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OCR B · H022 / H422 Papers 1–3 A-level Biology Maths skills

OCR B A-level Biology Maths — Worked Solutions

OCR B A-level Biology (Advancing Biology) (H022 / H422) 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 B 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 B 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 1 (A2), 2020, Q32(b)(ii)Statistics4 marks

A dihybrid cross (RrYy × RrYy) gave 320 offspring, expected in a 9:3:3:1 ratio. Observed: tall purple 175, dwarf purple 57, tall white 63, dwarf white 25. Using χ² = Σ[(O − E)² ÷ E], calculate χ².

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The method

  1. Work out the expected numbers from the 9:3:3:1 ratio (16 parts of 320): one part = 320 ÷ 16 = 20 tall purple = 9 × 20 = 180 dwarf purple = 3 × 20 = 60 tall white = 3 × 20 = 60 dwarf white = 1 × 20 = 20
  2. For each class find (O − E)², then ÷ E: (175 − 180)² ÷ 180 = 25 ÷ 180 = 0.139 (57 − 60)² ÷ 60 = 9 ÷ 60 = 0.15 (63 − 60)² ÷ 60 = 9 ÷ 60 = 0.15 (25 − 20)² ÷ 20 = 25 ÷ 20 = 1.25
  3. Sum for χ²: χ² = 0.139 + 0.15 + 0.15 + 1.25 = 1.69
Answer: χ² = 1.69. With 3 degrees of freedom (4 classes − 1) this is below the p = 0.05 critical value of 7.81, so the deviation from 9:3:3:1 is not significant.
The step students miss is the expected numbers: a 9:3:3:1 dihybrid ratio splits 320 into 16 parts (÷16 = 20 per part), giving 180/60/60/20 — not an even split. Then it’s the standard routine: difference, square, divide by that class’s E, add. OCR awards a mark for each stage (E values, (O−E)², (O−E)²÷E, Σ), so lay it out as a full table. The dwarf-white cell (1.25) dominates the sum — the smallest expected class usually contributes most.
2

Spearman’s rank correlation coefficient

Paper 3 (A2), 2020, Q2(a)(ii)Statistics2 marks

Ivy leaf area and petiole length were measured at ten sites. Ranking each variable separately gives a sum of squared rank differences of Σd² = 15.50 with n = 10. Using rₛ = 1 − [6Σd² ÷ n(n² − 1)], calculate rₛ to four decimal places.

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The method

  1. Numerator 6Σd²: 6 × 15.50 = 93
  2. Denominator n(n² − 1): 10 × (10² − 1) = 10 × 99 = 990
  3. Divide: 93 ÷ 990 = 0.0939
  4. Subtract from 1: rₛ = 1 − 0.0939 = 0.9061
Answer: rₛ = 0.9061. Close to +1, signalling a strong positive correlation — larger ivy leaves have longer petioles.
Spearman’s measures correlation between two ranked variables. Build the formula in pieces: 6Σd² on top, n(n²−1) on the bottom, then 1 minus the quotient. The two common errors are forgetting the “1 −” (leaving 0.094, which wrongly suggests no correlation) and mis-computing n(n²−1) as 10³ instead of 10 × 99 = 990. Watch the required precision — four decimal places (0.9061) to match the critical-value table; compare with the n = 10 critical value to judge significance.
3

Student’s t-test — calculating t

Paper 3 (A2), 2017, Q3(b)(iii)Statistics3 marks

Mean metabolic rate of a fish was 86.0 at 10 °C and 131.6 at 16 °C (mg O₂ kg⁻¹ hour⁻¹), variance s² = 68.89 for both, n = 10 each. Using t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂), calculate t to three decimal places.

