CELLTRANSPORT

AQA GCSE Biology · Lessons 1–5
⇩ Student Booklet ⇩ Teacher Booklet

Key Terms — Cell Transport

Specification

Who needs this topic?All of it. Cell transport is AQA specification section 4.1.3, and it is core content in both GCSE Biology (8461) and GCSE Combined Science: Trilogy (8464), at Foundation and Higher tier alike. Nothing in this topic is marked “biology only” (separate science) or “HT only”, so there is no Triple-only or Higher-only content to skip. One difference worth knowing: the osmosis investigation is Required Practical 3 in Biology, but Required Practical 2 in Combined Science.

All Terms

01
Cell membrane
DefinitionA selectively permeable membrane around the cell that controls the entry and exit of materials.
02
Concentration
DefinitionHow much of a substance is dissolved in water; higher concentration means more particles.
03
Concentration gradient
DefinitionThe difference in concentration between two areas. A steeper gradient gives a faster rate of diffusion.
04
Diffusion
DefinitionThe net movement of particles from an area of higher concentration to an area of lower concentration. A passive process.
05
Net movement
DefinitionThe overall movement in one direction, once movement both ways is taken into account.
06
Partially permeable
DefinitionAllows water and small molecules through, but not larger molecules such as starch.
07
Surface area
DefinitionThe outside area of an object, e.g. the area around the outside of a cell. Unit = cm².
08
Volume
DefinitionThe amount of space inside an object. Unit = cm³.
09
Surface area to volume ratio
DefinitionSurface area divided by volume. Small organisms have a large SA:V; large organisms have a small SA:V and need exchange surfaces and a transport system.
10
Exchange surface
DefinitionA surface adapted for fast diffusion: thin, large surface area, and a good blood or air supply.
11
Osmosis
DefinitionThe diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
12
Solute
DefinitionThe dissolved substance in a solution, e.g. sugar or salt.
13
Solvent
DefinitionThe liquid in which the solute dissolves to form a solution.
14
Solution
DefinitionA mixture formed by a solute and a solvent, e.g. salt water.
15
Dilute
DefinitionA solution with a low solute concentration (and so a high water concentration).
16
Concentrated
DefinitionA solution with a high solute concentration (and so a low water concentration).
17
Isotonic
DefinitionA solution with the same water concentration as the cells or tissues, so there is no net movement of water.
18
Turgid
DefinitionEnlarged and swollen with water. A plant cell whose vacuole has swollen because of water gained by osmosis.
19
Flaccid
DefinitionSoft and floppy, lacking stiffness. The opposite of turgid.
20
Plasmolysed
DefinitionA plant cell in which the vacuole has shrunk and the membrane has pulled away from the cell wall, because of extreme water loss by osmosis.
21
Cell wall
DefinitionThe outer structure that supports a plant cell and stops it bursting when water enters by osmosis. Made of cellulose in plants.
22
Cellulose
DefinitionA carbohydrate. It forms the cell wall in plant cells.
23
Vacuole
DefinitionA space within the cytoplasm of a plant cell that contains cell sap.
24
Cell sap
DefinitionThe liquid in the vacuole of a plant cell: water with dissolved sugars, salts and amino acids.
25
Percentage change in mass
Definition(change in mass ÷ starting mass) × 100. Used so cylinders that did not start at the same mass can be compared fairly.
26
Active transport
DefinitionThe movement of a substance from a lower concentration to a higher concentration, against the concentration gradient. It requires energy from respiration.
27
Passive transport
DefinitionTransport that happens by the random movement of particles, giving a net movement from higher to lower concentration. No energy is required.
28
Aerobic respiration
DefinitionRespiration that requires oxygen. It releases the energy used for active transport.
29
Root hair cell
DefinitionA specialised cell that increases the surface area of the root epidermis to improve the uptake of water and mineral ions. It has many mitochondria for active transport.

Lesson 1 · Diffusion

Do Now

Q1
What is the cell membrane?
Model AnswerA selectively permeable membrane around the cell that controls what enters and leaves.
Q2
Name a gas that your cells need to take in.
Model AnswerOxygen.
Q3
What does the word 'concentration' mean?
Model AnswerHow much of a substance is dissolved in water - the higher the concentration, the more particles.

Part 1 · Defining diffusion

Read the passage.

Particles in a gas or solution move in a random and passive way - no energy is needed for them to move. Diffusion is the spreading out of particles: they diffuse from an area of high concentration to an area of low concentration. Although all particles move randomly, there is always an overall (net) movement in this direction.

In biology, diffusion often happens across a partially permeable (semi-permeable) membrane. These membranes only let some substances pass through (e.g. small molecules such as oxygen, carbon dioxide, urea and glucose) but not others (large molecules such as proteins and starch).

Fig 1.1 - Diffusion of particles across the cell membrane, from high to low concentration, over time.
Fig 1.1 — Diffusion of particles across the cell membrane, from high to low concentration, over time.

