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Eduqas Components 1–3 A-level Biology Maths skills

Eduqas A-level Biology Maths — Worked Solutions

Eduqas A-level Biology expects you to handle statistics like the t-test and chi-squared alongside the standard biology maths — and these questions cost marks every year. Below are fully worked examples spanning Components 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 Eduqas Components 1–3: the t-test and chi-squared, Hardy–Weinberg, Simpson’s Diversity Index, microscope calibration, standard deviation, logarithmic growth, Rf values, magnification and percentage calculations. Click any example to expand the full method. Video walk-throughs are being added.

1

The t-test — calculating and interpreting t

Component 3, Summer 2023, Q2(b)Statistics7 marks

The number of stomata per field of view was compared for full-sun and shade leaves. Full sun: mean = 40, variance = 11.6, n = 10. Shade: mean = 27, variance = 13.3, n = 10. Using t = (x̄₁ − x̄₂) ÷ √(s₁²/n₁ + s₂²/n₂), calculate t. Then, given the critical value at the 5% level for the correct degrees of freedom, state whether to accept or reject the null hypothesis and explain your conclusion.

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The method

  1. Difference in means (the numerator): 40 − 27 = 13
  2. Each variance ÷ sample size, then add and root: 11.6/10 = 1.16 13.3/10 = 1.33 √(1.16 + 1.33) = √2.489 = 1.578
  3. Divide for t: t = 13 ÷ 1.578 = 8.24
  4. Degrees of freedom & decision: df = (10−1) + (10−1) = 18; critical value = 2.101. 8.24 > 2.101 → reject the null hypothesis
Answer: t = 8.24; since 8.24 > 2.101 (p = 0.05, df = 18), reject the null hypothesis — there is a significant difference between sun and shade leaves.
Two traps in the calculation: the values given are variances (s²) already, so don’t square them again — just divide each by n; and take the square root of the whole bracket before dividing. For the test, df for an unpaired t-test is (n₁−1) + (n₂−1) = 18 (not 20 or 10). Decision rule: calculated > critical → reject the null, then state the biological meaning too.
2

Chi-squared — calculating χ² and degrees of freedom

Component 2, Summer 2023, Q6(b)Statistics5 marks

A test cross was expected to give a 1 : 1 : 1 : 1 ratio of four phenotypes. The observed numbers were: glossy green 64, glossy striped 12, rough green 11, rough striped 73. Using χ² = Σ[(O − E)² ÷ E], calculate χ², state the degrees of freedom, and (using a critical-value table) give the probability that the results differ significantly from expected.

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The method

  1. Expected value (1:1:1:1 splits the total of 160 equally): E = 160 ÷ 4 = 40
  2. For each phenotype, (O−E)² ÷ E: (64−40)²/40 = 14.4 (12−40)²/40 = 19.6 (11−40)²/40 = 21.03 (73−40)²/40 = 27.23
  3. Sum for χ²: 14.4 + 19.6 + 21.03 + 27.23 = 82.25
  4. df and probability: df = categories − 1 = 3; 82.25 > 11.34 (the p = 0.01 value), so p < 0.01
Answer: χ² = 82.25; df = 3; p < 0.01 — a highly significant difference, so reject the null hypothesis (the genes turn out to be linked).
Work expected first (a 1:1:1:1 ratio splits the total into four 40s), then for each class do (O−E)² ÷ E and sum — a sign error in O−E disappears when you square. Degrees of freedom = number of categories − 1 = 3. A χ² of 82 vastly exceeds the table value, so the probability the difference is due to chance is less than 0.01.
3

Hardy–Weinberg — frequency of heterozygotes

Component 2, Summer 2025, Q5(c)(ii)Genetics4 marks

In a blood-donor sample, 49% are blood group O (the recessive homozygote IᵒIᵒ). Using the Hardy–Weinberg equations p + q = 1 and p² + 2pq + q² = 1 (where q is the frequency of the recessive Iᵒ allele), calculate the percentage of people who are heterozygous (carry one dominant and one recessive allele).

