. The independent variable is the salt solution concentration. The dependent variable is what you measure: the length, or the mass of potato.

Keep constant the volume of solution, the starting length, the diameter and the soaking time. The solution giving a change closest to 0 mm has a concentration closest to that of the solutes in the potato cells.

With dialysis tubing, measure the starting and final volume of the sugar solution in the tubing.

change in length = final length - starting length

Examiner insight

“Most candidates correctly stated that the mass of potato was the dependent variable in the investigation. Many gave incomplete answers such as just potato or size, instead of mass of potato.” “Using gloves to cut the potato or cutting it carefully were not accepted responses.”

Worked example 12 marks

You are going to investigate osmosis in cylinders of potato tissue placed in different concentrations of salt solution.

Step 3: Place the potato cylinders on a white tile and cut each cylinder to a length of exactly 40 mm.

Identify one hazard in step 3 and describe an appropriate safety precaution.

1
hazard: use of a knife / cutting [1]
2
precaution: cut away from the body / fingers / hand, or cut downwards on a tile / flat surface [1]

9. Plant tissue in different solutions

Plant tissue in distilled water takes in water by osmosis. It gains mass and its length increases.

Tissue in a concentrated solution loses water by osmosis. It loses mass and its length decreases.

It is the water that moves, not the solution and not the sugar or salt. Only water moves into or out of the potato cells.

Examiner insight

“Many candidates incorrectly thought that the potato would increase in mass in the concentrated sugar solution and lose mass in distilled water.” “Common misconceptions included linking ‘hardness’ of the potato stick to water loss, or more commonly, thinking that the potato would go soft when it soaked up water, as well as statements relating to the uptake of the solution rather than the water in the solution.”

Worked example 11 mark

Identical pieces of potato were placed in two test-tubes.

Distilled water was added to test-tube R and concentrated salt solution was added to test-tube S.

Which row shows the expected change in mass of the potato?

	test-tube R	test-tube S
A	mass decreased	mass decreased
B	mass decreased	mass increased
C	mass increased	mass decreased
D	mass increased	mass increased

1
C: R in distilled water gains mass, S in concentrated solution loses mass [1]

10. Water pressure supports plants

Plants are supported by the pressure of water inside the cells pressing outwards on the cell wall. This pressure is called turgor pressure.

The cell wall also stops a plant cell from bursting as it takes in water.

Examiner insight

“Only a small proportion of candidates understood that turgor pressure is the pressure that helps to provide support for plants.” “This highlighted a misconception as candidates should realise that turgor can only occur in cells that have a cell membrane as well as a cell wall.”

Worked example 11 mark

Some red blood cells and some plant cells were placed in distilled water for 10 minutes.

After 10 minutes the red blood cells had burst. The plant cells increased in volume but did not burst.

Suggest a reason why the plant cells do not burst when placed in distilled water.

1
they have a cell wall [1]

11. Osmosis and water potential

Osmosis is the net movement of water molecules from a region of higher water potential (dilute solution) to a region of lower water potential (concentrated solution), through a partially permeable membrane.

Examiner insight

“Many correctly described the movement of water molecules from a high water potential to a low water potential and then contradicted themselves by stating that it was down a water concentration gradient.” “Some candidates ignored the instruction in the question and wrote about the concentrations of sucrose and water instead of using water potential terminology.”

Worked example 12 marks

A dilute sugar solution and a concentrated sugar solution are separated by a partially permeable membrane. Describe the net movement of water molecules.

1
from higher water potential (dilute) to lower water potential (concentrated) [1]
2
through the partially permeable membrane [1]

12. Turgid, flaccid and plasmolysed cells

In distilled water, a plant cell has a lower water potential than the water, so it gains water by osmosis. Vacuole volume and turgor pressure increase. The cell is turgid: its contents push against the cell wall.

In a solution with a lower water potential than the cell, it loses water. Turgor pressure decreases and the cell becomes flaccid. Plasmolysis: the cell membrane pulls away from the cell wall and the cell is plasmolysed.

Examiner insight

“Some candidates incorrectly believed that a plant cell gains water by osmosis and becomes plasmolysed.” “Few candidates referred to the lack of turgor pressure in their responses.”

Worked example 16 marks

A sample of plant tissue was immersed in a highly concentrated salt solution. Another sample of plant tissue was immersed in distilled water.

Describe and explain the expected appearance of the cells in the two samples of plant tissue after 60 minutes of immersion.

