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Edexcel International · IAL XBI11 / YBI11 Units 1–6 Jan · June · Oct Maths skills

Edexcel International A-level Biology Maths — Worked Solutions

Edexcel International A-level Biology (Pearson’s IAL, XBI11 / YBI11) sits in three series a year — January, June and October — and examines a remarkably wide range of statistics: the Student’s t-test, Spearman’s rank, chi-squared AND the Mann–Whitney U test all appear, mostly in the practical-skills units (Unit 3 at AS, Unit 6 at A2), which are the maths-heaviest of the six. Below are fully worked examples spanning Units 1–6, 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 Edexcel International Units 1–6: all four of its statistical tests (the Student’s t-test, Spearman’s rank, chi-squared and the Mann–Whitney U test), the Hardy–Weinberg equation, the index of diversity, the Michaelis–Menten enzyme equation, the bacterial growth-rate constant, the Q10 temperature coefficient and the volume of a sphere. Units 3 and 6 (the practical-skills exams) carry the heaviest maths load. Click any example to expand the full method. Video walk-throughs are being added.

1

The Student’s t-test — calculating t

Unit 6 (A2), October 2020, Q2(d)(i)Statistics3 marks

Leaf lengths were compared for insecticide-treated and water-treated thistles (n = 15 each). The means are x̄₁ = 12.6 cm and x̄₂ = 9.8 cm, and the two variance terms give a standard error of the difference of 0.272. Using t = |x̄₁ − x̄₂| ÷ √[(S₁²÷n) + (S₂²÷n)], calculate the value of t.

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The method

  1. Top line — difference between the means: | 12.6 − 9.8 | = 2.8
  2. Bottom line — the standard error of the difference (built from the two variances, each ÷ n, added and square-rooted): √[(S₁²÷15) + (S₂²÷15)] = 0.272
  3. Divide top by bottom for t: t = 2.8 ÷ 0.272 = 10.3
Answer: t = 10.3 (10.31). This is then compared with the critical value (df = n₁ + n₂ − 2 = 28; critical value at p = 0.05 is 2.048) — 10.31 far exceeds it, so reject the null hypothesis.
The t-test compares two means. Build it in two halves: the top line is the difference between the means (2.8); the bottom line is the standard error of the difference — √[(S²/n) added for each group] (0.272). Then divide. A large t (10.3) signals a big, clear difference. Decision rule: if calculated t ≥ critical value, reject H₀.
2

Spearman’s rank correlation coefficient

Unit 6 (A2), October 2021, Q3(c)(i)Statistics2 marks

Marram-grass leaf length was ranked against position on a sand dune for n = 7 sampling points, giving Σd² = 8. Using rₛ = 1 − [6Σd² ÷ n(n² − 1)], calculate rₛ.

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The method

  1. Denominator n(n² − 1): 7 × (49 − 1) = 7 × 48 = 336
  2. Numerator 6Σd²: 6 × 8 = 48
  3. Divide, then subtract from 1: rₛ = 1 − (48 ÷ 336) = 1 − 0.143 = 0.857
Answer: rₛ = 0.857 (0.86). The critical value at n = 7 (p = 0.05) is 0.786; 0.857 exceeds it, so reject the null hypothesis — a significant positive correlation (longer leaves higher up the dune).
Spearman’s measures correlation between two ranked variables. Compute the two parts separately (n(n²−1) = 336; 6Σd² = 48), divide, then subtract from 1 — the easy step to forget. A value near +1 (0.857) means a strong positive correlation; compare its magnitude with the critical value to judge significance.
3

Chi-squared — calculating χ²

Unit 1 (AS), October 2021, Q8(c)(ii)–(iii)Statistics3 marks

Chick plumage was tested against a 2:1:1 ratio. Observed: speckled 243, white 125, black 112 (total 480). Expected: speckled 240, white 120, black 120. Using χ² = Σ[(O − E)² ÷ E], calculate χ².

