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GCSE Biology · GCSEChapter 3 · Diffusion, osmosis and active transport
GCSE · Chapter 3

Diffusion, osmosis and active transport

By the end of this chapter you will be able to

  • Explain diffusion using a concentration gradient, and say how temperature, gradient, surface area and distance change its rate
  • Explain osmosis and predict what happens to a plant or animal cell in a dilute or concentrated solution
  • Explain why active transport needs energy, and give two examples
  • Calculate surface area to volume ratios, and explain how exchange surfaces are adapted
  • Plan and interpret the potato osmosis practical, including the isotonic concentration

You should already know

  • Particles in solids, liquids and gases (KS3 Chemistry and Biology)
  • Cells and their parts (Chapter 2)
  • Calculating percentage change (Chapter 1)

Where this leads

  • Gas exchange in the lungs (Chapter 7)
  • Absorption in the digestive system (Chapter 5)
  • Water and mineral uptake by roots (Chapter 8)

GCSE Biology. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Describe diffusion and the factors that affect its rate, Describe osmosis across a partially permeable membrane, and interpret results from potato tissue, [H] Explain water potential, Explain active transport against a concentration gradient, using energy from respiration, Explain how surface area to volume ratio and exchange surfaces help substances move in and out of cells.

Exercise P  Do you remember?

1
FWhich way do particles move in diffusion: from high to low concentration, or from low to high? Write the words.
2
FWhat is 12 divided by 100, written as a decimal? Write the number.
3
FA potato cylinder is 5 g. It gains 0.6 g. Find the mass gained, as a fraction of the starting mass. Write the number.
4
FName the organelle that releases energy by aerobic respiration. Write the word.
5
FWhat is 5 multiplied by 6? Write the number.
3.1

Diffusion

GCSE
Describe diffusion as movement down a concentration gradient, Explain how temperature, the gradient, surface area and distance affect diffusion
Video lesson: DiffusionComing soon
Context. A bottle of perfume is opened at the front of a classroom. Within a short time, students at the back can smell it. Nobody stirred the air. The particles have spread out by themselves.

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. The difference in concentration is called the concentration gradient. Diffusion happens in gases and in liquids, and it does not need energy. It is passive, because the particles move because they have kinetic energy, not because a cell uses energy to push them.

Substances that move by diffusion in living things include oxygen and carbon dioxide across cell membranes, and some small molecules such as glucose across membranes.

high concentrationlow concentrationnet movement of particles, down the concentration gradient
Diffusion: particles move from high to low concentration

Key ideas

  • Diffusion is net movement from high to low concentration.
  • It is passive: it does not need energy from respiration.
  • A steeper concentration gradient makes diffusion faster.
  • Higher temperature, larger surface area and a shorter distance all speed up diffusion.
Worked example 3.1.1
Name the process that moves particles from high to low concentration with no energy needed. Write the word.

Method. Recall the process.

  1. Recall. The movement down a concentration gradient, with no energy, is called diffusion.
    no energy needed
  2. Name it. So the process is diffusion.
    diffusion
  3. Write the answer.
    diffusion

Check. Active transport is the process that needs energy. ✓

Worked example 3.1.2
Explain why diffusion is faster at a higher temperature. Write one sentence.

Method. Think about the particles.

  1. Think. At higher temperatures the particles have more kinetic energy.
    more kinetic energy
  2. Conclude. So they move faster and spread out faster.
    move faster
  3. Write the answer.
    Particles have more kinetic energy at a higher temperature, so they move faster and diffuse faster.

Check. Temperature changes the speed of the particles. ✓

Worked example 3.1.3
Name two factors that increase the rate of diffusion. Accept any two of: a steeper concentration gradient, a higher temperature, a larger surface area, a shorter distance.

Method. Recall the factors that speed diffusion up.

  1. Recall. A steeper gradient, a higher temperature, a larger surface area and a shorter distance all speed up diffusion.
    speed it up
  2. Choose two. Any two of these answer the question.
    any two
  3. Write the answer.
    Accept any two of: a steeper concentration gradient, a higher temperature, a larger surface area, a shorter distance.

Check. Each factor works in the same direction. ✓

Watch out

Watch out: diffusion goes from high to low concentration, not the other way. It is the net movement that matters, because particles move in both directions.

