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Edexcel B · 9BI0 / 8BI0 Papers 1–3 A-level Biology Maths skills

Edexcel B A-level Biology Maths — Worked Solutions

Edexcel B A-level Biology (Pearson’s “Biology B”, 9BI0) puts real statistics in the exam — the Student’s t-test, Spearman’s rank and chi-squared — alongside the standard biology maths, and these questions cost marks every year. Paper 3 is the maths-heaviest of the three. Below are fully worked examples spanning Papers 1–3, taught step by step the way an examiner wants to see them, plus a searchable database of every maths question in the past papers.

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Worked examples

A spread of the maths skills examined across Edexcel B Papers 1–3: the Student’s t-test, Spearman’s rank correlation and chi-squared, the Hardy–Weinberg equation, the index of diversity, standard deviation, magnification, the surface area of a sphere, efficiency of energy transfer and the bacterial growth rate constant. Paper 3 carries the heaviest maths load. Click any example to expand the full method. Video walk-throughs are being added.

1

The Student’s t-test — calculating t

Paper 3 (A2), 2025, Q8(a)(ii)Statistics3 marks

Tar-spot counts on leaves were compared at two sites. Near the road: mean x̄₁ = 12.93, S₁ = 6.78, n₁ = 15. Parkland: mean x̄₂ = 17.07, S₂ = 4.06, n₂ = 15. Using t = |x̄₁ − x̄₂| ÷ √[(S₁²÷n₁) + (S₂²÷n₂)], calculate the value of t.

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The method

  1. Top line — difference between the means: 17.07 − 12.93 = 4.14
  2. Work out each S²÷n term: S₁²÷n₁ = 6.78² ÷ 15 = 45.97 ÷ 15 = 3.06 S₂²÷n₂ = 4.06² ÷ 15 = 16.48 ÷ 15 = 1.10
  3. Add them and square-root (the bottom line): √(3.06 + 1.10) = √4.16 = 2.04
  4. Divide top by bottom for t: t = 4.14 ÷ 2.04 = 2.03
Answer: t = 2.03. This then feeds into the critical-value comparison (df = n₁ + n₂ − 2 = 28; critical value at p = 0.05 is 2.05).
The t-test is Edexcel B’s signature test for comparing two means. Build it in two halves: the top line is the difference between the means (4.14); the bottom line is √[(S²/n) added for each group]. Work the two S²/n terms separately — square the SD, then ÷ n — add, square-root (2.04), then divide. Don’t round until the very end.
2

Spearman’s rank correlation coefficient

Paper 3 (A2), 2024, Q8(b)(i)Statistics3 marks

Lesser celandine abundance was ranked against distance from a path for 9 sampling points. The sum of the squared rank differences is Σd² = 236 and n = 9. Using rₛ = 1 − [6Σd² ÷ n(n² − 1)], calculate rₛ to three significant figures.

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The method

  1. Numerator 6Σd²: 6 × 236 = 1416
  2. Denominator n(n² − 1): 9 × (9² − 1) = 9 × 80 = 720
  3. Divide: 1416 ÷ 720 = 1.967
  4. Subtract from 1: rₛ = 1 − 1.967 = −0.967
Answer: rₛ = −0.967. The negative result signals a strong negative correlation — abundance falls as distance from the path rises.
Spearman’s measures correlation between two ranked variables. Compute the two parts of the fraction separately (6Σd² = 1416; n(n²−1) = 720), divide (1.967), then subtract from 1 — that final step is easy to forget. To judge significance you compare the magnitude (0.967) with the critical value (here 0.700 for n = 9): being above it means the correlation is significant.
3

Chi-squared — calculating χ²

Paper 2 (A2), 2018, Q10(a)(ii)–(iii)Statistics2 marks

A dihybrid fruit-fly cross was tested against a 9:3:3:1 ratio. For “normal wings, red eyes”: observed O = 885, expected E = 900. The other three phenotype classes give (O − E)²÷E values of 1.61, 0.65 and 0.49. Complete the first row, then calculate the total χ² using χ² = Σ[(O − E)² ÷ E].

