Diffusion, osmosis and active transport (RP3) – GCSE Biology exam questions with mark scheme

1Grade 6–75 marks

A student investigated osmosis in potato cylinders placed in sugar solutions for 24 hours.

Concentration in mol/dm³	Start mass in g	End mass in g	Percentage change in mass
0.0	2.50	2.85	+14.0
0.2	2.48	2.66	+7.3
0.4	2.52	2.55	+1.2
0.6	2.50	2.38	X
0.8	2.46	2.23	−9.3

(a) Calculate value X. [2]

(b) The graph shows the results with a line of best fit. Use the graph to estimate the concentration of the solution inside the potato cells. [1]

(a) change in mass = 2.38 − 2.50 = −0.12 (g) and (−0.12 ÷ 2.50) × 100 (allow 2.38 ÷ 2.50 × 100 = 95.2 then − 100)(1)
(a) −4.8 (%) (minus sign needed for this mark; correct answer with no working scores 2 marks)(1)
(b) concentration where the line crosses 0% change: 0.46 mol/dm³ (accept 0.43–0.49)(1)
(c) water moves into the potato cells by osmosis(1)
(c) because the water outside is more dilute than the (cell sap / cytoplasm in the) cells, so water moves from a dilute to a more concentrated solution (through the partially permeable cell membrane); allow higher water concentration outside the cells; ignore 'water moves down a concentration gradient' without saying which is more dilute(1)
Answer: (a) −4.8% (b) about 0.46 mol/dm³ (c) water enters cells by osmosis from dilute solution into more concentrated cell sap
Examiner tip: Percentage change uses the START mass as the denominator and a decrease must keep its minus sign.

2Grade 6–74 marks

(a) Give two differences between active transport and diffusion. [2]

(b) Root hair cells contain many mitochondria. Explain why. [2]

(a) active transport moves substances against a concentration gradient / from a dilute to a more concentrated solution, but diffusion is down a concentration gradient(1)
(a) active transport requires energy (from respiration), but diffusion does not / diffusion is passive(1)
(b) mitochondria are the site of (aerobic) respiration, which releases / transfers energy; do not accept 'makes / produces energy'(1)
(b) energy is needed for active transport of mineral ions from the (dilute) soil water into the root hair cell; allow (to absorb) mineral ions / named ion (e.g. nitrate, magnesium) against a concentration gradient; reject active transport of water(1)
Answer: (a) AT goes against the gradient; AT needs energy (b) respiration releases energy for active transport of mineral ions
Examiner tip: Never say mitochondria 'make energy' — energy is released (transferred) by respiration.

3Grade 6–75 marks

A student cut cubes of agar jelly containing an indicator and placed them in acid. The table shows the cube sizes.

Side length in cm	Surface area in cm²	Volume in cm³
1	6	1
2	?	?
3	54	27

(a) Calculate the surface area to volume ratio of the 2 cm cube. [2]

(b) A single-celled organism does not need a specialised gas exchange surface, but a large multicellular organism does.

Explain why. [3]

(a) surface area = 24 (cm²) and volume = 8 (cm³) (allow 24 : 8 for this mark)(1)
(a) 3 : 1 (allow 3; correct answer with no working scores 2 marks)(1)
(b) the single-celled organism has a large surface area to volume ratio; allow converse: the large organism has a small surface area to volume ratio(1)
(b) the single-celled organism has a short diffusion distance (to all parts of the cell); allow converse: in the large organism the diffusion distance to cells inside the body is long(1)
(b) so in the large organism diffusion across the body surface is too slow / cannot supply enough oxygen (or remove carbon dioxide) for all its cells (so it needs an exchange surface); allow converse: in the single-celled organism diffusion across the cell membrane is enough to meet its needs(1)
Answer: (a) 3 : 1 (b) single-celled organism: large SA:V and short diffusion distance, so diffusion through the membrane is enough; large organism: small SA:V and long diffusion distance, so diffusion across the surface is too slow to supply all cells
Examiner tip: Give the ratio as x : 1 — writing 24 : 8 without simplifying can lose the mark.

