Cambridge O Level Biology 5090: Movement into and out of cells (5090) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Topic 3, Movement into and out of cells, of the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028: sections 3.1 (Diffusion and osmosis) and 3.2 (Active transport). The syllabus is not tiered, so every question applies to every candidate.

Revise first with the study guide and the revision notes. The course hub and the printable checklist list the other topics.

Questions

1. Define the term diffusion. [2]

2. Describe the role of water as a solvent in digestion, excretion and transport in humans. [3]

3. Diffusion and osmosis are passive processes.

(a) State the source of energy for osmosis. [1] (b) Explain why oxygen diffuses faster at a higher temperature. [2]

4. Two blocks of pink indicator agar are put into the same dilute acid. Block P is 2 cm × 2 cm × 2 cm; block Q is 1 cm × 1 cm × 8 cm. The colour disappears as acid diffuses in.

(a) Calculate the surface area to volume ratio of each block. Show your working. [3] (b) Predict which block will go colourless first, and explain your answer. [2]

5. A student places a drop of dye in the centre of a dish of agar and measures how far the colour spreads in 30 minutes, at 15 °C, 25 °C, 35 °C and 45 °C.

(a) State the independent variable and the dependent variable. [2] (b) State two variables she should keep constant. [2] (c) Suggest how she could make her results more reliable. [1]

6. Red blood cells and onion epidermis cells are each placed in distilled water.

(a) Define the term osmosis. [2] (b) Describe and explain what happens to the red blood cells. [3] (c) Explain why the onion cells do not burst. [2]

7. Cylinders of carrot were weighed, left in sucrose solutions for 40 minutes, blotted and reweighed.

Sucrose / mol per dm³	Start mass / g	End mass / g	Change in mass / %
0.0	3.20	3.52	+10.0
0.1	3.00	3.15	+5.0
0.2	3.10	3.10	0.0
0.3	3.00	2.85
0.4	3.20	2.88	−10.0
0.5	3.00	2.58	−14.0

(a) Calculate the percentage change in mass at 0.3 mol per dm³. [2] (b) Use the results to estimate the concentration of sucrose solution that has the same water potential as the carrot cells. Explain your answer. [2] (c) Explain the result at 0.0 mol per dm³. Use the terms turgid and turgor pressure. [3] (d) Explain the result at 0.5 mol per dm³. Use the terms flaccid and plasmolysis. [3] (e) Explain why the results are given as percentage change rather than change in mass. [1] (f) Suggest why the cylinders were blotted before reweighing. [1]

8. Dialysis tubing filled with concentrated sucrose solution is attached to a capillary tube and placed in distilled water. The height of liquid in the capillary tube is recorded.

Time / min	0	4	8	12	16	20
Height / mm	0	10	19	27	34	40

(a) Calculate the mean rate of rise over the 20 minutes. Give the unit. [2] (b) Explain why the liquid rises up the capillary tube. [3] (c) The rate of rise falls during the experiment. Suggest why. [2] (d) Describe a suitable control for this experiment. [1]

9. A potted bean plant is left without water on a hot, sunny windowsill. After two days its leaves and young stem droop. Explain why the plant droops, with reference to its cells. [4]

10. Two sets of barley roots were kept in the same potassium ion solution, one with air bubbled through and one with nitrogen. Potassium ion uptake was measured.

Time / h	0	1	2	3	4
Uptake with air / arbitrary units	0	12	24	36	48
Uptake with nitrogen / arbitrary units	0	3	4	5	6

(a) Calculate the percentage reduction in uptake after 4 hours when nitrogen is used instead of air. [2] (b) Explain the difference between the two sets of results. [4] (c) Suggest why there is some uptake in nitrogen during the first hour. [1]

11. A root hair cell takes in water and nitrate ions from the soil. Compare the ways in which these two substances enter the cell. [6]

Answers

1. The net movement of molecules or ions from a region of their higher concentration to a region of their lower concentration [1], down a concentration gradient, as a result of their random movement [1]. [2] Examiner insight: “Net” and “random movement” are what separate a two-mark definition from a one-mark one; “high to low” alone usually earns only the first mark.

2. Digestion: food molecules and enzymes are in water, and the soluble products dissolve so they can be absorbed [1]. Excretion: urea (and excess salts) are removed dissolved in water, as urine [1]. Transport: plasma, which is mostly water, carries dissolved substances such as glucose, amino acids or carbon dioxide [1]. [3] Examiner insight: One mark is given per process, so each needs a named example of something dissolved; three general statements about “carrying things” earn one mark at most.

3. (a) Kinetic energy of the random movement of (water) molecules [1]. (b) At a higher temperature molecules have more kinetic energy [1], so they move faster and more of them move down the concentration gradient in a given time [1]. Examiner insight: “Particles move faster” scores only if it is linked to kinetic energy; answers that mention enzymes or respiration for diffusion score nothing.