Questions — Defining diffusion

Q1
How do particles that are suspended in a liquid or gas move? (1 mark)
Model AnswerRandomly (and passively).
Q2
Define 'diffusion'. (2 marks)
Model AnswerThe net movement of particles from an area of higher concentration to an area of lower concentration.
Q3
What is meant by 'net movement'? (1 mark)
Model AnswerThe overall movement (in one direction).

Part 2 · Factors affecting the rate of diffusion

Read the passage about the factors.

The rate of diffusion is affected by several factors.

Factors that affect the rate of diffusion include:

  • Temperature: at higher temperatures, particles have more kinetic energy and so move around faster, increasing the rate of diffusion.
  • Concentration: particles diffuse down a concentration gradient, so useful substances that are being used up are replaced, and waste products do not build up in the cell.

A concentration gradient is kept steep when a substance is constantly used up or carried away. In the lungs, oxygen is at a high concentration in the alveolar sac and diffuses into the blood; the blood then carries the oxygen away, so its concentration in the blood stays low and oxygen keeps diffusing in.

Fig 1.2 - Oxygen diffuses from a high concentration in the alveolar sac into the blood, where blood flow keeps the concentration low.
Fig 1.2 — Oxygen diffuses from a high concentration in the alveolar sac into the blood, where blood flow keeps the concentration low.
Substances all animal cells need to removeSubstances plant cells also need to removeSubstances all cells need to take inSubstances plant cells also need to take in
Carbon dioxide, urea, waterOxygenOxygen, water, glucoseCarbon dioxide

Questions — Factors affecting the rate of diffusion

Q4
State three factors that affect the rate of diffusion. (3 marks)
Model AnswerTemperature, concentration difference (gradient) and surface area.
Q5
How does the concentration difference affect the rate of diffusion? (2 marks)
Model AnswerA bigger concentration difference (steeper gradient) gives a faster rate of diffusion.
Q6
Explain why a higher temperature results in faster diffusion. (2 marks)
Model AnswerParticles gain more kinetic energy and move faster, so they diffuse more quickly.
Q7
Name a useful substance that diffuses into our cells. (1 mark)
Model AnswerOxygen or glucose.
Q8
Name a waste substance that diffuses out of our cells. (1 mark)
Model AnswerCarbon dioxide (or urea).

Part 3 · Surface area and diffusion

Read the passage about exchange surfaces.

Increasing the surface area of a membrane increases the amount of diffusion that can happen across it, because there is more space for particles to move through.

Dividing an object into smaller pieces increases its total surface area even though the volume stays the same, so smaller units have a larger surface area for exchange.

Fig 1.3 - One large cube has a smaller total surface area than the same volume split into 27 small cubes.
Fig 1.3 — One large cube has a smaller total surface area than the same volume split into 27 small cubes.

Exchange surfaces where substances diffuse are adapted for fast diffusion.

These adaptations include:

  • being thin, to reduce the diffusion distance
  • having a large surface area, to increase the amount of membrane
  • having a good supply of blood or air, to maintain the concentration gradient
Fig 1.4 - The alveolar sac as an exchange surface: thin permeable walls, a large surface area, and a good blood supply that maintains the concentration gradient.
Fig 1.4 — The alveolar sac as an exchange surface: thin permeable walls, a large surface area, and a good blood supply that maintains the concentration gradient.

Questions — Surface area and diffusion

Q9
Explain how each adaptation of exchange surfaces helps speed up diffusion. (6 marks)
Model AnswerBeing thin reduces the diffusion distance, so substances diffuse across more quickly. A large surface area increases the amount of membrane where diffusion can happen, so more diffuses at once. A good blood (or air) supply transports substances to and from the surface, maintaining the concentration gradient so diffusion stays fast.