4Grade 1–3Foundation4 marks

(a) Complete the sentence. Choose answers from the box.

active

higher

lower

partially permeable

same

Diffusion is the net movement of particles from an area of ______ concentration to an area of ______ concentration. [2]

(b) Which substance diffuses out of body cells into the blood plasma so that it can be removed by the kidneys?

Tick (✓) one box.

☐ glucose ☐ oxygen ☐ starch ☐ urea [1]

Give one other factor that affects the rate of diffusion. [1]

(a) higher (first gap)(1)
(a) lower (second gap)
both gaps must be in this order; 'lower … higher' scores 0 for both(1)
(b) urea (only)
if more than one box ticked, 0 marks(1)
(c) any one from:
• (difference in) concentration / concentration gradient
• surface area (of the membrane)
allow distance / thickness of membrane
ignore 'amount of particles' unless linked to concentration
ignore temperature (given in the question)(1)
Answer: (a) higher; lower (b) urea (c) concentration gradient (or surface area of membrane)
Examiner tip: Diffusion always goes from higher to lower concentration — learn the exact definition including the word 'net', and do not give the factor already named in the question.

5Grade 4–55 marks

A student set up the apparatus shown in the diagram.

The Visking tubing bag was filled with concentrated sugar solution and placed in a beaker of water. After 30 minutes the level of liquid in the capillary tube had risen.

(a) Visking tubing lets water molecules through but not sugar molecules.

What type of membrane is Visking tubing? [1]

(b) Explain why the level of liquid in the capillary tube rose. [3]

Predict what would happen to the level of liquid in the capillary tube. [1]

(a) partially permeable
accept selectively permeable / semi-permeable
ignore 'permeable' on its own(1)
(b) water moves (into the bag) by osmosis(1)
(b) (water moves) from a dilute solution / water (in the beaker) to a more concentrated solution (in the bag)
allow from high water concentration to low water concentration
do not accept 'from high concentration to low concentration' unless water concentration is clearly meant(1)
(b) through the partially permeable membrane / Visking tubing
or sugar (molecules) cannot move out of the bag (so the volume in the bag increases and liquid is pushed up the tube)
do not accept sugar moves into the bag / water moves out of the bag(1)
(c) the level would fall / go down
ignore reasons(1)
Answer: (a) partially permeable (b) water enters the bag by osmosis, from the dilute water to the concentrated sugar solution, across the partially permeable tubing; sugar cannot leave, so the volume increases (c) level falls
Examiner tip: Say which way the WATER moves and name both solutions (dilute → concentrated); 'osmosis moves from high to low concentration' is ambiguous and loses the mark.

6Grade 6–76 marks

A student placed a drop of purple dye in the centre of a dish of agar jelly. She measured how far the purple colour had spread from the centre every 10 minutes.

She did the experiment at 20 °C and at 40 °C. The graph shows her results.

(a) Use the graph to give the distance moved by the dye at 40 °C after 40 minutes. [1]

(b) Calculate the mean rate of diffusion of the dye at 40 °C over the 40 minutes.

Give your answer in mm per minute. [2]

(d) At both temperatures the rate of diffusion slowed down as time went on.

Suggest why. [1]

(a) 16 (mm)(1)
(b) 16 ÷ 40
allow their (a) ÷ 40(1)
(b) 0.4 (mm per minute)
allow correct answer from their (a) ÷ 40
an answer of 0.4 with no working scores 2 marks(1)
(c) higher temperature increases the rate of diffusion
allow dye moves further / faster at 40 °C(1)
(c) (because) particles / molecules have more (kinetic) energy or particles / molecules move faster
ignore 'particles vibrate more'(1)
(d) the concentration gradient becomes smaller / less steep
allow the dye becomes more spread out / diluted so the difference in concentration is less
allow the dye has to spread over a larger area / volume
ignore 'dye runs out'(1)
Answer: (a) 16 mm (b) 0.4 mm per minute (c) higher temperature gives faster diffusion because particles have more kinetic energy / move faster (d) the concentration gradient gets less steep
Examiner tip: In 'describe and explain', the description (faster at higher temperature) and the explanation (particles have more kinetic energy) are separate marks — give both.

7Grade 6–76 marks

A student wants to find the concentration of salt solution that is the same as the concentration of the solution inside carrot cells.

