Plant tissues, transport and transpiration – GCSE Biology exam questions with mark scheme

1Grade 6–76 marks

A student used a potometer to measure the rate of water uptake by a leafy shoot. The air bubble moved 36 mm along the capillary tube in 5 minutes. The radius of the capillary tube is 0.5 mm.

(a) Calculate the rate of water uptake in mm³ per minute.

Use: volume = $\pi r^2 l$

Give your answer to 2 significant figures. [4]

(b) The student then put the potometer in a warmer room. Explain why the rate of water uptake increased. [2]

(a) $\pi \times 0.5^2 \times 36$; alternative route: $36 \div 5$ (= 7.2 mm per minute) scores this mark(1)
(a) = 28.3 (mm³); allow 28.27 / 28.274 / 9π; alternative route: $\pi \times 0.5^2 \times 7.2$ scores this mark(1)
(a) 28.3 ÷ 5 = 5.65 or 5.66 (either route); allow ecf from an incorrect volume(1)
(a) 5.7 (mm³/min) to 2 s.f.; a correct answer of 5.7 with no working scores 4 marks; 5.65 or 5.66 (not rounded) scores 3 marks(1)
(b) (higher temperature so) water evaporates faster / water vapour diffuses out of the stomata faster (because particles have more kinetic energy)(1)
(b) so the rate of transpiration increases, so more water is drawn up the xylem (from the potometer) to replace it; ignore references to photosynthesis(1)
Answer: (a) 5.7 mm³/min (b) faster evaporation and diffusion of water vapour from stomata, so more water is drawn up
Examiner tip: Use the RADIUS, not the diameter, and do not round 28.27 too early.

2Grade 6–75 marks

A student counted stomata on the lower surface of a leaf using a microscope. There were 21 stomata in a circular field of view with a diameter of 0.4 mm.

(a) Calculate the number of stomata per mm².

Use: area of a circle = $\pi r^2$

Give your answer to 3 significant figures. [3]

(b) Give two differences between xylem and phloem. [2]

(a) area = $\pi \times 0.2^2$ = 0.126 mm² (uses radius 0.2); allow 0.1257 / 0.04π(1)
(a) 21 ÷ 0.126 (correct method); allow ecf from an incorrect area(1)
(a) 167 (per mm²) to 3 s.f.; correct answer with no working scores 3 marks; an unrounded answer (e.g. 166.7 or 167.1) scores 2 marks(1)
(b) first difference, any one from: xylem transports water and mineral ions, phloem transports dissolved sugars / food (translocation); xylem is hollow tubes / made of dead cells, phloem is made of living (elongated) cells; xylem walls are strengthened by lignin, phloem walls are not; xylem transports in one direction only (upwards), phloem in both directions; phloem cells have pores in their end walls (sieve plates), xylem has no end walls(1)
(b) a second, different difference from the list above; each difference must refer to both tissues or be clearly comparative(1)
Answer: (a) 167 per mm² (b) e.g. xylem carries water/minerals, phloem carries sugars; xylem dead/lignified, phloem living
Examiner tip: Halve the diameter before squaring — using 0.4 as the radius gives an answer four times too small.

3Grade 1–3Foundation6 marks

The diagram shows a cross-section through part of a leaf.

(a) Name the tissues labelled A and B. [2]

(b) Cell C is one of a pair of cells found around a small pore in the lower surface of the leaf.

(i) Name cell C. [1]

(ii) Describe the function of cell C. [1]

Name tissue D. [1]

(d) New cells for growth are made in a tissue found at the tips of roots and shoots.

Name this tissue. [1]

(a) A: palisade mesophyll
allow palisade (layer / cells)(1)
(a) B: spongy mesophyll
allow spongy (layer / cells); do not accept ‘mesophyll’ alone for A or for B(1)
(b)(i) guard cell
do not accept stoma / stomata(1)
(b)(ii) opens and closes the stoma / controls the size of the stoma
allow controls gas exchange or controls water loss (from the leaf)(1)
(c) xylem
do not accept phloem(1)
(d) meristem (tissue)
ignore stem cells(1)
Answer: (a) A palisade mesophyll, B spongy mesophyll (b)(i) guard cell (ii) opens and closes the stoma to control gas exchange and water loss (c) xylem (d) meristem
Examiner tip: ‘Stoma’ is the pore, not the cell around it — naming the gap instead of the guard cell loses the mark.

4Grade 1–3Foundation5 marks

Root hair cells absorb water and mineral ions from the soil.