Exam-style questions

Q1
The diagram shows four ways (A-D) molecules may move into and out of a respiring cell; the dots show the concentration. (3 marks)
Fig 1.5 - Four possible directions of movement (A-D) for a respiring cell.
Fig 1.5 — Four possible directions of movement (A-D) for a respiring cell.
(a)(i) Which arrow shows the movement of oxygen?(a)(ii) Which arrow shows the movement of carbon dioxide?(b) Name the process by which these gases move into and out of the cell.
Mark Scheme(a)(i) The arrow showing oxygen moving into the cell (down its gradient).(a)(ii) The arrow showing carbon dioxide moving out of the cell.(b) Diffusion.
Q2
What is diffusion? (2 marks)
Mark SchemeThe net movement of particles from a higher to a lower concentration (a passive process).
Q3
The diagram shows an animal cell. (3 marks)
Fig 1.6 - An animal cell surrounded by oxygen molecules.
Fig 1.6 — An animal cell surrounded by oxygen molecules.
(a) How can you tell it is an animal cell and not a plant cell? Give two reasons.(b) Oxygen will diffuse into the cell. Explain why, using information from the diagram.
Mark Scheme(a) It has no cell wall and no chloroplasts (accept: no large permanent vacuole).(b) The oxygen concentration is higher outside the cell than inside, so oxygen diffuses in down the gradient.
Q4
The diagram shows a cell from the lining of the lung, specialised to let gases pass through quickly. (5 marks)
Fig 1.7 - A thin cell from the lung lining (label A, B and C).
Fig 1.7 — A thin cell from the lung lining (label A, B and C).
(a) Label structures A, B and C.(b)(i) Which feature lets oxygen pass quickly: it is thin / has a large nucleus / has many mitochondria?(b)(ii) Oxygen passes through this cell by diffusion / osmosis / respiration.
Mark Scheme(a) Accept the labelled parts (e.g. cell membrane, cytoplasm, nucleus).(b)(i) It is thin.(b)(ii) Diffusion.
Q5
Glucose is absorbed into the blood in the small intestine, mostly by diffusion. (3 marks)(a) How does the glucose concentration in the blood compare with that in the small intestine?(b) Which two features help absorption of glucose: villi have a cell wall / are covered in thick mucus / give a large surface area / have many blood capillaries?
Mark Scheme(a) The concentration in the blood is lower (so glucose diffuses in).(b) Villi give a large surface area; villi have many blood capillaries.
Q6
The diagrams show four cells (A-D) containing and surrounded by oxygen molecules. Oxygen can move into or out of the cells. (2 marks)
Fig 1.8 - Four cells (A-D) with different oxygen concentrations inside and outside.
Fig 1.8 — Four cells (A-D) with different oxygen concentrations inside and outside.
(a) Into which cell (A, B, C or D) will oxygen move the fastest?(b) Oxygen is taken into cells by the process of diffusion / osmosis / respiration.
Mark Scheme(a) Cell A - it has the steepest concentration gradient (most oxygen outside, least inside), so oxygen diffuses in fastest.(b) Diffusion.

Lesson 2 · Surface Area to Volume Ratio

Do Now

Q1
What are the units of surface area?
Model Answercm² (square centimetres).
Q2
What are the units of volume?
Model Answercm³ (cubic centimetres).
Q3
What is diffusion?
Model AnswerThe net movement of particles from a higher to a lower concentration.

Part 1 · Calculating SA:V ratio

Read the method and the worked examples.

Key wordMeaning
DiffusionThe movement of molecules from an area of higher concentration to an area of lower concentration.
Surface areaRefers to the outside area of an object, e.g. it is the area around the outside of a cell. Unit = cm².
VolumeRefers to the amount of space inside of the object, e.g. it is the space.

An object's surface area to volume ratio is like a way of describing how close every internal part of it is to its surface. It is worked out by dividing an object's surface area by its volume: SA:V ratio = surface area / volume.

Small Cube: side length = 1 cm. Volume = 1 cm x 1 cm x 1 cm = 1 cm³. Surface area = 6 x (1 cm x 1 cm) = 6 cm². SA:Vol ratio = 6 / 1 = 6.

Key ideaA high SA:V ratio shows that every part of an object is quite close to the edge. This means there is only a short distance from the edge to the middle so things can get in and out quickly.

Large Cube: side length = 10 cm. Volume = 10 cm x 10 cm x 10 cm = 1000 cm³. Surface area = 6 x (10 cm x 10 cm) = 600 cm². SA:Vol ratio = 600 / 1000 = 0.6.

Key ideaA low SA:V ratio means that there are parts of an object that are a long way from the edge. This means there is a longer distance from the edge to the middle so things take a long time to get in and out.

Questions — Calculating SA:V ratio

Q1
Calculate the SA:V ratio for each cube. (10 marks)
CubeSide LengthVolume / cm³Surface Area / cm²SA:Vol Ratio
A2???
B15???
C0.5???
D7.5???
E5???
Model AnswerA: V=8, SA=24, ratio=3. B: V=3375, SA=1350, ratio=0.4. C: V=0.125, SA=1.5, ratio=12. D: V=421.875, SA=337.5, ratio=0.8. E: V=125, SA=150, ratio=1.2.
Q2
What is the relationship between size and SA:Vol ratio? (2 marks)
Model AnswerAs the size (side length) of an object increases, its SA:V ratio decreases.
Q3
What trend do you notice as the size of the cells increases? (2 marks)
Model AnswerThe surface area to volume ratio gets smaller - larger cells have relatively less surface area for their volume.
Q4
Calculate the surface area to volume ratio of the organisms in the table. (5 marks)
OrganismLengthSA / m²Vol / m³SA / Vol
Bacterium1 um6 x 10⁻¹²10⁻¹⁸?
Amoeba100 um6 x 10⁻⁸10⁻¹²?
Housefly10 mm6 x 10⁻⁴10⁻⁶?
Dog1 m6 x 10⁰10⁰?
Whale100 m6 x 10⁴10⁶?
Model AnswerBacterium: 6x10^6. Amoeba: 6x10^4. Housefly: 6x10^2. Dog: 6. Whale: 0.06. (Each found by dividing SA by volume - SA:V falls sharply as organisms get bigger.)