She is given:

a large carrot
a cork borer, a knife and a ruler
a 1.0 mol/dm³ salt solution and distilled water
boiling tubes, measuring cylinders and paper towels
a balance

Describe a method the student could use.

Your method should give valid results. [6]

use cork borer (and knife / ruler) to cut carrot cylinders (all of the) same diameter and same length
allow same size; remove the skin / use same carrot(1)
blot cylinders dry with paper towel and record the (starting) mass of each with the balance(1)
make a range of concentrations (at least 5) by diluting the 1.0 mol/dm³ salt solution with distilled water, e.g. 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³(1)
put the same volume of each solution (e.g. 10 cm³) in a labelled boiling tube, measured with a measuring cylinder(1)
put one cylinder in each tube; leave all for the same time (e.g. 24 hours) at the same temperature(1)
remove cylinders, blot dry (to remove surface solution) and reweigh(1)
calculate percentage change in mass for each concentration(1)
repeat at each concentration (e.g. three cylinders each) and calculate a mean(1)
plot percentage change in mass against concentration; the concentration where the line crosses 0 % (no change in mass) is the concentration of the cell contents(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent method that could be followed by someone else to produce valid results. Includes a range of concentrations, control of cylinder size, volume and time, mass measured before and after with blotting, and a way of using the results (percentage change and the point of no change in mass) to find the concentration.
Level 2	3–4	A method with the main steps in a sensible order (cut cylinders, weigh, place in different concentrations, reweigh) and some control of variables, but with steps missing or not explained, or no clear way of finding the concentration from the results.
Level 1	1–2	Simple, relevant statements about the method, e.g. putting carrot pieces in salt solutions and weighing them, but the steps do not link into a workable method.
No marks	0	No relevant content.

Answer: Cut equal-sized carrot cylinders, blot and weigh; make a range of salt concentrations by dilution; same volume in each tube, same time and temperature; blot and reweigh; calculate % change in mass; repeat for means; plot % change against concentration and read off where the line crosses 0 %.
Examiner tip: Top-level answers say how the result is found: plot percentage change in mass against concentration and read where the line crosses zero — many methods stop at 'reweigh'.

8Grade 8–9Higher6 marks

A student placed cylinders of beetroot in different concentrations of salt solution for 18 hours.

The table shows the results.

Concentration of salt solution in mol/dm³	Percentage change in mass
0.0	+11.0
0.2	+4.0
0.4	−2.0
0.6	−7.5
1.0	−12.5

(a) The cylinder in 1.0 mol/dm³ salt solution had a starting mass of 3.20 g.

Calculate its mass at the end of the experiment. [2]

(b) Assume the percentage change in mass changes at a constant rate between 0.2 mol/dm³ and 0.4 mol/dm³.

Calculate the concentration of salt solution that would cause no change in mass.

Give your answer to 2 significant figures. [2]

(c) The student blotted each cylinder dry before weighing it at the start, but forgot to blot the cylinders before weighing them at the end.

Explain how this would affect the concentration calculated in part (b). [2]

(a) 3.20 × 0.875 or 3.20 − (0.125 × 3.20) or 0.40 (g lost)(1)
(a) 2.80 (g)
allow 2.8 (g)
correct answer with no working scores 2 marks(1)
(b) 0.2 + (4 ÷ 6) × 0.2 or 4 ÷ 6 of the way from 0.2 to 0.4
allow equivalent, e.g. 0.4 − (2 ÷ 6) × 0.2(1)
(b) 0.33 (mol/dm³)
an unrounded answer (0.333…) or an answer not to 2 s.f. (e.g. 0.3) scores the method mark only
correct answer with no working scores 2 marks(1)
(c) the end masses would be too high (because extra solution is left on the surface)
allow percentage gains would be too big / percentage losses too small(1)
(c) so the calculated concentration (for no change in mass) would be too high
allow the line would cross 0 % at a higher concentration
this mark depends on the direction of the change in end mass being given(1)
Answer: (a) 2.80 g (b) 0.33 mol/dm³ (c) end masses too high, so % changes are too high and the calculated concentration would be too high
Examiner tip: For (c), follow the error all the way through: wet cylinders → end masses too high → % change too high → line crosses zero at a higher concentration.

