AQA GCSE Biology Biology Paper 1 (Higher), June 2025: Question 8

9 marks · Standard Demand difficulty · Extended Answer

Answer questions comparing bacterial and yeast cell structures, comparing antibiotic effectiveness from a growth curve graph, and estimating bacterial population density.

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Question

Question 8 includes diagrams of a rod-shaped bacterial cell with plasmids and a loop of DNA, and a yeast cell showing a nucleus, vacuole, and mitochondria. Sub-question 8.1 is a multiple-choice question on prokaryotic versus eukaryotic cells. Sub-question 8.2 asks for two features of a yeast cell not found in a bacterial cell. Figure 11 presents a line graph showing the number of bacteria per cm³ over 16 hours when treated with Antibiotics A, B, and C. Sub-question 8.3 asks to compare the effectiveness of the antibiotics based on the graph. Sub-question 8.4 asks how to use ten 0.02 cm³ samples to estimate the bacterial population per cm³ when cells are too crowded to count directly.
Question text

08 Figure 10 shows a bacterial cell and a yeast cell.

Figure 10

08.1 Which statement about bacterial cells and yeast cells is correct?

[1 mark]

Tick ( ) one box.

Bacterial cells and yeast cells are eukaryotic.

Bacterial cells and yeast cells are prokaryotic.

Bacterial cells are eukaryotic and yeast cells are prokaryotic.

Bacterial cells are prokaryotic and yeast cells are eukaryotic.

08.2 Give two features of a yeast cell that are not found in a bacterial cell.

Use Figure 10.

[2 marks]

A scientist investigated the effect of different antibiotics on one type of bacterium.

The scientist:

• used three samples of culture medium

• added a different antibiotic to each sample

• added equal numbers of bacteria to each sample.

Figure 11 shows the results.

Figure 11

08.3 Compare the effectiveness of the antibiotics shown in Figure 11.

[4 marks]

08.4 It was not possible for the scientist to count the total number of bacteria per cm3 of

culture medium.

The cells were too close together to be counted accurately.

The scientist took 10 samples from the culture medium.

Each sample had a volume of 0.02 cm3.

Describe how the scientist could use the samples to estimate the number of bacteria

in 1 cm3 of culture medium.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 8 with a total of 9 marks: 08.1 accepts 'bacterial cells are prokaryotic and yeast cells are eukaryotic' (1 mark). 08.2 accepts any two from nucleus, vacuole, mitochondria, chromosomes, or large ribosomes (2 marks). 08.3 is a 4-mark leveled response comparing Antibiotics A (least effective, slowest drop, highest peak), B (intermediate), and C (most effective, earliest drop, reduces to zero). 08.4 awards 1 mark for diluting each sample and counting under a microscope, and 1 mark for calculating the mean and scaling up by multiplying by 50 (or total by 5).

Question 8

AO /

Question Answers Extra information Mark

Spec Ref.

08.1 bacterial cells are prokaryotic 1 AO1

and yeast cells are eukaryotic 4.1.1.1

AO /

Spec Ref.

08.2 any two from: 2 AO3

• nucleus 4.1.1.1

• vacuole 4.1.1.2

• mitochondria

allow chromosomes

allow large ribosomes

if no other mark awarded allow 1

mark for membrane bound sub-

cellular structures

AO /

Question Answers Mark

Spec Ref.

08.3 Level 2: Scientifically relevant features are identified; the way(s) in 3-4 AO3

which they are similar / different is made clear and (where 4.1.1.6

appropriate) the magnitude of the similarity / difference is noted.

Level 1: Relevant features are identified and differences noted. 1-2

No relevant content. 0

Indicative content:

• all three antibiotics have similar / no effect in the first 4 / 5 hours

antibiotic A

• A is least / less effective

o because A starts to reduce the number of bacteria latest

o because A reduces the number of bacteria the slowest

o because the (maximum) number of bacteria using A is the

highest

o because A does not reduce the number of bacteria as much

as C and / or B

o because with A there is still a (much) larger number of

bacteria left at 16 hours

antibiotic B

• B is more effective than A

• B is less effective than C

o because B starts to reduce the number of bacteria after C

o because B starts to reduce the number of bacteria before A

o because B reduces the number of bacteria more than A

o because B reduces the number of bacteria quicker than A

o because B reduces the number of bacteria slower than C

antibiotic C

• C is most / more effective

o because C starts to reduce the number of bacteria soonest

o because C reduces the number of bacteria the fastest

o because the (maximum) number of bacteria with C is the

lowest

o because C is the (only) antibiotic to reduce the number of

bacteria to zero

For Level 2, all three antibiotics must be compared.

AO /

Spec Ref.

08.4 dilute each sample and count 1 AO1

the number of bacteria (using a 4.1.1.6

microscope) 4.2.2.5

find the mean number of allow find the mean number of 1

bacteria and scale / multiply up bacteria and multiply by 50

allow find the total number of

bacteria and multiply by 5

Total Question 8 9

How to answer it

Cell Structure and Investigating Antibiotic Effectiveness

📋 WHAT THIS QUESTION TESTS

Core GCSE Biology Specifications: 4.1.1.1, 4.1.1.2, 4.1.1.6, and 4.2.2.5

  • Cell Biology: Distinguishing between eukaryotic and prokaryotic cell structures using diagrams.
  • Data Analysis (AO3): Interpreting and comparing line graphs showing bacterial population changes over time when treated with antibiotics.
  • Quantitative Sampling Techniques: Explaining serial dilution and the mathematical scaling required to estimate microbial populations.
QUESTION 08.1 • 1 MARK

Classifying Bacteria and Yeast

Cell classification: prokaryotic vs eukaryotic

✅ Correct Answer

Bacterial cells are prokaryotic and yeast cells are eukaryotic.

awarded 1 mark for correctly ticking the fourth box.

