AQA A-Level Biology Paper 2, June 2025: Question 7

11 marks · Medium difficulty · Practical Techniques & Data Analysis

Evaluate an investigation on Gammarus pulex response to light intensity in a choice chamber, identify the taxis behaviour, formulate a null hypothesis, and interpret chi-squared results.

Practise this question

Question

Question 07 shows an investigation into the effect of light intensity on the movement of Gammarus pulex using a four-quadrant choice chamber. Figure 3 shows a photograph of G. pulex. Figure 4 shows the choice chamber divided into quadrants A (0%), B (25%), C (50%), and D (100% light intensity). Subquestion 07.1 asks for three features that ensure validity (3 marks). Table 3 provides the count of shrimps after 10 minutes: A: 15, B: 8, C: 4, D: 3. Subquestion 07.2 asks for the type of behaviour and an advantage (2 marks). Subquestion 07.3 asks for a null hypothesis and justification for the chi-squared test (2 marks). Table 4 provides critical values of chi-squared across degrees of freedom 1 to 5 and P values 0.95 to 0.005. Subquestion 07.4 asks for conclusions drawn from the calculated value of 11.87 using Table 4, referencing probability and chance (4 marks).
Question text

07.1 Figure 3 shows Gammarus pulex, commonly known as a freshwater shrimp.

Figure 3

A student used a choice chamber to investigate the effect of light intensity on the

movement of G. pulex.

The student:

• collected 30 shrimps from a freshwater stream

• added freshwater and gravel substrate from the same stream to each segment

• used layers of black paper on the lid of the choice chamber to reduce the

percentage light intensity in 3 segments of the choice chamber as shown in

Figure 4

• placed 30 shrimps at the centre of the choice chamber

• waited 10 minutes before counting the number of shrimps in each segment.

Figure 4

The student stated that ‘the procedure used in this investigation helps to ensure the

validity of any conclusions made’.

Suggest and explain three features of this investigation that justify this statement.

[3 marks]

07.2 Table 3 shows the student’s results.

Table 3

Segment of choice Percentage light Number of G. pulex counted

chamber intensity after 10 minutes

A 0 15

B 25 8

C 50 4

D 100 3

The student observed directional movement by G. pulex.

Name the type of behaviour shown by G. pulex and suggest an advantage to G. pulex

of this behaviour.

Use the information in Table 3 in your answer.

[2 marks]

Type of behaviour

Advantage

The student analysed the results in using the chi-squared test (χ2) and

07.3 Table 3

calculated a χ2 value of 11.87

shows the critical values of χ2 at different probability ( ) values for different

Table 4 P

degrees of freedom.

Table 4

Critical values of χ2 at different values

P

Degrees

P = 0.95 P = 0.5 P = 0.05 P = 0.01 P = 0.005

of freedom

10.00 0.46 3.84 6.64 7.88

20.10 1.39 5.99 9.21 10.60

30.35 2.37 7.82 11.35 12.84

40.71 3.36 9.49 13.28 14.86

51.15 4.35 11.07 15.09 16.75

State a null hypothesis for this investigation and explain why the χ2 test was the

appropriate statistical test to use.

[2 marks]

Null hypothesis

Explanation for using a χ2 test

Using the calculated χ2 value of 11.87 and the information in , what can you

07.4 Table 4

conclude from this investigation?

You should include the terms probability and chance in your answer.

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for question 07 outlining four parts. 07.1 gives 3 marks for: same stream water/gravel to keep conditions natural or control variables; 30/large sample so representative/reliable; placed at centre so equidistant/no bias; 10 minute delay to acclimatise or allow enough time to respond. 07.2 awards 2 marks for negative photo taxis and predation less likely/less visible to predators. 07.3 awards 2 marks for stating no significant difference in shrimp distribution across light intensities, and explaining data are discrete/categorical or comparing observed vs expected. 07.4 gives 4 marks (Route A) for stating calculated chi-squared 11.87 is greater than 7.82/11.35; probability of difference due to chance is less than 0.05/0.01; significant difference; rejecting the null hypothesis.

Question Marking Guidance Mark Comments

1. First 3 options can

1. Same/stream water/gravel so natural/usual/

refer to same

normal habitat/environment

stream/water/gravel or

stream water/gravel

OR

1. Accept ‘so light is

Same/stream water/gravel so normal/usual the only variable’ for

behaviour ‘so variables

controlled’

OR

1. 2. Accept 30/large

Same/stream water/gravel so variables sample so reduces

controlled/same effect of anomalies

OR 2. Ignore ‘so reliable

mean’ obtained

Same/stream water so oxygen/pH controlled/

same; 4. Accept ‘get used to

the conditions’ for

2. 30/large sample so representative/reliable; 3 max acclimatise

07.1 (3 x

3. (Placed at) centre so same distance to each AO3) 4. Enough time to

segment/light intensity move ‘without

qualification’ is

OR insufficient for mark

(Placed at) centre so no bias;