(a) Complete the sentence. Choose answers from the box.

active transport diffusion osmosis respiration translocation

Root hair cells take up water by ____________ and take up mineral ions by ____________. [2]

(b) Each root hair cell has a long, thin extension that grows out into the soil.

Explain how this helps the cell to absorb water. [1]

Explain why. [2]

(a) osmosis(1)
(a) active transport
both answers must be in the correct order(1)
(b) (gives a) large(r) surface area (for absorption / osmosis)
ignore ‘bigger’ unqualified(1)
(c) for (aerobic) respiration (to release energy)
do not accept ‘to make / produce energy’(1)
(c) (energy) is needed for active transport (of mineral ions)
allow (energy) needed to move mineral ions against the concentration gradient(1)
Answer: (a) osmosis; active transport (b) increases the surface area for absorption (c) mitochondria carry out respiration, releasing energy needed for active transport of mineral ions
Examiner tip: Never write that respiration ‘makes energy’ — say it ‘releases’ (or transfers) energy, or the mark is lost.

5Grade 4–57 marks

A student investigated how different conditions affect water loss from plants.

She used four plants of the same species, each in a pot.
She covered the soil surface of each pot with a plastic bag.
She weighed each plant and pot, left it in different conditions for 5 hours, then weighed it again.

Plant	Conditions	Mass at start in g	Mass after 5 hours in g
A	light, 20 °C, still air	245.6	241.1
B	light, 30 °C, still air	238.2	230.7
C	light, 20 °C, still air, very humid air	251.0	249.8
D	dark, 20 °C, still air	243.4	242.5

(a) Suggest why the student covered the soil surface with a plastic bag. [1]

(b) Calculate the mean rate of mass loss of plant B.

Give your answer in g per hour. [2]

(d) Explain why plant D lost the least mass. [2]

(a) so that (the only) water lost is from the leaves / plant
allow to stop water evaporating from the soil; ignore ‘to make it a fair test’ unqualified(1)
(b) 238.2 − 230.7 = 7.5 (g)(1)
(b) 7.5 ÷ 5 = 1.5 (g per hour)
allow ecf from an incorrect mass loss; correct answer with no working scores 2 marks(1)
(c) (in humid air) there is more water vapour in the air (around the leaf)
allow the concentration of water vapour outside the leaf is higher(1)
(c) so the concentration gradient (of water vapour) is less steep / smaller so less water vapour diffuses out of the stomata
allow (so) less evaporation / transpiration(1)
(d) (in the dark) stomata close / are closed
allow guard cells close the stomata; do not accept ‘no photosynthesis’ alone(1)
(d) so less water vapour can diffuse / evaporate out of the leaf
allow so the rate of transpiration is lower(1)
Answer: (a) so water is only lost from the plant, not from the soil (b) 1.5 g per hour (c) humid air reduces the water vapour concentration gradient, so less diffuses out of stomata (d) stomata close in the dark, so less water vapour diffuses out
Examiner tip: In (c) and (d) the second mark needs the consequence — say less water vapour diffuses out of the stomata, not just ‘it is humid’ or ‘it is dark’.

6Grade 4–56 marks

A student investigated the distribution of stomata on a leaf.

She painted a thin layer of clear nail varnish on the upper surface and on the lower surface of the leaf and let it dry.
She peeled off each layer with sticky tape and stuck it onto a microscope slide.
She counted the stomata in five different fields of view for each surface.

Surface	Number of stomata in field of view
Surface	1	2	3	4	5
Upper	3	2	4	3	3
Lower	24	27	9	25	26

(a) Identify the anomalous result. [1]

(b) Calculate the mean number of stomata in a field of view on the lower surface.

Do not include the anomalous result. [2]

(d) Explain why having most stomata on the lower surface is an advantage to the plant. [2]

(a) 9 (lower surface, field of view 3)(1)
(b) (24 + 27 + 25 + 26) ÷ 4
allow 102 ÷ 4(1)
(b) 25.5
allow 25.5 rounded to 26 only if 25.5 is seen; correct answer with no working scores 2 marks; 22.2 (anomaly included) scores 1 mark(1)
(c) stomata are not evenly spread (across the leaf) so a mean is more representative
allow to calculate a mean / to make results more representative / to reduce the effect of an anomalous result; ignore ‘more accurate’ / ‘fair test’(1)
(d) the lower surface is cooler / shaded / out of direct sunlight
allow less exposed to the sun(1)
(d) so less water is lost (by evaporation / transpiration) (but gas exchange can still take place)
allow reduces the risk of wilting(1)
Answer: (a) 9 (b) 25.5 (c) stomata are not evenly distributed, so a mean of several fields is more representative (d) the lower surface is cooler / shaded, so less water is lost by transpiration while gas exchange still happens
Examiner tip: For (d) ‘to stop water loss’ is wrong — stomata on the lower surface reduce water loss; they do not prevent it.

