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OCR A · H020 / H420 Papers 1–3 A-level Biology Maths skills

OCR A-level Biology Maths — Worked Solutions

OCR A-level Biology A (H020 / H420) examines real statistics in the exam — the chi-squared test, Spearman’s rank correlation and the Student’s t-test — alongside the standard biology maths such as magnification, surface area to volume ratio, Simpson’s Index of Diversity and the Hardy–Weinberg equation, and these questions cost marks every year. Below are fully worked examples spanning the AS and A-level papers (2016–2024), 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 OCR A Papers 1–3: the chi-squared test, Spearman’s rank correlation and the Student’s t-test, the Hardy–Weinberg equation, Simpson’s Index of Diversity, magnification, the volume and surface area of a sphere, the surface area to volume ratio and rates from a tangent. OCR A is distinctive in setting genuine statistical tests in the exam. Click any example to expand the full method. Video walk-throughs are being added.

1

Chi-squared — calculating χ²

Paper 2 (A2), 2024, Q20(a)(i)Statistics3 marks

Wood sorrel was counted within a 3 m radius of three tree species: Ash 44, Oak 56, Sycamore 20. The null hypothesis is that tree species has no effect on wood sorrel numbers. Using χ² = Σ[(O − E)² ÷ E], calculate the value of χ².

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The method

  1. Work out the expected value E under the null hypothesis — an even split: Total = 44 + 56 + 20 = 120 E = 120 ÷ 3 = 40 for each tree
  2. For each tree, find O − E, square it, then ÷ E: Ash: (44 − 40)² ÷ 40 = 16 ÷ 40 = 0.4 Oak: (56 − 40)² ÷ 40 = 256 ÷ 40 = 6.4 Sycamore: (20 − 40)² ÷ 40 = 400 ÷ 40 = 10.0
  3. Sum the last column to get χ²: χ² = 0.4 + 6.4 + 10.0 = 16.8
Answer: χ² = 16.8. With 2 degrees of freedom (3 categories − 1) this exceeds the p = 0.05 critical value of 5.991, so the difference is significant.
The make-or-break step is the expected value. Under “no effect” the expectation is an even split, so E = total ÷ number of categories = 120 ÷ 3 = 40 — not the first observed value (44) or the total (120). Each category contributes (O−E)²÷E (always positive, because −20 squares to +400), and χ² is the sum of that column, not a single row.
2

Spearman’s rank correlation coefficient

Paper 3 (A2), 2021, Q5(c)(i)Statistics3 marks

Soil water content and rate of photosynthesis were measured at six sites for Z. mays. Ranking each variable separately gives a sum of squared rank differences of Σd² = 2 with n = 6. Using rₛ = 1 − [6Σd² ÷ n(n² − 1)], calculate rₛ to three significant figures.

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The method

  1. Numerator 6Σd²: 6 × 2 = 12
  2. Denominator n(n² − 1): 6 × (6² − 1) = 6 × 35 = 210
  3. Divide: 12 ÷ 210 = 0.0571
  4. Subtract from 1: rₛ = 1 − 0.0571 = 0.943
Answer: rₛ = 0.943. Close to +1, signalling a strong positive correlation — soil water content and photosynthesis rise together in maize.
Spearman’s measures correlation between two ranked variables. Rank each column independently, find d for each pair, square and sum (Σd²), then compute the two parts of the fraction separately — the commonest slip is miscomputing n(n²−1) as 6³ instead of 6 × 35 = 210. Divide, then subtract from 1: that final step is easy to forget. To judge significance, compare 0.943 with the critical value at n = 6 (0.829) — being above it means the correlation is significant.
3

Student’s t-test — interpreting the result

Paper 3 (A2), 2017, Q2(b)(v)Statistics1 mark

A Student’s t-test comparing yeast respiration at 30 °C and 35 °C gave a calculated t = 2.200. The critical value at p = 0.05 is 2.306. State the conclusion.

