Edexcel International GCSE Biology 4BI1: Movement of substances into and out of cells – 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 sub-topic 2(d), Movement of substances into and out of cells (points 2.15–2.17), of the Pearson Edexcel International GCSE Biology (4BI1) specification, Issue 3. The qualification is untiered and no point here has a B reference, so every question suits Paper 1 and Paper 2. A calculator may be used on both papers. They apply to the June and November series examined under Issue 3.

Learn the content first with the study guide. For cell structure and enzymes, use the cell structure practice questions. Course hub: Edexcel IGCSE Biology. Checklist: printable 4BI1 checklist.

Questions

1. State what is meant by the term diffusion. [2]

2. Give two differences between osmosis and active transport. [2]

3. Root hair cells take up nitrate ions from soil water. The concentration of nitrate ions in the soil water is lower than in the root hair cells.

(a) Explain why the uptake of nitrate ions by these cells must be by active transport. [2] (b) A field becomes waterlogged, so the soil contains very little oxygen. Explain how this affects nitrate uptake by the root hair cells. [2]

4. A student cuts two blocks from agar containing an indicator. Block A is a cuboid 1 cm × 2 cm × 4 cm. Block B is a cube of side 8 cm.

(a) Calculate the surface area to volume ratio of block A. [2] (b) Calculate the surface area to volume ratio of block B. [2] (c) Both blocks are placed in acid at the same time. The indicator changes colour when acid reaches it. Explain which block will change colour completely first. [2]

5. A student places 1 cm agar cubes containing an indicator in acid at different temperatures and records the time taken for each cube to change colour completely.

Temperature / °C	Time / s
20	420
30	330
40	270

(a) The student calculates rate as 1000 ÷ time. Calculate the rate at 30 °C and at 40 °C, to 3 significant figures. [2] (b) Describe and explain the effect of temperature on the rate of diffusion shown by these results. [3] (c) Give two variables the student should control. [2]

6. A length of Visking tubing is filled with concentrated sucrose solution, tied at both ends and placed in a beaker of distilled water. Its mass is 12.0 g at the start and 13.8 g after 30 minutes.

(a) Calculate the percentage change in mass. [2] (b) Explain why the mass of the tubing changed. [2] (c) Predict the change in mass if the beaker had contained the same sucrose solution as the tubing. [1]

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

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

8. A student investigates osmosis in carrot tissue. Carrot cylinders are blotted, weighed, left in sucrose solutions for one hour, then blotted and reweighed.

Sucrose / mol/dm³	Start mass / g	End mass / g	% change
0.0	3.20	3.52	+10.0
0.1	3.00	3.18	+6.0
0.2	3.10	3.19
0.3	2.90	2.84
0.4	3.00	2.82	−6.0
0.5	3.20	2.88	−10.0

(a) Calculate the percentage change in mass at 0.2 mol/dm³ and at 0.3 mol/dm³. Give each to 1 decimal place. [2] (b) Estimate the concentration of sucrose solution that would cause no change in mass. Explain how you arrived at your answer. [2] (c) Explain the result at 0.5 mol/dm³. [3] (d) Suggest why the student calculated percentage change rather than change in mass. [1] (e) Suggest one way to make the estimate in (b) more precise. [1]

9. After a meal, glucose is absorbed from the small intestine into the cells lining it. Hours later, gut glucose concentration is lower than in these cells, but glucose is still absorbed. Explain how glucose is absorbed at each of these times. [3]

10. A single-celled organism lives in pond water and takes in oxygen across its cell membrane by diffusion. Explain how each of the following would affect the rate at which oxygen enters the cell:

• a rise in the temperature of the water
• the oxygen concentration of the water falling
• the cell growing much larger while keeping the same shape. [6]

11. Pieces of red onion epidermis are placed in sucrose solutions for 20 minutes. A student counts how many of 50 cells in each piece are plasmolysed.

Sucrose / mol/dm³	Plasmolysed cells (out of 50)
0.2	0
0.3	4
0.4	23
0.5	41
0.6	50

(a) Calculate the percentage of cells plasmolysed at 0.4 mol/dm³ and at 0.5 mol/dm³. [2] (b) Estimate the sucrose concentration at which 50% of the cells would be plasmolysed. Show your working. [2] (c) Explain why all the cells are plasmolysed at 0.6 mol/dm³. [3]

Answers

1. The net movement of particles [1] from a region of higher concentration to a region of lower concentration / down a concentration gradient [1]. [2] Examiner insight: “Net” and “concentration” are both expected; “particles move from high to low” without the word concentration does not get the second mark.

2. Osmosis moves only water, active transport moves ions or other solutes [1]; osmosis uses no energy, active transport uses energy from respiration [1]. Also credit: osmosis is down a gradient, active transport against it; active transport needs carrier proteins. [2] Examiner insight: Each difference must give both sides; “active transport needs energy” alone is not a comparison.

3. (a) The ions move from a lower concentration (soil water) to a higher concentration (cell) / against the concentration gradient [1], so diffusion cannot do this and energy is needed [1]. (b) Less oxygen means less aerobic respiration in the root cells [1], so less energy is released for active transport and nitrate uptake falls [1]. Examiner insight: In (b) the chain oxygen → respiration → energy must be complete; “the roots cannot breathe” earns nothing.