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The method

  1. Group O is the recessive homozygote = q²: q² = 0.49 q = √0.49 = 0.7
  2. Find p (p + q = 1): p = 1 − 0.7 = 0.3
  3. Heterozygotes = 2pq: 2 × 0.3 × 0.7 = 0.42
  4. As a percentage: 0.42 × 100 = 42%
Answer: 42%. Marks are awarded line by line: q = 0.7, p = 0.3, 2pq = 0.42, and 42%.
The idea that unlocks it: group O is the recessive homozygote, so its frequency IS q² — take the square root to get q. The classic errors are using 0.49 as q itself (forgetting the square root) or forgetting the factor of 2 in 2pq. Show every line, because each correct line scores even if the final % slips.
4

Simpson’s Diversity Index

Component 2, Summer 2025, Q6(a)(iii)Ecology3 marks

A survey of one site recorded nine species, giving a total of N = 82 organisms and a sum of n(n−1) across all species of 886. Using Simpson’s Diversity Index D = 1 − [Σn(n−1) ÷ N(N−1)], calculate D.

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The method

  1. Σn(n−1) (sum of the column): Σn(n−1) = 886
  2. N(N−1) (the denominator): 82 × 81 = 6642
  3. Divide and subtract from 1: 886 ÷ 6642 = 0.1334 D = 1 − 0.1334 = 0.87
Answer: D = 0.87. 1 mark for Σn(n−1) = 886 or N(N−1) = 6642; 2 marks for 0.13 (forgetting the “1 −”).
Standard Simpson’s routine: sum n(n−1), work out N(N−1), divide, then subtract from 1. A species with n = 0 contributes nothing — a zero count is fine. The most common lost mark is forgetting the “1 −” step, leaving 0.13. This index runs 0–1, where closer to 1 = higher diversity (greater evenness, not just richness).
5

Microscope calibration — the eyepiece graticule

Component 3, Summer 2023, Q1(c)(i)Microscopy2 marks

An eyepiece graticule is calibrated against a stage micrometer viewed at ×40, where one stage micrometer division = 0.01 mm. The scales line up so that one eyepiece unit equals about 0.37 of a micrometer division. Calculate the value of one eyepiece unit in micrometres (µm).

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The method

  1. Real length of one eyepiece unit in mm: 0.37 × 0.01 mm = 0.0037 mm
  2. Convert mm → µm (×1000): 0.0037 × 1000 = 3.7 µm
Answer: 1 eyepiece unit = 3.7 µm (accept 3.3–3.8). 1 mark for the value in mm (0.0037), or for the fraction-of-a-division working.
Graticule calibration: line up the eyepiece scale against the stage micrometer (whose divisions are a known real size), find how much of a division one eyepiece unit equals, then convert. The mm→µm step (×1000) is the one students drop — the answer is wanted in µm. Once calibrated, you can measure any specimen in eyepiece units and convert.
6

Standard deviation

Component 1, Summer 2020, Q6(d)(i)Statistics3 marks

Five repeat times (in seconds) for a meniscus to travel 10 mm were 254, 246, 255, 253 and 252, giving a mean of 252. Using SD = √[Σ(x − x̄)² ÷ (n − 1)], calculate the standard deviation.

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The method

  1. Square each deviation from the mean: (254−252)²=4 (246−252)²=36 (255−252)²=9 (253−252)²=1 (252−252)²=0
  2. Sum the squared deviations: 4 + 36 + 9 + 1 + 0 = 50
  3. Divide by (n−1) and square-root: √(50 ÷ 4) = √12.5 = 3.54
Answer: SD = 3.54. 2 marks for √(50 ÷ 4); 1 mark for Σ(x−x̄)² = 50.
Find each deviation, square it (removing the sign), sum (Σ = 50), divide by n−1 (= 4, not 5 — the sample SD), then square-root. The two classic errors are dividing by n instead of n−1, and forgetting the final square root. Getting Σ(x−x̄)² = 50 alone scores a mark, so set out the deviation table.
7

Logarithmic growth — time to reach a population

Component 1, Summer 2025, Q6(b)(ii)Logarithms3 marks

An elephant population has a generation (doubling) time of G = 22 years. Using G = t ÷ [3.3 × log₁₀(Nₜ₋ₙₐₜ ÷ Nₛₜₐₙₜ)], calculate the time t for the population to grow from 10 to 1000, to 2 significant figures.

