AQA AS Level Biology Paper 1, June 2022: Question 8
7 marks · Medium difficulty · Practical Techniques & Data Analysis
Calculate the concentration of undiluted bacterial cells and sketch a bar chart showing the effects of antimicrobial substances J and K on E. coli growth.
Practise this questionQuestion
Question text
08 A student investigated the effect of two antimicrobial substances, J and K, on the
growth of E. coli bacteria.
She transferred E. coli cells using a sterilised pipette to make three identical cultures,
1, 2, and 3. She then added:
• no antimicrobial substance to culture 1
• antimicrobial substance J to culture 2
• antimicrobial substance K to culture 3.
She incubated the cultures for 24 hours, after which she determined the number of
cells per mm3 in each culture.
08.1 The student used a sterilised pipette to transfer E. coli into each culture.
Suggest why the number of E. coli cells per mm3 in each culture after 24 hours might
have been lower if the student had not used a sterilised pipette. Explain your answer.
[2 marks]
08.2 The student diluted 3 cm3 of culture 1 with 12 cm3 of water. She observed a sample
of this diluted mixture using an optical microscope and counted 24 cells in
0.000 25 mm3 of the diluted mixture.
Use this information to calculate the number of cells per mm3 in undiluted culture 1.
[2 marks]
19 3
Number of cells = per mm
08.3 After 24 hours, the student compared the number of cells per mm3 in cultures 1, 2
and 3. She found:
• substance J killed 80% of the cells
*18* • substance J killed twice as many cells as substance K.
Using the axes shown in Figure 9, sketch a bar chart showing the results the student
obtained from cultures 1, 2 and 3.
Do not draw a grid on the chart.
Do not include figures for the number of cells per mm3
[3 marks]
Figure 9
Mark scheme
Show the mark scheme
Question Marking Guidance Mark Comments
08.1 1. Unknown/new/different 1. Ignore chemical
microorganisms/pathogens/microbes/bacteria contaminant
(introduced);
2. (these bacteria) use food/space 2 2. Accept description
(2 x AO2) of competition for
OR other resources
(these bacteria) produce toxins;
08.2 Correct answer, for 2 marks = 480 000;; 4.8 x 105 = 2 marks
Accept for 1 mark,
–3
96 000 (correct number mm )
120 (correct number in 0.000 25 mm3 undiluted)
OR any one of: (2 x AO2)
0.2 ×5 (correct dilution factor)
OR
– LOGY – – JUNE 2022
Evidence of dividing by 0.000 25
08.3 Correct answer for 3 marks, 3 bars given For labelling of x axis
1. Correct labelling of x axis; Accept ‘no substance’
for culture 1
eg culture 1, culture 2 and culture 3
Accept ‘substance J’
OR for culture 2
Accept ‘substance K’
1,2,3 and (bacterial) culture;
for culture 3
2. Height of culture 2 bar about one fifth height
Accept ‘substance’ for
of culture 1 bar;
culture
3. Height of culture 3 bar higher than culture 2
bar and lower than culture 1 bar;
If 2 bars given OR if histogram given OR if 3
graph given, accept for 2 marks, (3 x AO2)
1. Correct labelling of x axis;
2. Correct relative height of 2 bars/coordinates;
eg height of culture 2 about one fifth height of
culture 1
14 OR
height culture 3 above height of culture 2
If no bar chart/histogram/graph given, accept for
1 mark,
1. Correct labelling of x axis;
How to answer it
Investigating Antimicrobial Substances on E. coli Growth
What this question tests
This question assesses your ability to apply microbiological techniques, perform multi-step dilution and cell concentration calculations, and visually interpret experimental data through scientific sketching. You are tested on understanding aseptic technique rationale (preventing contamination), dilution factors, concentration formulas, and proportional bar chart representation.
Aseptic Techniques and Population Dynamics
Suggest why the number of E. coli cells per mm³ might be lower if a sterilised pipette was not used, and explain your answer. [2 marks]
✅ Correct Answer
- Point 1: Introduces unknown/new/different microorganisms, pathogens, microbes, or bacteria.
- Point 2: These contaminant bacteria use up food/space (resources) OR produce toxins.
💡 Key Knowledge
- Aseptic technique prevents the introduction of unwanted microorganisms from the environment, skin, or equipment.
- Contaminants introduce interspecific competition within the culture broth.
🧠 Exam Technique
- This is a "suggest and explain" question. You must first state what enters (contaminants) and then give the mechanism of how it lowers cell yield (competition for resources or toxin production).
- Note: Ignore generic references to "chemical contaminants"; the focus must be on living biological agents.
❌ Common Errors
- Vaguely stating "contamination" without specifying that foreign microorganisms are introduced.
- Failing to link the presence of foreign bacteria to a limiting factor such as competition for nutrients or space.
Dilution and Cell Concentration Calculations
Calculate the number of cells per mm³ in undiluted culture 1. [2 marks]
✅ Correct Answer
480 000
or
4.8 × 10⁵
(Award 1 mark for correct intermediate cell concentration in the diluted mixture: 96 000, or for correct identification of the dilution factor/division by 0.00025).
📐 Step-by-Step Calculation
- Find total volume of diluted mixture: 3 cm³ (culture) + 12 cm³ (water) = 15 cm³ total volume.
- Calculate the dilution factor: Dilution factor = Total volume / Culture volume = 15 / 3 = 5 (meaning it is diluted 1-in-5, or multiplied by 5 to go back).
- Calculate cell concentration in the diluted mixture: Cells per mm³ = Counted cells / Volume observed = 24 / 0.00025 mm³ = 96 000 cells mm⁻³.
- Scale up to find the concentration in the undiluted culture: 96 000 × 5 = 480 000 cells mm⁻³ (or 4.8 × 10⁵ ).
❌ Common Calculation Traps
- Using 12 cm³ instead of 15 cm³ as the denominator when calculating the dilution ratio.
- Multiplying by 0.00025 instead of dividing by it when working out cells per mm³.
- Forgetting to multiply by the dilution factor at the final step, leaving the answer as 96 000.
Data Presentation and Graph Sketching
Sketch a bar chart showing the results obtained from cultures 1, 2 and 3. [3 marks]
✅ Correct Answer Structure
- Mark 1: Correct labelling of the x-axis (e.g., Culture 1, Culture 2, Culture 3, or 'No substance', 'Substance J', 'Substance K').
- Mark 2: Height of Culture 2 bar is about one-fifth (20%) the height of Culture 1 bar.
- Mark 3: Height of Culture 3 bar is higher than Culture 2 bar, but lower than Culture 1 bar.
💡 Key Knowledge & Data Breakdown
- Culture 1 (Control): Baseline growth (maximum height).
- Culture 2 (Substance J): Killed 80% of cells, meaning 20% remain alive. Therefore, bar 2 must be 1/5th the height of bar 1.
- Culture 3 (Substance K): Substance J killed twice as many cells as Substance K. This means Substance K only killed 40% of cells (leaving 60% alive). Therefore, bar 3 must be higher than bar 2, but lower than bar 1.
🧠 Exam Technique
- Read negative constraints carefully: Do not draw a grid on the chart, and do not include numerical figures on the y-axis.
- Ensure bars are distinct and separated (as it is a bar chart, not a histogram).
Topics
Biology · Practical skills · Required Practicals · 3.2 Cells · Data analysis · AS practicals (1–6)
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.