4. (10 minute) delay so shrimps can

acclimatise/equilibrate;

OR

(10 minute) delay so shrimps enough time to

respond/react/distribute

OR

(10 minute) delay so shrimps not stressed–; A-LEVEL BIOLOGY – 7402/2 –

2. Accept ‘consumers’

1. (Negative photo) taxis;

for ‘predators’

2. Predation less likely

2. Accept ‘stops

OR predation’

2. Accept ‘hides from

Less visible to predators/prey

predator/prey’

OR

07.2 (2 x

AO2) 2. Accept descriptions

Less likely to be prey of reduced predation

e.g. ‘fewer are eaten’

OR

2. Accept ‘move to

Move away from heat cooler place

OR /conditions’

– A-LEVEL BIOLOGY – 7402/2 –

Move towards food;

1. There is no (significant) difference in the 1.Accept ‘shrimps

distribution/ movement /behaviour/number of show no preference for

shrimps at different light intensities (different) light

intensity/intensities

OR

1. Accept correlation

(Different) light intensity/intensities

for mark point 1 but

has/have no effect/relationship on/with the

2 ignore for mark point

distribution/movement/behaviour of

07.3 (2 x 2.

shrimp(s);

AO2)

2. The data are discrete/categoric

OR

(To determine) if there is a difference between the

observed and expected

(results/data/values/frequencies);

Note. All 4 marks can

Route A – 4 max

only be obtained via

2 route A.

1. (The calculated / χ (value)/11.87) is

There are 2 max

greater than 7.82 / 11.35;

marks for route B

ignoring mark points

2. The probability of the difference being

1 and 4.

due to chance is less than 5% / 0.05 /

Route B allows credit

1% / 0.01

for understanding of

probability although

OR

context incorrect.

16 Use the route

The probability of the difference not

providing most

being due to chance is more than 95% /

marks.

0.95 / 99% / 0.99;

2. Reject 0.005% /

3. There is a significant difference; 4

0.05% / 50% / 0.001% /

07.4 (4 x

0.01% / 10%

4. Reject the null hypothesis; AO3)

2. Accept P for

Route B – 2 max probability

Ignore mark point 1 and 4

2 and 3. if no reference

to difference is the only

2. The probability of the difference being due to error in both 2 and 3

chance is more than 5% / 0.05 / 1% / 0.01 allow one mark

4. Accept Ho for null

OR

hypothesis

The probability of the difference not being due to

chance is less than 95% / 0.95 / 99% / 0.99;

3. There is no significant difference;

How to answer it

Choice Chambers, Simple Animal Responses and Chi-Squared Analysis

📋 What this question tests

This multi-part question assesses practical methodology, simple organism responses (taxes vs kineses), and statistical interpretation using the chi-squared (χ²) test:

  • Experimental design & internal validity: Identifying control variables, habituation periods, and unbiased starting setups in choice-chamber experiments.
  • Innate animal behaviours: Distinguishing directional responses (taxes) from non-directional movements (kineses) and rationalising their selective survival value.
  • Statistical hypothesis testing: Formulating a rigorous null hypothesis and explaining why categorical count data demands a χ² test.
  • Interpreting critical values: Determining degrees of freedom, comparing test statistics to critical cut-offs, and drawing standard A-Level biological conclusions using precise statistical wording.
Question 07.1

Experimental Validity in Choice Chambers

Suggest and explain three features of this investigation that justify the student's statement regarding validity (3 marks)

✅ Mark Scheme Credit (Any 3 pairs of Suggest + Explain)

  • Feature 1: Using stream water / gravel substrate from the same stream...
    → Explanation: Keeps conditions natural / ensures light is the only variable affecting movement (controls abiotic factors like pH and O₂ concentration).
  • Feature 2: Large sample size (30 shrimps)...
    → Explanation: Ensures data is representative / reduces the impact of anomalous individual behaviours.
  • Feature 3: Placing shrimps in the centre at the start...
    → Explanation: Shrimps are equal distance from each light segment, preventing initial positional bias.
  • Feature 4: 10-minute waiting delay before recording...
    → Explanation: Allows animals time to acclimatise, calm down after handling stress, and freely distribute.

🧠 Exam Technique: Suggest & Explain

This is a paired question: simply quoting the method gets 0 marks. Every feature you extract from the bullet points must have a biological justification:

  • Never just say "it was fair". State exactly which confounding variable was controlled.
  • Never say "so it was reliable" on its own. For a sample size of 30, specify that it is representative or reduces the effect of anomalies.
  • If mentioning the 10-minute delay, state that it permits acclimatisation or recovery from handling stress.

