AQA GCSE Combined Science: Trilogy Biology Paper 2 (Higher), June 2025: Question 3

15 marks · Standard Demand difficulty · Short Answer

Investigate the distribution and percentage cover of buttercup plants at different distances from a tree using gridded quadrats and point quadrats.

Practise this question

Question

Question 3 consists of six sub-questions about sampling buttercup plants using quadrats. Figure 2 shows two 5 by 5 grid quadrats representing buttercups at 1 metre and 10 metres from a tree. Figure 3 compares Quadrat A, a 5 by 5 grid of width 0.25 metres, to Quadrat B, a 10 by 10 grid of width 0.5 metres. Figure 4 illustrates a 10-pin point quadrat of width 0.5 metres on a frame. The sub-questions ask for independent variables, percentage cover calculation, sampling considerations, advantages of quadrat B using area calculations, multiple choice advantages of point quadrats, and a rearrangement calculation to find the number of pin touches from given percentage cover data.
Question text

03 Buttercups are small plants.

Students investigated the distribution of buttercup plants at different distances from

a tree.

Figure 2 shows quadrats being used to sample two areas.

Figure 2

03.1 What was the independent variable in this investigation?

[1 mark]

03.2 Students counted every small square in the quadrat that was at least half covered

by buttercup plants.

Determine the percentage cover of buttercup plants 10 metres away from the tree.

Use Figure 2.

[3 marks]

Percentage cover = %

03.3 The students placed:

• several quadrats 1 metre away from the tree

• several quadrats 10 metres away from the tree.

Give two factors the students should consider when deciding how many times to

place a quadrat at each distance.

Do not refer to time available.

[2 marks]

03.4 Figure 3 shows two different quadrats.

Figure 3

Explain two advantages of using quadrat B instead of quadrat A in the investigation.

You should include calculations in your answer.

[4 marks]

A point quadrat can also be used to estimate the percentage cover of plants in

an area.

Figure 4 shows a point quadrat.

Figure 4

When using a point quadrat, the pins are pushed down to the soil.

The number of pins that touch a buttercup plant is counted.

03.5 What are two advantages of using a point quadrat compared with using

a square quadrat?

[2 marks]

Tick ( ) two boxes.

With a point quadrat, a plant that is smaller than half a square

is more likely to be counted.

With a point quadrat, each species of plant is easier to identify.

With a point quadrat, no judgement of cover is needed.

With a square quadrat, a buttercup plant covering other plants cannot

be counted.

With a square quadrat, the results are less likely to be biased.14

03.6 The percentage cover of buttercup plants can be estimated using the equation:

number of times a pin touched a buttercup plant

*13percentage cover = * × 100

total number of pins used

30 students each collected results from 50 pins.

The students then put all their results together.

Buttercup plants had a percentage cover of 7%.

Calculate the number of times a pin touched a buttercup plant.

[3 marks]

Number of times a pin touched a buttercup plant =

Mark scheme

Show the mark scheme Mark scheme for Question 3 with a total of 15 marks. Question 03.1 awards 1 mark for distance from the tree. Question 03.2 awards 3 marks for identifying 7 squares, showing 7/25 x 100, and calculating 28%. Question 03.3 awards 2 marks for sample representativeness, quadrat size, number of students, or physical obstacles. Question 03.4 awards 4 marks for comparing sample areas (0.25 m² vs 0.0625 m²) and precision/resolution (1% vs 4%). Question 03.5 awards 2 marks for ticking options regarding small plants being counted and no judgement of cover needed. Question 03.6 awards 3 marks for substituting into the formula, rearranging for n, and obtaining 105 touches.

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 distance from the tree allow distance 1 AO2

4.7.2.1

RPA7

AO /

Spec. Ref.

03.2 7 (squares) 1 AO2

× 100 allow 0.28 × 100

allow correct working shown

from an incorrect attempt at 1 AO3

counting squares (in the range 6

to 10) covered by buttercups

28 (%) allow a correct calculation from 1 AO2

an incorrect attempt at counting

squares (in the range 6 to 10)

covered by buttercups 4.7.2.1

RPA7

alternative route

(percentage cover of one small

square)

( =) 4

7 × 4

allow correct working shown

with incorrectly calculated

percentage cover of one small

square

28 (%)

allow a correct calculation from

an incorrect attempt at counting

squares (in the range 6 to 10)

covered by buttercups

AO /

Spec. Ref.