Part 2 · Exchange structures as an adaptation

Read the passage about exchange structures.

The rate of exchange of substances depends on the surface area in contact with the surroundings. As organisms get bigger their SA:V ratio gets smaller, so larger organisms have more difficulty exchanging materials with their surroundings.

Large SA:V (small organisms)Small SA:V (large organisms)
Short diffusion distance, so diffusion is fastLong diffusion distance, so diffusion from outside is too slow
Advantage: can exchange materials directly with the environmentAdvantage: lose less heat, can survive in cold environments
Disadvantage: lose heat and dry out quicklyDisadvantage: need internal transport systems to reach inner cells

Questions — Exchange structures as an adaptation

Q5
What does the rate of exchange of substances depend on? (1 mark)
Model AnswerThe surface area in contact with the surroundings.
Q6
Give one advantage of a large SA:V. (1 mark)
Model AnswerCan exchange materials directly with the environment.
Q7
Give one disadvantage of a large SA:V. (1 mark)
Model AnswerLoses heat and dries out quickly.
Q8
Give one advantage of a small SA:V. (1 mark)
Model AnswerLoses less heat, so can survive in cold environments.
Q9
Give one disadvantage of a small SA:V. (1 mark)
Model AnswerNeeds internal transport systems to reach the inner cells.
Q10
Why do large organisms have difficulty exchanging materials? (2 marks)
Model AnswerThey have a small SA:V and a long diffusion distance, so diffusion alone is too slow.
Q11
Explain why mammals need a circulatory system but unicellular organisms do not. (3 marks)
Model AnswerMammals are large with a small SA:V and long diffusion distances, so they need a transport system to carry substances to all cells; unicellular organisms have a large SA:V and can exchange everything directly by diffusion across their surface.

Exam-style questions

Q1
The diagram shows two model cubic cells (sides 0.5 cm and 1 cm). (6 marks)
Fig 2.2 - Two model cells with sides of 0.5 cm and 1 cm.
Fig 2.2 — Two model cells with sides of 0.5 cm and 1 cm.
(a) Describe how the surface area to volume ratio changes as the length of the side of the model cell increases. Include calculations in your answer.(b) Explain why a bacterium can rely on diffusion for gas exchange, but animals need a transport system.
Mark Scheme(a) As the side increases the SA:V ratio decreases, e.g. a 1 cm cube is 6:1, a 2 cm cube is 3:1, a 3 cm cube is 2:1.(b) A bacterium is small with a large SA:V and short diffusion distance, so diffusion is fast enough; an animal is large with a small SA:V and long diffusion distance, so diffusion is too slow and it needs a transport system.
Worked Answer — VESSU
V
Variables cube side length = 2 cm · surface area = 6 x (2 x 2) = 24 cm² · volume = 2 x 2 x 2 = 8 cm³
E
Equation SA:V ratio = surface area / volume
S
Substitute SA:V = 24 / 8
S
Solve SA:V = 248 = 3, so 3 : 1
U
Units a ratio (no units)
Answer: 3 : 1
Q2
The image shows an epithelial cell from the lining of the small intestine. Suggest how the highly folded cell surface helps the epithelial cell to absorb soluble food. (1 mark)
Fig 2.3 - An epithelial cell from the lining of the small intestine.
Fig 2.3 — An epithelial cell from the lining of the small intestine.
Mark SchemeThe folds increase the surface area, so more food can be absorbed (faster).

Lesson 3 · Osmosis Theory

Do Now

Q1
What is diffusion?
Model AnswerThe net movement of particles from a higher to a lower concentration.
Q2
What is a partially permeable membrane?
Model AnswerA membrane that lets water and small molecules through but not larger molecules.
Q3
Name the dissolved substance in a solution.
Model AnswerThe solute.

Part 1 · Defining osmosis

Read the passage.

Osmosis is the diffusion of water - the net movement of water molecules from a dilute solution to a concentrated solution, through a partially permeable membrane.

Dilute solutions have a high concentration of water molecules and a low solute concentration. Concentrated solutions have a lower concentration of water molecules and a high solute concentration. During osmosis, water moves from pure water or a dilute solution to a more concentrated solution. It is a passive process and does not require energy.

Fig 3.1 - Osmosis across a partially permeable membrane: water moves from the dilute solution to the concentrated solution.
Fig 3.1 — Osmosis across a partially permeable membrane: water moves from the dilute solution to the concentrated solution.

