AQA AS Level Biology Paper 1, June 2022: Question 4

9 marks · Medium difficulty · Practical Techniques & Data Analysis

Conclude the oxygen uptake across a lugworm's body using partial pressure data, calculate dissolved oxygen volume from saturation and equation, and describe how to use a colorimeter calibration curve to determine pO2.

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Question

Question text

04.1 Lugworms create tubes in the sand on seashores. The tubes are filled with seawater.

A scientist measured the partial pressure of dissolved oxygen (pO2) in seawater at

different places in a tube with a lugworm inside.

Figure 3 shows her results.

Figure 3

The pO2 of dissolved oxygen in lugworm blood is < 2.7 kPa

Using the data in Figure 3, what can you conclude about the uptake of oxygen over

the entire body of the lugworm?

[4 marks]

04.2 Figure 4 shows the oxyhaemoglobin dissociation curve for a lugworm.

Figure 4

The oxygen saturation in the blood of a lugworm is 92%

The lugworm has 0.2 cm3 of blood.

Calculate the volume of dissolved oxygen in the blood of this lugworm using this

equation

CdO2

pO2 =

0.000 031

CdO2 is the concentration of dissolved oxygen in the blood, with units

cm3 oxygen per cm3 of blood.

Show your working.

[3 marks]

Answer cm3

04.3 The intensity of the red colour in blood is affected by the pO2 of the blood.

The intensity of the colour in a solution is measured using a colorimeter.

The scientist used a colorimeter to measure the intensity of red colour in samples of

lugworm blood with different pO2 values. She prepared a calibration curve with this

information.

Describe how the scientist will use information from the colorimeter and her calibration

curve to determine the pO2 in a sample of lugworm blood.

[2 marks]

Mark scheme

Show the mark scheme

Question Marking Guidance Mark Comments

04.1 1. Enters by diffusion; 1. Reject facilitated

diffusion

1 and 2 ‘down a

2. Down a concentration gradient diffusion gradient’ = 2

marks

2. Reject ‘along’ for

From high to low pO2; ‘down’

2. Accept description

4 max of O2 is always

3. More/most across parts of body with gills; (4 x AO3) higher in the water

than the lugworm

4. Gills provide a larger surface area (for 4. Accept Gills

absorption); increase SA:volume

ratio

5. 8.8 (kPa) over gills;

6. 2.4 (kPa - rest of body surface) / 1.9 (kPa -

front end before gills) / 0.5 (kPa - rear end after – LOGY – – JUNE 2022

gills);

04.2 Correct answer for 3 marks, 9.3 × 10–6 /

Accept correct

rounding of 9.3 x 10–6

0.000 0093;;;

MP1 – correct reading from graph (1.5)

MP2 – correct rearrangement of equation 3

(CdO2 = 0.000 031 × their pO2) (3 x AO2)

MP3 – their CdO2 × 0.2

their CdO2 ÷ 5

04.3 1. (Measure light) absorption/transmission; 1. Accept

2. Interpolate/draw line to curve/line then to pO2 ‘absorbance’ for

absorption

OR (2 x AO3)

Read off (pO2 figure) against

absorbance/transmission value obtained;

How to answer it

Gas Exchange and Transport in Lugworms Study Guide

What this question tests

This multi-part exam question assesses your ability to apply diffusion principles to biological data (AO3), manipulate and interpret physiological equations involving graphs (AO2/AO3), and understand practical biochemistry techniques such as colorimetry and calibration curves (AO3).

Question 0.4.1

Oxygen Uptake Along the Lugworm's Body

✅ Correct Answer Structure (Any 4 points)

Oxygen enters by diffusion down a concentration gradient (high to low partial pressure).
Most uptake occurs across the parts of the body with external gills.
Gills provide a larger surface area for absorption.
Data citation: External gills sit where seawater pO₂ is highest ( 8.8 kPa ), whereas the rest of the body absorbs oxygen at lower external pO₂ values ( 2.4 kPa / 1.9 kPa / 0.5 kPa ).

💡 Key Knowledge

Gases move across exchange surfaces passively via simple diffusion.
External gills drastically improve the surface area to volume ratio, enhancing diffusion rates according to Fick's Law.

🧠 Exam Technique

Always support comparative statements with manipulated numerical data from the figure.
State the mechanism clearly ( diffusion ) and specify the direction ( down a concentration gradient ).

❌ Common Errors

Writing "facilitated diffusion" instead of simple diffusion.
Using vague phrases like "along a gradient" instead of "down a concentration gradient" or "high to low pO₂".

Mark allocation: 4 marks available (4 × AO3). Max 4 marks for referencing mechanism, surface area adaptation, and data manipulation.

Question 0.4.2

Calculating Dissolved Oxygen Volume

✅ Correct Answer

Final Answer: 9.3 × 10⁻⁶ cm³ (or 0.0000093 cm³)

📐 Step-by-Step Calculation

Step 1 (Read Graph): Locate 92% saturation on the y-axis of Figure 4, read across to the curve, and read down to find pO₂ = 1.5 kPa .
Step 2 (Rearrange Equation): The equation given is pO₂ = CdO₂ / 0.000031 . Rearrange to solve for concentration: CdO₂ = pO₂ × 0.000031 .
Step 3 (Calculate CdO₂): 1.5 × 0.000031 = 0.0000465 cm³ O₂ per cm³ blood .
Step 4 (Total Volume): Multiply concentration by total blood volume ( 0.2 cm³ ): 0.0000465 × 0.2 = 9.3 × 10⁻⁶ cm³ .

🧠 Exam Technique

Show all intermediate working clearly. Even if your final evaluation contains a minor rounding slip, process marks (MP1, MP2, MP3) can still be secured.
Pay close attention to standard form representation and decimal places.

❌ Common Errors

Reading the wrong axis values from the dissociation curve.
Inverting the equation rearrangement (dividing instead of multiplying by 0.000031).

Mark allocation: 3 marks (3 × AO2). MP1: Correct reading from graph (1.5). MP2: Rearranging/applying equation. MP3: Multiplying by blood volume (0.2).

Question 0.4.3

Using a Colorimeter and Calibration Curve

✅ Correct Answer Structure (2 marks)

Measure light absorption (or transmission/absorbance) of the sample using the colorimeter.
Interpolate / draw a line from the obtained reading on the calibration curve to determine the corresponding pO₂ value.

💡 Key Knowledge

Colorimetry relies on the principle that the intensity of a solution's colour (in this case, red blood colour tied to oxyhaemoglobin states) is proportional to the concentration of the colored species or its correlating factor.

🧠 Exam Technique

Use precise experimental terminology: mention "measuring absorbance/transmission" and "interpolating using the calibration curve".

❌ Common Errors

Stating "extrapolate" instead of "interpolate" when reading values from within the range of a calibration curve.

Mark allocation: 2 marks (2 × AO3). Mark 1: Measuring absorption/transmission. Mark 2: Using the calibration curve to read off the pO₂ value.

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 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.