AQA A-Level Biology Paper 3, June 2025: Question 1

8 marks · Medium difficulty · Short Answer

Describe binary fission in bacteria, calculate percentage increase in bacterial population from log10 data, and explain how an ATP synthase inhibitor kills mycobacteria.

Practise this question

Question

Question 01 consists of three parts. 01.1 asks to describe binary fission in bacteria (3 marks). 01.2 presents Figure 1, a bar chart showing Log10 number of bacteria against Temperature (°C) for two cultures: at 20 °C the bar reaches 4.1, and at 24 °C it reaches 5.2. Students must calculate the percentage increase to the nearest whole number using the antilog (10^x) values (2 marks). 01.3 explains that bedaquiline inhibits ATP synthase in mycobacteria by stopping proton movement, and asks to explain how it kills these bacteria (3 marks).
Question text

01.1 Describe binary fission in bacteria.

[3 marks]

01.2 Microbiologists investigated the effect of two temperatures on the population growth of

a species of bacterium.

They grew separate bacterial cultures in identical nutrient media at 20 °C and 24 °C

for 8 hours.

Figure 1 shows the estimated final population size of each culture.

Figure 1

The bacterial population at 24 °C is larger than that at 20 °C

Calculate this difference as a percentage increase.

The actual number of bacteria can be calculated from the log10 value by using the

10x button on your calculator.

Give your answer to the nearest whole number.

Show your working.

[2 marks]

Percentage increase =

01.3 Bedaquiline is a drug that inhibits ATP synthase in mycobacteria by stopping

hydrogen ion (proton) movement.

Explain how bedaquiline can kill these bacteria.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for questions 01.1 to 01.3. 01.1 awards up to 3 marks for: replication of circular DNA; replication of plasmids; and division of cytoplasm or new membrane forming. 01.2 awards 2 marks for the correct answer 1159, with 1 mark for evidence of intermediate values (e.g. 10^5.2 - 10^4.1, or 1158 without rounding, or 145900 increase). 01.3 awards 3 marks for: fewer/no protons diffuse through ATP synthase; less/no energy to add phosphate to ADP or form ATP / less oxidative phosphorylation; and bacteria die due to lack of ATP for cellular or metabolic processes.

Question Marking Guidance Mark Comments

1. Replication of (circular) DNA; 1. and 2. Accept

doubling for replication

2. Replication of plasmids;

1. Reject

3. (New) membrane forms (dividing cell) chromosome

1. Reject mitosis

OR 1. Reject single

strand of DNA

Division of cytoplasm (to produce daughter 3

01.1 cells); (3 x 2. Ignore (named)

AO1) organelles

replicate

3. Accept

descriptions of

either alternative

3. Accept cytokinesis

(occurs)

Correct answer of 1159 = 2 marks;;

Evidence of 1158 (incorrect rounding) = 1 mark

OR

Evidence of 1158 followed by any number of

decimal places (not to nearest whole number)

= 1 mark

OR

Evidence of 1259 (answer that failed to subtract

100) = 1 mark

OR

Evidence of 12 (not multiplied by 100) = 1 mark

OR

Evidence of 145 900(.0651) (calculation of

increase in number of bacteria) = 1 mark

OR

Evidence of 92 (dividing by 105.2 instead of 104.1) =

1 mark

OR 2

01.2 Evidence of 27 (percentage increase of log values (2 x

read from graph) = 1 mark AO2)

OR 5

Evidence of the following working (calculation

method including correct powers of 10) = 1 mark;

105.2 − 104.1

4.1 × 100

OR

105.2

( 4.1 × 100) − 100

OR

158489(.32) − 12589(.25)

× 100

12589(.25)

OR

145900(.07)

× 100

12589(.25)

OR

158489(.32)

× 100) − 100

12589(.25)

+ 1. Accept no/less

1. Fewer/no protons/hydrogen ions/H

chemiosmosis

diffuse/move through ATP synthase;

1. Reject reference to

protons/hydrogen

ions/H+ diffusing if

in mitochondrion

1. Accept

hydrogen/H for

protons/hydrogen

ions/H+ (as told

hydrogen ion in

stem of question)

2. Less/no energy to add phosphate to ADP 2. or 3. Reject less/no

energy produced once

OR

2. Accept P , Pi or

PO 3– for

Less/no energy for phosphorylation of ADP 4

phosphate

OR 3 2. Reject

01.3 (3 x phosphorous/P

Less/no energy to form ATP AO2)

3. Accept named

OR process e.g.

synthesis of an

Less/no oxidative phosphorylation; organic molecule,

binary fission,

DNA replication,

3. (Bacteria die due to) less/no ATP/energy for active transport,

cellular/metabolic processes; etc.

3. Reject ‘less/no

energy/ATP for

respiration’ unless

qualified e.g.

glucose to glucose

phosphate

3. Reject incorrect

metabolic process

for a bacterium,

e.g. muscle

contraction

How to answer it

Bacterial Growth, Binary Fission & Chemiosmosis

What This Question Tests

This question evaluates foundational prokaryotic biology, mathematical data processing with logarithms, and the biochemical mechanism of ATP synthesis:

  • Prokaryotic Cell Division: Stages of binary fission (DNA/plasmid replication and cytokinesis).
  • Mathematical Data Handling: Converting logarithmic data ( log₁₀ ) to linear numbers and calculating percentage increase.
  • Biochemistry & Respiration: Mechanism of chemiosmosis across bacterial membranes, ATP synthase function, and cellular consequences of halting ATP generation.
Part 01.1 • 3 Marks

Binary Fission in Bacteria

Describe binary fission in bacteria.

