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WJEC Units 1–5 A-level Biology Maths skills

WJEC A-level Biology Maths — Worked Solutions

WJEC A-level Biology expects you to handle statistics like the t-test and chi-squared as well as the standard biology maths — and these questions cost marks every year. Below are fully worked examples spanning Units 1–5, 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 WJEC Units 1–5: the t-test and chi-squared, Hardy–Weinberg, Simpson’s Diversity Index, microscope calibration, surface area and volume formulae, logarithmic growth, the inverse-square law and standard deviation. Click any example to expand the full method. Video walk-throughs are being added.

1

The t-test — accept or reject the null hypothesis

Unit 3, Summer 2024, Q3(b)(i)Statistics4 marks

A t-test compared the mean NO₂ concentration before and after a motorway junction was closed. The calculated t-value is 1.769 with 28 degrees of freedom; the critical value at p = 0.05 is 1.701. The null hypothesis is that there is no significant difference between the two means. Decide whether to accept or reject the null hypothesis, and explain your decision.

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The method

  1. Read the critical value at p = 0.05 for df = 28: critical value = 1.701
  2. Compare calculated vs critical: 1.769 > 1.701 — the calculated value exceeds the critical value.
  3. Apply the rule. Calculated > critical → reject the null hypothesis.
  4. State the meaning: there is a significant difference between the means — the difference is not due to chance.
Answer: reject the null hypothesis, because 1.769 > 1.701 (p = 0.05), so there is a significant difference in mean NO₂ concentration. ECF applies if a different critical value is read but reasoned from correctly.
Learn the four-step ritual: choose p = 0.05, find the critical value at the right df, compare, then state the decision and what it means. The decision rule for a t-test: calculated > critical → reject the null hypothesis. The common errors are reading the wrong row/column, or stopping at “reject” without saying it means a significant difference. Note 1.769 only just exceeds 1.701 — significant, but marginally.
2

Chi-squared — calculating χ² and testing significance

Unit 4, Summer 2022, Q4(b)–(c)Statistics5 marks

Brown unbanded snails were counted in three habitats: woodland 28, hedgerow 2, meadow 3 (total 33). Under the null hypothesis of no difference, the snails would be spread equally between habitats. Using χ² = Σ[(O − E)² ÷ E], calculate χ², then (with 2 degrees of freedom, critical value 5.99 at p = 0.05) decide whether to accept or reject the null hypothesis.

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The method

  1. Expected value (total shared equally): E = 33 ÷ 3 = 11 per habitat
  2. For each habitat: (O−E)² ÷ E: Wood: (28−11)²/11 = 289/11 = 26.27 Hedge: (2−11)²/11 = 81/11 = 7.36 Meadow: (3−11)²/11 = 64/11 = 5.82
  3. Sum for χ²: 26.27 + 7.36 + 5.82 = 39.45
  4. Compare with the critical value (2 df, p = 0.05): 39.45 > 5.99 → reject the null hypothesis
Answer: χ² = 39.45; since 39.45 > 5.99, reject the null hypothesis — the distribution between habitats is significant, not due to chance.
The expected value is the total divided equally (33 ÷ 3 = 11). Then work column by column: O−E, square it (negatives become positive), divide by E, and sum. Degrees of freedom = categories − 1 = 2. Same decision rule as the t-test (calculated > critical → reject), but a different df formula. Finish with the meaning, not just “reject”.
3

Hardy–Weinberg — genotype frequencies

Unit 4, Summer 2024, Q4(b)(i)Genetics3 marks

In a population of 900 wild horses, 891 have a dun coat (dominant, D) and 9 are bay (recessive, dd). Using the Hardy–Weinberg equations p + q = 1 and p² + 2pq + q² = 1, calculate the frequencies of the three genotypes (dd, DD and Dd).

