AQA A-Level Biology Paper 1, June 2023: Question 8
10 marks · Hard difficulty · Practical Techniques & Data Analysis
Analyze the effects of the cancer-treating drug MiTMAB on cell division, enzyme inhibition, and cell culture growth using graphical and microscopic data.
Practise this questionQuestion
Question text
08.1 Scientists investigated a drug called MiTMAB as a treatment for cancer. MiTMAB
inhibits cytokinesis.
Figure 8 shows drawings of cancer cells seen with an optical microscope from a:
• sample treated with MiTMAB
• control sample.
Figure 8
A B
The cells in drawing A can be identified as those treated with MiTMAB.
Explain why.
[2 marks]
08.2 MiTMAB acts as a non-competitive inhibitor of an enzyme called dynamin.
Suggest how MiTMAB can cause dynamin to become inactive.
[3 marks]
When active, dynamin has two functions:
• it stimulates cytokinesis
• it inhibits cell death.
The scientists treated actively growing cultures of cancer cells with MiTMAB.
They incubated:
• one sample of 2500 cells without MiTMAB as a control
• eight samples, each with 2500 cells and a different concentration of MiTMAB.
After 72 hours, the scientists measured the number of cells in each sample.
Figure 9 shows the scientists’ results.
A negative value for proportion of control growth means that fewer than 2500 cells
were counted after 72 hours.
Figure 9
08.3 Use all the information given to explain the results shown in Figure 9.
[3 marks]
08 4 3
. 0.01 dm of MiTMAB solution was added to the treated cells.
Calculate the increase in mass of MiTMAB (in μg) added to the cells to reduce the cell
growth from equal to the control to 0.0 of the control.
Show your working.
[2 marks]
Answer μg
Mark scheme
Show the mark scheme
Question Marking Guidance Mark Comments
1. 2 nuclei (in cells)
OR
Cells (stopped) at telophase;
2. Cytokinesis prevented
08.1 (2 x
OR 2. Accept cell
AO3)
membrane not
Stopped (new) cell membrane forming
dividing/splitting/
OR pinching (in the cell)
Stopped cytoplasm dividing;
1. (MiTMAB) binds (to dynamin) other than the 1. Accept (MiTMAB)
active site; binds to dynamin at
an allosteric OR
2. Changes the shape of (dynamin) active site
inhibitor site
OR
2. Accept denature for
Changes the tertiary structure (of dynamin/
08.2 (1 x ‘change in shape’
enzyme);
AO1, 2
3. Not complementary so substrate does not bind x AO2) 3. Accept ES complex
(to active site) in this instance
OR 3. Ignore ESC
3. Accept fit OR
Not complementary so no/fewer enzyme- attach for bind
substrate complexes (form);
1. (At) lowest concentrations (all) dynamin is not
1. Accept graph
inhibited
readings in range 30
OR to 70 for lowest;
(At) lowest concentrations (MiTMAB) does not 1. Accept ‘has no
cause cell death/inhibit cytokinesis; effect’ for ‘cause cell
death/inhibit
2. (As MiTMAB) concentration increases more
cytokinesis’
dynamin is inhibited/inactive
1. and 3. Accept
OR
‘prevents inhibition
(As MiTMAB) concentration increases cell death of cell death’ for
increases ‘causes cell death’
OR 1, 2 and 3. Accept
(As MiTMAB) concentration increases ‘cell replication’ OR
3 mitosis for
cytokinesis decreases
08.3 (3 x cytokinesis
OR
AO2)
No change in cell number at 2000 (μg dm-3)
OR
No change in cell number at 0.0 (on y axis);
3. (At) highest (MiTMAB) concentrations all
dynamin is inhibited 3. Accept graph
readings >2000 to
OR 8000 for highest
(At) highest concentrations (MiTMAB) causes
cell death
OR
(At) highest concentrations (MiTMAB) inhibits
cytokinesis;
Correct answer in range 19.3 to 19.7 = 2 marks;; Correct answer
19.3 (is obtained from
70 and 2000)
Accept for 1 mark evidence of
19.7 (is obtained from
70 and 2000 (correct readings from the graph) /
30 and 2000)
30 and 2000 (correct readings from the graph)
OR Accept for 1 mark,
08.4 (2 x
any value in range 30
1930 (correct increase in MiTMAB, 70 to 2000) / AO2) to 70 and 2000
1970 (correct increase in MiTMAB, 30 to 2000)
Accept for 1 mark,
OR any value in the range 21
1930 to 1970
Division by 100/multiplication by 0.01 (correct
conversion to mass in 0.01 dm3);
How to answer it
A-Level Biology Study Guide: Enzyme Inhibition & Cell Division
What this question tests
This exam question tests your understanding of the cell cycle (specifically cytokinesis and telophase), enzyme kinetics (non-competitive inhibition), data interpretation from complex logarithmic graphs, and multi-step mass-concentration calculations.
