OCR GCSE Biology A: Transport in plants: water uptake, transpiration and translocation (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 plant transport in the OCR GCSE (9-1) Biology A specification (J247), Version 4.0: statements B2.2f to B2.2j, none of which is Higher Tier only. Use the study guide or the revision notes first. Course hub: OCR GCSE Biology. Printable checklist. Find gaps with the free diagnostics.

Questions

1. Describe two ways in which a root hair cell is adapted to take in water and mineral ions. For each, say how it helps. [4]

2. The concentration of nitrate ions is 0.3 mmol per dm³ in soil water and 4.5 mmol per dm³ inside a root hair cell. The cell still takes up nitrate.

(a) Name the process by which nitrate enters the cell. [1] (b) Explain why diffusion could not move the nitrate in. [2] (c) After heavy rain the soil is waterlogged and contains very little oxygen. Explain what happens to nitrate uptake. [2]

3. A gardener puts far too much fertiliser on a potted plant. Within a day the plant wilts. Use osmosis to explain why. [4]

4. The xylem and phloem are both transport tissues.

(a) Copy and complete the table.

	Xylem	Phloem
Substances carried
Direction of movement
Cells living or dead

[3] (b) Explain how lignin in the walls of xylem vessels helps the plant. [2] (c) State the job of a companion cell. [1]

5. Describe how water moves from the xylem in a leaf to the air outside the plant. [5]

6. A leaf impression was made from each surface of a leaf. Stomata were counted in five fields of view, each 0.40 mm in diameter.

Surface	Counts in five fields
Upper	2, 3, 1, 2, 2
Lower	11, 9, 13, 12, 10

(a) Calculate the area of one field of view, to 3 significant figures. [2] (b) Calculate the stomatal density of the lower surface in stomata per mm², to the nearest whole number. [2] (c) Suggest why the upper surface has fewer stomata. [2]

7. A ring of bark containing the phloem was cut from one branch of a tree, leaving the xylem. After some weeks the sugar concentration in bark tissue was measured just above and just below the ring, and at the same positions on an unringed branch.

Position	Sugar in bark (mg per g)
Ringed branch, above ring	42
Ringed branch, below ring	6
Unringed branch, above	24
Unringed branch, below	25

(a) Calculate how many times greater the sugar concentration is above the ring than below it, and state what the unringed branch shows. [2] (b) Explain the results for the ringed branch. [2] (c) The leaves above the ring did not wilt. Explain why. [2]

8. (a) Explain why a rise in temperature increases the rate of transpiration. [2] (b) Explain why a fan blowing on a plant increases the rate of transpiration. [2] (c) Name one other environmental factor, not given in (a) or (b), that changes the rate of transpiration. [1]

9. A potometer has a capillary tube of internal radius 0.50 mm. The distance moved by the bubble in 5.0 minutes was measured three times at each light level.

Light level	Distance moved (mm)
Bright	40, 42, 41
Dim	20, 22, 21

(a) Calculate the mean rate of bubble movement in bright light, in mm per minute. [2] (b) Use the radius to find the volume of water absorbed each minute in bright light, in mm³ per minute, to 2 significant figures. [2] (c) The mean rate in dim light is 4.2 mm per minute. Calculate the percentage decrease in rate from bright to dim light, to 2 significant figures. [2]

10. Describe how you would use a potometer to investigate the effect of air movement on the rate of water uptake by a leafy shoot. [6]

11. Two similar leafy shoots, A and B, were each placed in a tube of water with a layer of oil on top. The lower surfaces of the leaves of shoot B were covered in petroleum jelly. Each tube was weighed at the start and after 4.0 hours in the same conditions.

Shoot	Start mass (g)	Final mass (g)
A	150.0	146.4
B	148.0	147.2

(a) Calculate the percentage loss in mass of A. [2] (b) Calculate the percentage loss in mass of B, to 2 significant figures. [2] (c) What do the results suggest about where most water is lost from a leaf? Explain. [2]

12. A potometer was used outdoors on one day, giving these rates of bubble movement.

Time	06:00	09:00	12:00	15:00	18:00	21:00
Rate (mm per minute)	0.8	3.2	5.6	6.4	2.4	0.4

(a) At what time was the rate highest? Calculate the percentage increase in rate from 09:00 to that time. [2] (b) Calculate the mean rate of change of the rate between 09:00 and 12:00, in mm per minute per hour. [2] (c) Explain the rise in rate from 06:00 to 15:00. [2] (d) Explain the fall in rate from 15:00 to 21:00. [2]

Answers

1. Long thin extension, giving a large surface area for absorbing water and ions [1] [1]. Many mitochondria, releasing energy for active transport of mineral ions [1] [1]. (Other valid pairs: a thin wall giving a short diffusion distance, or a large vacuole keeping the water concentration low.) [4] Examiner insight: A feature earns little without its function, so add “which means”.

2. (a) Active transport [1]. [1] (b) Diffusion moves particles from high to low concentration [1]. The nitrate concentration is lower in the soil than in the cell, so diffusion would move it out, not in [1]. [2] (c) Uptake falls [1], because with little oxygen there is less respiration, so less energy for active transport [1]. [2] Examiner insight: Quote both concentrations in a gradient argument. Saying only “against the gradient” is a weaker answer than showing which way diffusion would go.