03.3 any two from: 2 AO3

• whether the sample is 4.7.2.1

representative RPA7

• the size of the quadrat

• the number of students allow the number of quadrats

(available)

• how many quadrats fit around allow size of tree

the tree at 1 m allow if the same (quadrat) area

is being sampled repeatedly

• factors such as buildings or

other tree(s)

AO /

Spec. Ref.

03.4 allow converse if clearly AO3

referring to quadrat A 4.7.2.1

RPA7

ignore references to time

(quadrat B) samples a larger allow (quadrat B) is larger / 1

area wider

(because) 0.25 (m2) rather than 1 2 1 2 1

allow (m ) rather than (m )

0.0625 (m2) 4 16

allow 2500 cm2 rather than 625

cm2

allow by 4 times

(quadrat B) gives more accurate allow (quadrat B) is more 1

/ precise results representative

allow (quadrat B) has greater

resolution

allow need to use quadrat A

4 times (as many times) to have

the same accuracy / precision

as quadrat B

ignore (quadrat B) is more

reliable / valid / repeatable

(because) to the nearest 1% (of allow (because) quadrat B has 1 15

the quadrat) rather than nearest 100 (small) squares rather than

4% 25 (small) squares

do not accept quadrat B has

smaller squares (within quadrat)

AO /

Spec. Ref.

03.5 with a point quadrat, a plant that 1 AO3

is smaller than half a square is 4.7.2.1

more likely to be counted RPA7

with a point quadrat, no 1

judgement of cover is needed

16 AO /

Spec. Ref.

03.6 n n 1 AO2

7 = × 100 allow 7 = × 100

30 × 50 1500 4.7.2.1

7 allow n = 0.07 × 1500 1

n = × 1500

100 allow n = 7 × 15

allow a correct rearrangement

using 50 pins as total number of

pins used

allow correct rearrangement

with an incorrect calculation of

number of pins used

n = 105 allow a correct calculation using 1

50 pins as total number of pins

used

allow correct calculation with an

incorrect calculation of number

of pins used

allow 2 marks for:

( × 50 =) 3.5

Total Question 3 15

How to answer it

Sampling Plant Populations: Quadrats & Point Frames

📋 What This Question Tests (Required Practical 7)

This question assesses practical ecology skills from Topic 4.7 (Ecology):

  • Identifying experimental variables (independent, dependent, control) in fieldwork.
  • Estimating percentage ground cover using gridded square quadrats based on standard counting rules.
  • Evaluating fieldwork sampling design, sample size, and factors affecting quadrat placement.
  • Comparing sampling apparatus (calculating area and grid resolution of different quadrats).
  • Comparing square quadrats with point quadrats and rearranging equations to calculate frequencies from pooled class data.
Question 03.1 • 1 Mark

Identifying the Independent Variable

Identifying what is changed by the investigator

✅ Correct Answer

Distance from the tree (or simply distance).

Award [1 mark] for stating distance from the tree.

💡 Key Knowledge

  • Independent Variable: The factor deliberately changed by the experimenter (here: 1 m vs 10 m).
  • Dependent Variable: What is measured (buttercup distribution / % cover).
  • Control Variables: Factors kept constant to ensure valid results.
Question 03.2 • 3 Marks

Calculating Percentage Cover in a Gridded Quadrat

Counting squares "at least half covered" and converting to a percentage

📐 Step-by-Step Calculation

  1. Count qualifying squares: Inspect the 10 m quadrat (5 × 5 grid = 25 total squares). Count squares that are at least half (≥ 50%) covered by buttercups.
    Count = 7 squares
  2. Calculate fraction of total area:
    (7 ÷ 25) × 100 or each square is worth 100% ÷ 25 = 4% , so 7 × 4% .
  3. Calculate final percentage:
    Percentage cover = 28%
[1 mark] for 7 squares identified.
[1 mark] for correct formula: (7 ÷ 25) × 100.
[1 mark] for 28%.