1
osmosis: water moves across a partially permeable membrane [1]
2
salt solution: cells have a higher water potential than the solution, so water moves out [1]
3
vacuole / cell sap decreases in volume and turgor pressure decreases [1]
4
cells become flaccid or plasmolysed: the cell membrane pulls away from the cell wall [1]
5
distilled water: cells have a lower water potential than the water, so water moves in [1]
6
turgor pressure increases: cells become turgid, contents push against the cell wall [1]

13. Water potential in organisms

Root hair cells take in water by osmosis. Nitrate ions in the root cells decrease their water potential, which enables the uptake of water.

If less water is absorbed, plant cells become flaccid and the plant wilts.

Red blood cells burst in distilled water. Plant cells take in water too, but their cell wall stops them bursting.

Examiner insight

“Most candidates knew that active transport was used for ion uptake in root hair cells, but a significant number confused it with water absorption by osmosis.” “The best answers explained that less water would be absorbed, and cells would become flaccid, and the plant would wilt.”

Worked example 13 marks

Scientists investigated the effect of four different concentrations of salt solution on the volume of red blood cells.

Complete the sentences to describe the data in Table 2.1.

As the percentage concentration of salt solution increases, the final mean volume of the red blood cells ____.

The final mean volume of the red blood cells is greater than the initial volume in ____ % salt solution and ____ % salt solution.

The mean volume of the red blood cells in the 1.6% salt solution changes by ____ arbitrary units.

percentage concentration of the salt solution	initial mean volume of the red blood cells / arbitrary units	final mean volume of the red blood cells / arbitrary units
0.4	100	139
0.8	100	107
1.2	100	71
1.6	100	30

1
decreases [1]
2
0.4% and 0.8%: the only two with a final volume above 100 [1]
3
100 - 30 = 70 arbitrary units [1]

Recap

Water is a solvent: substances dissolve in it.
Water diffuses through partially permeable membranes by osmosis. Only water moves.
Potato in distilled water gains mass. Potato in concentrated solution loses mass.
In the investigation, the salt concentration is the independent variable. The change closest to 0 mm shows the cell solute concentration.
Plants are supported by the pressure of water inside the cells pressing outwards on the cell wall.
Extended: osmosis is net movement of water from higher to lower water potential.
Extended: distilled water makes a plant cell turgid. A concentrated solution makes it flaccid or plasmolysed.

14. Moving against the concentration gradient

Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration, using energy from respiration.

This is movement against a concentration gradient. Diffusion goes the other way.

Examiner insight

“Some candidates incorrectly opted for the diagram showing particles moving from a region of higher concentration to a region of lower concentration.”

Worked example 13 marks

Some substances are transported across cell membranes against a concentration gradient.

(i) State the name of this type of transport.

(ii) State the name of the process that releases energy for this type of transport.

(iii) State the name of the cell structure where the process named in(ii) occurs.

1
(i) active transport [1]
2
(ii) respiration [1]
3
(iii) mitochondria [1]

15. Energy from respiration

Active transport needs energy. The energy comes from respiration.

If respiration stops, active transport slows down because the energy it needs is no longer released.

Examiner insight

“Many candidates did not appreciate that since cyanide is a poison that stops respiration, it must also decrease active transport. Active transport requires energy from respiration.”

Worked example 11 mark

Cyanide is a chemical that prevents the release of energy in flowering plants.

Which process will slow down if cyanide is added?

A active transport · B diffusion · C osmosis · D transpiration

1
A: active transport needs energy from respiration, which cyanide prevents. [1]

16. Why cells need it: ion uptake by root hairs

Active transport moves molecules or ions across membranes. A plant uses it for ion uptake by root hairs.

A root hair cell has a higher concentration of nitrate ions than the soil, so the ions enter against the concentration gradient. Say mineral ions or name the ion, not "nutrients".

Examiner insight

“Most candidates knew that active transport was used for ion uptake in root hair cells, but a significant number confused it with water absorption by osmosis.”

Worked example 11 mark

What is an example of active transport?

A absorption of nitrate ions from the soil

B evaporation of water from leaf mesophyll cells

C movement of oxygen from an alveolus into a blood capillary

D movement of water from the small intestine into the blood

1
A: nitrate ions are taken up by root hairs by active transport. [1]

17. Protein carriers

Protein carriers in the cell membrane move molecules or ions across it during active transport.

The carrier moves the ions. It does not cross the membrane and does not supply the energy: it uses energy from respiration.

Examiner insight

“Most candidates understood active transport, but some did not realise that it requires a protein carrier in the cell membrane.” “A common misconception was to refer to the proteins, rather than the ions, moving across the membrane.”

