AQA AS Level Biology Paper 2, November 2021: Question 2
7 marks · Medium difficulty · Practical Techniques & Data Analysis
Describe features of the alveolar epithelium adapted for gas exchange, calculate the FEV1:FVC ratio from a volume-time graph for a person with emphysema, and explain how reduced tidal volume affects carbon dioxide exchange.
Practise this questionQuestion
Question text
02.1 Describe and explain one feature of the alveolar epithelium that makes the epithelium
well adapted as a surface for gas exchange. Do not refer to surface area or moisture
in your answer.
[2 marks]
Doctors measure the health of lungs by calculating the FEV1:FVC ratio.
• FEV1 is the maximum volume of air exhaled in one second.
• FVC is the maximum volume of air exhaled in one breath.
The minimum FEV1:FVC ratio of healthy lungs is 0.7:1
A man with the lung disease emphysema inflated his lungs fully. He then exhaled as
much of this air as quickly as possible in one breath. Figure 3 shows how the volume
of exhaled air changed during this breath.
Figure 3
02.2 Use the information provided to determine the FEV1:FVC ratio of this man’s lungs.
Go on to determine how many times greater the minimum ratio of healthy lungs is than
his ratio.
[2 marks]
FEV1:FVC ratio of man’s lungs =
6How many times greater?
02.3 Tidal volume is the volume of air inhaled and exhaled during a single breath when a
person is resting. The tidal volume in a person with emphysema is reduced compared
with the tidal volume in a healthy person.
Suggest and explain how a reduced tidal volume affects the exchange of carbon
*05* dioxide between the blood and the alveoli.
[3 marks]
Mark scheme
Show the mark scheme
Question Marking Guidance Mark Comments
02.1 Mark in pairs: 1 and 2 OR 3 and 4
1. Flattened cells 1. Reject thin cell
wall/membrane
OR 1. Accept thin cells
1. Accept ‘one cell
Single layer of cells; 2 thick’
2. Reduces diffusion distance/pathway;
3. Permeable; 4. Ignore gas
exchange
4. Allows diffusion of oxygen/carbon dioxide;
02.2 Correct answer for 2 marks = 1.10–1.15;;
Accept for 1 mark, 2
0.6(1) : 1 (correct FEV1 : FEC ratio)
02.3 1. Less carbon dioxide exhaled/moves out (of lung)
OR
More carbon dioxide remains (in lung);
2. (So) reduced diffusion/concentration gradient
(between blood and alveoli); 3
3. Less/slower movement of carbon dioxide out of
blood
OR
More carbon dioxide stays in blood;
TOTAL 7
How to answer it
Gas Exchange and Lung Function Study Guide
What this question tests
This exam sequence evaluates your understanding of gas exchange surfaces, your ability to interpret spirometer-style graphs to calculate lung volume ratios, and your application of diffusion principles to explain how conditions like emphysema disrupt gas exchange gradients in the alveoli.
Alveolar Epithelium Adaptations
✅ Correct Answer Structure
Marks are awarded in linked pairs (1 and 2, OR 3 and 4):
- Point 1: Flattened cells OR single layer of cells.
- Point 2: Reduces diffusion distance/pathway.
- OR Point 3: Permeable.
- Point 4: Allows diffusion of oxygen/carbon dioxide.
💡 Key Knowledge
Fick's Law relies heavily on short diffusion pathways. The alveolar epithelium consists of squamous epithelium (single layer of flat cells) which minimizes the distance gases must travel between alveoli and capillaries.
🧠 Exam Technique
Always read the negative constraints carefully! The question explicitly banned mentioning surface area or moisture. Writing about moisture would instantly lose you credit.
❌ Common Errors
Students frequently write "thin cell membrane" or "thin cell wall". Examiners strictly reject this; you must refer to thin cells or "one cell thick".
Calculating FEV₁:FVC Ratios
📐 Step-by-Step Calculation
Step 1: Determine FEV₁ (max volume in 1 second)
Looking at Figure 3 at time = 1 s, the volume exhaled is approximately 2.25 dm³ (accept 2.2 to 2.3).
Step 2: Determine FVC (maximum total volume exhaled)
The graph plateaus at approximately 3.6 dm³ (accept 3.5 to 3.6).
Step 3: Calculate the ratio (FEV₁ : FVC)
2.25 / 3.6 = 0.625 (Accept range 0.61 to 0.64 : 1 , or simplified to 0.6).
Step 4: Calculate how many times greater healthy lungs are
Minimum healthy ratio = 0.7
Man's ratio = 0.61 (using middle of accepted range)
Calculation: 0.7 / 0.61 = 1.10 to 1.15 times greater.
✅ Final Marked Answers
FEV₁:FVC ratio of man's lungs: 0.61 : 1 (or correct value based on graph readings)
How many times greater: 1.10 to 1.15
❌ Common Calculation Traps
A 1-mark partial credit is available if the student calculates the man's ratio incorrectly but correctly divides the healthy threshold (0.7) by their incorrect ratio. Always show working out!
Effect of Reduced Tidal Volume on Carbon Dioxide Exchange
✅ Correct Answer Breakdown
Awarded via marking points:
- Point 1: Less carbon dioxide exhaled / moves out of lung OR more carbon dioxide remains in the lung.
- Point 2: Reduces the concentration/diffusion gradient between the blood and the alveoli.
- Point 3: Less/slower movement of carbon dioxide out of the blood OR more carbon dioxide stays in the blood.
💡 Key Knowledge
Tidal volume is the volume of air moved in and out per breath. If tidal volume drops, stale air rich in carbon dioxide is not fully replaced by fresh atmospheric air during ventilation, maintaining a higher partial pressure of CO₂ in the alveoli.
🧠 Top-Level Exam Expression
To secure all 3 marks, you must link the physical change (less air expelled) to the driving force of transport (the concentration/diffusion gradient) and finally to the net consequence (reduced gas movement out of blood).
Topics
Biology · Practical skills · 3.3 Organisms exchange substances with their environment · Data analysis
Question and mark scheme from the AQA AS Level Biology examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.