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The method

  1. Complete the first row — (O−E), then square, then ÷E: O − E = 885 − 900 = −15 (O − E)² = (−15)² = 225 (O − E)² ÷ E = 225 ÷ 900 = 0.25
  2. Add the (O−E)²÷E values for all four classes: 0.25 + 1.61 + 0.65 + 0.49 = 3.0
Answer: first row (O−E)²÷E = 0.25; χ² = 3.0. With 3 degrees of freedom (4 classes − 1), this is below the critical value 7.82, so the results fit the expected 9:3:3:1 ratio.
Each phenotype class contributes (O−E)²÷E; remember (O−E)² is always positive (the −15 squares to +225). Sum the column to get χ². Degrees of freedom = number of classes − 1 = 3. Decision rule: if χ² is smaller than the critical value, the difference from expected is not significant.
4

Hardy–Weinberg — heterozygote frequency (2pq)

Paper 2 (A2), 2024, Q9(a)(ii)Genetics2 marks

In zoo populations of Asiatic lions the recessive allele has a frequency of 0.30 (i.e. q = 0.30). Using p² + 2pq + q² = 1, determine the frequency of heterozygous lions (2pq).

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The method

  1. Find p (p + q = 1): p = 1 − 0.30 = 0.70
  2. Heterozygotes = 2pq: 2 × 0.70 × 0.30 = 0.42
Answer: 2pq = 0.42 (= 42%).
Watch the wording. Here the allele frequency q is given directly (0.30), so you do not square-root anything — just find p = 1 − q = 0.70, then 2pq = 2 × 0.70 × 0.30 = 0.42. In other questions you’re given q² (the recessive homozygote frequency) and must take the square root first — always check which one you’ve been handed.
5

Index of diversity (the Edexcel B formula)

Paper 2 (A2), 2024, Q5(b)(i)Ecology3 marks

Seven years after a fire, tree counts in a plot were: Aleppo pine 45, Silver fir 30, Beech 14, Oak 12, Chestnut 18, Plane 11. Using the index of diversity D = N(N − 1) ÷ Σn(n − 1), calculate D to two decimal places.

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The method

  1. Total all counts for N: 45 + 30 + 14 + 12 + 18 + 11 = 130
  2. Numerator N(N−1): 130 × 129 = 16 770
  3. Σn(n−1) — each species’ n(n−1), then add: (45×44)+(30×29)+(14×13)+(12×11)+(18×17)+(11×10) = 1980 + 870 + 182 + 132 + 306 + 110 = 3580
  4. Divide: 16 770 ÷ 3580 = 4.68
Answer: D = 4.68.
This is the Edexcel B index of diversity D = N(N−1) ÷ Σn(n−1) — the reciprocal form, so a higher value means more diverse. It is NOT the “1 − …” Simpson’s version some boards use, so don’t subtract from 1. Total all counts for N (130), then each n(n−1) for the denominator (3580), and divide.
6

Standard deviation

Paper 3 (A2), 2022, Q11(d)(i)Statistics3 marks

At 3 hours, five cell counts were 4, 9, 9, 3, 7, with a mean x̄ = 6.4 and n = 5. Using s = √[Σ(x − x̄)² ÷ (n − 1)], calculate the standard deviation.

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The method

  1. Each deviation from the mean, squared: (4 − 6.4)² = 5.76 (9 − 6.4)² = 6.76 (9 − 6.4)² = 6.76 (3 − 6.4)² = 11.56 (7 − 6.4)² = 0.36
  2. Sum the squared deviations: 5.76 + 6.76 + 6.76 + 11.56 + 0.36 = 31.2
  3. Divide by (n − 1) = 4 (the variance): 31.2 ÷ 4 = 7.8
  4. Take the square root: √7.8 = 2.8
Answer: s = 2.8.
Standard deviation in four moves: subtract the mean and square each deviation, sum them (= 31.2), divide by n − 1 (= 4, the sample correction — not n), then square-root. The two classic errors are dividing by n instead of n−1, and forgetting the final root. The three marks here are for Σ(x−x̄)², the ÷(n−1) step, and the final SD.
7

Magnification (image ÷ actual), in standard form

Paper 2 (A2), 2024, Q1(b)Microscopy2 marks

On an electron micrograph of a root-tip cell, a structure with an actual width of 36 µm measures about 50 mm across the image. Calculate the magnification, giving your answer in standard form.