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The method

  1. State the decision rule. For a t-test, the difference between the means is significant only when the calculated t exceeds the critical value.
  2. Compare the calculated t with the critical value: calculated t = 2.200 critical value = 2.306 2.200 < 2.306 → below the critical value
  3. Apply the rule. Because the calculated t is below the critical value, the difference between the means is not significant.
Answer: the difference between the means is not significant (it could be due to chance) at p = 0.05; accept the null hypothesis. Do not say “reject the null hypothesis” here.
The whole mark is in the direction of the comparison. The fixed rule is “calculated > critical = significant”. Here 2.200 < 2.306, so the result is not significant and the means could differ just by chance — so you accept the null hypothesis. Students most often flip the inequality and wrongly reject H₀; pair the calculation with the comparison so the test is only “finished” once the decision about H₀ is stated.
4

Hardy–Weinberg — heterozygote percentage

Paper 3 (A2), 2017, Q3(b)Genetics3 marks

About 1 in 10 000 Bengal tiger births is white (the recessive homozygote). Using p + q = 1 and p² + 2pq + q² = 1, calculate the percentage of tigers that are heterozygous for this gene, to 1 significant figure.

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The method

  1. q² = frequency of the recessive phenotype: q² = 1 ÷ 10 000 = 0.0001
  2. q = √q²: q = √0.0001 = 0.01
  3. p = 1 − q: p = 1 − 0.01 = 0.99
  4. Heterozygotes 2pq, then convert to a percentage: 2 × 0.99 × 0.01 = 0.0198 = 1.98% ≈ 2% (1 s.f.)
Answer: about 2% (to 1 s.f.). 1.98% scores full method; round to the stated 1 s.f.
Standard Hardy–Weinberg: the recessive phenotype (white, 1 in 10 000) gives you q² directly (0.0001), then square-root for q (0.01), 1 − q for p (0.99), and 2pq for the heterozygotes. The classic error is forgetting the square root. Note the population-genetics insight: a very rare recessive phenotype (1 in 10 000) still gives a noticeable carrier frequency of about 2%.
5

Simpson’s Index of Diversity (the 1 − … form)

Paper 2 (A2), 2022, Q20(b)(i)Ecology3 marks

Mean numbers per quadrat for seven grassland species were: Creeping buttercup 3, Daisy 7, Dandelion 1, Grass 26, Red clover 4, Ribwort plantain 3, White clover 6. Using D = 1 − Σ(n/N)², calculate Simpson’s Index of Diversity.

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The method

  1. Total all counts for N: 3 + 7 + 1 + 26 + 4 + 3 + 6 = 50
  2. For each species work out (n/N)²: Grass: (26/50)² = 0.52² = 0.2704 Daisy: (7/50)² = 0.14² = 0.0196 White clover: (6/50)² = 0.12² = 0.0144 … and so on for all seven species
  3. Sum the squared proportions: Σ(n/N)² = 0.3184
  4. Subtract from 1: D = 1 − 0.3184 = 0.6816
Answer: D = 0.6816 (accept 0.68 / 0.682).
This is the OCR A Simpson’s Index of Diversity D = 1 − Σ(n/N)² — the 1 − … form, bounded 0–1, so a higher value (approaching 1) means more diverse. It is NOT the reciprocal N(N−1) ÷ Σn(n−1) version some boards use, so don’t divide and don’t skip the final “1 −”. A sense-check: grass (26 of 50) dominates the Σ term — (26/50)² = 0.27 is over half of 0.3184 — which is exactly why one abundant species pulls diversity down.
6

Magnification — rearrangement + unit conversion

Paper 3 (A2), 2022, Q2(a)(iii)Microscopy2 marks

A blood smear micrograph has a magnification of ×800. A blood cell measures about 10 mm across the image with a ruler. Calculate the actual diameter of the cell, in µm.

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The method

  1. Rearrange: actual size = image size ÷ magnification: 10 mm ÷ 800 = 0.0125 mm
  2. Convert mm to µm (×1000), because the answer is wanted in µm: 0.0125 mm × 1000 = 12.5 µm ≈ 13 µm
Answer: 13 µm (accept 12.5 µm).
The magnification staple: actual = image ÷ magnification. The image is what you measure (10 mm), so divide by 800, then convert mm → µm (×1000). A sanity check: a blood cell is only a few to ~12 µm across, so 12.5 µm is right and any answer in the hundreds means you multiplied instead of dividing. Keep the working in clear steps so the method mark is bankable even if the rounding slips.
7

Volume of a spherical cell (from a scale bar)

Paper 1 (A2), 2017, Q17(b)(i)Geometry3 marks

On a micrograph of a Kupffer cell, a scale bar represents 4 µm and the cell’s image diameter measures 9.1 cm. Assuming the cell is spherical, and using volume of a sphere = (4/3)πr³, calculate the actual volume of the cell to 4 significant figures.