Questions — Defining osmosis

Q1
Define 'osmosis'. (2 marks)
Model AnswerThe net movement of water across a partially permeable membrane from a higher to a lower water concentration.
Q2
Define 'dilute'. (1 mark)
Model AnswerA solution with a low solute concentration (high water concentration).
Q3
Define 'concentrated'. (1 mark)
Model AnswerA solution with a high solute concentration (low water concentration).
Q4
Give two similarities and two differences between diffusion and osmosis. (4 marks)
Model AnswerSimilarities: both are passive processes (no energy needed); both involve net movement of particles/molecules from a higher to a lower concentration. Differences: osmosis is only the movement of water, whereas diffusion is the movement of any particles; osmosis must occur across a partially permeable membrane, whereas diffusion does not have to.

Part 2 · Osmosis in animal and plant cells

Look at the diagram showing both cell types in each solution.

Fig 3.2 - Animal and plant cells in dilute, isotonic and concentrated solutions.
Fig 3.2 — Animal and plant cells in dilute, isotonic and concentrated solutions.

Questions — Osmosis in animal and plant cells

Q5
What is an isotonic solution? (2 marks)
Model AnswerA solution with the same water concentration as the cell, so there is no net movement of water.
Q6
If a cell is in a dilute solution, will water enter or leave the cell? (1 mark)
Model AnswerWater will enter the cell.
Q7
What will happen to an animal cell in a dilute solution? (1 mark)
Model AnswerIt swells and may burst.
Q8
What will happen to an animal cell in a concentrated solution? (1 mark)
Model AnswerIt shrivels (shrinks) as it loses water.
Q9
Why do animal cells burst in pure water? (2 marks)
Model AnswerWater moves in by osmosis and the cell membrane is not strong enough to withstand the pressure, so it bursts.
Q10
If a cell is in a concentrated solution, will water enter or leave the cell? (1 mark)
Model AnswerWater will leave the cell.
Q11
How do cell walls protect plant cells from the effects of osmosis? (2 marks)
Model AnswerThe strong cellulose cell wall stops the cell bursting when water enters, and supports the cell so it becomes turgid rather than splitting.
Q12
What will happen to a plant cell in a dilute solution? (1 mark)
Model AnswerIt becomes turgid (swollen) but does not burst.
Q13
What will happen to a plant cell in an isotonic solution? (1 mark)
Model AnswerNo net movement of water - it stays the same.
Q14
What will happen to a plant cell in a concentrated solution? (1 mark)
Model AnswerIt loses water; the vacuole shrinks and it becomes plasmolysed.
Q15
What does 'plasmolysed' mean? (2 marks)
Model AnswerThe vacuole has shrunk and the membrane has pulled away from the cell wall due to extreme water loss by osmosis.

Exam-style questions

Q1
Substances can move into and out of cells. (5 marks)
Fig 3.3 - Plant cells after one hour in sugar solutions A and B.
Fig 3.3 — Plant cells after one hour in sugar solutions A and B.
(a)(i) Complete the sentence: 'Water moves into and out of cells by ______ (active transport / osmosis / reabsorption).'(a)(ii) Complete the sentence: 'The water moves through a ______ membrane (freely permeable / non-permeable / partially permeable).'(b)(i) Students put plant cells into two sugar solutions, A and B. Describe two ways the cell in solution B is different from the cell in solution A.(b)(ii) A student put red blood cells into water. Suggest what would happen.
Mark Scheme(a)(i) osmosis(a)(ii) partially permeable(b)(i) Accept two of: the cell in B is smaller/plasmolysed, the vacuole has shrunk, the membrane has pulled away from the wall (B was the more concentrated solution).(b)(ii) The red blood cells take in water by osmosis and burst.
Q2
What is osmosis? (3 marks)
Mark SchemeThe net movement of water molecules from a region of higher water concentration to a region of lower water concentration, across a partially permeable membrane.
Q3
A plant cell is placed in distilled water. (3 marks)
Fig 3.4 - A plant cell swelling and becoming turgid in distilled water.
Fig 3.4 — A plant cell swelling and becoming turgid in distilled water.
(a) Explain why the cell swells and becomes turgid, and name the process involved.(b) Give one feature of the cell wall that allows the cell to become turgid.
Mark Scheme(a) Water moves into the cell by osmosis (higher water concentration outside); the vacuole expands and the cell becomes turgid.(b) It is strong/elastic, so it resists the pressure without bursting.
Q4
A plant cell is put into water and swells up. (6 marks)
Fig 3.5 - Osmosis apparatus: Visking tubing and a capillary tube (for part b).
Fig 3.5 — Osmosis apparatus: Visking tubing and a capillary tube (for part b).
(a) Explain why the cell swells up, as fully as you can.(b)(i) Students set up osmosis equipment with a capillary tube. What will happen to the water level in the capillary tube?(b)(ii) Explain why.
Mark Scheme(a) There is a higher water concentration outside the cell, so water moves in by osmosis across the partially permeable membrane; the vacuole expands and the cell becomes turgid.(b)(i) The water level rises.(b)(ii) Water moves by osmosis into the more concentrated solution, increasing its volume and pushing the level up the tube.