4. (a) Surface area = 2(1×2 + 2×4 + 1×4) = 28 cm² and volume = 8 cm³ [1]; ratio = 3.5 : 1 [1]. (b) Surface area = 6 × 8² = 384 cm² and volume = 8³ = 512 cm³ [1]; ratio = 0.75 : 1 [1]. (c) Block A [1], because it has a larger surface area to volume ratio and its centre is a shorter distance from the surface (0.5 cm compared with 4 cm), so acid reaches all of it sooner [1]. Examiner insight: A ratio written upside down loses the accuracy mark, but correct surface area and volume still earn the method mark.

5. (a) 1000 ÷ 330 = 3.03 [1]; 1000 ÷ 270 = 3.70 [1]. (b) As temperature increases, the rate of diffusion increases / time taken decreases [1]; particles (acid) have more kinetic energy [1] so they move faster and reach the centre of the cube sooner [1]. (c) Any two from: size (volume) of cube [1]; concentration of acid [1]; volume of acid; concentration of indicator in the agar. [2] Examiner insight: In (a), 3.7 does not meet the 3 significant figure instruction; write 3.70. In (c), “same amount of acid” is too vague.

6. (a) (13.8 − 12.0) ÷ 12.0 × 100 [1] = +15% [1]. (b) Water moved into the tubing by osmosis, from the dilute solution (water) to the concentrated sucrose solution [1], through the partially permeable tubing; sucrose molecules are too large to pass out [1]. (c) No (net) change in mass [1]. Examiner insight: Adding “sucrose moved out” to (b) contradicts the mass gain and loses the mark.

7. (a) Water enters the cells by osmosis [1] because the water outside is more dilute than the cell contents [1]; the cells swell and burst because they have no cell wall [1]. (b) The cell wall is strong / rigid [1] and resists further expansion as the vacuole fills, so the cell becomes turgid instead of bursting [1]. Examiner insight: “The cell wall stops it bursting” restates the question; the second mark needs the wall resisting expansion or the cell becoming turgid.

8. (a) (3.19 − 3.10) ÷ 3.10 × 100 = +2.9% [1]; (2.84 − 2.90) ÷ 2.90 × 100 = −2.1% [1]. (b) About 0.26 mol/dm³ (accept 0.25–0.27) [1], because the percentage change goes from positive at 0.2 to negative at 0.3, so it crosses zero between them [1]. (c) The solution is more concentrated than the carrot cell contents [1]; water leaves the cells by osmosis through the partially permeable membrane [1]; so the cylinder loses mass / cells become flaccid [1]. (d) The cylinders had different starting masses, so percentage change allows a fair comparison [1]. (e) Use more concentrations between 0.2 and 0.3 mol/dm³ / repeat each concentration and use a mean [1]. Examiner insight: In (a) the sign is part of the answer; a percentage change of “2.1%” for a mass loss does not earn the mark.

9. Soon after the meal, glucose is absorbed by diffusion, down the concentration gradient from the gut into the cells [1]. Later, when the gut concentration is lower, glucose is absorbed by active transport, against the concentration gradient [1], using energy from respiration [1]. [3] Examiner insight: Naming the two processes without linking each to the direction of the gradient earns only one mark.

10. Temperature: rate increases [1] because oxygen molecules have more kinetic energy and move faster [1]. Lower oxygen concentration in the water: rate decreases [1] because the concentration gradient between the water and the cell is less steep [1]. Larger cell: oxygen supply per unit volume decreases / rate is not enough for the cell’s needs [1] because the surface area to volume ratio falls and the distance to the centre of the cell increases [1]. [6] Examiner insight: Each effect needs its own reason; three effects with no reasons score three at most.

11. (a) 23 ÷ 50 × 100 = 46% [1]; 41 ÷ 50 × 100 = 82% [1]. (b) 50% lies between 46% (0.4) and 82% (0.5): 0.4 + 0.1 × (50 − 46) ÷ (82 − 46) [1] = 0.41 mol/dm³ (accept 0.41–0.42 from a graph) [1]. (c) The solution is more concentrated than the cell sap in every cell [1]; water leaves the cells (vacuoles) by osmosis [1]; the cytoplasm and cell membrane pull away from the cell wall [1]. Examiner insight: In (c) “the cell shrinks” is not enough; plasmolysis needs the membrane pulling away from the wall.

Where marks are usually lost

• Omitting net or concentration from a definition of diffusion.
• Reversing the direction of osmosis (it is dilute solution to concentrated solution).
• Writing that active transport needs energy without linking it to respiration, or without saying it is against the gradient.
• Dropping the sign on a percentage change in mass.
• Giving surface area to volume ratio upside down or without “: 1”.
• Describing a trend without explaining it (no kinetic energy, no gradient).
• Describing plasmolysis as “the cell shrinks” instead of the membrane pulling away from the wall.

Next steps

• Revision notes for this sub-topic
• Study guide with full explanations
• Cell structure revision notes
• Edexcel IGCSE Biology course hub
• Printable 4BI1 checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Biology (4BI1) specification, Issue 3, published by Pearson Education Limited: Topic 2, Structures and functions in living organisms, sub-topic (d) Movement of substances into and out of cells, points 2.15–2.17.