AQA GCSE Biology Biology Paper 2 (Higher), June 2022: Question 7

11 marks · High Demand difficulty · Short Answer

Explain mutations, count cells using a counting grid, calculate the undiluted cell concentration from sample dimensions and dilution factor, and evaluate the haemocytometer method.

Practise this question

Question

Question 7 introduces an investigation of blue algae in pond water caused by a mutation. Part 07.1 asks to define a mutation. A 4-step method describes using a special counting slide with a 0.1 mm depth and 0.2 mm by 0.2 mm grid after diluting 2.5 cm³ pond water to 10 cm³. Figure 8 shows a side cross-section of the slide and coverslip. Figure 9 shows the microscope view of the 0.2 mm square containing algal cells. Part 07.2 asks to count the cells using inclusion/exclusion rules for edges. Part 07.3 is a 6-mark calculation asking for the number of algal cells in 1.0 mm³ of undiluted pond water based on a repeat count of 14 cells. Part 07.4 asks why the water was diluted. Part 07.5 asks to explain how using a thin coverslip that pulls downwards affects the results.
Question text

07 A scientist found a polluted pond which had a new type of blue algae in the water.

The blue colour of the algae was caused by a mutation.

07.1 What is a mutation?

[1 mark]

The scientist measured the number of blue algal cells in a sample of the pond water.

The scientist used a special slide which has a counting grid.

This is the method used.

1. Dilute 2.5 cm3 of pond water to a volume of 10 cm3 with distilled water.

2. Place a drop of the diluted pond water on the special slide, as shown in Figure 8.

3. Place a thick coverslip over the diluted pond water to give a depth of 0.1 mm of

pond water.

4. Use a microscope to count the number of algal cells in a 0.2 mm × 0.2 mm square

on the counting grid.

Figure 8 shows a side view of the special slide.

Figure 8

Figure 9 shows the view of the counting grid through a microscope.

Figure 9

07.2 How many algal cells are in the 0.2 mm × 0.2 mm square in Figure 9?

*26* Use the following procedure:

• Count all cells that are completely within the 0.2 mm × 0.2 mm square in the

counting grid.

• Count cells that are touching the left side or the lower side of the square.

• Do not count cells that are touching the right side or the top side of the square.

[1 mark]

Number of algal cells in the 0.2 mm × 0.2 mm square =

07.3 One week later the scientist repeated the test and counted 14 cells on the

0.2 mm × 0.2 mm counting grid.

Calculate the number of algal cells in 1.0 mm3 of undiluted pond water.

Use the scientist’s second count of 14 cells.

[6 marks]

Number of algal cells in 1.0 mm3 of undiluted pond water =28

07.4 Suggest why the scientist diluted the pond water before placing it on the special slide.

[1 mark]

07.5 A student repeated the scientist’s method.

The student used a thin coverslip over the diluted pond water instead of the

thick coverslip.

The liquid pulled the thin coverslip downwards slightly.

Explain how the use of the thin coverslip would affect the results for the cell count.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 7 shows: 07.1 accepts a change in DNA, base code, base, gene/allele, or chromosome for 1 mark. 07.2 awards 1 mark for 16. 07.3 breaks down 6 marks: volume = 0.004 mm³ (1 mark), 14 ÷ 0.004 (1 mark), 3500 (1 mark), dilution factor x4 (1 mark), 3500 x 4 (1 mark), final answer 14 000 or 1.4 x 10⁴ (1 mark) with error carried forward allowed. 07.4 awards 1 mark for 'to make it easier to count'. 07.5 awards 1 mark for smaller volume and 1 mark for fewer cells / lower cell count. Total 11 marks.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 any one from: 1 AO1

4.6

a change in

• DNA 4.6.1.5

• base code or nucleotide

sequence

• a base (in DNA)

• a gene / allele

• part of a chromosome

• number of chromosomes

• genetic code / material

ignore genetic information

AO /

Spec. Ref.

07.2 16 / sixteen 1 AO3

4.7.2.1

RPA9

AO /

Spec. Ref.

07.3 volume of sample in mm3 AO2

0.004 1 4.7.2.1

RPA9

number of cells in 1 mm3

diluted pond water

14 ÷ 0.004 allow 14 ÷ (0.2 × 0.2 × 0.1) 1

allow use of an incorrectly

calculated volume of 0.04

3 500 allow ecf from answer to q.07.2 1

for number of algal cells

22 correct dilution factor allow dilution = ×4 1

¼ or 4 times

number of cells in 1 mm3

undiluted pond water allow a calculation based on a

3500 × 4 dilution factor of 5 1

14 000 or 1.4 × 104 1

AO /

Spec. Ref.

07.4 to make it easier to count ignore easier to see or more 1 AO3

spread out 4.7.2.1

ignore quicker to count RPA9

AO /

Spec. Ref.