Exam tip

Exam tip. When you explain a rate, say which factor changes, then say what that does to the particles.

Try it (with your teacher). Put a few drops of food colouring in a beaker of warm water and in a beaker of cold water, at the same time, with a teacher. Watch how quickly the colour spreads. Describe the effect of temperature.

Exercise 3A  Diffusion

1
FIn diffusion, which way do particles move? Write the words.
2
FDoes diffusion need energy from respiration? Write yes or no.
3
FName the difference in concentration between two regions. Write the words.
4
FWhich gas moves into a leaf by diffusion for photosynthesis? Write the words.
5
FWhich of these makes diffusion faster? Write a or b. (a) a shorter distance (b) a longer distance
6
FWhich makes diffusion faster: a higher temperature or a lower temperature? Write the words.
7
FA cell has a large surface area. Does this make diffusion faster or slower? Write the word.
8
PSA cell uses oxygen, so the oxygen concentration inside it is kept low. Is the oxygen concentration outside the cell higher or lower than inside? Write the word.
9
PSA student heats a beaker of water and the food colouring spreads faster. Write one reason why. Accept any sentence that says the particles have more kinetic energy at a higher temperature.
10
PSCarbon dioxide leaves a leaf by diffusion. Name one factor that makes this faster. Accept any one: a steeper gradient, a higher temperature, a larger surface area, or a shorter distance.
11
RChallenge: explain why a fish gill, with a very large surface area and thin walls, takes in oxygen faster than a flat sheet of the same thickness. Accept any sentence that says the large surface area gives more space for oxygen to diffuse, and the thin walls shorten the distance.
3.2

Osmosis, and water potential [H]

GCSE
Describe osmosis across a partially permeable membrane, Predict what happens to plant and animal cells in dilute and concentrated solutions, [H] Explain water potential
Video lesson: Osmosis, and water potential [H]Coming soon
Context. Ruth places a plant cell in pure water and an animal cell in pure water. "The plant cell will swell, but the animal cell may burst," her teacher says. To understand why, we need to look at how water moves.

Osmosis is the net movement of water molecules across a partially permeable membrane, from a dilute solution (a high concentration of water) to a concentrated solution (a low concentration of water). Sugar molecules are too large to pass through the membrane, but water molecules can pass through it.

In a plant cell, water entering by osmosis makes the cell turgid (firm), which supports the plant. In a concentrated solution, water leaves the cell and the cell becomes flaccid (limp). Animal cells have no cell wall, so in pure water they can swell and burst, and in concentrated solution they shrink.

partially permeable membranedilute: high water potentialconcentrated: low water potential
Osmosis: water moves across a partially permeable membrane from the dilute side to the concentrated side

Key ideas

  • Osmosis is the net movement of water across a partially permeable membrane.
  • Water moves from a dilute solution to a concentrated solution.
  • A plant cell in pure water becomes turgid; in concentrated solution it becomes flaccid.
  • [H] Pure water has the highest water potential (0); solutions have lower (negative) water potentials.
Worked example 3.2.1
Which way does water move by osmosis: from the dilute solution into the concentrated, or the other way? Write the words.

Method. Recall the direction of osmosis.

  1. Recall. Water moves from a dilute solution to a concentrated solution.
    dilute to concentrated
  2. Name the direction. So water moves from the dilute solution into the concentrated solution.
    dilute into concentrated
  3. Write the answer.
    from the dilute solution into the concentrated solution

Check. Water moves down the water potential gradient. ✓

Worked example 3.2.2
A plant cell is placed in pure water. Describe what happens to the cell. Write one sentence.

Method. Think about where water is more concentrated.

  1. Think. Pure water is dilute, so water moves into the cell by osmosis.
    water moves into the cell
  2. Conclude. The cell swells and becomes turgid.
    becomes turgid
  3. Write the answer.
    The cell takes in water by osmosis, so it swells and becomes turgid.

Check. The cell wall stops the cell bursting. ✓

Worked example 3.2.3
[H] Which has the higher water potential: pure water or a sugar solution? Write the words.

Method. Compare the amount of water in each.

