OCR GCSE Biology A: Transport across membranes, mitosis and stem cells (J247) – 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 statements B2.1a to B2.1f of the OCR GCSE (9-1) Biology A Gateway Science specification (J247), Version 4.0, for first assessment in 2018. No statement in this unit is Higher Tier only. Revise first with the study guide and revision notes. Course links: OCR GCSE Biology hub and printable checklist.

Questions

1. Name the process by which each of the following occurs. (a) Oxygen passes from the lungs’ air sacs into the surrounding capillaries. [1] (b) Water enters a root hair cell from the soil. [1] (c) Mineral ions enter a root hair cell although their concentration is higher inside the cell than in the soil. [1]

2. A strip of courgette has a mass of 4.00 g. After 90 minutes in a salt solution its mass is 3.52 g. (a) Work out the percentage change in the strip’s mass. [2] (b) Explain the change using the term water potential. [3]

3. Apple cylinders were left in sucrose solutions of different concentrations.

Sucrose (mol/dm³)	Initial mass (g)	Final mass (g)
0.1	2.00	2.16
0.3	2.50	2.45
0.5	2.20	1.98

(a) Calculate the percentage change in mass at each concentration. [3] (b) Use the results to estimate the sucrose concentration where the apple would neither gain nor lose mass. [2] (c) Explain what this concentration tells you about the apple cells. [1]

4. The uptake of ions by a root is 6.0 units per hour. When a chemical that stops respiration is added, uptake falls to 1.5 units per hour. (a) Calculate the percentage change in uptake. [2] (b) Explain why the chemical reduced uptake. [2]

5. (a) Describe what happens to the chromosomes when DNA replicates in the cell cycle. [2] (b) Describe what happens to the chromosomes during mitosis. [2]

6. A body cell of a fruit fly has 8 chromosomes. (a) State the number of chromosomes in each daughter cell after mitosis. [1] (b) State the number of chromatids in the cell just after DNA replication. [1] (c) In a root tip, a student counts 180 cells, of which 27 are in mitosis. Calculate the percentage of cells in mitosis. [2] (d) Explain why the daughter cells are genetically identical. [2]

7. Plan an investigation using potato chips and sucrose solutions. Include the variable you change, the variable you measure, two variables you control, and how you would compare results when the starting masses differ. Say how you would find the concentration giving no net change. [6]

8. (a) Describe how a sperm cell is adapted to its function. [2] (b) Describe how a root hair cell is adapted to its function. [2] (c) State one reason why cell differentiation is important in a multicellular organism. [1]

9. (a) State where stem cells are found in (i) a growing plant and (ii) an adult animal. [2] (b) Describe one difference between embryonic and adult stem cells. [2]

10. A gardener grows a new plant from a tiny piece of shoot tip. Explain how the piece of tissue can grow into a whole plant. [4]

11. A child’s mass rises from 18.0 kg to 20.0 kg over a year. (a) Calculate the percentage gain in mass, to 3 significant figures. [2] (b) In a group of 40 children of the same age, 30 are lighter than this child. State the child’s percentile. [2] (c) Explain how mitosis and stem cells contribute to the growth of the child. [3]

Answers

1. (a) Diffusion [1] (b) Osmosis [1] (c) Active transport [1] Examiner insight: One precise word is enough for each part, but writing “diffusion” for part (b) usually gains nothing, because the question depends on telling the three processes apart.

2. (a) Change = 3.52 − 4.00 = −0.48 g [1]; −0.48 ÷ 4.00 × 100 = −12.0% [1] (b) Outside the cells the water potential is lower than inside [1]. Water leaves the courgette cells by osmosis [1], crossing the membrane down the water potential gradient, which lowers the mass [1]. Examiner insight: Showing the division by the initial mass is how a method is visible; a bare −12 can be read as correct, but a wrong answer with no working leaves nothing to credit.

3. (a) 0.1: (2.16 − 2.00) ÷ 2.00 × 100 = +8.0% [1]; 0.3: (2.45 − 2.50) ÷ 2.50 × 100 = −2.0% [1]; 0.5: (1.98 − 2.20) ÷ 2.20 × 100 = −10.0% [1] (b) The change crosses zero between 0.1 and 0.3 [1]. The fall from +8.0 to −2.0 is 10.0 points, of which 8.0 occur before zero: 0.1 + 0.2 × (8.0 ÷ 10.0) = 0.26 mol/dm³ (a plotted line gives about the same) [1]. (c) At that concentration the solution and the cell contents have the same water potential, so there is no net movement of water [1]. Examiner insight: Raw mass changes cannot be compared when starting masses differ, so percentage change is the quantity to use; a graph read-off in the right region usually earns the estimate even if your decimals differ slightly.