Lesson 4 · Osmosis Practical

Do Now

Q1
What is osmosis?
Model AnswerThe net movement of water across a partially permeable membrane from high to low water concentration.
Q2
What is the independent variable in an experiment?
Model AnswerThe variable that you change.
Q3
How do you calculate a percentage?
Model AnswerDivide the part by the whole and multiply by 100.

Part 1 · Aim and variables

Read the lesson objectives, the aim and the variables.

Lesson objectivesBy the end of this lesson you should be able to: predict how osmosis will affect the mass of a piece of plant tissue in different solutions; write a method to show how you could measure the effect of osmosis on a piece of plant tissue; calculate % change in mass and use a line graph to estimate the cytoplasm concentration of a piece of tissue.

Aim: to investigate the effect of the concentration of sugar solution on the mass of aubergine tissue.

The variables in this experiment are:

  • Independent variable (what we change): the concentration of the sugar solution.
  • Dependent variable (what we measure): the mass of the tissue.
  • Control variables (what we keep the same): the type of aubergine, the length of the cylinders, the temperature of the solutions, and the volume of the cylinders.

Part 2 · The method

Read the method.

Method:

  • Use a cork borer to cut aubergine cylinders of the same diameter, making sure all the skin is removed (it has a layer of waterproof membrane that could affect results).
  • Use a ruler to trim the cylinders so they are the same length (3 cm) and therefore have the same volume. HAZARD - be careful when using a sharp knife to cut the aubergine.
  • Dry the aubergine carefully by blotting it with a paper towel, to remove any liquid left on the outside of the plant tissue which would affect the mass.
  • Accurately measure and record the length and mass of each aubergine cylinder.
  • Measure 10 cm³ of 1.0 M, 0.75 M, 0.5 M and 0.25 M sugar solution into boiling tubes, labelled to show the concentration.
  • Measure 10 cm³ of distilled water into a fifth boiling tube.
  • Add one aubergine cylinder into each boiling tube - you must know which cylinder is in which solution!
  • Leave the tubes for 20 minutes.
  • Remove the cylinders, carefully blotting dry with a paper towel to remove extra solution.
  • Re-measure the length and mass of each cylinder. If the cylinders have gained water by osmosis, they will have increased in mass; if water has left the cylinders by osmosis, they will have decreased in mass.
  • Calculate the change in mass of each cylinder by subtracting the final mass from the initial mass - cylinders that have decreased in mass will have negative numbers.
  • Calculate the percentage change in mass for each cylinder. This lets you compare the effect of the sugar concentration on cylinders that didn't have the same initial mass.
Fig 4.1 - The osmosis required practical: potato cores are measured, placed in a dilution series of sugar solution and water, then re-measured after a set time.
Fig 4.1 — The osmosis required practical: potato cores are measured, placed in a dilution series of sugar solution and water, then re-measured after a set time.

Questions — The method

Q1
Why does the potato skin need to be removed before putting the cylinders into the solutions? (2 marks)
Model AnswerThe skin has a waterproof (membrane) layer that would stop water moving in or out and affect the results.
Q2
Why must we dry the chips before weighing them? (2 marks)
Model AnswerTo remove liquid left on the outside, which would otherwise add to the mass and affect the results.

Part 3 · Calculating and graphing the results

Read about processing the results.

Calculate the change in mass (final mass minus starting mass) for each cylinder; a decrease gives a negative number.

Then calculate the percentage change in mass, which lets you compare cylinders that did not start with the same mass:

% change in mass = change in massstarting mass x 100

Plot a graph of % change in mass (y axis) against concentration (x axis) and draw a line of best fit. The point where the line crosses the x axis (no change in mass) gives the concentration that is isotonic with the tissue.

Questions — Calculating and graphing the results

Q3
How do we calculate % change in mass? (2 marks)
Model Answerchange in massstarting mass x 100.
Q4
A cylinder's mass changed from 10.0 g to 13.0 g. Calculate the percentage change in mass. (2 marks)
Model Answer((13.0 - 10.0) / 10.0) x 100 = 30% gain.
Worked Answer — VESSU
V
Variables starting mass = 10.0 g · final mass = 13.0 g · change in mass = 13.0 - 10.0 = 3.0 g
E
Equation % change in mass = (change in mass / starting mass) x 100
S
Substitute = (3.0 / 10.0) x 100
S
Solve = 3.010.0 x 100 = 30
U
Units percent (%)
Answer: 30% gain in mass
Q5
Why use percentage change rather than change in mass? (2 marks)
Model AnswerIt lets you fairly compare cylinders that did not have the same starting mass.
Q6
What does the point where the line of best fit crosses the x axis tell you? (2 marks)
Model AnswerThe concentration of solution that causes no change in mass - the concentration that is the same as (isotonic with) the tissue.