  1. Compare. Pure water contains only water, so it has the higher water potential.
    higher water potential
  2. Name it. A sugar solution has less free water, so its water potential is lower.
    lower water potential
  3. Write the answer.
    pure water

Check. Water moves from the higher to the lower water potential. ✓

Watch out

Watch out: osmosis is about water, not sugar. The sugar molecules do not move across the membrane; the water does.

Exam tip

Exam tip. Name the solution as dilute or concentrated, then say which way water moves, then say what happens to the cell.

Try it (with your teacher). With your teacher, look at a labelled diagram or a slide of plant cells in pure water and in concentrated salt or sugar solution. Describe what has changed in each cell.

Exercise 3B  Osmosis, and water potential [H]

1
FWhat is osmosis? Write the words.
2
FWater moves into a cell from a dilute solution. Write the word for this type of movement. Write the word.
3
FCan sugar molecules pass through a partially permeable membrane? Write yes or no.
4
FIn which solution does a plant cell become turgid: pure water or concentrated salt solution? Write the words.
5
FIn a concentrated sugar solution, does water leave a plant cell? Write yes or no.
6
FWhat is the word for a plant cell that has lost water and become limp? Write the word.
7
FAn animal cell is placed in pure water. Does it swell and burst? Write yes or no.
8
F[H] Which has the higher water potential: a dilute solution or a concentrated solution? Write the words.
9
PS[H] Explain why water moves from a higher water potential to a lower water potential. Write one sentence. Accept any sentence that says water moves down a water potential gradient, from the less concentrated side to the more concentrated side.
10
PSRuth places an animal cell in concentrated salt solution. Predict what happens to the cell. Write the word. Accept shrink.
11
PSA plant cell is placed in a dilute sugar solution, and water enters it. Write what happens to the cell. Accept any sentence that says the cell swells and becomes turgid.
12
RChallenge: explain in one sentence why a plant cell in pure water does not burst, but an animal cell in pure water can. Accept any sentence that says the plant cell wall is strong and resists the pressure of the water, while the animal cell has no cell wall.
3.3

Active transport

GCSE
Explain active transport against a concentration gradient, Give examples of active transport in plants and animals
Video lesson: Active transportComing soon
Context. Ali’s roots take up mineral ions from the soil. The soil has fewer mineral ions than the root hair cells, so the ions must move against the gradient. “How does the plant do this when diffusion would move them the other way?” Ali asks.

Active transport moves particles from a region of lower concentration to a region of higher concentration, against the concentration gradient. It needs energy from respiration. Proteins in the cell membrane act as carrier proteins, which bind the particles and move them across.

Two examples are: mineral ions moving into root hair cells from the soil, and glucose moving from the gut into the blood when the gut contains less glucose than the blood.

root hair cellsoil waterions move into the cell against the gradient, using energy from respiration
Active transport of mineral ions into a root hair cell

Key ideas

  • Active transport moves particles against the concentration gradient.
  • It needs energy from respiration, so cells with many mitochondria can do more.
  • Carrier proteins in the membrane move the particles.
  • Examples: mineral ions into root hair cells; glucose from the gut into the blood.
Worked example 3.3.1
Name the type of transport that moves particles against a concentration gradient. Write the words.

Method. Recall the process that needs energy.

  1. Recall. Moving particles against the gradient needs energy, so it is active transport.
    needs energy
  2. Name it. The answer is active transport.
    active transport
  3. Write the answer.
    active transport

Check. Diffusion moves particles down the gradient. ✓

Worked example 3.3.2
Explain why active transport needs energy. Write one sentence.

Method. Think about the direction of the movement.

  1. Think. The particles move against the concentration gradient.
    against the gradient
  2. Conclude. Energy from respiration is needed to push them against the gradient.
    energy from respiration
  3. Write the answer.
    Particles are moved against the gradient, so energy from respiration is needed.

Check. Energy comes from aerobic respiration in the mitochondria. ✓

Worked example 3.3.3
Mineral ions are at 2 units in the soil and 10 units in the root hair cell. Which way do they move by active transport? Write the words.

Method. Compare the two concentrations.