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The method

  1. Start from the recessives (bay = q², the only single-genotype phenotype): q² = 9 ÷ 900 = 0.01
  2. Square-root for q, subtract for p: q = √0.01 = 0.1 p = 1 − 0.1 = 0.9
  3. Each genotype frequency: dd = q² = 0.01 DD = p² = 0.9² = 0.81 Dd = 2pq = 2 × 0.9 × 0.1 = 0.18
Answer: dd = 0.01, DD = 0.81, Dd = 0.18 (check: they sum to 1.00). A mark is available for finding q = 0.1 and p = 0.9.
Always start from the recessives — bay (dd) is the one phenotype you can read straight off as q², because dun could be DD or Dd. From q² take the square root for q, subtract for p, then three quick substitutions. The built-in check: the three frequencies sum to 1. The classic mistake is treating the 891 dun horses as p² (they’re actually p² + 2pq).
4

Simpson’s Diversity Index

Unit 5 Analysis Task, Summer 2024, Q1(c)(i)Ecology3 marks

Kick-sampling of a river recorded 11 invertebrate species, giving N = 212 organisms in total and a sum of n(n−1) across all species of 5756. Using Simpson’s Diversity Index D = 1 − [Σn(n−1) ÷ N(N−1)], calculate the index to two decimal places.

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The method

  1. Σn(n−1) (sum of the column): Σn(n−1) = 5756
  2. N(N−1) (the denominator): 212 × 211 = 44 732
  3. Divide and subtract from 1: 5756 ÷ 44 732 = 0.1287 D = 1 − 0.1287 = 0.8713
  4. Round to 2 dp: 0.8713 → 0.87
Answer: D = 0.87 (to 2 dp). 1 mark for either N(N−1) = 44 732 or Σn(n−1) = 5756; 2 marks for 0.8713 or for 0.13 (forgetting the “1 −”).
Careful bookkeeping: sum the n(n−1) column, work out N(N−1), divide, then subtract from 1. The classic errors are forgetting the “1 −” step (leaving 0.13) and the rounding. This version of the index runs 0–1, where closer to 1 = higher diversity — which matters for the conclusion. Keep full precision until the final 2-dp rounding.
5

Microscope calibration — the eyepiece graticule

Unit 5 Analysis Task, Summer 2024, Q2(a)(ii)Microscopy2 marks

An eyepiece graticule is calibrated against a stage micrometer: 100 stage micrometer units line up with 40 eyepiece units, and each stage micrometer unit is 0.01 mm. Calculate the length of one eyepiece unit in micrometres (µm).

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The method

  1. Real length of 100 stage units (the known scale): 100 × 0.01 mm = 1.0 mm = 1000 µm
  2. That equals 40 eyepiece units (viewed together): 40 eyepiece units = 1000 µm
  3. Divide for one eyepiece unit: 1000 ÷ 40 = 25 µm
Answer: 1 eyepiece unit = 25 µm. 1 mark for (100 × 0.01 × 1000) ÷ 40, or the intermediate 0.025 mm.
Calibration asks “how many real micrometres does one eyepiece division represent?” The stage micrometer gives the true length (100 units × 0.01 mm = 1 mm = 1000 µm), and that real distance covers 40 eyepiece units, so each = 1000 ÷ 40 = 25 µm. The trap (partial answer 0.025) is stopping in mm instead of converting to µm. Drill mm→µm (×1000).
6

Surface area of a sphere (4πr²)

Unit 1, Summer 2024, Q5(b)(i)Geometry2 marks

HIV is assumed to be spherical, with a diameter across the viral envelope of 120 nm. Using surface area of a sphere = 4πr² with π = 3.14, calculate the surface area in nm².

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The method

  1. Diameter → radius first (the formula uses r): r = 120 ÷ 2 = 60 nm
  2. Substitute into 4πr²: 4 × 3.14 × 60² = 4 × 3.14 × 3600
  3. Calculate: = 45 216 nm²
Answer: 45 216 nm² (45 239 using the calculator’s π). 1 mark for 4 × 3.14 × 60². Using 120 as the radius gives 180 864 — worth 1 mark.
The decisive step is the first: the question gives the diameter (120 nm) but the formula needs the radius (60 nm). Plug 120 straight in and your answer is four times too big — the mark scheme lists exactly that wrong value. Drill “diameter ÷ 2 = radius before you touch the formula”, then square the radius first, then ×4 and ×π.
7

Logarithmic growth — generations per hour

Unit 3, Summer 2024, Q5(b)(i)Logarithms3 marks

On a log-scaled growth curve, a bacterial culture has 10⁴ cells cm⁻³ at 1.5 hours and 10⁷ cells cm⁻³ at 3.5 hours. Using generations per hour = (log₁₀Xₜ − log₁₀X₀) ÷ (0.301 × t), where t is the growth period in hours, calculate the number of generations per hour to the nearest whole number.