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The method

  1. Difference of the means (top line): |86.0 − 131.6| = 45.6
  2. Denominator — s²/n for each group, add, then square-root: 68.89 ÷ 10 = 6.889 (each group) 6.889 + 6.889 = 13.78 √13.78 = 3.712
  3. Divide: t = 45.6 ÷ 3.712 = 12.285
Answer: t = 12.285. With df = n₁ + n₂ − 2 = 18, this hugely exceeds the p = 0.05 critical value (2.101), so the difference is significant — reject the null hypothesis.
The unpaired t formula: difference of means on top, and on the bottom the square root of (variance ÷ n) added for both groups. Keep the full variance (68.89) all the way through — rounding early shifts the answer. Watch the “three decimal places” instruction (12.285, not 12.28). Even a wrong final value earns method marks for correct substitution, so always show the working. For an unpaired test df = n₁ + n₂ − 2 = 18, not n − 1.
4

Hardy–Weinberg — number of carriers

Paper 3 (A2), 2024, Q2(b)(i)Genetics3 marks

NGLY1 deficiency is a recessive disease. 75 people worldwide had it in 2022; world population = 7 880 000 000. Using p + q = 1 and p² + 2pq + q² = 1, estimate the number of heterozygous carriers, to 4 significant figures.

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The method

  1. q² = frequency of affected (homozygous recessive) people: q² = 75 ÷ 7 880 000 000 = 9.95 × 10⁻⁹
  2. q = √q²: q = √(9.95 × 10⁻⁹) = 9.756 × 10⁻⁵
  3. p = 1 − q: p = 1 − 0.00009756 = 0.99990244
  4. Carrier frequency 2pq, then × population: 2pq = 2 × 0.99990244 × 0.00009756 = 0.000195098 carriers = 0.000195098 × 7 880 000 000 = 1 537 372
  5. Round to 4 significant figures: 1 537 372 → 1 537 000
Answer: about 1 537 000 carriers (4 s.f.). One mark each for q, for 2pq, and for the final population figure to 4 s.f.
The classic Hardy–Weinberg chain: the recessive phenotype gives you q² directly, then square-root for q, 1 − q for p, 2pq for the carrier frequency, and × population for the number. Students lose marks skipping the square root. When q is tiny, p ≈ 1, so 2pq ≈ 2q — a useful check, but do the full calculation. Watch the significant figures: OCR asks for 4 s.f., so 1 537 372 becomes 1 537 000. Show every intermediate value so each mark is banked even if the rounding slips.
5

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

Paper 1 (A2), 2017, Q35(b)(i)Ecology3 marks

Bird counts in field M were: Dunnock 3, Song thrush 40, Reed bunting 23, Meadow pipit 12, Willow warbler 4, Common redstart 18. 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 + 40 + 23 + 12 + 4 + 18 = 100
  2. For each species work out (n/N)²: Song thrush: (40/100)² = 0.16 Reed bunting: (23/100)² = 0.0529 Common redstart: (18/100)² = 0.0324 … and so on for all six species
  3. Sum the squared proportions: Σ(n/N)² = 0.2622
  4. Subtract from 1: D = 1 − 0.2622 = 0.7378
Answer: D = 0.7378 (accept 0.74 / 0.738).
This is the OCR B 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. Three stages: total the counts for N (= 100), square every proportion, sum them, then 1 − that sum. The two common slips are forgetting the final “1 −” (leaving 0.2622) and dropping a species from the sum. A D of 0.74 here is higher than the 0.54 of the comparison field, so field M is the more diverse.
6

Magnification of a TEM (image ÷ actual)

Paper 1 (AS), 2018, Q23(d)(i)Microscopy2 marks

On a TEM of a bacterium, the actual length between two points is 4 µm and the image measures about 43 mm. Calculate the magnification, to 3 significant figures.

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The method

  1. Put image and actual sizes in the same units: 43 mm = 43 000 µm
  2. Magnification = image ÷ actual: 43 000 µm ÷ 4 µm = 10 750
  3. Round to 3 significant figures: 10 750 → 10 800
Answer: ×10 800 (accept 10 750; a 42–43 mm read is allowed).
Magnification = image ÷ actual, with both in the same units — convert 43 mm to 43 000 µm before dividing by 4 µm. Students slip by not converting (43 ÷ 4 = 10.75, a thousand times too small) or by wrong significant figures (10 750 → 10 800 to 3 s.f.). Show the 43 000 ÷ 4 working to bank the method mark, and remember magnification has no units.
7

Surface area of a sphere (4πr²) as a scale-up

Paper 1 (A2), 2024, Q33(b)Geometry2 marks

An apple is modelled as a sphere of diameter 9 cm with 1.2 lenticels per cm². Using surface area of a sphere = 4πr², calculate the mean number of lenticels on one apple.

