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AQA 7402 A-level Biology Maths skills

AQA A-level Biology Maths – Worked Solutions

At least 10% of the marks in AQA A-level Biology test maths skills, and they catch students out every year. Below are fully worked examples covering the full range of skills examined – 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 at A-level: genetics, statistics, microscopy, geometry, percentages, ecology indices, exponential growth, logarithms and surface area to volume. Click any example to expand the full method. Video walk-throughs are being added.

1

Hardy–Weinberg — finding an allele frequency

Paper 2, June 2023, Q04.4Genetics2 marks

In a population of cats, 36% showed white fur, which is produced by the dominant F allele (f is recessive). Using the Hardy–Weinberg equations, calculate the frequency of the recessive f allele.

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What the question gives you

  • 36% of cats had the F allele and showed white fur (F dominant, f recessive).
  • Hardy–Weinberg: p + q = 1 and p² + 2pq + q² = 1
  • Find: the frequency of the recessive f allele (q).

The method

  1. Read what “36% had F and white fur” means. White fur is the dominant phenotype, so 36% show the dominant trait (FF or Ff). That leaves 64% as ff — the only genotype with no F allele.
  2. Identify the term: q² = ff = 0.64
  3. Square-root to get q: q = √0.64 = 0.8
Answer: q = 0.8 (80% accepted). 1 mark for showing q² = 0.64 or ff = 64%.
The classic error is writing p = 0.36 straight away — but 36% is a phenotype, not an allele frequency. Rule: a recessive phenotype % = q² (square-root it); a dominant phenotype % = p² + 2pq (subtract from 1 first to get q², then square-root). p, q are allele frequencies; p², 2pq, q² are genotype frequencies.
2

Standard deviation — testing for a significant difference

Paper 1, June 2023, Q05.5Statistics2 marks

Clear-zone diameters were measured for an antimicrobial (cinnamon oil) and a positive control. Cinnamon oil had a mean of 17 mm (SD 2.4); the positive control had a mean of 13 mm (SD 2.2). Given that mean ± 2 standard deviations includes over 95% of the data, evaluate whether the difference between the two means is likely to be due to chance.

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What the question gives you

  • Mean ± 2 standard deviations includes >95% of the data.
  • Cinnamon oil: mean 17 mm, SD 2.4. Positive control: mean 13 mm, SD 2.2.
  • Find: is the difference in means likely due to chance (do the ranges overlap)?

The method

  1. Range for cinnamon oil (mean ± 2SD): 17 − (2 × 2.4) = 12.2 17 + (2 × 2.4) = 21.8So 12.2–21.8 mm.
  2. Range for positive control: 13 − (2 × 2.2) = 8.6 13 + (2 × 2.2) = 17.4So 8.6–17.4 mm.
  3. Check overlap. Cinnamon starts at 12.2; control reaches 17.4. Since 12.2 < 17.4, the ranges overlap.
Answer: the ranges overlap (12.2–17.4 mm), so the difference between the means is likely due to chance — no significant difference.
A small SD means data cluster near the mean; a large SD means they’re spread out. Mean ± 2SD captures ~95% of the data — if two such ranges overlap, you can’t be confident the means truly differ. Say “the difference is likely due to chance,” not “the results are due to chance.” Draw both ranges on a number line.
3

Magnification, mean & unit conversion

Paper 1, June 2024, Q03.4Microscopy2 marks

Eight alveolus diameters were measured from an image (in mm): 4, 2, 5, 1, 2, 3, 5, 2. The image magnification is ×40. Calculate the mean actual diameter in µm, showing your working.

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What the question gives you

  • Eight alveolus diameters measured off the image (mm): 4, 2, 5, 1, 2, 3, 5, 2.
  • Magnification of the image = ×40.
  • Find: the mean actual diameter in µm.

The method

  1. Mean the image diameters first (convert once at the end): (4+2+5+1+2+3+5+2) ÷ 8 = 24 ÷ 8 = 3 mm
  2. Divide by magnification (actual = image ÷ magnification): 3 ÷ 40 = 0.075 mm
  3. Convert mm → µm (×1000): 0.075 × 1000 = 75 µm
Answer: 75 µm. 1 mark for any correct stage — rearranging, ÷40, the mean of 3000 µm, ×1000, or right digits/wrong units (0.075 mm).
Use the “I = A × M” triangle: Image = Actual × Magnification. The most common error is not matching units — check both are the same before dividing. Memorise 1 mm = 1000 µm, 1 µm = 1000 nm. Magnification has no units.
4

Volume of a cylinder — rearranging the formula

Paper 1, June 2023, Q05.2Geometry2 marks

A well cut into agar is a cylinder with a diameter of 6 mm. It must hold 50 mm³ of cinnamon oil. Using volume = πr²l and π = 3.14, calculate the minimum depth of the well.