  1. Compare. The ions are at a higher concentration inside the cell than in the soil.
    higher inside
  2. Name the direction. By active transport they move into the cell, from the soil.
    into the cell
  3. Write the answer.
    into the cell

Check. Moving them in is against the gradient, so energy is used. ✓

Watch out

Watch out: active transport is not the same as diffusion. Diffusion needs no energy and goes down the gradient; active transport needs energy and goes up it.

Exam tip

Exam tip. Say “against the concentration gradient” and “energy from respiration” together for full marks.

Try it (with your teacher). Think about why a root hair cell needs many mitochondria. Write one sentence, and check it with your teacher. No practical is needed.

Exercise 3C  Active transport

1
FWhat is the process that moves particles against the concentration gradient? Write the words.
2
FActive transport needs energy from which process? Write the word.
3
FWhich way do particles move by active transport: down the gradient or up it? Write the words.
4
FName the proteins in the cell membrane that help move particles in active transport. Write the words.
5
FName one example of active transport in a plant. Write the words.
6
FName one example of active transport in an animal. Write the words.
7
FMineral ions are at 2 units in the soil and 10 units in the root hair cell. Is the movement into the cell against the gradient? Write yes or no.
8
PSTwo things are needed for active transport: energy from respiration and carrier proteins. Write one more thing that is not needed. Accept any sensible answer: a concentration gradient in the same direction, or no energy.
9
PSA cell that uses a lot of active transport has many mitochondria. Explain why, in one sentence. Accept any sentence that says mitochondria release the energy from respiration that active transport needs.
10
RChallenge: a student says active transport is faster than diffusion because it moves particles up a gradient. Explain why this is wrong. Accept any sentence that says active transport is slow in comparison, and it moves particles up a gradient only because energy is used to do this, not because it is faster.
3.4

Exchange surfaces and surface area to volume ratio

GCSE
Calculate surface area to volume ratio, Explain how exchange surfaces are adapted
Video lesson: Exchange surfaces and surface area to volume ratioComing soon
Context. A single-celled organism is a very small blob, so every part of its body is close to the outside. A large animal is much bigger, and most of its cells are deep inside. “How does a big animal get enough oxygen for all its cells?” asks Priya.

The surface area to volume ratio (SA:V) of an object tells us how much surface there is for each unit of volume. A small object has a large SA:V, so substances can reach all its cells quickly. As an object gets bigger, its SA:V gets smaller, so larger organisms need specialised exchange surfaces.

An exchange surface has a large surface area, a thin layer (a short distance), and a good supply of blood or air to keep the gradient steep.

Cube sideSurface area (cm2)Volume (cm3)Surface area : volume
1 cm616 : 1
2 cm2483 : 1
3 cm54272 : 1
Surface area to volume ratio of cubes of different sizes
Exchange surfaceWhereHow it is adapted
villismall intestinemany folds and projections; thin walls; blood supply
root hair cellsrootslong, thin extensions give a large surface area
alveolilungsvery many tiny sacs; thin walls; good blood supply
Exchange surfaces in plants and animals

Key ideas

  • SA:V = surface area divided by volume.
  • A smaller object has a larger SA:V.
  • Exchange surfaces have a large surface area and thin walls.
  • Villi, root hair cells and alveoli are all exchange surfaces.
Worked example 3.4.1
A cube has a side of 2 cm. Its surface area is 24 cm2 and its volume is 8 cm3. Find the surface area to volume ratio.

Method. Divide the surface area by the volume.

  1. Divide. SA:V = 24 / 8.
    24 / 8
  2. Simplify. So the ratio is 3 : 1.
    3 : 1
  3. Write the answer.
    3 : 1

Check. 3 x 8 = 24. ✓

Worked example 3.4.2
Explain why a villus has many folds. Write one sentence.

Method. Think about the surface area.

  1. Think. The folds increase the surface area.
    increase the surface area
  2. Conclude. A larger surface area means more food can be absorbed by diffusion.
    more food absorbed
  3. Write the answer.
    The folds increase the surface area, so more food can be absorbed.

Check. Large surface area speeds up absorption. ✓

Worked example 3.4.3
State how the thin walls of an alveolus help gas exchange. Write one sentence.

Method. Think about the distance.

  1. Think. The walls are one cell thick.
    one cell thick
  2. Conclude. So the gases have a short distance to diffuse.
    short distance
  3. Write the answer.
    The walls are thin, so gases diffuse over a short distance.