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The method

  1. Rearrange for t: t = G × [3.3 × log₁₀(Nₜ₋ₙₐₜ ÷ Nₛₜₐₙₜ)]
  2. Ratio inside the log: 1000 ÷ 10 = 100
  3. Take the log (10² = 100): log₁₀(100) = 2
  4. Substitute and round to 2 sf: t = 22 × (3.3 × 2) = 22 × 6.6 = 145.2 → 150 years
Answer: 150 years. 2 marks for 145 / 145.2 (not rounded to 2 sf); 1 mark for the full substitution.
Two things make this approachable: the log of 100 is just 2 (10² = 100 — no calculator needed if you spot it), and G is the doubling time. The real trap is the 2 sig fig instruction: 145.2 rounds to 150, not 145 or 140. Rearrange first, simplify the log, then round last.
8

Rf value from a chromatogram

Component 1, Summer 2024, Q4(c)(i)Chromatography2 marks

On a paper chromatogram of leaf pigments, a pigment spot has travelled about 52 mm from the origin and the solvent front has travelled about 72 mm. Using Rf = distance travelled by the pigment ÷ distance travelled by the solvent front, calculate the Rf value of the pigment.

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The method

  1. Measure both distances from the origin (to the centre of the spot): pigment = 52 mm solvent front = 72 mm
  2. Divide pigment distance by solvent-front distance: 52 ÷ 72 = 0.72
Answer: Rf ≈ 0.72 (accept 0.70–0.75). 1 mark for showing the two measured distances. An Rf of ~0.72 matches phaeophytin (reference 0.70).
Rf is always distance moved by the spot ÷ distance moved by the solvent front, both from the origin — so Rf is always between 0 and 1 (the spot can’t outrun the solvent). Measure to the centre of the spot. Rf depends on the solvent and paper, so it only identifies a pigment reliably against like-for-like reference values.
9

Actual size from a scale bar

Component 3, Summer 2025, Q2(a)(ii)Microscopy2 marks

On a micrograph of red blood cells, the mean diameter measured on the image is 27 mm. A scale bar representing 5 µm measures 20 mm on the same image. Calculate the actual (real) mean diameter of the cells.

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The method

  1. Use the scale bar to set the conversion (20 mm image = 5 µm real): real µm = image mm ÷ 20 × 5
  2. Apply it to the mean diameter: 27 ÷ 20 × 5 = 1.35 × 5 = 6.75 µm
Answer: 6.7–6.8 µm (≈ 6.75). 1 mark for the correct method “image ÷ 20 × 5”. ECF if the mean was wrong but used correctly.
A scale-bar conversion: the bar gives the real length of a known image length, so scale the measurement in the same proportion — real size = measured × (bar’s real length ÷ bar’s image length). Sanity-check: the real cell (µm) is far smaller than the image (tens of mm), so the answer must be a small number of µm. A real erythrocyte is ~7 µm, so 6.7–6.8 µm is spot on.
10

Percentage inhibition from a formula

Component 1, Summer 2025, Q5(d)(ii)Percentages2 marks

An enzyme’s rate without inhibitor (V₀) is 79 µmol min⁻¹ mg⁻¹, and with 50 µmol NADH present the rate (V) falls to 51. Using percentage inhibition I% = [(V₀ − V) ÷ V₀] × 100, calculate the percentage inhibition at 50 µmol NADH.

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The method

  1. Find the drop in rate, V₀ − V: 79 − 51 = 28
  2. Divide by V₀ and ×100: (28 ÷ 79) × 100 = 35.44%
  3. Round: ≈ 35%
Answer: 35%. 1 mark for 35.4 (un-rounded), or for showing (79−51)/79 × 100.
A percentage-change calculation with the formula given — the only skill is substituting the right two numbers. The trap is which rate is V₀: it’s the rate at zero inhibitor (79), not the row you’re working on. Sanity-check: 50 µmol sits between the 25 µmol (19%) and 100 µmol (51%) values, so 35% is sensibly in between.

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