3. The soil water becomes a concentrated solution, so its water concentration falls below that inside the root hair cells [1]. Water moves by osmosis [1] out of the root cells, through the partially permeable membrane, down the concentration gradient [1]. The cells lose water and become flaccid, so the plant wilts [1]. [4] Examiner insight: Say which has the higher water concentration, soil or cell. “Water moves to where there is more salt” gives no gradient.

4. (a) Xylem: water and mineral ions; up only; dead. Phloem: dissolved sugars; up and down; living. One mark for each correct row [1] [1] [1]. [3] (b) Lignin makes the walls strong and waterproof [1], so the vessels do not collapse inward when water is pulled up, and they give support [1]. [2] (c) It supplies energy (it has many mitochondria) to load sugars into the sieve tube [1]. [1] Examiner insight: Write the exact substance, such as dissolved sugars. “Food” is vague.

5. Water evaporates from the surface of mesophyll cells [1] into the air spaces in the leaf [1]. The water vapour diffuses out through open stomata [1] down a concentration gradient into the drier air [1]. More water is pulled up the xylem to replace it [1]. [5] Examiner insight: Use “evaporates” for the liquid-to-vapour step and “diffuses” for the movement out.

6. (a) Radius = 0.20 mm [1]. Area = π × 0.20² = 0.126 mm² [1]. [2] (b) Mean = (11 + 9 + 13 + 12 + 10) ÷ 5 = 11 [1]. Density = 11 ÷ 0.1257 = 88 per mm² [1]. [2] (c) The upper surface is more exposed to sun and warm air [1], so fewer stomata means less water lost [1]. [2] Examiner insight: Keep the unrounded area until the end. Rounding it to 0.126 first gives 87, so you can lose accuracy from early rounding.

7. (a) 42 ÷ 6 = 7 times [1]. The unringed branch shows about the same concentration above and below, so the ring causes the difference [1]. [2] (b) Sugar made in the leaves travels down in the phloem [1]. The ring removed the phloem, so the sugar builds up above the cut and little reaches the tissue below [1]. [2] (c) The xylem was left intact [1], so water and mineral ions still moved up to the leaves [1]. [2] Examiner insight: Use the unringed branch as the comparison.

8. (a) Water evaporates faster from the mesophyll cells [1] and vapour diffuses out faster, so the rate of transpiration increases [1]. [2] (b) Draughts remove vapour from beside the stomata [1], so the concentration gradient stays steep and diffusion out is quicker [1]. [2] (c) Light intensity (or humidity) [1]. [1] Examiner insight: A reason must say what happens at the leaf. “Because it is hotter” restates the question.

9. (a) Mean distance = (40 + 42 + 41) ÷ 3 = 41 mm [1]; 41 ÷ 5.0 = 8.2 mm per minute [1]. [2] (b) Area = π × 0.50² = 0.785 mm² [1]; 8.2 × 0.785 = 6.4 mm³ per minute [1]. [2] (c) (8.2 − 4.2) ÷ 8.2 × 100 [1] = 49% [1]. [2] Examiner insight: Percentage change uses the starting value, here the bright light rate.

10. Cut the shoot while submerged and connect it to the potometer underwater, with airtight joints [1]. Introduce an air bubble and record its start position [1]. Time the bubble over a fixed interval, such as 5 minutes, and record its distance [1]. Change one variable, such as the distance of a fan from the shoot [1], and keep temperature, light and the shoot the same [1]. Repeat each setting, calculate the mean rate (distance ÷ time) [1]. [6] Examiner insight: A plan needs the independent variable (fan distance), the measurement and the controls. “Keep it fair” names no control.

11. (a) Loss = 150.0 − 146.4 = 3.6 g [1]; 3.6 ÷ 150.0 × 100 = 2.4% [1]. [2] (b) Loss = 148.0 − 147.2 = 0.8 g [1]; 0.8 ÷ 148.0 × 100 = 0.54% [1]. [2] (c) Most water is lost from the lower surface [1], because blocking it cut the loss from 2.4% to 0.54%, and the lower surface has most of the stomata [1]. [2] Examiner insight: Back the conclusion with numbers taken from the table.

12. (a) 15:00 [1]; (6.4 − 3.2) ÷ 3.2 × 100 = 100% [1]. [2] (b) (5.6 − 3.2) ÷ 3 [1] = 0.8 mm per minute per hour [1]. [2] (c) Light intensity and temperature increase through the morning [1], so the stomata open wider and evaporation and diffusion speed up [1]. [2] (d) Light fades and temperature drops [1], so stomata close and the gradient becomes smaller, slowing water loss [1]. [2] Examiner insight: A rate of change needs the time interval and units with “per hour”.

Where marks are usually lost

• Writing osmosis for the uptake of mineral ions.
• Describing a feature without its function.
• Saying plants lose water through the xylem instead of through the stomata.
• Mixing up xylem and phloem direction and cell type.
• Calculating percentage change from the final value.
• Forgetting the cross-sectional area in a volume rate.
• Leaving out units, or omitting “per hour” in a rate of change.
• Naming no control variables in a potometer method.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Course hub: OCR GCSE Biology.
• Tick off topics on the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Biology A (Gateway Science) J247 Specification, Version 4.0 (August 2026), first assessment 2018, statements B2.2f to B2.2j.