Check. A short distance speeds up diffusion. ✓

Watch out

Watch out: a bigger object has a bigger surface area, but a smaller SA:V. Do not write that a large animal has a large SA:V.

Exam tip

Exam tip. Give the ratio as a : 1 when a ratio is asked for. Divide the larger number by the smaller number if you need to simplify.

Try it (with your teacher). With your teacher, cut three cubes of jelly or sponge (with a grown-up or teacher only), of sides 1 cm, 2 cm and 3 cm. Work out the surface area to volume ratio of each cube. Describe the pattern.

Exercise 3D  Exchange surfaces and surface area to volume ratio

1
FWhat is the surface area of a cube with a side of 1 cm? Write the number, in cm2.
2
FWhat is the volume of a cube with a side of 2 cm? Write the number, in cm3.
3
FA cube has a surface area of 24 cm2 and a volume of 8 cm3. Find the surface area to volume ratio. Write the number.
4
FA cube has a side of 3 cm. Its surface area is 54 cm2 and its volume is 27 cm3. Find the surface area to volume ratio. Write the number.
5
FAs the size of a cube increases, does the SA:V ratio increase or decrease? Write the word.
6
FWhich has the larger SA:V: a small single-celled organism or a large animal? Write the words.
7
FName the structure in the small intestine that increases the surface area for absorption. Write the word.
8
FName the structures on roots that increase the surface area for absorbing water and minerals. Write the words.
9
FName the tiny sacs in the lungs where gas exchange takes place. Write the word.
10
PSGive one feature of an exchange surface that helps diffusion, other than a large surface area. Accept any one: thin walls, a good blood supply, or a steep gradient maintained.
11
PSA cube has a side of 4 cm. Find its surface area in cm2. Write the number.
12
RChallenge: a cube has a side of 4 cm, so its volume is 64 cm3. Find the surface area to volume ratio, and explain how this compares with the 2 cm cube. Accept 1.5 : 1 and a sentence that says the ratio is smaller for the bigger cube.
3.5

Required practical: osmosis in potato tissue

GCSE
Plan and carry out an investigation into osmosis in potato tissue, Calculate percentage change in mass and interpret a graph
Video lesson: Required practical: osmosis in potato tissueComing soon
Context. Ruth cuts potato cylinders of the same size and places one in each of several sugar solutions. After an hour she weighs each cylinder. “Some gain mass, some lose mass, and one stays the same,” she writes. “Which solution is the same as the inside of the potato?”

In this practical the independent variable is the concentration of the sucrose solution. The dependent variable is the percentage change in mass of each potato cylinder. Control variables include the size of the cylinders, the volume of solution, the time in solution and the temperature.

The cylinders are blotted dry before each weighing, so that only the water taken up by osmosis is measured. The concentration at which the mass does not change is the isotonic concentration: the potato cells and the solution have the same water potential.

Percentage change in mass = (final mass - starting mass) / starting mass x 100.

Sucrose (mol/dm3)Percentage change in mass (%)
0.0+12
0.1+7
0.2+3
0.30
0.4-4
0.5-8
0.6-11
Percentage change in mass of potato cylinders in sucrose solutions
Potato mass change in sucrose solution-15-10-50510150.00.10.20.30.40.50.6sucrose concentration (mol/dm3)percentage change in mass (%)
Line graph of percentage change in mass against sucrose concentration

Key ideas

  • Independent variable: sucrose concentration. Dependent variable: percentage change in mass.
  • Keep the cylinder size, volume, time and temperature the same.
  • A gain in mass means water entered by osmosis; a loss means water left.
  • The isotonic concentration is where the percentage change is zero.
Worked example 3.5.1
A potato cylinder has a starting mass of 5.0 g and a final mass of 5.6 g. Find the percentage change in mass.

Method. Use percentage change = (final - starting) / starting x 100.

  1. Change. The change is 5.6 - 5.0 = 0.6 g.
    0.6 g
  2. Divide. Divide the change by the starting mass: 0.6 / 5.0 = 0.12.
    0.6 / 5.0 = 0.12
  3. Multiply. Multiply by 100 to get a percentage.
    0.12 x 100 = 12
  4. Write the answer.
    12%

Check. 12% of 5.0 g is 0.6 g. ✓

Worked example 3.5.2
Read the graph. At which sucrose concentration does the potato cylinder show no change in mass? Write the number, in mol/dm3.

