AQA A-Level Biology Paper 1, June 2022: Question 6

10 marks · Medium difficulty · Practical Techniques & Data Analysis

Explain the rate of transpiration, calculate percentage increase from a graph, and describe an experiment to investigate water potential in mangrove root cells.

Practise this question

Question

Question text

06 Mangrove trees grow near the sea. Sea water surrounds the lower parts of the trees

at high tide.

Scientists investigated the rate of transpiration in a mangrove tree.

Figure 9 shows the scientists’ results.

Figure 9

06.1 Explain the rate of transpiration between 5 am and midday shown in Figure 9.

[4 marks]

06.2 Use Figure 9 to calculate the percentage increase in the rate of transpiration from

1 pm to 2 pm.

*20* [2 marks]

Percentage increase in rate of transpiration %

06.3 The higher rate of transpiration at high tide shows that the mangrove tree is absorbing

water from the sea water surrounding its roots.

Describe an experiment that you could do to investigate whether the mangrove root

cells have a lower water potential than sea water.

You are given:

• a piece of fresh mangrove root

• sea water

• access to laboratory equipment.

[4 marks]

Mark scheme

Show the mark scheme

Question Marking Guidance Mark Comments

1. (Rate of) transpiration/evaporation increases due 4 Ignore reference to

to increased temperature tide

(4 x

OR AO2) 1 and 2 Reject tide

affecting transpiration/

(Rate of) transpiration/evaporation increases due

water potential/

to increased light intensity

OR humidity

(Rate of) transpiration/evaporation increases due

1 and 2 Correct link

to decreased humidity

needed between

factor affecting

(Rate of) transpiration/evaporation increases due

transpiration and the

to increased wind/air movement;

explanation

2.(So) increased kinetic energy (causing more

06.1 water loss)

(So) increased water potential gradient (so more

water lost)

(So) increased (water) diffusion gradient (so more

water lost);

3.Stomata open (at sunrise/after 5 am) allowing

gas exchange

Stomata open (at sunrise/after 5 am) allowing 4 Accept at 11 am as

carbon dioxide to enter; the time when

stomata close

4. (Some) stomata close at midday/after 11 am

(reducing transpiration);

Correct answer for 2 marks, 6.6̇, 6.67 – 7 (%);; 2

Accept for 1 mark, (2 x

AO2)

0.05 (correct difference in transpiration rate)

6.6 (correct calculation, but incorrect rounding)

06.2

6.25/6.3 (correct calculation using incorrect

denominator)

666/667 correct number sequence but decimal

place in wrong place eg 66.7/0.0667

0.75 as denominator – A-LEVEL BIOLOGY – –

18Question Marking Guidance Mark Comments

Mark in groups, either 1 to 4 OR 5 to 8 4 Accept ‘weight’ for

‘mass’.

1. Record mass/length before and after; (4 x

2. Place in sea water for (specified/equal) time; AO3) Accept ‘diameter’ for

‘length’.

3. Method to remove surface water;

4. Increase in mass/length shows water has been 2. Ignore period of

absorbed by osmosis time

OR 2. Accept seawater in

Increase in mass/length shows cells have lower a dilution series

water potential; Ignore blot dry before

initial mass

OR measurement

5. Put tissue/cells on (microscope/glass) slide; Reject ‘size’ once

6. Add seawater (and leave) then allow ECF.

06.3 7. Observe under (optical) microscope; 3. Accept eg use

8. If cells become flaccid they do not have a lower tissue paper to dry

water potential than seawater OR blot dry

OR 4. Accept

(If cells become) turgid cells show water is root/mangrove for

absorbed by osmosis cells

(If cells become) turgid cells show cells have a

lower water potential

OR 8. Accept description

(If cells are) not flaccid/plasmolysis cells show of turgid (cells)

water is not lost by osmosis

(Determine) percentage plasmolysis;

How to answer it

Mangrove Transpiration and Osmosis

What this question tests

This question assesses your ability to interpret biological line graphs involving environmental factors and transpiration, perform percentage increase calculations accurately, and design a practical investigation to test water potential using osmosis principles.

Part 06.1 (4 Marks)

Explaining Transpiration Trends from Data

✅ Correct Answer Framework

Factor (1): Transpiration rate increases from 5 am to 11 am due to increased temperature, increased light intensity, decreased humidity, or increased air movement.
Mechanism (2): Higher kinetic energy increases water evaporation, or a steeper water potential (diffusion) gradient is established.
Stomatal Opening (3): Stomata open at sunrise / after 5 am, allowing gas exchange or carbon dioxide uptake.
Stomatal Closure (4): Stomata close around midday / after 11 am, reducing the rate of transpiration.

💡 Key Knowledge

Transpiration is driven by environmental conditions that affect the evaporation of water from mesophyll cells and the diffusion gradient out through the stomata. Light triggers stomatal opening, while excessive heat or water stress can trigger midday closures.

🧠 Exam Technique

Link physical environmental changes directly to the biological consequences. Do not just quote graph numbers; explain why the trend occurs (e.g., mention kinetic energy, water potential gradients, or stomatal behavior).

❌ Common Errors

Students frequently lost marks by erroneously attributing changes directly to the tide level shown on the secondary axis. Remember: tide height does not directly alter atmospheric humidity or stomatal resistance in the canopy.

Mark scheme guidance: 4 marks available via AO2. Requires matching an environmental factor or stomatal change with a valid physiological explanation.

Part 06.2 (2 Marks)

Calculating Percentage Increase

📐 Step-by-Step Calculation

Step 1: Read values from Figure 9.
At 1 pm = 0.75 cm³ hr⁻¹
At 2 pm = 0.80 cm³ hr⁻¹
Step 2: Find the absolute increase:
0.80 - 0.75 = 0.05
Step 3: Apply the percentage formula (Increase ÷ Original Value × 100):
(0.05 / 0.75) × 100 = 6.666...
Step 4: Round appropriately to 6.7% (or 6.6 / 6.67).

❌ Common Calculation Traps

Using the final value ( 0.80 ) as the denominator instead of the starting value ( 0.75 ).
Rounding errors or failing to show working when numbers repeat indefinitely (e.g., 6.6 recurring).

Mark scheme guidance: 2 marks for the correct answer ( 6.6, 6.67 - 7 (%) ). 1 mark awarded for correct working/difference ( 0.05 ) if final evaluation is wrong.

Part 06.3 (4 Marks)

Designing an Osmosis Investigation

✅ Correct Experimental Design (Option 1: Mass/Length)

Record initial mass or length of a fresh mangrove root piece.
Immerse the root tissue in sea water for a specified, equal period of time.
Remove and blot dry surface water before recording final mass/length.
An increase in mass or length indicates water has been absorbed by osmosis, proving the root cells have a lower water potential than sea water.

✅ Alternative Design (Option 2: Microscopy)

Place root tissue/cells on a microscope slide with sea water.
Observe under an optical microscope.
If cells become turgid (or do not undergo plasmolysis/flaccidity), water has not left the cells, confirming root cells possess a lower water potential than the surrounding sea water.

🧠 Exam Technique

For experimental design questions, make sure to include controls or measurable quantitative variables (e.g., "record mass before and after"). Precision matters: explicitly state how surface liquid is removed (blotting).

Mark scheme guidance: 4 marks available via AO3. Marks grouped into actions (recording mass, timing, blotting, linking directional water movement via osmosis to water potential conclusions).

Topics

Biology · Practical skills · 3.3 Organisms exchange substances with their environment · 3.1 Biological molecules · Data analysis · Experimental design

Question and mark scheme from the AQA A-Level Biology examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.