07.5 smaller volume allow (some) liquid / cells would 1 AO3

leak out (from under the cover 4.7.2.1

slip) RPA9

so fewer cells or lower cell allow this mark only if there is 1

count an attempt at an explanation

Total Question 7 11

How to answer it

Counting Algal Cells Using a Hemocytometer Grid

📌 What this question tests

This 11-mark question assesses fundamental genetics recall (mutations) combined with practical microscopy techniques (counting grid sampling rules, dilution factors, 3D volume calculations), and evaluating experimental errors in cell suspension measurements.

Part (a) / Question 07.1 1 Mark

Definition of a Mutation

✅ Correct Answers (Any one)

  • A change in DNA
  • A change in the base sequence (or base code / nucleotide sequence)
  • A change in a gene or allele
  • A change in a chromosome (or number/part of chromosomes)

❌ Common Errors & Examiner Trap

  • "Change in genetic information" is explicitly marked as IGNORE — it is too vague.
  • Writing "a disease" or "an adaptation" rather than stating the biological definition at the molecular/genetic level.

Part (b) / Question 07.2 1 Mark

Applying the Hemocytometer Inclusion/Exclusion Rule

✅ Correct Answer

16 (or sixteen)

Award 1 mark for exactly 16.

🧠 Exam Technique: North/West vs South/East

The question gave strict instructions:

  • INCLUDE: All cells entirely inside + cells touching the Left (West) or Bottom (South) grid lines.
  • EXCLUDE: Any cells touching the Right (East) or Top (North) grid lines.
  • Tip: Cross off cells as you count them, marking touching cells with an "X" or a tick to avoid duplicate counts.

Part (c) / Question 07.3 6 Marks

Multi-Step Quantitative Calculation (Volume & Dilution Factor)

📐 Step-by-Step Calculation

  1. Calculate the volume of the counting area:
    Volume = length × width × depth
    Volume = 0.2 mm × 0.2 mm × 0.1 mm = 0.004 mm³
    Mark 1: for finding sample volume = 0.004 mm³
  2. Find the number of cells per 1.0 mm³ of DILUTED pond water:
    There are 14 cells in 0.004 mm³.
    Cells in 1.0 mm³ (diluted) = 14 ÷ 0.004 = 3500 cells
    Mark 2: expression (14 ÷ 0.004)  |  Mark 3: 3500
  3. Determine the dilution factor:
    2.5 cm³ was diluted to a total volume of 10 cm³.
    Dilution factor = 10 ÷ 2.5 = 4 (or pond water is 1/4 of the total volume).
    Mark 4: correct dilution factor of 4 (or 1/4)
  4. Scale up to UNDILUTED pond water:
    Since the sample was diluted 4-fold, multiply by 4:
    Cells in 1.0 mm³ (undiluted) = 3500 × 4 = 14 000
    Mark 5: expression (3500 × 4)  |  Mark 6: 14 000 (or 1.4 × 10⁴)

❌ Common Errors

  • Mixing units: The grid dimensions are already given in millimetres ( mm ). Do NOT convert them to cm or μm!
  • Wrong dilution factor: Calculating 10 ÷ 2.5 = 4 , but dividing by 4 instead of multiplying at the final step, giving an impossible answer of 875.
  • Using 07.2 count: The question explicitly says: "Use the scientist's second count of 14 cells". Using 16 loses direct marks (though error carried forward is permitted).

💡 Key Mathematical Insight

Alternatively, calculate the total multiplier in one go:

Factor = (1.0 ÷ 0.004) × (10 ÷ 2.5) = 250 × 4 = 1000

Then: 14 × 1000 = 14 000 cells per mm³ .

Part (d) / Question 07.4 1 Mark

Purpose of Dilution in Practical Cell Counting

✅ Correct Answer

To make it easier to count (the cells).

❌ Insufficient / Rejected Answers

  • IGNORE "easier to see" (the microscope handles magnification, not dilution).
  • IGNORE "spread out" (too vague on its own).
  • IGNORE "quicker to count".
  • Must link explicitly to avoiding overlap so cells can actually be counted accurately.

Part (e) / Question 07.5 2 Marks

Evaluating Experimental Error: Thin Coverslip Effect

✅ Marking Points

  • Mark 1 (Cause): The depth decreases, resulting in a smaller volume of liquid in the grid (allow: liquid/cells leak out from under the coverslip).
  • Mark 2 (Effect): There will be fewer cells counted / a lower cell count.

🧠 Examiner Insight: Cause and Effect

This is an "Explain" question worth 2 marks. To score both:

  • State the physical change to the chamber: pulling downward reduces the 0.1 mm depth → smaller volume.
  • Link this to the outcome: a smaller volume over the same 0.2 × 0.2 mm square contains fewer cells.
  • Note: Mark 2 is dependent on an attempt to explain the reason (Mark 1).

Topics

Biology · Required Practicals · B6: Inheritance, Variation and Evolution · B7: Ecology · Required Practicals

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 2 (Higher), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.