7Grade 6–78 marks

A student used a potometer to measure the water uptake of a leafy shoot.

She recorded the distance moved by an air bubble along the capillary tube every 2 minutes.

For the first 10 minutes the shoot was in still air. At 10 minutes she switched on a fan pointed at the shoot.

The graph shows her results.

(a) Use the graph to calculate the rate of movement of the bubble in still air.

Give your answer in mm per minute. [2]

(b) Calculate the percentage increase in the rate of movement of the bubble after the fan was switched on. [3]

(d) Give one variable the student should have kept constant during the investigation. [1]

(a) 16 (mm) ÷ 10 (min)
allow any correct pair of readings from the first section, e.g. 8 ÷ 5(1)
(a) 1.6 (mm per minute)
correct answer with no working scores 2 marks(1)
(b) rate with fan = (56 − 16) ÷ 10 = 4.0 (mm per minute)
allow correct readings from any part of the second section(1)
(b) (4.0 − 1.6) ÷ 1.6 × 100
allow ecf from (a) and from the rate with the fan(1)
(b) 150 (%)
correct answer with no working scores 3 marks; 250 (%) scores 1 mark (for the rate of 4.0 with the fan)(1)
(c) moving air removes water vapour from around the leaf / stomata(1)
(c) (so) maintains a steep concentration gradient (of water vapour) so water vapour diffuses out of the stomata faster / transpiration rate increases
allow so more water is drawn up the xylem to replace water lost(1)
(d) any one from: temperature; light intensity; humidity (of the air); same shoot / same leaf area
ignore ‘amount of water’; do not accept wind speed / air movement(1)
Answer: (a) 1.6 mm per minute (b) 150% (c) the fan removes water vapour from around the leaf, keeping a steep concentration gradient, so water vapour diffuses out of the stomata faster (d) e.g. temperature
Examiner tip: Percentage increase = change ÷ ORIGINAL rate × 100, so (4.0 − 1.6) ÷ 1.6 × 100 = 150% — writing 4.0 ÷ 1.6 × 100 = 250% forgets to subtract the original rate.

8Grade 8–9Higher7 marks

A gardener removed a complete ring of bark from the trunk of a young tree. The ring of bark contained the phloem. The xylem was not removed.

After several weeks:

the bark just above the ring had swollen
the leaves had not wilted.

(a) Explain why the bark just above the ring swelled. [3]

(b) Explain why the leaves did not wilt. [2]

Suggest why. [2]

(a) phloem transports / translocates (dissolved) sugars (made in the leaves)
allow sucrose / glucose; do not accept sugars carried in the xylem(1)
(a) (sugars) cannot pass the ring / cannot move down to the roots (because the phloem has been removed)(1)
(a) (so) sugars build up above the ring
allow (sugars) used for growth of cells above the ring / water enters (cells above the ring) by osmosis(1)
(b) the xylem is still intact(1)
(b) so water (and mineral ions) can still be transported from the roots to the leaves
allow the transpiration stream continues; allow water lost by transpiration is replaced(1)
(c) roots receive no sugars / glucose (for respiration)
allow roots are starved(1)
(c) (so) root cells cannot respire / release energy so (die and) cannot absorb water or mineral ions
allow no energy for active transport of mineral ions (into root hair cells)(1)
Answer: (a) phloem translocates dissolved sugars from the leaves; they cannot pass the missing ring so accumulate above it (b) xylem is intact so water and mineral ions still reach the leaves (c) roots get no sugars, cannot respire, so cannot take up water and mineral ions and die
Examiner tip: Keep xylem and phloem straight: sugars go in the phloem (which was removed), water and mineral ions in the xylem (which was not).

9Grade 6–76 marks

Water moves from the soil, through a plant and into the air.