9Grade 6–76 marks

A flatworm is a thin, flat animal that lives in ponds. It has no gas exchange organs. A scientist modelled a flatworm as a cuboid, as shown in the diagram.

(a) Calculate the surface area to volume ratio of the model flatworm.

Give your answer in the form x : 1. [3]

(b) Fish are much larger than flatworms, so they need gills to absorb oxygen.

Gills have many thin filaments covered in lamellae. Each lamella has a thin wall and a good blood supply. Water is pumped over the gills all the time.

Explain how these features allow fish to absorb enough oxygen. [3]

(a) surface area = 2(10 × 2) + 2(10 × 0.5) + 2(2 × 0.5) = 52 (mm²)(1)
(a) volume = 10 × 2 × 0.5 = 10 (mm³)(1)
(a) 5.2 : 1
allow their surface area ÷ their volume correctly evaluated, as x : 1
correct answer with no working scores 3 marks(1)
(b) any three from:
• many filaments / lamellae give a large surface area (for diffusion)(1)
• thin walls give a short diffusion distance / path(1)
• good blood supply carries oxygen away, keeping a (steep) concentration gradient
• water flowing over the gills (ventilation) brings in more oxygen, keeping a (steep) concentration gradient
for the concentration-gradient marks the gradient must be linked to the blood supply or the water flow
ignore 'large surface area' unless linked to the filaments / lamellae(1)
Answer: (a) SA = 52 mm², V = 10 mm³, ratio 5.2 : 1 (b) many filaments/lamellae → large surface area; thin walls → short diffusion distance; blood supply and constant water flow keep a steep oxygen concentration gradient
Examiner tip: Link each feature to its effect: 'thin' only scores when you say it gives a short diffusion distance, and 'good blood supply' only scores when it keeps the concentration gradient steep.

10Grade 8–9Higher7 marks

Scientists investigated how barley roots absorb potassium ions.

They grew roots in a dilute solution of potassium ions. They bubbled air through one container and nitrogen through another. Nitrogen contains no oxygen. All other conditions were kept the same.

	Air bubbled through	Nitrogen bubbled through
Concentration of potassium ions in the solution in mmol/dm³	0.6	0.6
Concentration of potassium ions in the root cells after 6 hours in mmol/dm³	18.0	2.1

(a) Calculate how many times greater the concentration of potassium ions in the root cells was than in the solution when air was bubbled through. [1]

(b) Explain how the results show that potassium ions are absorbed by active transport. [3]

(c) When nitrogen was bubbled through, the roots still took up a small amount of potassium against the concentration gradient.

Suggest why. [1]

(d) In the small intestine, glucose is absorbed into the blood by diffusion and by active transport.

Explain why active transport is needed as well as diffusion. [2]

(a) 30 (times)(1)
(b) (with air) potassium ions moved into the root cells from a lower to a higher concentration / against the concentration gradient
allow concentration in roots (18.0) is higher than the solution (0.6)(1)
(b) less uptake / lower concentration in roots when there is no oxygen (2.1 compared with 18.0 mmol/dm³)(1)
(b) (because) without oxygen there is less (aerobic) respiration, so less energy is released / available for active transport
do not accept energy is made / produced(1)
(c) (some) anaerobic respiration releases a small amount of energy
allow some oxygen was left in the roots / solution for a short time(1)
(d) (when) the concentration of glucose in the gut becomes lower than in the blood, diffusion (into the blood) stops / glucose would diffuse back(1)
(d) active transport moves glucose against the concentration gradient so (almost) all the glucose is absorbed / none is wasted (in faeces)(1)
Answer: (a) 30 (b) ions moved against the concentration gradient (18.0 vs 0.6); much less uptake without oxygen; less aerobic respiration so less energy released for active transport (c) anaerobic respiration still releases some energy (d) once gut glucose is lower than blood glucose, diffusion cannot absorb it; active transport moves it against the gradient so it is all absorbed
Examiner tip: Active transport needs two ideas: movement against the concentration gradient AND energy released by respiration — and never write that respiration 'makes' energy.

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