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The method

  1. Read the two counts off the log axis: X₀ = 10⁴ (at 1.5 h) Xₜ = 10⁷ (at 3.5 h)
  2. Take logs (clean, as they’re powers of ten): log₁₀(10⁷) − log₁₀(10⁴) = 7 − 4 = 3
  3. Denominator (t = 3.5 − 1.5 = 2 h): 0.301 × 2 = 0.602
  4. Divide and round: 3 ÷ 0.602 = 4.98 → 5 generations per hour
Answer: 5 generations per hour. 2 marks for 4.98 (not rounded); 1 mark for the substitution (7−4) ÷ (0.301×2).
The log step collapses neatly because the graph values are clean powers of ten — log₁₀(10⁷) = 7, no calculator needed. (The 0.301 is log₁₀2, which is why this counts doublings.) Traps: reading the wrong points off the log axis, and the rounding — 4.98 rounds up to 5.
8

The inverse-square law — light intensity

Unit 3, Summer 2022, Q3(a)Formulae3 marks

In a photosynthesis experiment, the light intensity reaching a plant is given by Iₚ = Iₗ ÷ d², where Iₗ is the lamp intensity (4 W m⁻²) and d is the lamp-to-plant distance. For a distance of 0.10 m, calculate d² and the light intensity at the plant.

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The method

  1. Square the distance (the formula divides by d²): d² = 0.10² = 0.0100 m²
  2. Apply the inverse-square formula: Iₚ = 4 ÷ 0.0100 = 400 W m⁻²
Answer: d² = 0.0100 m²; light intensity at the plant = 400 W m⁻². 2 marks for 4 × (1 ÷ 0.10)²; 1 mark for the correct d² value.
“Inverse square” means divide by distance squared. The two steps are squaring the distance (0.10² = 0.01, not 0.1) and dividing. The pattern down the table confirms it — halving the distance quadruples the intensity. Students slip on the squaring; check the answer fits the increasing trend as the lamp gets closer.
9

Volume of a cylinder & rate of water uptake

Unit 2, Summer 2024, Q4(b)(i)Geometry & rates5 marks

In a potometer, an air bubble travels 200 mm along a capillary tube of internal diameter 1 mm. Using V = πr²h with π = 3.14, find the volume of water taken up in mm³. Then, given that the mean time for the bubble to travel 200 mm is 192 seconds, calculate the mean rate of water uptake in mm³ min⁻¹.

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The method

  1. Diameter → radius: r = 1 ÷ 2 = 0.5 mm
  2. Volume = πr²h: 3.14 × 0.5² × 200 = 3.14 × 0.25 × 200 = 157 mm³
  3. Rate = volume ÷ time, converting 192 s to minutes (×60): (157 ÷ 192) × 60 = 49.1 mm³ min⁻¹
Answer: volume = 157 mm³; rate = 49.1 mm³ min⁻¹. Using the diameter (1 mm) not radius gives 628 mm³ (1 mark). ECF carries a wrong volume into the rate.
Two skills chained: V = πr²h needs the radius (halve the diameter, and 0.5² = 0.25 — students often leave it as 0.5), then a rate with a seconds→minutes conversion (×60). ECF means a slip in the volume isn’t punished twice. Protect the 1-dp mark at the end.
10

Standard deviation

Unit 4, Summer 2017, Q9(c)(iii)Statistics2 marks

In a stickleback behaviour study, the silver-belly model gave a sum of squared deviations from the mean, Σ(x − x̄)², of 343.3 for N = 12 fish. Using SD = √[Σ(x − x̄)² ÷ (N − 1)], calculate the standard deviation.

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The method

  1. Divide by (N − 1): N − 1 = 12 − 1 = 11 343.3 ÷ 11 = 31.2
  2. Take the square root: √31.2 = 5.6
Answer: SD = 5.6 (5.59). 1 mark for 343.3 ÷ 11, or √(343.3 ÷ 11).
Divide the sum of squared deviations by (N − 1), then square-root. The two slips are using N (12) instead of N−1 (11), and forgetting the final square root. A teaching point for the wider question: if two data sets have high, overlapping SDs, that weakens any conclusion that their means differ.

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