Cambridge International AS & A Level Biology 9700: Cell membranes and transport (by subtopic) – 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 work through topic 4 of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, outcome by outcome: section 4.1 (Fluid mosaic membranes, outcomes 4.1.1 to 4.1.4) and section 4.2 (Movement into and out of cells, outcomes 4.2.1 to 4.2.6). This is AS Level content, examined on Paper 1 and Paper 2 and assumed on Paper 4. Questions 7, 9 and 10 are based on the 4.2 investigations, so they also practise Paper 3 and Paper 5 skills. Calculators are allowed.

These questions differ from the cell membranes practice questions. For teaching, use the study guide; for recall, the revision notes. Also see the Cambridge A Level Biology hub, the printable 9700 checklist and the 9700 AS Level diagnostic.

Questions

1. (4.1.1) Describe the fluid mosaic model. Explain how hydrophobic and hydrophilic interactions account for the bilayer and for the position of a protein that spans the membrane. [4]

2. (4.1.2, 4.1.3)

(a) Describe the arrangement of cholesterol, glycolipids and glycoproteins in a cell surface membrane. [3] (b) Explain how cholesterol affects membrane fluidity and permeability. [2] (c) Explain the roles of glycolipids and glycoproteins in membrane stability and cell recognition. [2]

3. (4.1.3) Compare channel proteins and carrier proteins. [3]

4. (4.1.4) Gland cells release a protein hormone. It reaches almost every cell in the body, but only kidney cells respond. Outline the stages of cell signalling in this example and explain why only kidney cells respond. [4]

5. (4.2.1) Name the process by which each substance crosses the cell surface membrane, with a reason.

(a) Carbon dioxide leaving a respiring muscle cell. [1] (b) Glucose entering a red blood cell through a transmembrane protein, down its concentration gradient, without ATP. [1] (c) Water entering a root hair cell from dilute soil water. [1] (d) Sodium ions leaving a nerve cell against their concentration gradient. [1] (e) Mucus, a glycoprotein, leaving a goblet cell. [1]

6. (4.2.1) Describe how a phagocyte takes in a bacterium, and state why the process needs energy. [3]

7. (4.2.2, 4.2.3) Visking tubing was filled with 1.0 mol dm⁻³ sucrose solution, fitted with a glass capillary tube and placed in distilled water. The height of liquid in the capillary above its starting level was recorded.

Time / min	0	5	10	15	20
Height / mm	0	16	30	42	52

(a) Calculate the mean rate of rise from 0 to 10 minutes and from 10 to 20 minutes. [2] (b) The capillary bore is a cylinder of internal radius 0.5 mm. Calculate the volume of water that entered in 20 minutes, to 3 significant figures. [2] (c) Explain why the liquid rose. [2] (d) Suggest why the rate of rise fell. [1]

8. (4.2.3)

(a) Calculate the surface area to volume ratio (SA:V) of a cube-shaped cell of side 4 µm and of one of side 8 µm. [2] (b) A flattened cell is a cuboid 16 µm × 4 µm × 1 µm, the same volume as the 4 µm cube. Calculate its SA:V. [2] (c) State the principle shown by your answers to (a). [1]

9. (4.2.4) Pink agar made with dilute sodium hydroxide and phenolphthalein was cut into cubes and placed in dilute hydrochloric acid. The time for each cube to become colourless was recorded.

Side of cube / cm	0.5	1.0	1.5	2.0
Time / min	3.0	12.5	27.0	48.5

(a) Calculate the SA:V of the 1.5 cm cube. Show your working. [2] (b) Describe and explain the relationship between SA:V and time. [3] (c) State two variables, other than cube size, to keep constant. [2] (d) Suggest one way to make the timing more reliable. [1]

10. (4.2.5, 4.2.6) Beetroot cylinders were blotted, weighed, left in solutions of known water potential for 2 hours at 25 °C, then blotted and reweighed.

Solution water potential / kPa	0	−250	−500	−750	−1000	−1250
Initial mass / g	2.40	2.36	2.44	2.38	2.42	2.40
Final mass / g	2.62	2.50	2.51	2.36	2.33	2.26
Change / %	+9.2	+5.9			−3.7	−5.8

(a) Calculate the two missing percentage changes, to 1 decimal place. [2] (b) Use them to estimate the water potential of the tissue. Show your working. [2] (c) Explain why the cylinders in −1250 kPa lost mass. [2] (d) Explain why percentage change is used. [1] (e) Suggest how to make the estimate more precise. [1]

11. (4.2.6) Red blood cells and onion epidermal cells were placed in distilled water and, separately, in concentrated sucrose solution. Describe and explain what happens to each type of cell in each liquid. [6]

Answers

1. A phospholipid bilayer in which molecules move sideways (fluid), with proteins scattered through it (mosaic) [1]. Hydrophilic phosphate heads interact with water, so they face the cytoplasm and tissue fluid [1]. Hydrophobic fatty acid tails are repelled by water, so they point inwards, forming a hydrophobic core that ions and polar molecules cannot easily cross [1]. A spanning protein has hydrophobic R groups where it meets the tails and hydrophilic R groups where it meets water [1]. Examiner insight: Each property needs its consequence; “heads are hydrophilic” without where they face does not score.

