Diffusion, osmosis and active transport – IGCSE Biology exam questions with mark scheme

1Grade 6–77 marks

A student investigated osmosis using potato cylinders. Each cylinder was weighed, placed in a sucrose solution for 30 minutes, blotted dry and reweighed.

Sucrose concentration / mol per dm³	Starting mass / g	Final mass / g	Percentage change in mass
0.0	2.50	2.80	+12.0
0.2	2.48	2.58	+4.0
0.4	2.52	2.42	-4.0
0.6	2.46	2.21
0.8	2.50	2.15	-14.0

(a) Calculate the percentage change in mass for the cylinder in 0.6 mol per dm³ sucrose solution. Give your answer to one decimal place. [2]

(b) Use the results to estimate the concentration of sucrose that would cause no change in mass. [1]

(d) Give two variables, other than the time, that the student should keep the same. [2]

(a) (2.21 − 2.46) ÷ 2.46 × 100 seen; accept change in mass = −0.25 g with ÷ 2.46; correct final answer with no working scores both marks(1)
(a) −10.2 (%); accept 10.2 (%) decrease; do not award −10.16 (not to 1 d.p.) for this mark(1)
(b) 0.3 (mol per dm³); accept any value from 0.28 to 0.32(1)
(c) water moves into the (potato) cells by osmosis(1)
(c) from a dilute solution / high water concentration (outside) to a more concentrated solution / lower water concentration (inside cells) through a partially permeable membrane; accept down a water concentration gradient / water potential gradient; accept cell sap / cytoplasm more concentrated than the water(1)
(d) any two from: size / length / diameter / surface area / starting mass of cylinder; volume of solution; temperature; same potato / variety; blotting method — first mark(1)
(d) second correct variable; ignore 'same amount' unqualified; do not credit time (excluded by the question) or concentration of sucrose (independent variable)(1)
Answer: (a) −10.2% (b) 0.3 mol per dm³ (c) water entered cells by osmosis from more dilute solution (d) e.g. cylinder size, volume of solution, temperature
Examiner tip: Percentage change must be divided by the STARTING mass, not the final mass — dividing by 2.21 gives −11.3% and loses the accuracy mark.

2Grade 6–75 marks

A student cut three cubes of agar jelly containing a pink indicator. The pink colour disappears when acid diffuses in. The cubes had sides of 1 cm, 2 cm and 3 cm and were placed in acid.

(a) Calculate the surface area to volume ratio of the 2 cm cube. Give your answer in the form n : 1. [2]

(b) The 3 cm cube took the longest time to turn completely colourless. Explain why. [2]

(c) Use your answer to (b) to explain why a large multicellular organism needs a transport system but an amoeba does not. [1]

(a) surface area = 24 (cm²) and volume = 8 (cm³)(1)
(a) 3 : 1(1)
(b) (largest cube has) smallest surface area to volume ratio / less surface area per unit volume(1)
(b) greater distance for acid to diffuse to the centre(1)
(c) amoeba has a large SA : V / short diffusion distance so diffusion alone is fast enough, but in large organisms diffusion is too slow to supply all cells; accept converse(1)
Answer: (a) 3 : 1 (b) smaller SA:V and longer diffusion distance (c) diffusion too slow over large distances / small SA:V
Examiner tip: Saying 'the cube is bigger' scores nothing — you must refer to the surface area to volume ratio or the diffusion distance.

3Grade 8–94 marks

The table shows the concentration of potassium ions in the cells of a plant root and in the soil water around it, in two different conditions.

Condition	Potassium ions in root cells / arbitrary units	Potassium ions in soil water / arbitrary units	Rate of uptake / arbitrary units
Oxygen bubbled through soil water	12.0	0.5	8.4
Respiratory inhibitor added	12.0	0.5	0.3

Explain how the data show that potassium ions are absorbed by active transport. [4]

ions move from low concentration (in soil water, 0.5) to high concentration (in root cells, 12.0) / against the concentration gradient(1)
so cannot be diffusion(1)
active transport requires energy from respiration; accept ATP(1)
inhibitor stops (aerobic) respiration so little / no energy released and uptake almost stops (falls from 8.4 to 0.3)(1)
Answer: Ions move against the concentration gradient, which needs energy from respiration; when respiration is inhibited uptake almost stops.
Examiner tip: Quote the data: the move from 0.5 to 12.0 and the drop from 8.4 to 0.3 are what earn the explanation marks.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro