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Edexcel A · SNAB Papers 1–3 A-level Biology Maths skills

Edexcel A (SNAB) A-level Biology Maths — Worked Solutions

Edexcel A-level Biology A (Salters-Nuffield / SNAB) 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. 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 A (SNAB) Papers 1–3: the Student’s t-test, Spearman’s rank correlation and chi-squared, the Hardy–Weinberg equation, the SNAB index of diversity, standard deviation, magnification, the volume of a sphere, the temperature coefficient Q₁₀ and percentage change. 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), June 2017, Q4(b)(iv)Statistics3 marks

Brine-shrimp lengths were compared for two salt solutions. Means: x̄₁ = 1.102 mm (3%) and x̄₂ = 0.949 mm (5%); variances s₁² = 0.0059 and s₂² = 0.0054; n = 10. Using t = (x̄₁ − x̄₂) ÷ √[(s₁² + s₂²) ÷ n], calculate the value of t.

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The method

  1. Top line — difference between the means: 1.102 − 0.949 = 0.153
  2. Bottom line — add the variances, ÷ n, then square-root the whole bracket: (0.0059 + 0.0054) ÷ 10 = 0.00113 √0.00113 = 0.0336
  3. Divide top by bottom for t: t = 0.153 ÷ 0.0336 = 4.55
Answer: t = 4.55 (accept 4.55–4.554). With df = (10−1) + (10−1) = 18, this far exceeds the 5% critical value, so the difference between the two means is significant.
Build it in two halves: the top line is just the difference of means (0.153); the bottom line is √[(s₁² + s₂²)/n]. The values given are already variances, so don’t square them again. Take the square root of the whole bracket before dividing, and don’t round until the very end.
2

Spearman’s rank correlation coefficient

Paper 3 (A2), June 2018, Q3(a)Statistics8 marks

For seven blowfly species, two temperatures were ranked: the temperature for 50% larval survival and the temperature of sand chosen for pupation. The sum of the squared rank differences is Σd² = 34 and n = 7. Using rₛ = 1 − [6Σd² ÷ n(n² − 1)], calculate rₛ, then say whether the correlation is significant (critical value at n = 7 is 0.786).

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The method

  1. Numerator 6Σd²: 6 × 34 = 204
  2. Denominator n(n² − 1): 7 × (7² − 1) = 7 × 48 = 336
  3. Divide, then subtract from 1: 204 ÷ 336 = 0.607 rₛ = 1 − 0.607 = 0.393
  4. Compare with the critical value: 0.393 < 0.786 → not significant
Answer: rₛ = 0.393; since 0.393 is below the critical value 0.786 (n = 7, p = 0.05), there is no significant correlation between the two temperatures.
Spearman’s measures correlation between two ranked variables: rank each variable separately, find the rank difference d for each pair and square it, sum the d² column, then substitute. Here n(n²−1) = 7 × 48 = 336. The decision rule is the reverse of the t-test’s wording: rₛ below the critical value means the correlation is not significant.
3

Chi-squared — calculating χ² and concluding

Paper 3 (A2), June 2023, Q6(c)(ii)–(iii)Statistics5 marks

A dihybrid zebrafish cross gave observed (O) and expected (E) numbers: stripes/long O = 270, E = 288; stripes/short O = 87, E = 96; spots/long O = 115, E = 96; spots/short O = 40, E = 32. Using χ² = Σ[(O − E)² ÷ E], calculate χ², then conclude using the critical value at 3 degrees of freedom (7.82).

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The method

  1. For each class, (O−E)² ÷ E: (270−288)²/288 = 324/288 = 1.125 (87−96)²/96 = 81/96 = 0.844 (115−96)²/96 = 361/96 = 3.760 (40−32)²/32 = 64/32 = 2.000
  2. Sum for χ²: 1.125 + 0.844 + 3.760 + 2.000 = 7.73
  3. Degrees of freedom = categories − 1: 4 − 1 = 3
  4. Compare with the critical value (3 df, p = 0.05): 7.73 < 7.82 → not significant
Answer: χ² = 7.73; df = 3; 7.73 is less than the critical value 7.82, so the results are not significantly different from expected — the two genes are inherited independently.
Lay the calculation out as a column table — (O−E), then (O−E)², then ÷E — and sum the last column. A sign error in O−E disappears when you square. Degrees of freedom = number of categories − 1 = 3. Decision rule: if χ² is smaller than the critical value, the difference is not significant. Quoting the exact critical value (7.82) earns the final mark.
4

Hardy–Weinberg — frequency of carriers

Paper 2 (A2), June 2019, Q3(a)Genetics3 marks

Cystic fibrosis is recessive and 1 in 2500 babies are born with it. Using the Hardy–Weinberg equations p + q = 1 and p² + 2pq + q² = 1, calculate the probability that a person is a carrier (heterozygous).