💡 Key Knowledge

  • Prokaryotes (e.g., bacteria): Lack a true nucleus; their genetic material is a single loop of free DNA and plasmids. They also lack membrane-bound organelles.
  • Eukaryotes (e.g., animals, plants, fungi): Yeast is a single-celled fungus, meaning it possesses a distinct nucleus and membrane-bound organelles.

❌ Common Misconceptions

Students often mistake yeast for a bacterium because both are microscopic single-celled organisms. Remember: yeast is a fungus, which places it firmly in the eukaryotic kingdom.

QUESTION 08.2 • 2 MARKS

Comparing Sub-Cellular Structures

Identifying features in yeast that are absent in bacteria using Figure 10

✅ Correct Answers (Any Two)

  • Nucleus (allow: nuclear membrane / chromosomes)
  • Vacuole (large central permanent vacuole visible in diagram)
  • Mitochondria (visible oval shapes with folds)
1 mark for each correct structure identified (max 2 marks). If neither is named, 1 mark could be awarded for identifying "membrane-bound organelles".

🧠 Exam Technique: Use the Diagram

The question specifies "Use Figure 10". Look closely at the yeast cell drawing:

  • The large round shaded circle is the nucleus.
  • The large empty clear circular space in the center is the vacuole.
  • The small sausage-shaped structures with inner lines are mitochondria.

❌ Common Errors

  • Naming Chloroplasts: Yeast cells are fungi, not plants—they never have chloroplasts.
  • Giving structures bacteria also have: Stating cell wall, cytoplasm, or cell membrane scores zero because bacteria also have them.
  • Stating "ribosomes": Both have ribosomes (though bacterial ribosomes are smaller, simply writing "ribosomes" is not creditworthy).
QUESTION 08.3 • 4 MARKS

Comparing Antibiotic Effectiveness

Extended response: interpreting population growth and death curves

💡 Marking Scheme Rubric (Level of Response)

  • Level 2 (3–4 marks): Scientifically relevant comparisons are made across all three antibiotics (A, B, and C). The ways they are similar and different are clear, and comparative/magnitude terms are used.
  • Level 1 (1–2 marks): Relevant features and simple differences are identified, but descriptions may be isolated or miss comparisons across all three.

✅ Model Comparative Response

  • Initial similarity: All three antibiotics show no significant effect for the first 4 to 5 hours, with bacteria multiplying at similar rates.
  • Antibiotic C is the most effective:
    • Starts reducing bacteria earliest (peaks at ~4.5 hours).
    • Reduces the population the fastest (steepest negative gradient).
    • Has the lowest peak bacterial count.
    • Is the only antibiotic that kills all bacteria (count drops to 0 at ~12 hours).
  • Antibiotic B is intermediate:
    • More effective than A, but less effective than C.
    • Starts reducing later than C (at ~8 hours) but before A.
    • Reduces the population to a very low number by 14 hours, but not to zero.
  • Antibiotic A is the least effective:
    • Takes the longest time to act (bacteria grow until ~8.5 hours).
    • Reduces bacteria the slowest.
    • Leaves the highest number of surviving bacteria at 16 hours.

🧠 Top-Grade Strategy

To secure a Level 2 (4/4 marks), follow this 3-step checklist:

  1. Mention all three: Rank them clearly (C is most effective, B is intermediate, A is least effective).
  2. Identify key graph phases:
    • Initial lag / delay (0–4 hours)
    • Time when numbers start falling (onset of bactericidal effect)
    • Final outcome at 16 hours (C = 0, A = high).
  3. Quote time milestones: e.g., "Antibiotic C reaches zero bacteria at around 12 hours".
QUESTION 08.4 • 2 MARKS

Estimating Bacterial Populations

Dilution and volume scaling calculation

✅ Correct Method (Mark Scheme)

  1. Dilute each sample and count the number of bacteria using a microscope [1 mark].
  2. Calculate the mean number of bacteria per sample and scale/multiply up to 1 cm³ [1 mark].

📐 Calculations & Scaling Factors

Two valid mathematical routes to scale up to 1 cm³:

Route 1: Per individual sample

Each sample volume = 0.02 cm³

Scaling factor = 1 cm³ ÷ 0.02 cm³ = 50

Total in 1 cm³ = Mean count per sample × 50

Route 2: Combining all 10 samples

Total volume counted = 10 × 0.02 cm³ = 0.2 cm³

Scaling factor = 1 cm³ ÷ 0.2 cm³ = 5

Total in 1 cm³ = Sum of 10 samples × 5

❌ Common Errors & Traps

  • Missing the dilution step: The question states the cells were "too close together to be counted accurately". You must explain how to fix this: by diluting the sample (serial dilution).
  • Multiplying by 10 instead of 50: 10 is the number of samples, not the volume conversion factor! 0.02 cm³ goes into 1 cm³ exactly 50 times.
  • Forgetting to find the mean: Taking repeat samples is only useful if you calculate an average to improve reliability.

Topics

Biology · B1: Cell Biology · B3: Infection and Response

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.