4. (a) P: surface area = 6 × (2 × 2) = 24 cm², volume = 8 cm³, SA:V = 3 : 1 [1]. Q: surface area = 2(1 × 1) + 4(1 × 8) = 2 + 32 = 34 cm², volume = 8 cm³ [1], SA:V = 34 : 8 = 4.25 : 1 [1]. (b) Q goes colourless first [1]. It has the larger surface area to volume ratio and no part of it is more than 0.5 cm from the surface, so acid has a shorter distance to diffuse [1]. Examiner insight: Show both surface area and volume; a wrong bare ratio with no working earns nothing.

5. (a) Independent: temperature [1]. Dependent: distance the colour has spread (in 30 minutes) [1]. (b) Two from: volume of dye drop, concentration of dye, depth of agar, time allowed [1] [1]. (c) Repeat each temperature (for example three times) and calculate a mean [1]. Examiner insight: Vague answers such as “same conditions” or “same amount” earn no credit for controlled variables.

6. (a) Net movement of water molecules from a region of higher water potential to a region of lower water potential [1], through a partially permeable membrane [1]. (b) Distilled water has a higher water potential than the cytoplasm of the red blood cells [1]. Water enters the cells by osmosis [1]. The cells swell and burst because they have no cell wall [1]. (c) Onion cells have a cellulose cell wall [1], which resists the pressure of the expanding vacuole, so the cells become turgid instead of bursting [1]. Examiner insight: Saying red blood cells burst “because of too much water”, without water potential and osmosis, gains only the description mark.

7. (a) (2.85 − 3.00) / 3.00 × 100 [1] = −5.0 % [1]. (b) 0.2 mol per dm³ [1], because there is no change in mass, so there is no net movement of water: the water potentials inside and outside are equal [1]. (c) The solution has a higher water potential than the carrot cells [1], so water enters by osmosis [1]. Vacuoles swell and push on the walls (turgor pressure); cells become turgid and the cylinder gains mass [1]. (d) The solution has a lower water potential than the carrot cells [1], so water leaves by osmosis [1]. Cells lose turgor pressure and become flaccid; in some, the membrane and cytoplasm pull away from the wall (plasmolysis) [1]. (e) The starting masses were different, so percentage change allows a fair comparison [1]. (f) To remove surface solution, so only water taken into or lost from the cells affects the mass [1]. Examiner insight: The minus sign is needed for the accuracy mark in (a); in (c) and (d) state which side has the higher water potential and use the named terms.

8. (a) 40 mm / 20 min [1] = 2.0 mm per min [1]. (b) Distilled water has a higher water potential than the sucrose solution [1]. Water enters the tubing by osmosis through the partially permeable membrane [1]. Water enters much faster than any sucrose leaves, so the volume inside increases and liquid is pushed up the tube [1]. (c) Water entering dilutes the sucrose solution [1], so the water potential gradient becomes less steep and osmosis slows [1]. (d) The same set-up with distilled water inside the tubing instead of sucrose solution [1]. Examiner insight: A rate needs a unit for the second mark; “2” alone loses it even with correct working.

9. The plant loses water from its leaves faster than it takes water up from the dry soil [1]. Its cells lose water by osmosis [1], lose turgor pressure and become flaccid [1]. Flaccid cells no longer press outwards on their walls, so cannot support the leaves and stem [1]. [4] Examiner insight: The link between turgor pressure and support is the key mark; answers that only say “the plant has no water” score at most one mark.

10. (a) (48 − 6) / 48 × 100 [1] = 87.5 % [1]. (b) Potassium ions are taken up by active transport [1], against a concentration gradient [1]. This needs energy released during respiration [1]. Without oxygen (in nitrogen) aerobic respiration stops, so little energy is released and uptake almost stops [1]. (c) Any one [1]: some ions diffuse in while the concentration inside the root is still lower than in the solution; some oxygen remains in the roots at first, so a little respiration continues. Examiner insight: “Energy” must be linked to respiration and to oxygen; a statement that nitrogen “is toxic” or “blocks the ions” gains no credit.

11. Water enters by osmosis; nitrate ions by active transport [1]. Water moves from higher to lower water potential [1]; nitrate ions move from lower to higher concentration, against a concentration gradient [1]. Osmosis uses only the kinetic energy of molecules [1]; active transport uses energy released during respiration [1]. Both cross the cell membrane, helped by the large surface area of the root hair [1]. [6] (Also accept: lack of oxygen reduces nitrate uptake but does not directly stop osmosis.) Examiner insight: A comparison question needs matched points for both substances; describing only one substance in detail caps the answer at about half the marks.

Where marks are usually lost

• Defining osmosis in terms of “water concentration” or “solution concentration” instead of water potential.
• Leaving out the partially permeable membrane in the osmosis definition.
• Leaving the minus sign off a percentage loss in mass.
• Giving a rate without a unit, or a ratio with no working.
• Saying a plant cell bursts in water, or that the cell wall shrinks in plasmolysis.
• Writing “energy” for active transport without “released during respiration”.
• Listing vague controlled variables (“same conditions”, “same amount”).

Next steps

• Revision notes for this topic
• Study guide for this topic
• Cell structure and organisation practice
• Cambridge O Level Biology course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4), Topic 3: Movement into and out of cells, published by Cambridge Assessment International Education.