AQA GCSE Biology Biology Paper 1 (Foundation), June 2024: Question 2

8 marks · Low Demand difficulty · Short Answer

Explain human skin defence against pathogens and analyse experimental data on the survival of bacteria at different pH levels.

Practise this question

Question

Question 02 begins with part 02.1 asking how the skin defends the human body against pathogens for 1 mark. A method is then described investigating the effect of acid on bacteria across pH 1, 2, 3, and 5 over 24 hours. Table 1 shows live bacteria counts at 0 hours (210, 210, 210, 216) and 24 hours (23, X, 63, 185). Question 02.2 asks for the fraction of bacteria surviving at pH 3 in simplest form for 2 marks. Question 02.3 asks candidates to calculate how many more bacteria were killed at pH 1 than pH 5 in 24 hours by completing three calculation steps for 3 marks. Question 02.4 states a student calculated value X as 43 and asks to suggest how this value was calculated for 2 marks.
Question text

02 Pathogens cause disease.

02.1 How does the skin defend the human body against pathogens?

[1 mark]

The stomach contains acid to kill pathogens.

A scientist investigated the effect of acid on the survival of bacteria.

This is the method used.

1. Prepare four test tubes each with 10 cm3 of culture solution.

2. Use acid to adjust the pH of the solutions to be pH1, pH2, pH3 and pH5

3. Add 1 cm3 of bacteria mixture to each test tube.

4. Take a 0.1 cm3 sample from each test tube and record the number of live bacteria.

5. Keep the test tubes at 37 °C for 24 hours.

6. Repeat step 4.

Table 1 shows some of the results.

Table 1

Number of live bacteria

Time in

hours

pH1 pH2 pH3 pH5

0 210 210 210 216

24 23 9 X 63 185

02.2 What fraction of the bacteria present at 0 hours for pH3 survived for 24 hours?

Give your answer in its simplest form.

[2 marks]

Fraction surviving =

02.3 How many more bacteria were killed at pH1 than at pH5 in 24 hours?

Complete the following steps.

[3 marks]

Calculate the number of bacteria killed at pH1

Calculate the number of bacteria killed at pH5

Calculate how many more bacteria were killed at pH1 than at pH5

Number =

02.4 A student calculated value X in Table 1 to be 43

Suggest how the student calculated this value.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 02 gives the answers: 02.1 accepts physical barrier or stops pathogens entering, or antimicrobial secretions/sebum/sweat (1 mark). 02.2 gives 1 mark for 63/210 and 1 mark for simplifying to 3/10 (or 0.3). 02.3 awards 1 mark for 187 killed at pH 1, 1 mark for 31 killed at pH 5, and 1 mark for 156. 02.4 awards 2 marks for stating the student calculated the midpoint or mean between 23 and 63, with alternative working shown as (63 - 23)/2 = 20 followed by addition or subtraction, or (23 + 63)/2 = 43.

Question 2

AO /

Question Answers Extra information Mark

Spec Ref.

02.1 allow named pathogen AO1

throughout 4.3.1.6

(physical) barrier 1

or

stops pathogens entering (blood

/ body)

allow produces antimicrobial

secretions

allow produces oil / sebum /

sweat

ignore reference to scabs / clots

AO /

Spec Ref.

02.2 63 1 AO2

210 4.3.1.6

4.2.2.1

3 allow 0.3 1

10 ignore 30%

if neither mark awarded allow

210 10

= for 1 mark

63 3

AO /

Spec Ref.

02.3 ignore negative symbol AO2

throughout 4.3.1.6

(at pH1) 187 (killed) 1 4.2.2.1

(at pH5) 31 (killed) 1

(187 – 31) = 156 allow correct subtraction using 1

(more bacteria killed) MARKincorrect calculation at pH1SCHEME– – –

and/or pH5

AO /

Spec Ref.

02.4 (the student) calculated the 1 AO3

midpoint 4.3.1.6

4.2.2.1

(between) 23 and 63 allow (between values at) pH1 1

and pH3

OR

any two from:

63 – 23

• = 20

• 20 + 23 = 43

• 63 – 20 = 43

23 + 63

allow 2 = 43 for 2 marks

allow plot data from Table 1 on

graph (1) then read off value for

pH2 (1)

allow other correct methods for

up to 2 marks

Total Question 2 8

How to answer it

Human Non-Specific Defences & Bacterial Survival at Different pH Values

📌 What this question tests

This question assesses knowledge of the body's primary non-specific defence systems (Topic 4.3: Infection and Response) and mathematical processing of experimental microbiology data.

  • Recall: How the skin acts as an external physical and chemical barrier against entry of pathogens.
  • Data Handling: Calculating fractions in their simplest form from experimental counts.
  • Multi-step Calculations: Determining differences in numbers of bacteria killed under acidic conditions.
  • Scientific Reasoning (AO3): Understanding how missing intermediate data points can be estimated (calculating a mean/midpoint).