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What the question gives you

  • A well cut in agar is a cylinder; diameter 6 mm; oil volume to fit 50 mm³.
  • Volume of a cylinder = πr²l, with π = 3.14.
  • Find: the minimum depth (l) of the well.

The method

  1. Diameter → radius (halve it): r = 6 ÷ 2 = 3 mm
  2. Rearrange for l: V = πr²l → l = V ÷ (πr²)
  3. Calculate πr²: 3.14 × 3² = 3.14 × 9 = 28.26 mm²
  4. Divide: l = 50 ÷ 28.26 = 1.77 mm
Answer: 1.77 mm (1.768–1.8 mm accepted).
Three steps in order: halve the diameter, square the radius before multiplying by π, and rearrange so the unknown is alone before substituting. Forgetting to halve gives ≈0.44 mm — worth 1 mark as a method slip.
5

Percentage increase

Paper 1, June 2022, Q06.2Percentages2 marks

The rate of transpiration was read from a graph as approximately 0.75 cm³ hr⁻¹ at 1 pm and 0.80 cm³ hr⁻¹ at 2 pm. Calculate the percentage increase in the rate of transpiration between 1 pm and 2 pm.

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What the question gives you

  • Transpiration rate at 1 pm ≈ 0.75 cm³ hr⁻¹; at 2 pm ≈ 0.80 cm³ hr⁻¹.
  • Find: the percentage increase from 1 pm to 2 pm.

The method

  1. Increase = new − old: 0.80 − 0.75 = 0.05
  2. Divide by the ORIGINAL (not the new value): 0.05 ÷ 0.75 = 0.0667
  3. ×100: 0.0667 × 100 = 6.67%
Answer: 6.67% (6.6̇ or 6.67–7% accepted).
% change = (new − old) ÷ old × 100. The denominator is always what you started from. Dividing by the new value (0.80) gives 6.25% — worth 1 mark only. Routine: difference, circle the old value as divisor, then ×100.
6

Simpson’s Index of Diversity

Paper 1, June 2022, Q01.4Ecology2 marks

The shoot biomass (g m⁻²) of seven plant species was recorded: 2, 4, 5, 7, 15, 36 and 51. Using the index of diversity d = 1 − Σ(n/N)², where n is each species’ biomass and N is the total biomass, calculate the index of diversity.

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What the question gives you

  • Formula: d = 1 − Σ(n/N)² where n = each species’ biomass, N = total biomass.
  • Seven species’ biomass (g m⁻²): 2, 4, 5, 7, 15, 36, 51.
  • Find: the index of diversity d.

The method

  1. Total, N: 2+4+5+7+15+36+51 = 120
  2. Each (n/N)², then sum. The two largest dominate: (36/120)² = 0.090 (51/120)² = 0.1806Sum of all seven ≈ Σ(n/N)² ≈ 0.2928
  3. Subtract from 1: d = 1 − 0.2928 = 0.71
Answer: d = 0.71 (0.707; range 0.7–0.71). 1 mark for Σ(n/N)² ≈ 0.29–0.30, or for N = 120.
Add a “(n/N)²” column to the table and fill it row by row, then sum and subtract from 1. Common slips: forgetting to square, or forgetting the final “1 −”. Values near 1 = high diversity; near 0 = dominated by few species.
7

Exponential growth — PCR amplification (2ⁿ)

Paper 2, June 2023, Q09.3Exponentials2 marks

A single short tandem repeat (STR) molecule is 12 base pairs long. It is amplified through 50 cycles of PCR. Calculate the total number of base pairs present after 50 cycles.

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What the question gives you

  • A single STR molecule of 12 base pairs, amplified through 50 PCR cycles.
  • Find: the total number of base pairs after 50 cycles.

The method

  1. Each cycle doubles the molecules, so after n cycles there are 2ⁿ: 2⁵⁰ = 1.1259 × 10¹⁵ molecules
  2. Multiply by base pairs per molecule (×12): 1.1259 × 10¹⁵ × 12 = 1.35 × 10¹⁶
Answer: 1.35 × 10¹⁶ base pairs (1.36 × 10¹⁶ accepted). 1 mark for writing 2⁵⁰ × 12 anywhere.
Doubling is 2ⁿ, not n × 2 — students often write 50 × 2 = 100. Do 2⁵⁰ first, then × 12. Show how to type 2 ^ 50 on the calculator. Even an unfinished calculation scores a mark if 2⁵⁰ × 12 is written.
8

Logarithms — bacterial generation time from a log graph

Paper 1, June 2021, Q09.2Logarithms2 marks

From a log-scale graph of bacterial population against time, the population was about 4 × 10⁶ cells cm⁻³ at −4 hours and about 3 × 10⁹ cells cm⁻³ at +4 hours. Using number of generations = log₁₀(pop₂ ÷ pop₁) ÷ log₁₀(2), calculate the generation time in hours.