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The method

  1. Convert the image width to the same unit (mm → µm, ×1000): 50 mm = 50 000 µm
  2. Magnification = image ÷ actual: 50 000 ÷ 36 = 1389
  3. Write in standard form: ≈ 1.4 × 10³
Answer: × 1.4 × 10³ (accept 1.36–1.42 × 10³). Magnification has no units.
Magnification = image size ÷ actual size. The two sizes must be in the same unit, so convert the 50 mm image to 50 000 µm before dividing by the 36 µm actual size, then express the answer in standard form (1.4 × 10³).
8

Surface area of a sphere (4πr²)

Paper 1 (A2), 2025, Q7(a)(i)Geometry3 marks

An alveolus has a diameter of 200 µm, and there are 300 million alveoli in each lung. Using surface area of one alveolus = 4πr², calculate the total surface area of a pair of lungs, in m².

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The method

  1. Radius = half the diameter, converted to metres: r = 100 µm = 1 × 10⁻⁴ m
  2. Surface area of one alveolus = 4πr²: 4 × π × (1 × 10⁻⁴)² = 1.26 × 10⁻⁷ m²
  3. Multiply by the number in two lungs (2 × 300 million): 1.26 × 10⁻⁷ × 6 × 10⁸ = 75.4 m²
Answer: ≈ 75.4 m² — about a tennis court’s worth of gas-exchange surface.
Three traps. (1) The radius is half the diameter (100 µm). (2) Convert µm to m before squaring (100 µm = 1 × 10⁻⁴ m). (3) It’s a pair of lungs, so multiply by 2 × 300 million = 6 × 10⁸. Then 4πr² × number gives ~75 m².
9

Efficiency of energy transfer

Paper 1 (A2), 2025, Q2(c)(i)Bioenergetics2 marks

Respiration yields 36 ATP per glucose. Oxidising one glucose releases 2 870 224 J, and one ATP stores 30 566 J of free energy. Calculate the efficiency of energy transfer, to two decimal places.

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The method

  1. Total free energy stored in 36 ATP: 30 566 × 36 = 1 100 376 J
  2. Divide by the energy released by glucose, ×100: (1 100 376 ÷ 2 870 224) × 100 = 38.34%
Answer: ≈ 38.34%.
Efficiency = (useful energy out ÷ total energy in) × 100. The useful energy is stored in all 36 ATP (30 566 × 36), and the input is the energy released by one glucose. The rest is lost as heat — which is why it is well below 100%. The same percentage-efficiency method is used for energy transfer between trophic levels.
10

Bacterial exponential growth rate constant (k)

Paper 3 (A2), 2025, Q11(a)(iii)Microbiology2 marks

During exponential growth a bacterial population rose from 400 per cm³ to 102 400 per cm³ over 3.4 hours. Calculate the growth rate constant k to two decimal places, using k = [log₁₀Nₜ − log₁₀N₀] ÷ [0.301 × t].

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The method

  1. Take log₁₀ of each population: log₁₀ 102 400 = 5.01 log₁₀ 400 = 2.60
  2. Subtract for the numerator: 5.01 − 2.60 = 2.41
  3. Denominator (0.301 × t): 0.301 × 3.4 = 1.023
  4. Divide: k = 2.41 ÷ 1.023 = 2.35
Answer: k ≈ 2.35.
Take log₁₀ of both populations and subtract (top line = 2.41), then divide by 0.301 × time (= 1.023). The 0.301 is log₁₀2 — it appears because the constant is defined in terms of doublings. This bacterial growth-constant calculation is a distinctive Edexcel B skill; quote the answer to 2 decimal places.

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