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The method

  1. Affected babies are the recessive homozygote = q²: q² = 1 ÷ 2500 = 0.0004
  2. Square-root for q: q = √0.0004 = 0.020
  3. Find p (p + q = 1): p = 1 − 0.020 = 0.980
  4. Carriers = 2pq: 2 × 0.980 × 0.020 = 0.0392 = 3.92%
Answer: 0.0392 (3.92%). Carriers (3.92%) hugely outnumber affected babies (0.04%).
Always start from the recessive homozygote, the only group that equals q²: 1/2500 = 0.0004. Take the square root to get q (0.020), then p = 1 − q, then 2pq for carriers. The classic errors are using 0.0004 as q (forgetting the root) and dropping the factor of 2 in 2pq.
5

Index of diversity (the SNAB formula)

Paper 1 (A2), June 2023, Q2(b)(ii)Ecology2 marks

A pond survey gave a total of N = 156 organisms and a sum of n(n−1) across all species of Σn(n−1) = 9110. Using the index of diversity D = N(N − 1) ÷ Σn(n − 1), calculate D for this pond.

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The method

  1. Numerator N(N−1): 156 × 155 = 24 180
  2. Divide by Σn(n−1): 24 180 ÷ 9110 = 2.65
Answer: D = 2.65 (accept 2.65–2.7).
This is the Edexcel / SNAB 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 other boards use, so don’t subtract from 1. Each n(n−1) is just n × (n−1); use the grand total N in the same pattern for the numerator.
6

Standard deviation

Paper 1 (A2), Autumn 2020, Q10(a)(ii)Statistics3 marks

Wheat-grain masses (tonnes per hectare) were recorded for 10 plots, with a mean of 2.489 and a sum of squared deviations Σ(x − x̄)² = 31.49. Using s = √[Σ(x − x̄)² ÷ (n − 1)], calculate the standard deviation.

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The method

  1. Sum of squared deviations (given): Σ(x − x̄)² = 31.49
  2. Divide by (n − 1) = 9 (this is the variance): 31.49 ÷ 9 = 3.499
  3. Take the square root: √3.499 = 1.87
Answer: s = 1.87 (tonnes per hectare).
Standard deviation in four moves: subtract the mean and square each deviation, sum them, divide by n − 1 (= 9, 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. A calculator’s “sₓ” / “σₙ₋₁” key does it in one step for checking.
7

Magnification from a measured image

Paper 1 (A2), November 2021, Q7(b)(iv)Microscopy3 marks

A line on a photograph represents the width of a vascular bundle whose actual width is 320 µm. The line measures about 36 mm on the image. Calculate the magnification.

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The method

  1. Convert the image measurement to the same unit (mm → µm, ×1000): 36 mm = 36 000 µm
  2. Magnification = image size ÷ actual size: 36 000 ÷ 320 = 112.5
Answer: × 112.5. Magnification has no units.
Magnification = image size ÷ actual size (the reverse of the “actual = image ÷ magnification” rearrangement). The two sizes must be in the same unit, so convert the 36 mm image measurement to 36 000 µm before dividing by the 320 µm actual size.
8

Volume of a sphere

Paper 3 (A2), June 2023, Q2(a)Geometry2 marks

A neutrophil has a diameter of 10 µm. Using the volume of a sphere V = ⁴⁄₃πr³, calculate the volume of one neutrophil in µm³.

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The method

  1. Radius = half the diameter: r = 10 ÷ 2 = 5 µm
  2. Cube the radius: 5³ = 125
  3. Multiply by ⁴⁄₃π: ⁴⁄₃ × π × 125 = 523.6 µm³
Answer: 523.6 µm³ (accept 523–524).
The classic slip is using the diameter in the formula instead of the radius — always halve first (r = 5, not 10). Then cube just the radius (5³ = 125) before multiplying by ⁴⁄₃π. Keep the unit cubed (µm³).
9

Temperature coefficient Q₁₀

Paper 1 (A2), June 2019, Q9(a)(i)Rates1 mark

Catalase produced a mean oxygen volume of 80 mm³ at 20 °C and 240 mm³ at 30 °C. Calculate the temperature coefficient Q₁₀ between 20 °C and 30 °C.

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The method

  1. Q₁₀ = rate at (T + 10 °C) ÷ rate at T: 240 ÷ 80 = 3
Answer: Q₁₀ = 3 — the rate triples for a 10 °C rise.
Q₁₀ = rate at the higher temperature ÷ rate at the temperature 10 °C lower (240 ÷ 80). A Q₁₀ of about 2–3 is typical for an enzyme-controlled reaction. It has no units — it’s a ratio — and the oxygen volume stands in for the rate.
10

Percentage increase from a graph

Paper 1 (A2), June 2025, Q8(a)(i)Percentages2 marks

A graph shows people diagnosed with cancer each year. Diagnoses read about 330 000 in 2009 and about 390 000 in 2019. Calculate the percentage increase from 2009 to 2019, to 3 significant figures.

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The method

  1. Find the increase (% increase uses the rise): 390 000 − 330 000 = 60 000
  2. Divide by the original (2009) value and ×100: (60 000 ÷ 330 000) × 100 = 18.18…
  3. Round to 3 significant figures: ≈ 18.2%
Answer: 18.2%.
Percentage increase divides the rise by the starting value (the 2009 figure), not the later one. Work out the difference first (60 000), then ÷ original × 100. Read both points off the graph carefully and respect the 3 sig fig instruction: 18.18… rounds to 18.2%.

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