4. (a) (1.5 − 6.0) ÷ 6.0 × 100 [1] = −75.0% [1] (b) Active transport needs energy [1]; the chemical stopped respiration, so less energy was released and less uptake occurred [1]. Examiner insight: Saying that “energy” is needed is weaker than saying the energy comes from respiration, which links the data to the process.

5. (a) Each chromosome is copied [1], forming two identical chromatids joined together [1]. (b) The chromosomes line up across the centre of the cell [1]; spindle fibres pull the chromatids to opposite ends [1]. Examiner insight: The words chromosome and chromatid are not interchangeable; answers that say “the chromosomes split in half” are imprecise, so name the chromatids.

6. (a) 8 [1] (b) 16 chromatids (8 chromosomes, each of two chromatids) [1] (c) 27 ÷ 180 [1] × 100 = 15.0% [1] (d) DNA is copied exactly during replication [1], and one chromatid of each chromosome goes to each daughter cell, so each receives the same genetic information [1]. Examiner insight: “Identical” needs a reason; stating that replication copies the DNA and that the copies are shared out evenly is the explanation, not the assertion.

7. Change: sucrose concentration (several values, including pure water) [1]. Measure: mass of the chip, before and after [1]. Controls, two of: chip size/surface, same plant tissue, volume of solution, time, temperature [1] [1]. Compare using percentage change, (final − initial) ÷ initial × 100 [1]. Plot percentage change against concentration; where the line crosses zero is the concentration with no net change [1]. Examiner insight: Naming controls without saying why they matter is usually enough here, but failing to blot the chips before weighing is a common source of error that is worth stating in a method.

8. (a) Tail, to swim [1]; many mitochondria, to release energy for movement (or enzymes in the head to digest the egg’s outer layer) [1]. (b) Long hair-like projection [1], which increases the surface for taking in water and mineral ions [1]. (c) Specialised cells carry out one job efficiently, so the organism as a whole works more efficiently [1]. Examiner insight: An adaptation only earns credit when linked to the function; listing “tail” without the swimming purpose is a frequent incomplete answer.

9. (a) (i) Meristems (root and shoot tips) [1]; (ii) bone marrow, for example [1]. (b) Embryonic stem cells can become any cell type [1], whereas adult stem cells can become only a limited range of cell types [1]. Examiner insight: A difference needs both sides stated; writing only that embryonic cells are “more flexible” can be too vague.

10. The shoot tip contains meristem (stem) cells [1] that divide by mitosis [1]. Daughter cells differentiate into the different tissues [1], so roots and shoots can all form, giving a whole plant [1]. Examiner insight: Linking the idea of meristem, division and differentiation in order is a clear structure; mentioning only that the plant “has stem cells” is usually too brief for four marks.

11. (a) Gain = 2.0 kg [1]; 2.0 ÷ 18.0 × 100 = 11.1% [1] (b) 30 ÷ 40 × 100 [1] = 75th percentile [1] (c) Mitosis produces genetically identical new cells so the body grows in cell number [1]; stem cells divide to supply cells [1]; they differentiate into specialised types such as bone and muscle cells that the larger body needs [1]. Examiner insight: A percentile is a position in a group, not a percentage gain, and mixing the two is a regular error; give the answer as an ordinal, such as 75th.

Where marks are usually lost

• Writing “high to low concentration” for osmosis instead of referring to water or water potential.
• Leaving out respiration as the source of energy in active transport.
• Using the final value as the divisor in a percentage change.
• Dropping the sign on a loss in mass.
• Confusing chromosomes with chromatids when counting after replication.
• Describing daughter cells as “similar” rather than genetically identical, and not explaining why.
• Listing a feature of a specialised cell without linking it to its function.
• Assuming all stem cells are embryonic and forgetting adult stem cells and meristems.

Next steps

• Review the revision notes and the study guide.
• Course hub: OCR GCSE Biology; printable checklist.
• Try all free 10-minute diagnostics.
• Want a tutor? Book a free trial class.

Official syllabus

OCR GCSE (9-1) Biology A (Gateway Science), specification J247, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Statements B2.1a to B2.1f.