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The method

  1. Use the scale bar to convert the image diameter to a real diameter: scaling the 9.1 cm image against the 4 µm bar gives a real diameter ≈ 18.1 µm
  2. Radius = half the diameter: r = 18.1 ÷ 2 = 9.05 µm
  3. Apply V = (4/3)πr³ — cube the radius, then × (4/3)π: r³ = 9.05³ = 741.2 V = (4/3) × π × 741.2 = 3157 µm³
  4. Give to 4 significant figures: V = 3157 µm³ = 3.157 × 10³ µm³
Answer: 3157 µm³ (3.157 × 10³ µm³). Full method scores 3; a correct value not to 4 s.f. (or missing/wrong units) scores 2; evidence of using (4/3)πr³ scores 1.
The sphere-volume routine has two traps. First, the scale bar: convert the image diameter to a real size before anything else. Second, cube the radius, not the diameter — so halve to r = 9.05 µm first (using the diameter, 18.1, would make the answer 8× too big). Then r³, ×(4/3)π (the factor is ≈ 4.19), and finish in standard form to the stated 4 s.f.
8

Surface area : volume ratio

Paper 2 (A2), 2024, Q9Cell biology1 mark

A cube has edges 4 cm long. Calculate its surface area to volume ratio (SA:VOL).

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The method

  1. Surface area — a cube has 6 identical square faces: one face = 4 × 4 = 16 cm² six faces = 6 × 16 = 96 cm²
  2. Volume — edge cubed: 4 × 4 × 4 = 64 cm³
  3. Divide surface area by volume: SA:VOL = 96 ÷ 64 = 1.5 (i.e. 1.5 : 1)
Answer: 1.5 (1.5 : 1).
Two things sink students here. First, “6 faces” — drill SA of a cube = 6 × (edge)² (computing one face and stopping gives 0.96). Second, the order of the ratio: SA:VOL means surface area on top, so divide SA by volume; the wrong-way-round answer (64 ÷ 96 = 0.67) is a classic trap. Write “SA over VOL” with SA physically above VOL so the fraction is unambiguous. This is the principle behind why small cells (high SA:V) exchange materials more efficiently than large ones.
9

Rate from a tangent (gradient) with units

Paper 1 (A2), 2024, Q21(b)(ii)Rates3 marks

A graph shows hydrogen peroxide decomposed (µg) against time (s) for a catalase reaction — a curve that flattens over time. Calculate the rate of the catalysed reaction at 30 s, and state the units.

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The method

  1. “Rate at a point on a curve” means the gradient of a tangent. The rate is changing, so you cannot just divide a y-value by 30. Draw a single straight tangent touching the curve at t = 30 s.
  2. Read a rise and a run off the tangent and divide: rise ≈ 170 µg over a run ≈ 46 s gradient = 170 ÷ 46 = 3.7 µg s⁻¹
  3. State the units from the axes: y-units per x-units = µg per second = µg s⁻¹
Answer: rate ≈ 3.7 µg s⁻¹ (accept 3.7 ± 0.6, because the tangent read varies). Units: µg s⁻¹.
The make-or-break skill is the tangent. Dividing the total peroxide decomposed by 30 gives an average rate, not the rate at 30 s — a curve has a different steepness at every point, so lay a ruler against the curve at 30 s and measure the slope of that. The three marks are: a correct straight tangent at 30 s, rise ÷ run from it, and the correct units copied straight off the axes (µg over s). Avoid false precision — quoting 3.71666 caps the marks.
10

Percentage increase from a graph

Paper 2 (A2), 2023, Q20(b)(i)Data handling2 marks

From a population graph, wolf numbers were about 20 in 1995 and about 90 in 2003. Calculate the percentage increase between 1995 and 2003.

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The method

  1. Find the increase (new − original): 90 − 20 = 70 wolves more
  2. Divide by the original (1995) value: 70 ÷ 20 = 3.5
  3. Multiply by 100: 3.5 × 100 = 350%
Answer: 350%.
A percentage increase of 350% surprises students who expect an answer under 100% — more-than-tripling really is a 350% increase. The formula is (change ÷ original) × 100, with the 1995 value (20) as the denominator; the error to avoid is dividing by the new value (90). Lay it out: subtract for the change, divide by the original year, ×100 — and read the two graph values accurately first.

Get the full worked solution packs

Complete step-by-step solutions to every maths question in the database — written as teaching scripts with common-mistake warnings — are available as a downloadable PDF pack covering all years, 2016–2024.

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