✅ Mark Scheme Criteria

  1. Replication of (circular) DNA [1 mark]
  2. Replication of plasmids [1 mark]
  3. Division of cytoplasm to produce daughter cells OR formation of a new cell membrane dividing the cell [1 mark]
Note: "Doubling" is accepted for replication. "Cytokinesis" is accepted for division of cytoplasm.

💡 Key Biological Principles

  • Bacteria are prokaryotes; they have no nucleus and contain a single, circular main DNA molecule alongside smaller variable loops called plasmids.
  • Plasmids may replicate multiple times, whereas the circular DNA molecule replicates once.
  • The cytoplasm pinches inward as new cell walls and membranes form, producing two genetically identical daughter cells.

❌ Critical Traps & Misconceptions

  • Mitosis: Never mention "mitosis" or "chromosomes". Mitosis only occurs in eukaryotes.
  • Single strand: Do NOT write "single strand of DNA" — bacterial DNA is double-stranded, just arranged in a closed loop.
  • Organelles: Claiming organelles replicate loses credit (bacteria lack membrane-bound organelles).

🧠 Exam Technique

Treat this as a sequential process with distinct structures. Mention both genetic elements independently (main circular DNA and plasmids) before stating cytoplasmic division.

Part 01.2 • 2 Marks

Logarithmic Data & Percentage Increase

Calculate the percentage increase in bacterial population from 20 °C to 24 °C.

📐 Step-by-Step Calculation

  1. Read the log values from Figure 1:
    Value at 20 °C = 4.1
    Value at 24 °C = 5.2
  2. Convert log₁₀ values to actual bacterial counts:
    At 20 °C: 10⁴·¹ = 12,589.25
    At 24 °C: 10⁵·² = 158,489.32
  3. Calculate absolute difference:
    158,489.32 − 12,589.25 = 145,900.07 [1 mark for correct method / working]
  4. Calculate percentage increase relative to the original (20 °C) population:
    Percentage increase = (145,900.07 ÷ 12,589.25) × 100 = 1158.93%
  5. Round to nearest whole number:
    1159% [2 marks for correct final value]

✅ Final Answer

1159%

Award 2 marks for 1159 with or without working. Award 1 mark if unrounded (e.g. 1158.9) or incorrectly rounded down to 1158.

❌ Common Calculation Pitfalls

  • Percentage of log values: Working directly with log values: (5.2 − 4.1) ÷ 4.1 × 100 = 26.8% (yields max 1 mark if method shown). You must take the antilog first!
  • Wrong denominator: Dividing by the larger value (10⁵·²) yields 92% (1 mark). Always divide by the initial/comparison value (20 °C).
  • Forgetting to subtract 100%: Calculating (10⁵·² ÷ 10⁴·¹) × 100 = 1259% without subtracting the original 100% (1 mark).
Part 01.3 • 3 Marks

Action of Bedaquiline (Inhibiting ATP Synthase)

Explain how bedaquiline can kill mycobacteria by stopping hydrogen ion (proton) movement.

✅ Mark Scheme Points

  1. Fewer / no protons (H⁺) diffuse / move through ATP synthase OR no / reduced chemiosmosis [1 mark]
  2. Less / no energy available to phosphorylate ADP (to add phosphate / Pᵢ to ADP / to form ATP) OR no oxidative phosphorylation [1 mark]
  3. Bacteria die because there is insufficient ATP / energy for vital cellular / metabolic processes [1 mark]

💡 Chemiosmosis in Bacteria

  • Bacteria generate a proton gradient by pumping H⁺ across their cell surface membrane into the periplasmic space.
  • Protons diffuse back down their electrochemical gradient into the cytoplasm through the channel in ATP synthase.
  • The flow of H⁺ provides the kinetic / electrical energy needed for the catalytic subunit of ATP synthase to combine ADP and inorganic phosphate (Pᵢ).

❌ Common Examiner Penalties

  • Mitochondria reference: Mentioning the mitochondrial inner membrane or matrix immediately loses Mark 1. Bacteria lack mitochondria!
  • Energy "produced": Writing "no energy is produced" is chemically incorrect (first law of thermodynamics) and rejected. Always write "energy released" or "ATP synthesized".
  • "Phosphorous" / "P": Using the single letter P or writing "phosphorous" instead of phosphate / Pᵢ loses Mark 2.
  • Inappropriate examples: Citing "muscle contraction" as a named metabolic process in bacteria loses Mark 3. Stick to active transport, protein synthesis, or binary fission.

🧠 Linking Cause to Effect

Structure your response logically: Direct molecular blocker (H⁺ movement through enzyme stopped) → Biochemical consequence (no ADP + Pᵢ phosphorylation) → Cellular fatality (lack of ATP halts vital metabolic reactions like active transport).

Topics

Biology · Practical skills · 3.2 Cells · 3.5 Energy transfers in and between organisms (A-level only) · Data analysis

Question and mark scheme from the AQA A-Level Biology examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.