2. (a) Cholesterol lies between phospholipids, in both layers [1]. Glycolipids are in the outer layer, carbohydrate chain outside the cell [1]. Glycoproteins carry carbohydrate chains on the outer surface only [1]. (b) It regulates fluidity: less fluid when warm, less rigid when cold [1]. It reduces permeability to ions and polar molecules [1]. (c) Their carbohydrate chains form hydrogen bonds with water outside the cell, stabilising the membrane [1]. The chains act as cell surface antigens, recognised by other cells such as lymphocytes [1]. Examiner insight: “Cholesterol makes the membrane more fluid” with no condition is marked wrong.

3. A channel is a water-filled pore; a carrier binds the substance and changes shape [1]. Channels allow only facilitated diffusion; carriers work in facilitated diffusion and active transport [1]. Both span the membrane and are specific [1]. Examiner insight: “Compare” needs a similarity as well as differences, or the answer cannot gain full marks.

4. Gland cells secrete the hormone, a specific ligand [1]. It is transported in the blood [1]. It binds to cell surface receptors on target cells [1]. Only kidney cells have receptors with a binding site complementary in shape to the hormone, so only they respond [1]. Examiner insight: Missing the transport stage loses a mark even when secretion and binding are well described.

5. (a) Simple diffusion: small, non-polar, passes through the bilayer down its gradient [1]. (b) Facilitated diffusion: down a gradient through a protein, no ATP [1]. (c) Osmosis: from higher to lower water potential across a partially permeable membrane [1]. (d) Active transport: against the gradient, needs carrier proteins and ATP [1]. (e) Exocytosis: too large to cross, so vesicles fuse with the membrane and release it [1]. Examiner insight: Each mark needs a correct name and reason; “osmosis because water is at high concentration” scores zero.

6. The cell surface membrane extends around the bacterium [1]. It fuses and pinches off a vesicle (phagocytic vacuole) containing it: endocytosis [1]. ATP is needed to move the membrane [1]. Examiner insight: Naming endocytosis alone does not earn the description marks.

7. (a) 30 ÷ 10 = 3.0 mm min⁻¹ [1]; (52 − 30) ÷ 10 = 2.2 mm min⁻¹ [1]. (b) V = πr²h = π × 0.5² × 52 [1] = 40.8 mm³ [1]. (c) The sucrose solution has a lower water potential than the water [1]. Water enters by osmosis through the partially permeable tubing; sucrose crosses far more slowly, if at all [1]. (d) Entering water dilutes the sucrose, so the water potential gradient falls [1]. Examiner insight: Using the diameter in πr² gives four times the volume; with no method shown, both marks are lost.

8. (a) 4 µm: 96 ÷ 64 = 1.5 : 1 [1]. 8 µm: 384 ÷ 512 = 0.75 : 1 [1]. (b) SA = 2(64 + 16 + 4) = 168 µm² [1]; 168 ÷ 64 = 2.63 : 1 [1]. (c) As size increases, SA:V decreases [1]. Examiner insight: Write ratios as “n : 1”; a bare “1.5” may not be credited.

9. (a) SA = 6 × 1.5² = 13.5 cm²; V = 1.5³ = 3.375 cm³ [1]; 4 : 1 [1]. (b) Larger SA:V, shorter time [1]. More surface for acid to diffuse in per unit volume [1], and a shorter distance to the centre [1]. (c) Any two: acid concentration [1]; temperature [1]; acid volume; agar batch. (d) Judge the end point against a white tile, or repeat and find a mean [1]. Examiner insight: “Amount of acid” is too vague; name volume or concentration.

10. (a) (2.51 − 2.44) ÷ 2.44 × 100 = +2.9 % [1]; (2.36 − 2.38) ÷ 2.38 × 100 = −0.8 % [1]. (b) Zero change is 2.9 ÷ (2.9 + 0.8) = 0.78 of the way from −500 to −750 kPa [1]; −500 − 0.78 × 250 = about −700 kPa [1]. (c) The solution’s water potential is lower than the tissue’s [1], so water leaves the cells by osmosis through partially permeable membranes [1]. (d) Starting masses differ, so it allows fair comparison [1]. (e) More solutions at smaller intervals between −500 and −750 kPa [1]. Examiner insight: The estimate must lie between −500 and −750 kPa, where the sign of the change flips; a value outside that range gets no accuracy mark.

11. In distilled water, water enters both cells by osmosis from the higher water potential outside [1]. The red blood cell swells and bursts, having no cell wall [1]. The onion cell becomes turgid [1]; the wall resists expansion, so it does not burst [1]. In sucrose, water leaves both by osmosis; the red blood cell shrinks and crinkles [1]. The onion protoplast shrinks and pulls away from the wall: plasmolysis [1]. Examiner insight: Solute and pressure potential are not expected; explain with water potential, and never say the plant cell bursts.

Where marks are usually lost

• Not saying which way phospholipid heads face.
• Putting carbohydrate chains inside the membrane.
• Saying channel proteins do active transport.
• Skipping the transport stage of cell signalling.
• Naming a process without a reason.
• Using the diameter in πr².
• Writing SA:V without “: 1”.
• Using mass change instead of percentage change.
• Saying plant cells “burst” or “shrink” instead of “turgid” and “plasmolysed”.

Next steps

• Revision notes
• Study guide
• Cell membranes practice questions
• Cambridge A Level Biology hub
• Printable 9700 checklist
• 9700 AS Level diagnostic, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International, Cambridge International AS & A Level Biology 9700 syllabus for 2025, 2026 and 2027 (Version 1, published September 2022), Cambridge University Press & Assessment. Topic 4, Cell membranes and transport: sections 4.1 and 4.2.