AQA AS Level Biology Paper 1, June 2023: Question 5
9 marks · Medium difficulty · Practical Techniques & Data Analysis
Calculate surface area to volume ratios, determine means from experimental diffusion data involving agar blocks, identify controlled variables, and explain gas exchange in single-celled versus large organisms.
Practise this questionQuestion
Question text
05 A student investigated the effect of changing surface area on the rate of diffusion of a
solution into the centre of agar blocks.
She used agar coloured by an indicator. The indicator is pink at pH > 8 and
colourless at pH 8 and pH < 8
She cut blocks in different shapes as shown in Figure 6.
Figure 6
05.1 Complete Table 2 to show the surface area and the surface area to volume ratio for the
two shapes.
Table 2
23 Surface area to volume
Shape Surface area / cm Volume / cm
ratio
C 64
: 1
D 64
: 1
[2 marks]
05.2 The student put the blocks into an acidic solution.
The acidic solution caused the blocks to gradually turn from pink to colourless.
She recorded the time taken for the blocks to turn completely colourless.
She repeated this three times.
Table 3 shows the student’s results.
Table 3
Time for block to turn colourless / s
Mean time for block to
Shape
turn colourless / s
Block 1 Block 2 Block 3 Block 4
C 3490 1200 3540 3530
D 1680 1500 1590 1610 1595
After collecting the data, the student noticed that shape C, block 2 was damaged.
Calculate the mean for shape C.
[1 mark]
05.3 Suggest what the student should have done when she saw that shape C, block 2 was
damaged.
[1 mark]
05.4 State three variables the student controlled in order to obtain valid results.
[2 marks]
05.5 Describe how gas exchange occurs in single-celled organisms and explain why this
method cannot be used by large, multicellular organisms.
[3 marks]
Mark scheme
Show the mark scheme
Question Marking Guidance Mark Comments
Mark as columns
Shape Surface Volume / Surface Allow ECF for ratio
area / cm2 cm3 area to
volume Allow 2.1:1 and
ratio 2
2.13:1 for D
05.1 (2 x
C 96 64 1.5 : 1 AO2)
Ignore fractions for
ratio
D 136 64 2.125 : 1
; ;
05.2 3520;
(AO2)
Accept ‘Replace with
Repeat with an undamaged (shape C) block;
an undamaged
1 (shape C) block and
05.3
(AO1b) repeat’
Allow ‘Repeat with a
new block (shape C)’
Any three for 2 marks;; Ignore
Any two for 1 mark; Mass, shape, volume
1. Temperature
2. Concentration of indicator 2 max
05.4 3. pH/concentration of solution/acid (2 x
AO3)
4. pH/concentration of alkali in blocks
5. Concentration/type of agar
6. Ensure total surface area of the block is in – LOGY – – JUNE 2023
contact with the solution/acid
Question Marking Guidance Comments
1. Diffusion (across the cell surface
membrane);
2. Large organisms have small(er) sa : vol ratio
10 OR
3 max
05.5 (3 x
Single-celled organisms have a large(r) sa : vol
AO1)
ratio;
3. Diffusion pathway would be too long 3. Must be in the
context of larger
OR
organisms
(Rate of) Diffusion too slow;
How to answer it
Surface Area, Volume, and Diffusion Study Guide
What this question tests
This exam question assesses your ability to calculate surface area and surface-to-volume ratios, process experimental data while handling anomalous or damaged data points, identify controlled variables to ensure validity, and apply core biological knowledge regarding diffusion limitations in single-celled versus large multicellular organisms.
Calculating Surface Area and Surface-to-Volume Ratios
✅ Correct Answers
- Shape C: Surface area = 96 cm² | SA:Vol ratio = 1.5 : 1
- Shape D: Surface area = 136 cm² | SA:Vol ratio = 2.125 : 1 (Accept 2.1 : 1 or 2.13 : 1 )
📐 Step-by-Step Calculation
- Shape C (Cube/Cuboid): Dimensions 4cm × 4cm × 4cm.
Area of one face = 4 × 4 = 16 cm².
Total surface area = 16 × 6 = 96 cm² .
Ratio = 96 ÷ 64 = 1.5 (written as 1.5 : 1 ). - Shape D (Cuboid): Dimensions 16cm × 2cm × 2cm.
Faces: (16 × 2) × 4 = 128 cm²; (2 × 2) × 2 = 8 cm².
Total surface area = 128 + 8 = 136 cm² .
Ratio = 136 ÷ 64 = 2.125 (written as 2.125 : 1 ).
Calculating a Mean with Anomalous Data Exclusion
✅ Correct Answer
3520 (seconds)
🧠 Exam Technique
Look closely at the data table provided in the prompt. Block 2 for Shape C recorded a time of 1200 s , whereas Blocks 1, 3, and 4 were 3490 s , 3540 s , and 3530 s respectively. The stem text explicitly tells you that shape C, block 2 was damaged.
Therefore, you must exclude Block 2 from your mean calculation and average only the remaining three valid values: (3490 + 3540 + 3530) ÷ 3 = 3520.
Handling Anomalous or Damaged Apparatus
✅ Correct Answer
Repeat the experiment with an undamaged (shape C) block / Replace with a new block and repeat.
💡 Key Knowledge
When an experimental unit is compromised or damaged during a trial, simply averaging around it or discarding the replicate entirely without replacement can reduce your dataset size. Standard practical protocol requires replacing the damaged sample and repeating that specific trial to maintain a full set of reliable replicates.
Controlling Variables for Valid Results
💡 Key Knowledge: Valid Variables
To ensure comparisons between shapes C and D are fair and valid, any factor affecting the rate of diffusion or colour change must be strictly controlled:
- Temperature of the solution
- Concentration of the indicator in the agar
- pH or concentration of the acidic solution
- Concentration or type of agar used
- Ensuring the total surface area of the block is fully submerged/in contact with the acid
❌ Common Errors
Do not list vague answers such as "mass", "shape", or "volume". Shape is the independent variable being deliberately changed, and mass/volume will naturally vary depending on the geometric layout chosen. Focus strictly on environmental and chemical conditions.
Gas Exchange and Organism Scale
✅ Expected Marking Points (Any 3)
- Point 1: Diffusion occurs directly across the cell surface membrane in single-celled organisms.
- Point 2: Large multicellular organisms have a small(er) surface area to volume ratio, whereas single-celled organisms have a large(r) surface area to volume ratio.
- Point 3: The diffusion pathway would be too long in large organisms, making the rate of diffusion too slow to meet metabolic demands.
🧠 Top-Level Response Strategy
To secure full marks, structure your answer comparatively:
- State how single-celled organisms exchange gases (direct membrane diffusion).
- Link this to their high surface area to volume ratio and short diffusion distance.
- Contrasting contrast this directly with large organisms, explicitly mentioning that their small SA:Vol ratio and large diffusion distance result in diffusion being too slow to sustain inner cells.
Topics
Biology · Practical skills · 3.3 Organisms exchange substances with their environment · Experimental design · Data analysis
Question and mark scheme from the AQA AS Level Biology examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.