AQA A-Level Biology Paper 1, June 2025: Question 8

11 marks · Medium difficulty · Practical Techniques & Data Analysis

Explain antibiotic resistance in hospitals using natural selection, calculate and interpret Spearman rank correlation coefficient for plasmid length and copy number, calculate total plasmid length, and analyze plasmid replication and binary fission during population growth phases.

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Question

Question text

08.1 Strains of bacteria with antibiotic resistance are common in hospitals.

Use your knowledge of selection to explain why.

[3 marks]

08.2 Antibiotic resistance alleles are located in plasmids.

There are 11 different types of plasmid inside cells of the bacterium

Bacillus thuringiensis.

A scientist determined the:

• length of each type of plasmid in kilobase pairs (kbp)

• mean plasmid copy number (mean number of each type of plasmid per cell).

Table 5 shows the scientist’s results for 5 of these plasmids.

Table 5

Rank of

Mean Squared

Rank of mean

Plasmid Plasmid plasmid difference

plasmid plasmid

identity length / kbp copy between

length copy 2

number ranks (d )

number

A 8 172 2 5 9

B 95 6 4 2 4

C 12 35 3 3 0

D 2 139 1 4 9

E 294 3 5 1 16

The Spearman rank correlation coefficient (rs) between plasmid length and mean

plasmid copy number is determined using this equation.

6 ∑ d2

rs = 1 − 2

n (n − 1)

where

Σd2 = the sum of all d2 figures

n = 5 (the highest rank number)

Use information in Table 5 and the formula to calculate the rs value for this set of data.

Show your working.

[2 marks]

24 r =

08.3 The scientist determined that the rs value between plasmid length and mean plasmid

copy number for all 11 types of plasmid is –0.94

*23What can you conclude from this r*s value?

[1 mark]

08.4 The length of the single circular (non-plasmid) DNA molecule in a B. thuringiensis cell

is 5400 kbp

The total length of all plasmid DNA molecules in a B. thuringiensis cell is 65% greater

than the length of the single circular DNA molecule.

Calculate the total number of kbp in all the plasmid molecules in a B. thuringiensis

cell.

Show your working.

[2 marks]

Answer26 kbp

08.5 A growing population of B. thuringiensis cells includes a:

• germination phase

• growth phase

• dormant phase.

A scientist incubated a culture of B. thuringiensis cells at 20 °C

At intervals, the scientist:

• measured the cloudiness of the culture

• determined the mean plasmid B copy number per cell.

Cloudiness is a measure of the number of cells in a culture.

Figure 11 shows the scientist’s results.

Figure 11

Using information in Figure 11, what can you conclude about binary fission and the

replication of plasmid B in the different phases of a growing population of

B. thuringiensis?

[3 marks]

Mark scheme

Show the mark scheme

Question Marking Guidance Mark Comments

1. Antibiotic use is common/regular (in hospitals)

2. and 3. Reject

OR immunity for

resistance only once

(Resistant) bacteria easily/readily/often spread

(in hospitals); Ignore reference to

2. Bacteria with resistant allele (survive), mutation

reproduce and pass on allele; 3

08.1 3. (So resistant) allele frequency increases (3 x 2 and 3. Penalise

AO1) gene for allele once

only

(So) number of bacteria with (resistant) allele

increases; 2. Accept

divide/replicate OR

binary fission for

reproduce

Accept for 2 marks,

Correct answer of – 0.9 = 2 marks;;

- 0.9 as a fraction

228 (correct numerator) = 1 mark

120 (correct denominator) = 1 mark

08.2 (2 x

AO2)

1.9 (correct calculated quotient) = 1 mark

0.9 (correct answer but positive integer) = 1 mark;

Strong/significant negative correlation; 1

08.3 – A-LEVEL BIOLOGY(1 x – 7402/1 –

AO3)

Correct answer of 8910;;

3510 (correct difference between total number of

plasmid kbp and non-plasmid kbp) = 1 mark

08.4 OR (2 x 15

x1.65 (correct multiplication factor for percentage AO3)

change) = 1 mark

Incorrect answer, but correct answer shown in

working = 1 mark;

Germination phase Each mark point must

be clearly linked to the

1. No DNA replication correct phase.

Accept correct time

OR references to indicate

each phase

No plasmid (B) replication

Accept references to

OR numbers to identify

each phase, eg first

phase for ‘germination

No binary fission;

phase’

Growth phase

3 max Accept ‘cell division’

08.5 2. Plasmid (B) replication occurs (3 x for ‘binary fission’

AO3)

OR Ignore ‘cloudiness’

and ‘number of cells’

(Rate of) plasmid (B) replication increases

Binary fission (only) occurs;

Dormant phase

3. Plasmid (B) replication occurs and no binary

fission;

How to answer it

Bacterial Selection, Plasmid Dynamics & Spearman's Rank

📌 What This Question Tests

This multi-step question assesses core biological understanding alongside quantitative and data analysis skills across five distinct sub-questions:

Natural selection in bacteria: Explaining antibiotic resistance using precise evolutionary terminology (selection pressure, survival, reproduction, allele frequency).
Statistical calculation (Spearman's rank correlation coefficient): Substituting values into the formula rₛ = 1 - (6Σd² / (n(n² - 1))) accurately.
Statistical interpretation: Drawing concise, scientifically valid conclusions from calculated correlation coefficients.
Percentage calculations: Calculating a value that is a given percentage greater than a baseline quantity.
Data analysis & biological processes: Interpreting population growth stages to draw conclusions about plasmid replication and binary fission.