Worked example 13 marks

Describe how a root hair cell takes up nitrate ions from the soil by active transport.

1
protein carriers are in the cell membrane of the root hair cell [1]
2
the carriers move the nitrate ions across the membrane, by changing shape [1]
3
the ions move against the concentration gradient, using energy from respiration [1]

Recap

Active transport: movement of particles through a cell membrane from lower to higher concentration, against a concentration gradient.
It uses energy from respiration. If respiration stops, active transport slows down.
Diffusion is the opposite direction: higher to lower concentration.
Extended: it moves molecules or ions across membranes, including ion uptake by root hairs.
Extended: protein carriers in the membrane move the ions. They use energy; they do not supply it.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

State what is meant by diffusion.1 mark
State the source of the energy for diffusion.1 mark
Oxygen moves into a red blood cell through the cell membrane by diffusion. State whether the cell uses energy from respiration for this movement, and give a reason.2 marks
Oxygen moves from the air in the alveoli into the blood. Name the process and state the direction of movement in terms of concentration.2 marks
Cells X and Y have the same volume. Cell X has a folded cell membrane and cell Y has a smooth one. State which cell takes in oxygen faster by diffusion and explain why.2 marks
A student placed identical agar cubes in hydrochloric acid of different concentrations and timed how long each took to turn red. State a conclusion about diffusion. | concentration of acid / mol per dm³ | time taken to turn red / s | | --- | --- | | 0.2 | 120 | | 0.4 | 75 | | 0.6 | 40 |2 marks
A drop of dye is added to cold water in one beaker and to hot water in another. After five minutes the dye has spread further in the hot water. Explain why.3 marks
A student investigates the effect of temperature on the diffusion of pigment out of identical beetroot cylinders into 10 cm³ of water. State the independent variable, the dependent variable and two variables that must be kept constant.4 marks
A student has two agar cubes containing universal indicator, each with sides of 20 mm. Cube P stays whole. Cube Q is cut into 8 equal cubes with sides of 10 mm. P and the 8 pieces of Q are put into separate beakers of hydrochloric acid and timed until they turn completely red. (a) Calculate the total surface area of the agar in each beaker. (b) Predict which beaker of agar turns completely red first. Give two reasons.5 marks
Three containers at the same temperature are each divided into halves X and Y by a membrane that allows glucose molecules through. The table gives the glucose concentration in each half at the start. (a) State the container in which glucose diffuses from X to Y fastest. Give a reason. (b) In container C glucose molecules still cross the membrane. Explain why there is no net movement. (c) Predict and explain how the rate of diffusion in container B changes as time passes. | container | glucose in X / g per dm³ | glucose in Y / g per dm³ | | --- | --- | --- | | A | 80 | 70 | | B | 60 | 20 | | C | 50 | 50 |6 marks
Salt dissolves in water. State the name given to water in this situation.1 mark
A student writes: ‘Osmosis is the movement of any substance through a membrane, using energy from respiration.’ Give two corrections.2 marks
Explain how water helps to support a plant.2 marks
A dialysis tubing bag containing concentrated sugar solution is placed in a beaker of distilled water. The volume of solution in the bag is measured at the start and after 30 minutes. State what happens to the volume and name the process.2 marks
A potato cylinder has a mass of 5.0 g. After one hour in a salt solution its mass is 4.2 g. Calculate the percentage change in mass.3 marks
Potato cylinders, each 40 mm long at the start, were placed in salt solutions for 30 minutes. (a) Calculate the change in length of each cylinder. (b) Suggest which concentration is closest to the concentration of solutes in the potato cells. | test-tube | final percentage concentration of salt solution | final length / mm | | --- | --- | --- | | A | 0.0 | 43 | | B | 0.5 | 41 | | C | 1.0 | 40 | | D | 2.0 | 38 | | E | 4.0 | 36 |3 marks
Describe how you would investigate the effect of the concentration of sugar solution on the mass of pieces of potato.4 marks
A gardener waters a plant with a very concentrated solution of fertiliser. The plant wilts. Explain why.5 marks
Four pieces of beetroot tissue were weighed, placed in sugar solutions for two hours, blotted dry and weighed again. (a) Calculate the percentage change in mass of the pieces in 0.2 and 0.4 moles per dm³ sugar solution. [2] (b) Explain why percentage change in mass is compared, not change in mass. [1] (c) Estimate the concentration of sugar solution in which the mass of the tissue would not change. [1] (d) Explain the result for the piece in 0.6 moles per dm³ sugar solution, and state the condition of its cells. Use the term water potential. [3] | concentration of sugar solution / moles per dm³ | starting mass / g | final mass / g | | --- | --- | --- | | 0.0 | 2.50 | 2.80 | | 0.2 | 2.40 | 2.52 | | 0.4 | 2.60 | 2.47 | | 0.6 | 2.50 | 2.20 |7 marks