❌ Common Errors

  • Vague timing: Writing "10 minutes gives them time to move" without qualifying that they need time to explore/acclimatise was rejected.
  • Confusing accuracy with validity: Saying "stream water makes it more accurate" instead of explaining that it prevents other variables (pH, salinity) from influencing the choice.
Mark allocation: 3 marks maximum (1 mark for each valid feature + explanation pair). Assesses AO3.
Question 07.2

Classifying Behaviour and Evolutionary Advantage

Name the type of behaviour shown and suggest an advantage to G. pulex (2 marks)

✅ Correct Answer

Type of behaviour:

  • (Negative photo)taxis (Accept: taxis / negative phototaxis)

Advantage:

  • Predation is less likely / less visible to predators (or less visible to prey).
  • Also accept: Moves away from heat / reaches cooler water / moves towards decaying food in dark crevices.

💡 Key Knowledge: Taxes vs Kineses

  • Taxis: A directional movement response to a directional stimulus. Shrimps moved away from the light source into the 0% light area, showing negative phototaxis.
  • Kinesis: A non-directional change in the speed of movement or rate of turning in response to stimulus intensity.
  • Survival value: Freshwater shrimps live under stones/gravel. Remaining in darkness hides them from sighted aquatic predators (fish/birds).

❌ Common Errors

  • Stating kinesis instead of taxis. The question specifically states: "observed directional movement", which is the definition of taxis.
  • Writing just "survival" or "protection" without describing how (e.g. avoiding detection by predators).
Mark allocation: 2 marks (1 mark for behaviour name, 1 mark for biological advantage). Assesses AO2.
Question 07.3

Formulating a Null Hypothesis & Choosing χ²

State a null hypothesis and explain why the χ² test was appropriate (2 marks)

✅ Model Answers

Null Hypothesis:

There is no significant difference in the distribution / number of shrimps between the different light intensities.

OR: Light intensity has no effect on the movement / distribution of shrimps.

Explanation for using χ²:

  • The data is categorical / discrete (counting frequencies/numbers of individuals in separate categories).
  • OR: To determine if there is a significant difference between observed and expected frequencies.

🧠 Exam Technique: Perfect Null Hypotheses

  • Always mention the word "significant" (e.g., "There is no significant difference...").
  • State both variables explicitly: number/distribution of shrimps and different light intensities.
  • Do not confuse difference with correlation: Chi-squared tests for differences in frequencies between categories, not correlation.

❌ Common Errors

  • Saying continuous data: Stating that chi-squared is used for continuous data will lose the mark. Chi-squared is strictly for discrete categorical / frequency count data.
  • Writing an alternative hypothesis: Saying "Light intensity will affect movement" instead of the null condition (no difference / no effect).
Mark allocation: 2 marks (1 for null hypothesis, 1 for justification of χ²). Assesses AO2.
Question 07.4

Interpreting Calculated Values and Drawing Conclusions

What can you conclude from calculated χ² = 11.87 and Table 4? Include "probability" and "chance" (4 marks)

📐 Step-by-Step Statistical Interpretation

  1. Step 1: Determine degrees of freedom (df):
    df = number of categories - 1 = 4 - 1 = 3
  2. Step 2: Identify the critical value:
    At p = 0.05 (standard significance level in biology), critical value for df = 3 is 7.82.
    (Notice calculated χ² 11.87 is also greater than 11.35 at p = 0.01).
  3. Step 3: Compare calculated value to critical value:
    Calculated χ² (11.87) is greater than the critical value of 7.82 (and 11.35).
  4. Step 4: Express probability (P) and chance:
    The probability of the differences being due to chance is less than 5% (p < 0.05) [or less than 1% / p < 0.01].
  5. Step 5: Conclude on significance and the null hypothesis:
    The difference is statistically significant; therefore, reject the null hypothesis.

✅ 4-Mark Model Answer (Route A)

1. The calculated χ² value of 11.87 is greater than the critical value of 7.82 (at p = 0.05, df = 3) [or greater than 11.35 at p = 0.01].
2. The probability that the difference in shrimp distribution is due to chance is less than 0.05 (less than 5%).
3. There is a statistically significant difference in shrimp distribution between light intensities.
4. Therefore, reject the null hypothesis.

❌ Critical Traps to Avoid

  • Wrong degrees of freedom: Using df = 4 (critical value 9.49) instead of 3. There are 4 segments (A, B, C, D), so df = 4 − 1 = 3.
  • Misconverting probability decimals: p = 0.05 is 5%, NOT 0.05%. The mark scheme explicitly states: Reject 0.05% or 0.005%.
  • Omitting required terms: The question commanded you to use both words: probability and chance. Omitting either forfeits mark point 2!
  • Stating results are "due to chance": The calculated value is higher than critical, so results are unlikely to be due to chance.
Mark allocation: 4 marks (1 mark per sequential step in Route A). Assesses AO3.

Topics

Biology · Practical skills · Required Practicals · 3.6 Organisms respond to changes in their internal and external environments (A-level only) · Experimental design · Data analysis · A-Level practicals (7–12)

Question and mark scheme from the AQA A-Level Biology examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.