Method. Find the concentration where the line crosses zero.

  1. Find zero. The line crosses the zero percentage change at one value.
    crosses zero
  2. Read the value. That value on the x-axis is 0.3 mol/dm3.
    0.3 mol/dm3
  3. Write the answer.
    0.3

Check. At 0.3 mol/dm3 the mass does not change. ✓

Worked example 3.5.3
Using the table, find the difference in percentage change in mass between 0.0 and 0.6 mol/dm3. Write the number of percentage points.

Method. Subtract the two values.

  1. Read. At 0.0 the change is +12%; at 0.6 it is -11%.
    +12 and -11
  2. Subtract. The difference is 12 - (-11) = 23.
    12 + 11 = 23
  3. Write the answer.
    23

Check. 23 percentage points between the two extremes. ✓

Watch out

Watch out: blot each cylinder dry before weighing, but do it gently and the same way every time. Skin oils or water left on the surface change the mass.

Exam tip

Safety. Cutting potato cylinders needs a teacher or a grown-up to supervise the cork borer and the scalpel. Keep the cutting board flat and use a dry surface.

Try it (with your teacher). With your teacher, cut potato cylinders of the same size, place them in sucrose solutions of two different concentrations, and weigh them before and after one hour. Calculate the percentage change in mass for each.

Exercise 3E  Required practical: osmosis in potato tissue

1
FIn the potato practical, which variable is the independent variable? Write the words.
2
FIn the potato practical, which variable is the dependent variable? Write the words.
3
FName one control variable in the potato practical. Accept any one: the size of the cylinders, the volume of solution, the time, or the temperature.
4
FWhy are the potato cylinders blotted dry before weighing? Write one sentence. Accept any sentence that says surface water would add mass that is not from osmosis.
5
FA potato cylinder has a starting mass of 4.0 g and a final mass of 4.8 g. Find the percentage change in mass. Write the number, with the percentage sign.
6
PSA potato cylinder has a starting mass of 6.0 g and a final mass of 5.4 g. Find the percentage change in mass. Write the number, with the percentage sign.
7
PSAt 0.2 mol/dm3, what is the percentage change in mass shown on the graph? Write the number, with the percentage sign.
8
PSAt which concentrations does the potato cylinder lose mass? Write the two values, in mol/dm3. Accept 0.4 and 0.5 and 0.6 (any of these three).
9
PSRuth says the potato cells have the same concentration as the 0.3 mol/dm3 solution. Write yes or no.
10
REvaluate the potato practical: name one improvement, and say why it helps. Accept any sensible improvement, for example more concentrations, or repeat readings for each concentration, to make the graph more reliable.
11
RChallenge: explain why the potato cylinders gain mass in pure water but lose mass in a 0.6 mol/dm3 solution. Write two sentences, using the words water potential and osmosis. Accept any answer that says water enters by osmosis in pure water, and leaves by osmosis in the concentrated solution because water potential is higher inside the potato.

Chapter summary

SkillMethod
DiffusionNet movement from high to low concentration; passive; faster with a steeper gradient, higher temperature, larger surface area and shorter distance.
OsmosisNet movement of water across a partially permeable membrane, from dilute to concentrated.
[H] Water potentialPure water has the highest water potential; water moves down the gradient.
Active transportMovement against the gradient; needs energy from respiration and carrier proteins.
Exchange surfacesLarge surface area, thin walls, and a good blood or air supply (villi, root hairs, alveoli).
SA:V ratioSurface area divided by volume; falls as an object gets bigger.
Potato practicalPercentage change in mass against sucrose concentration; zero change is the isotonic point.

Can you…?