A single-celled organism lives in fresh water. It has a cell membrane but no cell wall. Water enters it by osmosis all the time, and the organism pumps this water out again. Scientists placed the organism in different concentrations of sea water and measured the rate at which it pumped out water. Fresh water is 0% sea water. (a) Explain why water enters the organism in fresh water. Use the term water potential. [2] (b) A poison stops the organism pumping out water. Suggest what happens to a poisoned organism in fresh water, and explain why this does not happen to a plant cell. [2] (c) Describe and explain the trend in the results. [2] (d) State what the result for 12% sea water tells you about the water potential of the contents of the organism. [1] | percentage concentration of sea water | rate of pumping out water / arbitrary units | | --- | --- | | 0 | 12 | | 4 | 8 | | 8 | 4 | | 12 | 0 |7 marks
Describe active transport.3 marks
A cell contains a high concentration of a substance. The concentration outside the cell is lower. State whether the cell can take up more of the substance by diffusion or by active transport. Explain your answer.3 marks
A student writes: "Active transport happens anywhere in a cell, and needs no membrane." State what is wrong.1 mark
A poison stops respiration in a cell. State the effect on active transport in the cell and explain it.2 marks
The concentration of an ion is 4 units outside a cell and 9 units inside it. The cell takes up more of the ion. Respiration takes place in the mitochondria of the cell. (a) State the name of the process by which the ion enters the cell. Give a reason, using the figures. (b) Explain why a cell that takes up many ions in this way contains many mitochondria.4 marks
State the name of the process by which mineral ions enter root hair cells.1 mark
A student answers: "Root hair cells take up nutrients by active transport." Rewrite the answer so that it earns the mark.1 mark
Describe the role of protein in active transport.3 marks
A scientist measures the concentration of two ions, P and Q, in pond water and inside the cells of a pond plant. The table gives the results in arbitrary units. Both ions enter the cells. (a) Calculate how many times more concentrated ion P is inside the cells than in the pond water. (b) A poison that stops respiration is added to the water. Uptake of ion P stops. Uptake of ion Q continues. Explain these two results. | ion | in pond water | inside the cells | | --- | --- | --- | | P | 2 | 48 | | Q | 30 | 12 |6 marks
Barley roots are placed in a solution of nitrate ions. The concentration of nitrate ions is higher in the root hair cells than in the solution. Air is bubbled through one solution and nitrogen gas through another. Root cells use oxygen for respiration. The table gives the nitrate ions taken up in six hours, in arbitrary units. (a) Calculate the percentage decrease in uptake when nitrogen is used instead of air. (b) A student writes: "With nitrogen the roots take up less water, so fewer nitrate ions are carried in by osmosis." State what is wrong with this, and give the correct explanation of the results. | gas bubbled through | nitrate ions taken up | | --- | --- | | air | 80 | | nitrogen | 20 |7 marks

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Read the notes

1. What diffusion is and where its energy comes from#

2. Diffusion into and out of cells#

3. Concentration gradient and temperature#

4. Surface area and distance#

5. Investigating diffusion#

6. Water as a solvent#

7. Osmosis and partially permeable membranes#

8. Investigating osmosis#

9. Plant tissue in different solutions#

10. Water pressure supports plants#

11. Osmosis and water potential#

12. Turgid, flaccid and plasmolysed cells#

13. Water potential in organisms#

14. Moving against the concentration gradient#

15. Energy from respiration#

16. Why cells need it: ion uptake by root hairs#

17. Protein carriers#

End-of-chapter exercise

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Chapter at a Glance

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Related chapters

1. What diffusion is and where its energy comes from

2. Diffusion into and out of cells

3. Concentration gradient and temperature

4. Surface area and distance

5. Investigating diffusion

6. Water as a solvent

7. Osmosis and partially permeable membranes

8. Investigating osmosis

9. Plant tissue in different solutions

10. Water pressure supports plants

11. Osmosis and water potential

12. Turgid, flaccid and plasmolysed cells

13. Water potential in organisms

14. Moving against the concentration gradient

15. Energy from respiration

16. Why cells need it: ion uptake by root hairs

17. Protein carriers