AQA AS Level Biology Paper 1, June 2022: Question 6

7 marks · Medium difficulty · Practical Techniques & Data Analysis

Calculate the ratio of the mass of iron ions in the plasma of a person with haemochromatosis to that of a healthy person using given data and explain iron ion roles and transport mechanisms.

Practise this question

Question

An exam question with three parts. Part 06.1 asks to explain a property of iron ions enabling their role in red blood cells for 2 marks. Part 06.2 includes Figure 7 showing the hormone hepcidin hydrolysing ferroportin and controlling iron ion concentration from cell cytoplasm to blood plasma, and asks to use this to explain high iron plasma concentration in haemochromatosis for 3 marks. Part 06.3 provides numerical data on iron ion mass and concentration to calculate a ratio for 2 marks.
Question text

06.1 Explain a property of iron ions that enables these ions to carry out their role in red

blood cells.

[2 marks]

06.2 The hormone hepcidin controls the iron ion concentration in blood plasma. Hepcidin

affects ferroportin, the iron ion channel protein in cell-surface membranes.

Figure 7 shows how hepcidin controls the iron ion concentration in plasma.

Figure 7

People with the disease haemochromatosis do not produce hepcidin.

Use information in Figure 7 to explain why the iron ion concentration is higher in the

plasma of people with haemochromatosis.

[3 marks]

06.3 The mass of iron ions in the plasma of a person with haemochromatosis is 6104 μg

The iron ion concentration in the plasma of a healthy person is 50 μg dm–3

The volume of blood in each of these people is 4000 cm3

Calculate the ratio of the mass of iron ions in the plasma of the person with

haemochromatosis to the mass of iron ions in the plasma of the healthy person.

[2 marks]

Answer

Mark scheme

Show the mark scheme Mark scheme showing answers for questions 06.1, 06.2, and 06.3. Question 06.1 awards 2 marks for iron being charged/polar or part of haemoglobin to bind/transport oxygen. Question 06.2 awards 3 marks for points about reduced ferroportin hydrolysis, more ferroportin in cell membranes, and more iron transport into blood plasma. Question 06.3 awards 2 marks for the correct ratio of 30.52:1 or equivalent working out.

Question Marking Guidance Mark Comments

2+

06.1 1. (Is) charged/polar 1. Accept Fe OR

3+

Fe for ‘charged’

OR

(Is) part of haem(oglobin);

2. (So) binds/associates/loads (with) oxygen

OR (2 x AO1)

(So) forms oxyhaemoglobin

OR

2. Accept ‘carries’ for

(So) transports oxygen; transports

06.2 1. Less/no ferroportin hydrolysis/breakdown; 1. and 2. Accept

‘channel protein’ for

2. (So) more ferroportin (in cell-surface

3 ferroportin

membranes);

(3 x AO3)

2. and 3. Accept

3. (So) more iron (ion) transport from – LOGY – – JUNE 2022

‘many’ for more

cytoplasm/cell;

06.3 Correct answer for 2 marks = 30.52:1 / 30.5:1 /

31:1;;

Accept for 1 mark,

31 (ratio not given)

OR

30:1 (incorrect rounding)

OR

200 (correct mass in healthy person)

(2 x AO2)

OR

1526 (correct iron concentration in person with

haemochromatosis)

OR Accept for 1 mark any

correct ratio (not

simplified) e.g. 763:25 11

6104 : 200 (correct ratio, but not simplified)

or 1526:50

How to answer it

Iron Ions, Haemochromatosis, and Plasma Calculations

What this question tests

This exam sequence evaluates your understanding of biological molecules and ions (specifically inorganic iron ions in haemoglobin), your ability to interpret unfamiliar biochemical pathway flowcharts, and your quantitative problem-solving skills involving concentrations, unit conversions, and ratios.

Part 06.1 (2 Marks)

Function and Properties of Iron Ions

✅ Correct Answer

Any two of the following marking points:

  • Ions are charged / polar (Accept Fe²⁺ or Fe³⁺).
  • Ions are part of haem (haemoglobin).
  • Ions bind, associate, or load with oxygen (forming oxyhaemoglobin) to transport oxygen.

💡 Key Knowledge

Inorganic ions have specific roles depending on their properties. Iron ions (Fe²⁺) sit at the centre of the haem group in haemoglobin proteins, temporarily binding oxygen molecules in the lungs and releasing them in respiring tissues.

🧠 Exam Technique

Read command words carefully. Explain requires you to link a property of the ion directly to how it allows the red blood cell to perform its specific physiological role.

❌ Common Errors

Students often state general properties of water or ions without linking them back to red blood cells and oxygen transport, or they incorrectly refer to iron atoms rather than ions (Fe²⁺).

Part 06.2 (3 Marks)

Interpreting Biochemical Pathways (Haemochromatosis)

✅ Correct Answer

To gain all 3 marks using Figure 7:

  • Point 1: There is less or no hydrolysis/breakdown of ferroportin (due to absence of hepcidin).
  • Point 2: Therefore, more ferroportin remains in the cell-surface membranes.
  • Point 3: As a result, more iron ions are transported from the cell cytoplasm into the blood plasma.

💡 Key Knowledge

Figure-based questions test your ability to track directional arrows and relational blocks. Follow the pathway step-by-step: hormone missing → protein not broken down → increased transport across membranes.

🧠 Exam Technique

Always anchor your explanation using the specific names provided in the prompt and diagram (e.g., ferroportin , hydrolysis , cytoplasm , plasma ). Do not rely on outside biological knowledge not shown in the flowchart.

❌ Common Errors

Vague references like "protein is destroyed" instead of using the flowchart's specific term hydrolysed ferroportin , or missing the causal link between membrane protein levels and iron transport rates.

Part 06.3 (2 Marks)

Quantitative Calculations & Ratios

✅ Correct Answer

30.52 : 1 (or 30.5 : 1, or 31 : 1)

Award 2 marks for a correct final ratio. Award 1 mark for intermediate correct values if the final ratio is incorrect or unsimplified (e.g., 6104 : 200).

📐 Step-by-Step Calculation

  1. Mass in haemochromatosis patient: Given directly as 6104 µg .
  2. Calculate mass in healthy person: Concentration = 50 µg dm⁻³ , Volume = 4000 cm³ . Convert volume to dm³: 4000 cm³ = 4 dm³ .
  3. Mass = Concentration × Volume = 50 µg dm⁻³ × 4 dm³ = 200 µg .
  4. Calculate ratio: 6104 : 200 → divide both sides by 200 to simplify: 30.52 : 1 .

🧠 Exam Technique

Always show your intermediate working! If your final division or rounding goes wrong, working out individual masses (200 µg and 6104 µg) secures the fallback 1 mark.

❌ Error Traps to Avoid

  • Forgetting to convert cm³ to dm³ (failing to divide 4000 by 1000).
  • Failing to format as a ratio relative to 1 (e.g. writing just 30.52 without : 1 loses a mark if specified).

Topics

Biology · 3.1 Biological molecules

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.