Question 08.1 • 3 Marks

Natural Selection & Antibiotic Resistance in Hospitals

Explaining why resistant strains are common using selection

✅ Model Answer (3 Marks)

Selection pressure / transmission: Antibiotics are frequently or regularly used in hospitals (acting as a strong selective agent) OR resistant bacteria spread easily between patients. [1 mark]
Differential survival and reproduction: Bacteria carrying the resistant allele survive and reproduce (divide by binary fission), passing the resistant allele on to offspring. [1 mark]
Change in population: This increases the frequency of the resistant allele (or increases the number of resistant bacteria) in the hospital population. [1 mark]

❌ Common Errors & Pitfalls

"Immunity" vs. "Resistance": Writing that bacteria become "immune" to antibiotics. Immunity involves an immune system (antibodies, lymphocytes); bacteria are simple prokaryotes and possess resistance. Mentioning immunity loses credit.
"Gene" instead of "Allele": Referring to passing on the "gene" rather than the specific allele. AQA penalises this vocabulary slip.
Antibiotics "cause" mutations: Stating that antibiotics cause the resistance mutation. Mutations arise spontaneously; antibiotics only select for pre-existing mutations.

🧠 Exam Technique: Natural Selection Checklist

Always structure selection answers with this 4-step sequence:

Pressure (antibiotic use) → Survival (possess advantageous allele) → Reproduction (binary fission / passing allele on) → Frequency (allele frequency rises in gene pool).

Question 08.2 • 2 Marks

Calculating Spearman's Rank Correlation Coefficient (rₛ)

Applying the rank correlation formula using Table 5

📐 Step-by-Step Calculation

Formula: rₛ = 1 - [6Σd² / (n(n² - 1))]

Calculate Σd²: Sum the squared difference column:
Σd² = 9 + 4 + 0 + 9 + 16 = 38
Calculate the numerator:
6 × Σd² = 6 × 38 = 228 [1 mark if seen in working]
Calculate the denominator: Since n = 5 (5 plasmids):
n(n² - 1) = 5(5² - 1) = 5(25 - 1) = 5 × 24 = 120 [1 mark if seen in working]
Calculate the quotient:
228 / 120 = 1.9
Subtract from 1 to find rₛ:
rₛ = 1 - 1.9 = -0.9

Final Answer: rₛ = -0.9 (2 marks)

❌ Common Errors

Losing the minus sign: Writing +0.9 gives only 1 mark. Remember 1 - 1.9 = -0.9 .
Wrong n value: Using n = 11 because the stem mentioned 11 plasmid types in total. The table only has data for n = 5 .
Order of operations error: Calculating (1 - 228) / 120 instead of 1 - (228 / 120) .

Question 08.3 • 1 Mark

Interpreting the Correlation Coefficient

Drawing a conclusion from rₛ = -0.94

✅ Model Answer (1 Mark)

There is a strong (or significant) negative correlation between plasmid length and mean plasmid copy number.

🧠 Exam Technique: Two-Part Conclusions

When asked to conclude from a correlation coefficient, always specify both:

Direction: Negative (since the value is below zero).
Strength: Strong (since -0.94 is very close to -1.0 ).

Note: Do not state that plasmid length causes a lower copy number; correlation does not imply causation.

Question 08.4 • 2 Marks

Percentage Greater Calculation

Calculating total plasmid DNA from circular genomic DNA

📐 Step-by-Step Calculation

Single circular DNA length = 5400 kbp .
Total plasmid DNA is 65% greater than this.

Method 1 (Direct multiplier):

"65% greater" means the total is 100% + 65% = 165% of the original.
Multiplication factor = 1.65 [1 mark]
5400 × 1.65 = 8910 kbp [2 marks]

Method 2 (Calculate difference and add):

Find 65% of 5400: 5400 × 0.65 = 3510 kbp [1 mark]
Add to original: 5400 + 3510 = 8910 kbp [2 marks]

Final Answer: 8910 kbp

❌ Common Errors

Stopping at 3510 kbp: Calculating 65% of 5400 and forgetting to add it back to 5400. This earns only 1 mark.
Rounding incorrectly: The result is an exact whole number; do not round to significant figures unless instructed.

Question 08.5 • 3 Marks

Phases of Bacterial Population Growth & Plasmid Replication

Evaluating binary fission and plasmid B replication across phases

✅ Model Answer (3 Marks - 1 per phase)

Germination phase (Phase 1): No DNA replication occurs OR no plasmid (B) replication occurs OR no binary fission / cell division occurs. [1 mark]
Growth phase (Phase 2): Plasmid (B) replication occurs (rate of replication increases) OR binary fission (only) occurs. [1 mark]
Dormant phase (Phase 3): Plasmid (B) replication occurs AND no binary fission occurs. [1 mark]

🧠 Exam Technique: Structuring by Named Phase

Clear Headings: The mark scheme explicitly states that each mark point must be clearly linked to the correct phase (or exact time interval/phase number).
Address both processes: The question asks about both binary fission and plasmid B replication. In the dormant phase, distinguishing that plasmids replicate while cells do not divide is key to securing full marks.
Avoid irrelevant descriptors: Examiners instruct to ignore mentions of "cloudiness", "turbidity", or general "number of cells". Focus strictly on binary fission and plasmid replication.

Topics

Biology · Practical skills · 3.2 Cells · 3.4 Genetic information, variation and relationships between organisms · Data analysis

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