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The method

  1. Convert the diameter to a radius first: r = 9 ÷ 2 = 4.5 cm
  2. Surface area of the sphere = 4πr²: 4 × π × (4.5)² = 4 × π × 20.25 = 254.5 cm²
  3. Multiply the area by the lenticel density: 254.5 × 1.2 = 305 lenticels
Answer: 305 lenticels (305.36…). A correct surface area alone (254.47 cm²) scores 1.
Two make-or-break moves. First, use the radius: halve the 9 cm diameter to r = 4.5 (using 9 makes the area 4× too big). Second, multiply by 1.2 lenticels per cm², not divide — “per cm²” means “for each cm², multiply”. OCR still credits the surface-area step with 1 mark even if the density step slips, so always show 4πr² first. The chain: diameter → halve → 4πr² → × density.
8

Standard deviation

Paper 1 (A2), 2017, Q33(a)(ii)Statistics2 marks

Water-uptake rates across three replicates were 7.0, 7.0 and 7.2 mm min⁻¹ (mean 7.07). Calculate the standard deviation.

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The method

  1. Find each deviation from the mean, then square it: (7.0 − 7.07)² = 0.0049 (7.0 − 7.07)² = 0.0049 (7.2 − 7.07)² = 0.0169
  2. Sum and divide by (n − 1): Σ = 0.0267 0.0267 ÷ (3 − 1) = 0.01335
  3. Take the square root: √0.01335 = 0.116 ≈ 0.12
Answer: 0.12 (0.1158).
Standard deviation: deviations from the mean, squared, summed, ÷ (n − 1), square-rooted. The two slips are dividing by n (= 3) instead of n − 1 (= 2), and forgetting the final square root. Lay the squared deviations in a column, sum, ÷ 2, √. A small SD (0.12) relative to the mean (7.07) shows tight, consistent replicates.
9

Rate from a tangent (gradient) with units

Paper 3 (A2), 2017, Q1(a)Rates2 marks

A fetal growth chart plots biparietal diameter (mm) against gestation (weeks) as a curve that flattens over time. Calculate the rate of growth at 30 weeks on the 95th-percentile curve, 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 30 weeks.
  2. Read a rise and a run off the tangent and divide: gradient = change in BPD ÷ change in time ≈ 2.7 mm week⁻¹
  3. State the units from the axes: y-units per x-units = mm per week = mm week⁻¹
Answer: ≈ 2.7 mm week⁻¹ (accept 2.35–2.85, because the tangent read varies). Units: mm week⁻¹.
The make-or-break skill is the tangent. Dividing the total BPD by 30 gives an average rate, not the rate at 30 weeks — a curve has a different steepness at every point, so lay a ruler against the curve at 30 weeks and measure the slope of that. OCR explicitly refuses a single-value read (81 ÷ 30) here. The marks are a correct tangent and rise ÷ run with the units copied straight off the axes (mm per week). The wide accepted band (2.35–2.85) reflects that everyone’s tangent differs slightly.
10

Percentage increase from a graph

Paper 1 (A2), 2017, Q32(b)(ii)Data handling2 marks

On a glucose-tolerance graph, blood glucose was about 7 mmol dm⁻³ at time 0 and about 12 mmol dm⁻³ at 90 minutes. Calculate the percentage increase, to 2 significant figures.

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The method

  1. Find the increase (new − original): 12 − 7 = 5
  2. Divide by the original (time-0) value: 5 ÷ 7 = 0.714
  3. Multiply by 100: 0.714 × 100 = 71.4% ≈ 71%
Answer: 71% (71.4%).
Percentage increase = (change ÷ original) × 100, with the time-0 value (7) as the denominator; the error to avoid is dividing by the new value (12). Read the two graph values accurately first — on a glucose-tolerance graph the glucose and insulin curves share the axis, so use the correct one. Watch the 2-s.f. instruction: 71.43 → 71. Show the (12 − 7) ÷ 7 × 100 working to bank the method mark even if the rounding slips.

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.

A complete OCR B maths worked-solutions pack — every paper, 2016–2024.