Interpreting Cell Drawings & Cytokinesis
The cells in drawing A can be identified as those treated with MitMAB. Explain why. [2 marks]
✅ Correct Answers (Max 2 marks)
- Cells contain 2 nuclei (1 mark).
- Cells are stopped at telophase / cytokinesis is prevented / new cell membrane failing to form / cytoplasm stopped dividing (1 mark).
💡 Key Knowledge
- Cytokinesis is the final stage of cell division where the cytoplasm splits to form two separate daughter cells.
- MitMAB explicitly inhibits this process, meaning nuclear division (mitosis) finishes, but cellular division does not, leading to binucleate cells.
❌ Common Errors
- Vague descriptions like "cells stuck in mitosis" without specifying telophase or cytokinesis lose credit.
- Confusing nuclear division with cellular division.
Non-Competitive Inhibition Mechanism
MitMAB acts as a non-competitive inhibitor of an enzyme called dynamin. Suggest how MitMAB can cause dynamin to become inactive. [3 marks]
✅ Correct Answers (Max 3 marks)
- MitMAB binds to the enzyme at a site other than the active site (allosteric/inhibitor site) (1 mark).
- Causes a change in shape of the active site / changes the tertiary structure of the enzyme (1 mark).
- Active site is no longer complementary to the substrate, so fewer/no enzyme-substrate complexes form (1 mark).
🧠 Exam Technique
Always structure enzyme inhibition answers using the triad: Binding location → Tertiary structure/Active site change → Impact on enzyme-substrate complex formation.
Interpreting Logarithmic Graph Trends
Use all the information given to explain the results shown in Figure 9. [3 marks]
✅ Correct Answers (Max 3 marks)
- At lowest concentrations (30 to 70 µg dm⁻³), dynamin is not inhibited / cell death or cytokinesis inhibition does not occur (1 mark).
- As MitMAB concentration increases, more dynamin is inhibited, increasing cell death or decreasing cytokinesis (1 mark).
- At highest concentrations (>2000 up to 8000 µg dm⁻³), all dynamin is inhibited, causing maximum cell death / halted growth (1 mark).
💡 Key Knowledge
- Dynamin has two roles: stimulating cytokinesis and inhibiting cell death. Inhibiting it simultaneously blocks division and triggers apoptosis/death, explaining negative control growth values below the 0.0 baseline.
- Pay close attention to the logarithmic x-axis scale when extracting values.
Calculation: Mass and Concentration
0.01 dm³ of MitMAB solution was added to the treated cells. Calculate the increase in mass of MitMAB (in µg) added to the cells to reduce the cell growth from equal to the control to 0.0 of the control. Show your working. [2 marks]
📐 Step-by-Step Calculation
- Step 1: Read values from Figure 9.
Proportion of control growth at 1.0 (equal to control) corresponds to concentration range 30 to 70 µg dm⁻³.
Proportion of control growth at 0.0 corresponds to concentration 2000 µg dm⁻³. - Step 2: Calculate the change in concentration.
Using 70: 2000 - 70 = 1930 µg dm⁻³
Using 30: 2000 - 30 = 1970 µg dm⁻³ - Step 3: Convert concentration change to total mass.
Mass = Concentration × Volume
Using 1930: 1930 × 0.01 = 19.3 µg
Using 1970: 1970 × 0.01 = 19.7 µg
Answer: Range 19.3 to 19.7 µg
❌ Common Calculation Traps
- Forgetting to multiply by the volume ( 0.01 dm³ ), leaving the answer in concentration units instead of mass.
- Misreading the logarithmic scale on the x-axis for lower values (reading 10 or 100 incorrectly instead of the grid lines for 30 and 70).
Topics
Biology · Practical skills · 3.2 Cells · 3.1 Biological molecules · 3.8 The control of gene expression (A-level only) · Data analysis
Question and mark scheme from the AQA A-Level Biology examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.