OxfordAQA International AS and A-level Biology: Mass transport in plants: xylem, transpiration and phloem (9610) – 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 follow the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, section 3.2.8 (3.2.8.1 xylem and 3.2.8.2 phloem) and Required practical 6. Unit 2 is listed as part of both the International AS and the International A-level. Learn the content first from the study guide and the revision notes. The course page, the checklist and the free diagnostics help you plan. All data below are supplied for the questions.

Questions

1. Describe the pathway taken by water from the soil to the xylem in a dicotyledonous root. [4]

2. Water crosses the cortex of a root by the apoplastic pathway and the symplastic pathway.

(a) Distinguish between the two pathways. [2] (b) Explain why water must enter the symplast at the endodermis. [2] (c) Suggest one benefit of this to the plant. [1]

3. A student used a potometer. The capillary bore had a radius of 0.35 mm. The air bubble moved 33 mm, 35 mm and 31 mm in three separate 5.0 minute runs.

(a) Calculate the mean rate of bubble movement in mm min⁻¹. [2] (b) Find the volume of water taken up each minute, to 2 significant figures. [2]

4. A student measured water uptake by a leafy shoot in four conditions.

Condition	Uptake / mm³ min⁻¹
Still air, 20 °C	2.4
Still air, 30 °C	3.9
Fan, 20 °C	5.1
Still air, 20 °C, polythene bag over the leaves	1.1

(a) Calculate the percentage increase in uptake when the temperature rises from 20 °C to 30 °C. [2] (b) Calculate the percentage decrease caused by the polythene bag. [2] (c) Explain the effect of the fan. [2]

5. Cut stumps of a plant were left to exude sap for 2.0 hours. The volumes collected were: 46 mm³ at 20 °C with oxygen supplied, 8 mm³ at 20 °C with a respiratory inhibitor and 13 mm³ at 5 °C.

(a) Calculate the percentage reduction caused by the inhibitor. [2] (b) Use all the data to explain what the results suggest about root pressure. [3]

6. A student plans to test whether moving air changes how fast a leafy shoot takes up water, using a potometer. Describe the procedure. [6]

7. Explain the mass flow hypothesis for the movement of sucrose in the phloem. [6]

8. A ring of bark was removed from a stem. After 48 hours the sucrose concentration in the bark sap was measured.

Stem	Above the ring / mg cm⁻³	Below the ring / mg cm⁻³
Ringed	68	14
Control (no ring)	37	36

(a) Describe the results. [2] (b) Explain the results for the ringed stem. [2] (c) Explain the purpose of the control. [1]

9. Radioactive carbon dioxide was fed to a leaf. Detectors on the stem were 12.0 cm and 31.5 cm from the leaf. The label first reached them 13 minutes and 38 minutes after feeding. Calculate the rate of movement between the detectors in m h⁻¹. Give your answer to 2 significant figures. [4]

10. The diameter of a tree trunk was measured at three times in one day: 312.40 mm at 06:00, 311.86 mm at 14:00 and 312.31 mm at 22:00.

(a) Calculate the decrease between 06:00 and 14:00 as a percentage of the 06:00 diameter. [2] (b) Explain the change in diameter. [3]

11. Aphid stylets were used to collect sap from a plant. Sap flowed from the cut stylets. The sucrose concentration was 0.62 mol dm⁻³ near a leaf, 0.48 mol dm⁻³ in the mid stem and 0.21 mol dm⁻³ near the root tip.

(a) Explain how the flow from the stylets supports the mass flow hypothesis. [2] (b) Explain the pattern in concentration. [2] (c) Suggest two limitations of this evidence. [2]

Answers

1. Water enters the root hair cell by osmosis [1]. It crosses the cortex by the apoplastic pathway (walls) or symplastic pathway (cytoplasm) [1]. It reaches the endodermis, where it passes through cell membranes into the cytoplasm [1]. It then enters the xylem in the centre of the root [1]. [4] Examiner insight: Name the tissues in order. Answers that say only “it goes through the root” cannot be credited for each stage.

2. (a) The apoplastic pathway runs through cell walls and intercellular spaces [1], whereas the symplastic pathway runs through the cytoplasm via plasmodesmata [1]. (b) The endodermis walls contain a waterproof suberin band [1], which blocks the apoplastic route, so water must cross the surface membrane [1]. (c) The membrane controls which ions enter the xylem [1]. [5] Examiner insight: A link between the structure and its function earns the mark, so “waterproof” must be paired with “blocks the walls”. Saying “the endodermis stops water” is too vague.