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What the question gives you

  • Log-scale graph: at −4 h, ≈ 4 × 10⁶ cells cm⁻³; at +4 h, ≈ 3 × 10⁹ cells cm⁻³.
  • Formula: generations = log₁₀(pop₂ ÷ pop₁) ÷ log₁₀(2)
  • Find: the generation time (one binary fission) in hours.

The method

  1. Population ratio: (3 × 10⁹) ÷ (4 × 10⁶) = 750
  2. Number of generations: log₁₀(750) ÷ log₁₀(2) = 2.875 ÷ 0.301 = 9.55
  3. Generation time = time ÷ generations (8 h interval): 8 ÷ 9.55 = 0.84 hours
Answer: 0.84 hours (≈ 50 min; 0.8 h accepted). 1 mark for any correct stage given the tricky log axis.
The formula is just “2ⁿ = growth ratio” rearranged with logs. Reading a log axis: halfway between 10⁶ and 10⁷ is ≈ 3 × 10⁶, not 5 × 10⁶. Final trap: 9.55 ÷ 8 = 1.19 is generations per hour, not the generation time — invert it (8 ÷ 9.55). Sanity check: ~50 min doubling is sensible for bacteria.
9

Surface area : volume ratio from formulae

Paper 3, June 2024, Q03.2SA:V ratio3 marks

Person A is 181 cm tall with a mass of 90.90 kg; Person B is 149 cm tall with a mass of 62.62 kg. Surface area (m²) = √[(height × mass) ÷ 3600] and volume (m³) = mass ÷ 1010. Determine which person has the smaller surface area to volume ratio, giving the ratio to 3 significant figures.

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What the question gives you

  • Person A: 181 cm, 90.90 kg. Person B: 149 cm, 62.62 kg.
  • Surface area (m²) = √[(height × mass) ÷ 3600]; Volume (m³) = mass ÷ 1010.
  • Find: who has the smaller SA:V ratio, to 3 sig figs.

The method

  1. Surface areas: A: √[(181×90.90)/3600] = √4.570 = 2.14 m² B: √[(149×62.62)/3600] = √2.592 = 1.61 m²
  2. Volumes: A: 90.90/1010 = 0.0900 m³ B: 62.62/1010 = 0.0620 m³
  3. SA:V ratios: A: 2.14/0.0900 = 23.7 : 1 B: 1.61/0.0620 = 26.0 : 1
Answer: Person A, 23.7 : 1 (23.8 : 1 accepted) has the smaller ratio.
Four substitutions plus a square root — slip-prone. The bigger person (A) has the smaller SA:V, which is the biological point and your sanity check. “3 sig figs” and “express as X : 1” each carry a mark; keep full precision on the calculator and round only the final ratio.
10

Interpreting P-values against the 0.05 threshold

Paper 2, June 2024, Q06.2Statistical inference3 marks

A statistical test gave the following P-values for the association between four factors and the risk of atrial fibrillation: age 0.004; high blood pressure 0.001; high LDL 0.222; hyperthyroidism 0.018. State the conclusions that can be drawn from these results.

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What the question gives you

  • P-values for association with atrial fibrillation: Age 0.004; High blood pressure 0.001; High LDL 0.222; Hyperthyroidism 0.018.
  • Find: a conclusion from the table.

The method

  1. Apply the 0.05 threshold. Below 0.05 = significant: age (0.004), blood pressure (0.001), hyperthyroidism (0.018). Above 0.05 = not significant: high LDL (0.222).
  2. Rank the significant ones — smaller P = stronger evidence: blood pressure (0.001) strongest; hyperthyroidism (0.018) weakest of the three.
  3. Link to the null hypothesis. Reject the null for the three significant factors; accept it for high LDL.
Answer: age, blood pressure and hyperthyroidism show a significant association (P < 0.05); high LDL does not (P = 0.222). Blood pressure is the strongest.
Two things students get wrong: thinking a bigger P means a stronger link (it’s the opposite), and saying “the results are significant” instead of “the association is significant” (AQA rejects the vague version). Drill: “P < 0.05 means less than a 5% probability the association is due to chance, so we reject the null hypothesis.”

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