Exam-style questions

Q1
In fish and chip shops, potatoes are cut into chips several hours before they are cooked. The mass of water in the chips must be kept constant during this time. To keep the water in the chips constant, the chips are kept in sodium chloride solution. (11 marks)
Fig 4.2 - Apparatus and materials for part a): potato, cork borer, knife and tweezers; boiling tubes of solution in racks; a balance.
Fig 4.2 — Apparatus and materials for part a): potato, cork borer, knife and tweezers; boiling tubes of solution in racks; a balance.
Sodium chloride concentration (mol dm-3)00.20.40.60.81.0
Change in length of cylinders (mm)+4.1+1.5-1.4-3.6-4.6-5.2
(a) The drawing shows apparatus and materials (boiling tubes and a balance). In this question you will be assessed on using good English, organising information clearly, and using scientific terms where appropriate. Describe how you would use the apparatus and materials to find the concentration of sodium chloride in which to keep the chips so that the mass of water in the chips remains constant. You should include: the measurements you would make, and how you would make the investigation a fair test.(b) In a similar investigation a student investigated the effect of the concentration of sodium chloride solution on standard-sized cylinders cut from a potato. The table shows the student's results. Plot a graph of the results: add a suitable scale and label to the y axis, plot the points, and draw a line of best fit.
Fig 4.3 - Grid for plotting the graph in part b).
Fig 4.3 — Grid for plotting the graph in part b).
(c) Explain the change in length of the potato cylinders placed in the 1 mol dm-3 sodium chloride solution.
Mark Scheme(a) Cut equal-sized chips using a cork borer and trim to the same length; measure and record the initial mass of a chip for each of a range of sodium chloride concentrations (e.g. 0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm-3); place one chip in each concentration for a set, timed period; remove, blot dry and re-measure the mass of each chip; calculate the change (or % change) in mass for each concentration; the concentration at which there is no change in mass is the concentration to use. Fair test: keep volume of solution, temperature, and time in solution the same for every chip.(b) Suitable y-axis scale and label (change in length / mm); all six points plotted correctly; appropriate line of best fit, crossing the x axis at approximately 0.28 mol dm-3 (the isotonic point).
Model answer: change in length of cylinder (mm) against sodium chloride concentration (mol dm-3), with line of best fit and isotonic point (x-intercept) marked.
Model answer: change in length of cylinder (mm) against sodium chloride concentration (mol dm-3), with line of best fit and isotonic point (x-intercept) marked.
(c) The 1 mol dm-3 solution is more concentrated (lower water concentration) than the cell contents, so water leaves the cells by osmosis; the cells lose water and become flaccid, so the cylinders decrease in length.
Q2
Some students used osmosis to find the sucrose concentration in potato cell sap, using potato discs (about 10 g) in five beakers. After two hours the discs were removed, blotted and reweighed. (4 marks)
Fig 4.4 - Potato discs in five beakers (distilled water to 40% sucrose solution).
Fig 4.4 — Potato discs in five beakers (distilled water to 40% sucrose solution).
Beaker 1Beaker 2Beaker 3Beaker 4Beaker 5
Final mass (g)13.012.29.07.97.3
Initial mass (g)10.010.610.010.110.4
% change in mass+30%?-10%??
(a) Why did the students blot the potato before weighing it?(b) Calculate the % change in mass for beakers 2, 4 and 5.
Mark Scheme(a) To remove surface liquid that would otherwise add to the measured mass.(b) Beaker 2: 12.2 - 10.610.6 x 100 = +15.1% gain.
Beaker 4: 7.9 - 10.110.1 x 100 = -21.8% loss.
Beaker 5: 7.3 - 10.410.4 x 100 = -29.8% loss.
Worked Answer — VESSU
V
Variables starting mass = 10.0 g · final mass = 13.0 g · change in mass = 13.0 - 10.0 = 3.0 g
E
Equation % change in mass = (change in mass / starting mass) x 100
S
Substitute = (3.0 / 10.0) x 100
S
Solve = 3.010.0 x 100 = 30
U
Units percent (%)
Answer: 30% gain in mass

Lesson 5 · Active Transport and Review

Do Now

Q1
What is diffusion?
Model AnswerThe net movement of particles from a higher to a lower concentration.
Q2
What is osmosis?
Model AnswerThe net movement of water across a partially permeable membrane from high to low water concentration.
Q3
Does diffusion need energy?
Model AnswerNo - it is passive.

Part 1 · Defining active transport

Read the passage, and look at the diagram.

Diffusion and osmosis are passive processes - they do not need energy for particles to move down their concentration gradient.

Active transport is an active process because it moves particles from an area of a low concentration to an area of a high concentration (against a concentration gradient).