  • I can explain diffusion and the factors that change its rate
  • I can explain osmosis and predict what happens to plant and animal cells
  • I can explain why active transport needs energy, and give two examples
  • I can calculate a surface area to volume ratio and explain how exchange surfaces are adapted
  • I can plan the potato practical and interpret its graph, including the isotonic point
  • [H] I can explain water potential

Key terms

diffusion
the net movement of particles from high concentration to low concentration
concentration gradient
the difference in concentration between two regions
osmosis
the net movement of water across a partially permeable membrane
partially permeable membrane
a membrane that lets some particles through but not others
active transport
movement of particles against a concentration gradient, using energy from respiration
carrier protein
a protein in the membrane that moves particles across it
turgid
firm, because the cell has taken in water by osmosis
flaccid
limp, because the cell has lost water
isotonic
when a solution and a cell have the same water potential, so no net movement of water
surface area to volume ratio
the surface area of an object divided by its volume
exchange surface
a surface with a large area and thin walls, where substances are exchanged
[H] water potential
a measure of how likely water is to move; pure water has the highest value

Exercise 3Z  Check-up

1
Name the process that moves particles from high to low concentration without energy. Write the word.
2
Name the process of water moving across a partially permeable membrane. Write the word.
3
Name the process that moves particles against the concentration gradient, using energy. Write the words.
4
A plant cell is placed in pure water. Does it become turgid? Write yes or no.
5
Name one factor that makes diffusion faster. Accept any one: a steeper gradient, a higher temperature, a larger surface area, or a shorter distance.
6
A cube has a side of 1 cm. Find the surface area to volume ratio. Write the number.
7
Name the structure in the small intestine that increases the surface area for absorption. Write the word.
8
A potato cylinder has a starting mass of 5.0 g and a final mass of 4.5 g. Find the percentage change in mass. Write the number, with the percentage sign.
9
Mineral ions are at 2 units in the soil and 10 units in a root hair cell. Name the process that moves them into the cell. Write the words.
10
Explain in one sentence why active transport needs energy. Accept any sentence that says particles are moved against the concentration gradient, which needs energy from respiration.

Total 24 marks. Allow about 24 minutes.

Exercise 3X  Exam-style questions

1
FName the process that moves water molecules across a partially permeable membrane. Write the word.[1 mark]
2
FWhich way does water move by osmosis: from a dilute solution to a concentrated solution, or the other way? Write the words.[1 mark]
3
FName one factor that increases the rate of diffusion. Accept any one: a higher temperature, a steeper gradient, a larger surface area, or a shorter distance.[1 mark]
4
RExplain why the root hair cells need active transport to take in mineral ions from the soil. Write two sentences. Accept any answer that says the ions are at a higher concentration inside the cell, so they move against the gradient, which needs energy from respiration.[2 marks]
5
PSA potato cylinder has a starting mass of 5.0 g and a final mass of 5.8 g. Find the percentage change in mass. Write the number, with the percentage sign.[2 marks]
6
RA cube has a side of 3 cm, with a surface area of 54 cm2 and a volume of 27 cm3. Explain why a large animal needs an exchange surface, and use the SA:V ratio in your answer. Accept any answer that gives the ratio 2 : 1 and says a large animal has a smaller SA:V, so it needs specialised exchange surfaces.[3 marks]
7
FDescribe how the villi are adapted for absorbing food. Write three points. Accept any three of: many finger-like projections, large surface area, thin walls, and a good blood supply.[3 marks]
8
RA student places potato cylinders in sucrose solutions and finds the mass gain at 0.0 mol/dm3 is +12%, and the mass loss at 0.6 mol/dm3 is -11%. Explain why the potato cells gain and lose mass, using the words water potential and osmosis. Accept any answer that says water enters by osmosis in pure water and leaves in the more concentrated solution, because water potential inside the potato is between the two.[4 marks]
9
PSPlan an investigation into the effect of sucrose concentration on the mass of potato cylinders. Name the independent variable, the dependent variable, and two control variables, and say how you would calculate the percentage change. Accept any plan that names sucrose concentration, percentage change in mass, and two sensible controls, with the formula (final mass - starting mass) / starting mass x 100.[5 marks]
10
RExplain why a plant cell in pure water becomes turgid, using the word osmosis. Write two sentences. Accept any answer that says water enters the cell by osmosis, and the cell wall stops the cell bursting, so the cell becomes firm.[2 marks]
← Chapter 2: Cell structure and microscopyChapter 4: Mitosis, the cell cycle and stem cells →

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