Describe how water moves from the soil into the air around a leaf and explain how the plant can control how much water it loses. [6]

water enters root hair cells by osmosis(1)
because the soil water is more dilute than the cell contents / water moves from a dilute to a more concentrated solution across a partially permeable membrane(1)
root hairs give a large surface area for absorption(1)
water moves (across the root) into the xylem(1)
xylem is made of hollow tubes (strengthened by lignin) so water moves up through the stem in a continuous column(1)
water evaporates from (the surfaces of) mesophyll cells into air spaces in the leaf(1)
water vapour diffuses out of the leaf through the stomata (down a concentration gradient)(1)
loss of water from the leaf draws more water up the xylem: the transpiration stream(1)
guard cells control the size of the stomata(1)
stomata close (e.g. at night / in hot, dry conditions / when water is short) to reduce water loss(1)
most stomata are on the lower (cooler / shaded) surface of the leaf so less water is lost(1)
closing stomata also reduces carbon dioxide uptake, so photosynthesis slows(1)
waxy cuticle on the epidermis reduces water loss(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, logically sequenced account of water moving from soil to air (osmosis into root hair cells, transport in the xylem, evaporation and diffusion out through stomata) together with a clear explanation of how guard cells and the position of stomata control water loss.
Level 2	3–4	Describes most stages of the route with some correct processes named, and makes at least one relevant point about controlling water loss; OR a detailed route with no explanation of control. Some steps may be out of order or not linked.
Level 1	1–2	Simple, relevant statements, e.g. water goes in through the roots, up the xylem and out of the leaves, or stomata close. Little or no use of processes or links between steps.
No marks	0	No relevant content.

Answer: Water enters root hair cells by osmosis, crosses the root into the xylem and moves up the stem in the transpiration stream; it evaporates from mesophyll cells and diffuses out through the stomata. Guard cells close the stomata to reduce water loss (e.g. at night or in dry conditions); most stomata are on the shaded lower surface.
Examiner tip: Many answers describe the journey well but forget the second command — top marks need guard cells closing stomata explained as a way to control water loss.

10Grade 8–9Higher8 marks

A student compared water uptake by leafy shoots of two plant species, P and Q.

He found the total leaf area of each shoot by drawing round the leaves on 1 cm² squared paper.
He used a potometer to measure the volume of water taken up by each shoot in 40 minutes.

Species	Total leaf area in cm²	Volume of water taken up in 40 minutes in cm³	Rate of water uptake in cm³ per cm² per hour
P	64	0.48	X
Q	150	0.90	9.0 × 10⁻³

(a) Calculate value X.

Give your answer in standard form. [3]

(b) The student concluded: ‘Species Q loses water faster than species P because it took up more water.’

Evaluate this conclusion. [2]

(d) On a hot, dry day the stomata of both species close.

Explain why this reduces the rate of photosynthesis. [2]

(a) 40 minutes = 0.667 / ⅔ hour or 0.48 ÷ 64 = 0.0075 (cm³ per cm² in 40 min)
allow 0.48 ÷ 40 × 60 = 0.72 (cm³ per hour)(1)
(a) 0.0075 ÷ ⅔ or 0.72 ÷ 64 = 0.01125
allow 0.0113 / 0.011(1)
(a) 1.125 × 10⁻² (cm³ per cm² per hour)
allow 1.13 × 10⁻² / 1.1 × 10⁻²; correct answer with no working scores 3 marks; correct value not in standard form scores 2 marks(1)
(b) the conclusion is not correct / not supported: P has a higher rate (of water uptake) per cm² of leaf (1.1 × 10⁻² compared with 9.0 × 10⁻³)
allow ecf from (a)(1)
(b) Q only took up more water because it has a larger leaf area (so the rate must be compared per unit area)
allow water uptake is not the same as water loss (some water is used by the plant, e.g. for photosynthesis)(1)
(c) more stomata per cm² / higher density of stomata
allow larger stomata / thinner (waxy) cuticle / more stomata on the upper surface(1)
(d) less carbon dioxide diffuses into the leaf (through the stomata)(1)
(d) carbon dioxide is a reactant / raw material for photosynthesis, so it becomes a limiting factor
ignore ‘less water’ unless linked to stomata closing(1)
Answer: (a) X = 1.125 × 10⁻² (1.1 × 10⁻²) cm³ per cm² per hour (b) conclusion not valid: per cm² of leaf, P has the higher rate; Q took up more only because it has a larger leaf area (c) e.g. more stomata per cm² (d) less CO₂ diffuses into the leaf, and CO₂ is needed for photosynthesis so it becomes limiting
Examiner tip: Convert 40 minutes to hours (40 ÷ 60 = ⅔ h) before dividing — dividing by 40 gives a rate per minute, not per hour, and loses the last two marks.

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