Context: Investigating Acid and Bacteria

Scientists investigated how acid kills bacteria by monitoring culture samples over 24 hours at 37 °C across different pH levels:

Time in hours Number of live bacteria
pH1 pH2 pH3 pH5
0 210 210 210 216
24 23 X 63 185
Question 02.1

Skin Defences Against Pathogens

1 Mark • AO1 (Specification Reference: 4.3.1.6)

✅ Correct Answers (Any 1 of the following)

  • Forms a (physical) barrier.
  • Stops pathogens entering the blood or body tissues.
  • Produces antimicrobial secretions.
  • Produces oil, sebum, or sweat (which inhibit bacterial growth).

🧠 Exam Technique

The question asks how the skin defends against pathogens, not how damage is repaired. Focus on the intact skin's role as a first line of defence before pathogens can reach the bloodstream.

❌ Common Errors & Examiner Notes

  • Do NOT mention scabs or blood clotting: Clotting is a response to a cut or wound, not how intact skin normally functions. The mark scheme explicitly states: "ignore reference to scabs / clots".
  • Vague answers like "it protects the body" without stating how (e.g. by acting as a barrier) will not score.
Mark distribution: [1 mark] for correctly identifying that skin forms a barrier or produces antimicrobial substances.
Question 02.2

Simplest Fraction Calculation

2 Marks • AO2 (Specification Reference: 4.3.1.6 / 4.2.2.1)

📐 Calculation: Step-by-Step

  1. Identify data from Table 1 at pH3:
    Number present at 0 hours = 210
    Number surviving at 24 hours = 63
  2. Form initial fraction:
    63 / 210 [1 mark]
  3. Simplify by dividing numerator & denominator by common factors:
    Divide both by 7: 63 ÷ 7 = 9 and 210 ÷ 7 = 30 → 9 / 30
    Divide both by 3: 9 ÷ 3 = 3 and 30 ÷ 3 = 10 → 3 / 10 [1 mark]

✅ Final Answer

Fraction surviving = 3/10 (or decimal equivalent 0.3 )

Mark 1: Initial fraction 63 / 210 .
Mark 2: Fully simplified form 3 / 10 .

❌ Common Errors

  • Inverting the fraction: writing 210 / 63 = 10 / 3 scores only a maximum of 1 mark.
  • Writing a percentage: the mark scheme explicitly states "ignore 30%" because the question requested a fraction.
  • Failing to simplify down completely to single-digit numerator 3 / 10 .
Question 02.3

Comparing Bacteria Killed at pH1 vs pH5

3 Marks • AO2 (Specification Reference: 4.3.1.6)

📐 Three-Stage Structured Calculation

Step 1: Calculate bacteria killed at pH1

210 - 23 = 187

Step 2: Calculate bacteria killed at pH5

216 - 185 = 31

Step 3: Difference in bacteria killed

187 - 31 = 156

🧠 Exam Technique: Follow the Prompts

This question splits the calculation onto three distinct printed lines. Always write each sub-calculation clearly on its allocated line. Even if you make an arithmetic error in Step 1 or 2, you can still gain the final mark via Error Carried Forward (ECF) if your final subtraction matches your previous answers.

✅ Mark Scheme Allocation

  • Mark 1: 187 (bacteria killed at pH1)
  • Mark 2: 31 (bacteria killed at pH5)
  • Mark 3: 156 (difference)

Note: The examiner will ignore negative signs if numbers were subtracted in reverse (e.g. 23 - 210 = -187).

Question 02.4

Suggesting How Value X Was Calculated

2 Marks • AO3 (Analysis & Evaluation)

💡 What Was the Student Doing?

Value X represents live bacteria remaining at pH2 at 24 hours. The student does not have experimental data for pH2, so they estimated it from the surrounding values at pH1 (23) and pH3 (63).

✅ Acceptable Answers (Any 1 full method for 2 marks)

Method A (Mean / Midpoint):

  • Calculated the mean / midpoint / average [1 mark] between 23 and 63 (or values at pH1 and pH3) [1 mark].
  • Showing formula: (23 + 63) / 2 = 86 / 2 = 43 [2 marks].

Method B (Step Difference):

  • (63 - 23) / 2 = 20 [1 mark]
  • 23 + 20 = 43 OR 63 - 20 = 43 [1 mark]

Method C (Graphical):

  • Plotted known data points on a graph [1 mark], then read off the value at pH2 [1 mark].

❌ Common Errors

  • Simply stating "they guessed" or "they estimated" without explaining the mathematical basis.
  • Forgetting to state which numbers were averaged (must mention 23 and 63 or pH1 and pH3).

Topics

Biology · B3: Infection and Response · B2: Organisation

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 1 (Foundation), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.