3. (a) Mean distance = (33 + 35 + 31) ÷ 3 = 33.0 mm [1]. Rate = 33.0 ÷ 5.0 = 6.6 mm min⁻¹ [1]. (b) Area = π × 0.35² = 0.385 mm² [1]. Volume per minute = 6.6 × 0.385 = 2.5 mm³ min⁻¹ [1]. [4] Examiner insight: Using the diameter in place of the radius gives a volume four times too large. Write the area line, so that your working shows each step clearly.

4. (a) (3.9 − 2.4) ÷ 2.4 × 100 [1] = 62.5% (63%) [1]. (b) (2.4 − 1.1) ÷ 2.4 × 100 [1] = 54% [1]. (c) A fan removes the humid air around the stomata [1], so the water vapour gradient is steeper and transpiration, and therefore uptake, is faster [1]. [6] Examiner insight: Percentage change is measured against the starting value, not the final one. A bag lowers uptake because it traps humid air, and “it blocks the stomata” would not earn the mark.

5. (a) (46 − 8) ÷ 46 × 100 [1] = 83% [1]. (b) The inhibitor lowers respiration, so less ATP is made [1]. Active transport of ions into the xylem is then reduced, so less water enters by osmosis [1]. Cold gives a similar fall (72%), so the process depends on metabolism, and root pressure is not just passive [1]. [5] Examiner insight: Quote data from both treatments. A conclusion that the inhibitor “kills the cells” is not what the results show.

6. Cut the shoot under water at a slant [1]. Assemble the potometer under water so there are no air bubbles [1]. Dry the leaves and seal the joints with petroleum jelly [1]. Introduce one air bubble and record its position [1]. Measure the distance moved in a set time, first in still air and then with a fan at a fixed distance, repeating each to calculate a mean [1]. Keep the temperature, light and shoot unchanged, and allow the shoot to adjust [1]. [6] Examiner insight: A plan needs a control variable named and a repeat. The word “fair test” alone is not a variable.

7. Sucrose is actively loaded into the sieve tubes at the source [1], which lowers the water potential [1]. Water enters by osmosis from the xylem, raising the hydrostatic pressure [1]. At the sink, sucrose is taken out and used or stored [1], so water leaves and pressure falls [1]. The sap flows from high pressure at the source to low pressure at the sink [1]. [6] Examiner insight: The order is source, pressure difference, sink. Writing that sucrose moves “by osmosis” confuses the movement of sucrose with the movement of water.

8. (a) In the ringed stem the sucrose is much higher above the ring than below it (68 against 14) [1], but in the control the two are almost equal (37 and 36) [1]. (b) The ring removed the phloem [1], so sucrose from the leaves could not move down and built up above the ring [1]. (c) The control shows that the difference is due to the ring and not to the stem’s normal gradient [1]. [5] Examiner insight: Compare two numbers from the table in a description. A statement that “sucrose is high above” with no figures is incomplete.

9. Distance = 31.5 − 12.0 = 19.5 cm = 195 mm [1]. Time = 38 − 13 = 25 min [1]. Rate = 195 ÷ 25 = 7.8 mm min⁻¹ [1]. In m h⁻¹: 7.8 × 60 ÷ 1000 = 0.47 m h⁻¹ [1]. [4] Examiner insight: The rate uses the distance between detectors, not the distance from the leaf. Unit conversion is the usual slip.

10. (a) 312.40 − 311.86 = 0.54 mm [1]; 0.54 ÷ 312.40 × 100 = 0.17% [1]. (b) Transpiration is fastest around midday [1], so the water in the xylem is under greater tension, which pulls the walls inwards and reduces the diameter [1]. By evening transpiration falls and the tension relaxes, so the diameter recovers [1]. [5] Examiner insight: Link tension to the walls. Answers that say “the tree loses water” without tension do not explain the change in diameter.

11. (a) Sap flows out of a cut stylet, so the sieve tube contents are under positive pressure [1], as the hypothesis predicts [1]. (b) Sucrose is loaded at the leaf (the source), so it is concentrated there [1]. It is used or stored in the root, which lowers the concentration at that end [1]. (c) Any two from: the pressure was not measured; only three positions were tested; the stylet could change the flow [2]. [6] Examiner insight: A limitation must say what the data cannot show. “Not enough repeats” is credited only if you say what was not repeated.

Where marks are usually lost

• Giving the apoplast as “through the cell membrane” instead of walls and spaces.
• Not stating that the endodermis barrier is waterproof in the walls.
• Using the capillary diameter rather than the radius in πr².
• Dividing percentage change by the final value.
• Writing “temperature speeds it up” without evaporation or diffusion.
• Describing potometer results as transpiration instead of uptake.
• Describing data without quoting figures or comparing with the control.
• Writing that water is moved in the phloem by tension.

Next steps

• Review the revision notes and the study guide.
• Go to the OxfordAQA A-level Biology course page and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.8 and Required practical 6.