  • This process requires energy.
  • Energy comes from respiration in the mitochondria.
  • The rate of active transport depends on the rate of respiration.
  • Oxygen is therefore also required for active transport to occur.
Fig 5.1 - Diffusion (passive, no energy) compared with active transport (against the gradient, requires energy from ATP).
Fig 5.1 — Diffusion (passive, no energy) compared with active transport (against the gradient, requires energy from ATP).

Questions — Defining active transport

Q1
Define 'active transport'. (2 marks)
Model AnswerThe movement of a substance from a low concentration to a high concentration (against the gradient), using energy from respiration.
Q2
Why is active transport not a passive process? (2 marks)
Model AnswerIt moves particles against the concentration gradient, which requires energy.

Part 2 · Why active transport is important

Read the passage and examples.

Cells that carry out a lot of active transport often have many mitochondria to provide the energy from respiration.

Examples include:

  • Cells lining the small intestine (epithelial cells) - so glucose can be absorbed against the concentration gradient into the blood.
  • Root hair cells in plants - which actively transport mineral ions from the soil. Mineral ions are often in low concentration in the soil, so root hair cells use the energy from respiration to take them up against the gradient.

For healthy growth, plants need mineral ions, but they are often in low concentration in the soil. Plant roots have specialised cells which have many mitochondria that provide the energy needed from respiration for the active transport of mineral ions against the concentration gradient into the root hair cells.

Fig 5.2 - A root hair cell actively transporting mineral ions from a low concentration in the soil to a higher concentration inside the cell.
Fig 5.2 — A root hair cell actively transporting mineral ions from a low concentration in the soil to a higher concentration inside the cell.

Questions — Why active transport is important

Q3
Give two examples of where active transport occurs in organisms. (2 marks)
Model AnswerIn the epithelial cells lining the small intestine (absorbing glucose) and in root hair cells (taking up mineral ions).
Q4
Explain one adaptation of specialised root hair cells in plants. (2 marks)
Model AnswerThey have many mitochondria, which provide the energy from respiration needed for the active transport of mineral ions.

Part 3 · Comparing the three transport processes

Use the table to compare the processes.

DiffusionOsmosisActive transport
What movesAny small particlesWater onlyParticles/ions
DirectionHigh to low (down gradient)High to low water conc.Low to high (against gradient)
Needs energy?No (passive)No (passive)Yes (from respiration)
MembraneAcross a membranePartially permeable membraneAcross a membrane

Questions — Comparing the three transport processes

Q5
How does active transport differ from diffusion and osmosis? (2 marks)
Model AnswerActive transport moves particles against the gradient (low to high) and needs energy; diffusion and osmosis move down the gradient and are passive.
Q6
State two differences between diffusion and osmosis. (2 marks)
Model AnswerDiffusion moves any particles; osmosis moves only water. Osmosis must occur across a partially permeable membrane; diffusion does not have to.

Exam-style questions

Q1
The diagram shows four ways (A-D) molecules may move into and out of a cell; the dots show the concentration. Which arrow represents the active uptake of sugar molecules by the cell? Explain the reason for your answer. (2 marks)
Fig 5.3 - Four possible directions of movement (A-D) for a cell.
Fig 5.3 — Four possible directions of movement (A-D) for a cell.
Mark SchemeThe arrow showing sugar moving from low to high concentration (into the cell against the gradient) - because active transport moves substances up the concentration gradient using energy.
Q2
Some sugar molecules are absorbed from the small intestine into the blood by active transport. Explain why the rate of absorption can depend on the concentration of oxygen in the cells lining the small intestine. (3 marks)
Mark SchemeActive transport needs energy from aerobic respiration; more oxygen allows faster respiration, releasing more energy, so the rate of active transport (and absorption) increases.
Q3
Plants use active transport to move some substances from the soil into root hair cells. (3 marks)(a) Active transport needs energy. Which part of the cell releases most of this energy (mitochondria / nucleus / ribosome)?(b) Explain why active transport is necessary in root hair cells.
Mark Scheme(a) Mitochondria.(b) Mineral ions are in a lower concentration in the soil than in the cell, so they must be moved against the gradient, which needs active transport.
Q4
The image shows an epithelial cell from the lining of the small intestine. (6 marks)
Fig 5.4 - An epithelial cell from the lining of the small intestine (part A contains chromosomes).
Fig 5.4 — An epithelial cell from the lining of the small intestine (part A contains chromosomes).
(a) Part A contains chromosomes. What is the name of part A?(b) Suggest how the highly folded cell surface helps absorb soluble food.(c)(i) Name one food molecule absorbed by active transport.(c)(ii) Why is it necessary to absorb some food by active transport?(c)(iii) Suggest why epithelial cells have many mitochondria.
Mark Scheme(a) The nucleus.(b) The folds increase the surface area, so more food is absorbed.(c)(i) Glucose (or amino acids).(c)(ii) When the concentration in the gut is lower than in the blood, the food must be moved against the gradient by active transport.(c)(iii) To release lots of energy from respiration for the active transport of food against the gradient.
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