❌ Common Errors & Traps

  • Counting squares with tiny patches: The rule states at least half covered. Counting squares with only tiny tips (e.g. counting 9 or 10) loses mark 1, though error carried forward (ECF) is allowed for math working if in range 6–10.
  • Dividing by 100 instead of 25: The grid is 5 × 5 (25 squares total), not 100 squares!
Question 03.3 • 2 Marks

Fieldwork Planning & Sample Size

Factors to consider when deciding sample size (excluding time)

✅ Any TWO from the Mark Scheme

  • Whether the sample is representative (large enough sample size).
  • The size of the quadrat used.
  • The number of students (or number of quadrats available).
  • How many quadrats physically fit around the tree at 1 m (circumference / area limit).
  • Obstacles / environmental factors such as buildings, paths, or other trees.
Award [1 mark] for each valid factor up to a maximum of [2 marks].

🧠 Exam Technique: Read the Restrictions!

The question explicitly says: "Do not refer to time available."

Any mention of "not having enough time" or "how long the lesson lasts" scores 0 marks. Focus on biological and physical constraints (representativeness, spatial limits around the tree, or equipment).

Question 03.4 • 4 Marks

Comparing Quadrats: Area & Grid Resolution

Explaining two advantages of Quadrat B over Quadrat A using calculations

📐 Calculations Required

Advantage 1: Quadrat B covers a larger total area

  • Area of Quadrat A = 0.25 m × 0.25 m = 0.0625 m² (or 625 cm²)
  • Area of Quadrat B = 0.5 m × 0.5 m = 0.25 m² (or 2500 cm²)
  • Quadrat B samples 4 times larger area!

Advantage 2: Quadrat B gives higher precision / resolution

  • Quadrat A: 5 × 5 = 25 squares → each square = 4%
  • Quadrat B: 10 × 10 = 100 squares → each square = 1%
  • Quadrat B estimates cover to the nearest 1% instead of 4%.

✅ Mark Breakdown (4 Marks)

  • Mark 1: Quadrat B samples a larger area (or is more representative).
  • Mark 2: Calculation showing 0.25 m² vs 0.0625 m² (or 4 times larger).
  • Mark 3: Quadrat B gives more accurate / precise results (or higher resolution).
  • Mark 4: Calculation showing percentage resolution: to the nearest 1% rather than nearest 4% (or 100 squares vs 25 squares).

❌ Common Errors

  • Confusing terms: Do not claim Quadrat B is "more reliable" or "valid" by itself without linking it to sample area or precision.
  • Vague statements: Saying "it has smaller squares" without quoting numbers or calculating percentage resolution (1% vs 4%) loses mark 4.
Question 03.5 • 2 Marks

Point Quadrat vs Frame Quadrat

Evaluating the advantages of using pins

✅ Two Correct Statements (Tick Boxes)

  • ☑️ With a point quadrat, a plant that is smaller than half a square is more likely to be counted.
  • ☑️ With a point quadrat, no judgement of cover is needed.
Award [1 mark] for each correct box ticked. Maximum [2 marks].

💡 Why Are These True?

  • No subjective judgement: A pin either physically touches a leaf or it doesn't (objective binary data). With square quadrats, students must estimate whether a plant covers ≥ 50% of a square.
  • Detecting small plants: Tiny leaves that never fill half a square are ignored in standard square quadrats, but can easily touch a needle-thin pin.
Question 03.6 • 3 Marks

Formula Rearrangement: Point Quadrat Data

Rearranging the percentage cover formula to find the number of pin touches

📐 Step-by-Step Calculation

Given formula:
Percentage cover = (number of touches ÷ total pins used) × 100

  1. Find the total number of pins used:
    30 students each used 50 pins:
    Total pins = 30 × 50 = 1500 pins
    [1 mark] for substituting 1500 into equation
  2. Rearrange the formula to make "touches (n)" the subject:
    7 = (n ÷ 1500) × 100
    n = (7 ÷ 100) × 1500  →  n = 0.07 × 1500
    [1 mark] for correct rearrangement
  3. Calculate final answer:
    n = 105
    [1 mark] for final answer of 105

❌ Common Trap: Forgetting the 30 Students

Many students calculate 7% of 50 pins: (7 ÷ 100) × 50 = 3.5 .

The question states 30 students put ALL their results together. If you calculate 3.5, you only score 2 marks (partial credit for using 50 instead of 1500). Always underline total sample quantities!

Topics

Biology · B7: Ecology

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Biology Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.