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The method

  1. For each colour, (O − E), squared, ÷ E: speckled: (243 − 240)² ÷ 240 = 9 ÷ 240 = 0.0375 white: (125 − 120)² ÷ 120 = 25 ÷ 120 = 0.2083 black: (112 − 120)² ÷ 120 = 64 ÷ 120 = 0.5333
  2. Sum the three terms for χ²: 0.0375 + 0.2083 + 0.5333 = 0.78
Answer: χ² = 0.78 (0.779). With df = categories − 1 = 2, this is well below the critical value (5.99 at p = 0.05), so accept the null hypothesis — the results fit the 2:1:1 ratio.
Each category contributes (O−E)²÷E; (O−E)² is always positive (black’s −8 squares to +64). Sum the column for χ². Degrees of freedom = number of categories − 1 = 2. Decision rule: if χ² is smaller than the critical value, the difference from expected is not significant.
4

The Mann–Whitney U test

Unit 6 (A2), June 2021, Q2(d)(i)–(ii)Statistics3 marks

Biomass of two switchgrass varieties was compared using ranks. With the rank sums (R) and sample sizes given, use U = n₁n₂ + [n(n + 1) ÷ 2] − R for each group and take the smaller U. The calculated U = 19.5 and the critical value is 8 — state the conclusion.

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The method

  1. Compute U for each group from the formula, substituting that group’s rank sum R and the two sample sizes, then take the smaller U: U = 19.5
  2. Compare with the critical value — note the reversed rule: 19.5 > 8 (U is NOT ≤ critical value)
  3. Conclude: because U is greater than the critical value, accept the null hypothesis — no significant difference.
Answer: U = 19.5; since 19.5 > 8, accept H₀ — no significant difference between the two varieties’ biomass.
The Mann–Whitney U test compares two independent samples using their ranks — a non-parametric alternative to the t-test. Compute U for each group, take the smaller. The decision rule is the reverse of t and chi-squared: reject H₀ only if U is ≤ the critical value. Here 19.5 > 8, so the difference is not significant.
5

Hardy–Weinberg — allele frequencies

Unit 2 (AS), January 2021, Q6(c)Genetics3 marks

Of 610 individuals, 140 show the homozygous-recessive phenotype (frequency = q²). Using p² + 2pq + q² = 1 and p + q = 1, find the dominant allele frequency p to two decimal places.

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The method

  1. Find q² as a proportion, then q (square root): q² = 140 ÷ 610 = 0.23 q = √0.23 = 0.48
  2. Find p (p + q = 1): p = 1 − 0.48 = 0.52
Answer: p = 0.52 (q = 0.48; q² = 0.23).
Hardy–Weinberg always starts from the homozygous recessive, whose frequency is q². Turn the count into a proportion (140÷610 = 0.23), square-root for q (0.48), then p = 1 − q (0.52). The carrier frequency, if asked, is 2pq. Watch the wording: sometimes q is given directly and no square root is needed.
6

Index of diversity (the IAL formula)

Unit 2 (AS), June 2021, Q7(c)(i)Ecology3 marks

Species counts were A 21, B 2, C 4, D 13, E 54, F 15, G 6, H 32 (total N = 147). Using the index of diversity D = N(N − 1) ÷ Σn(n − 1), calculate D to one decimal place.

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The method

  1. Σn(n−1) — each species’ n(n−1), then add: (21×20)+(2×1)+(4×3)+(13×12)+(54×53)+(15×14)+(6×5)+(32×31) = 420 + 2 + 12 + 156 + 2862 + 210 + 30 + 992 = 4684
  2. Numerator N(N−1): 147 × 146 = 21 462
  3. Divide: D = 21 462 ÷ 4684 = 4.6
Answer: D = 4.6 (4.58).
This is the reciprocal index of diversity D = N(N−1) ÷ Σn(n−1) — a higher value means more diverse. It is NOT the “1 − …” version, so don’t subtract from 1. Total all counts for N (147), work out each species’ n(n−1) for the denominator (4684), then divide.
7

The Michaelis–Menten equation (enzyme rate)

Unit 1 (AS), January 2020, Q3(c)(ii)Enzymes2 marks

From the graph, Vmax = 50 and the Michaelis constant K = 1.9. At a substrate concentration S = 4, use V = (Vmax × S) ÷ (K + S) to find the rate V.

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The method

  1. Substitute Vmax, K and S: V = (50 × 4) ÷ (1.9 + 4)
  2. Work the top and bottom, then divide last: V = 200 ÷ 5.9 = 33.9
Answer: V ≈ 33.9 (34).
The Michaelis–Menten equation V = Vmax·S ÷ (K + S) gives the rate of an enzyme reaction. Read Vmax and K off the graph, substitute S, then do the division last — the whole top (Vmax × S) over the whole bottom (K + S). Keep the denominator as a single bracket so you don’t divide too early.
8

Bacterial growth-rate constant (from logs)

Unit 4 (A2), January 2020, Q9(d)Microbiology3 marks

During exponential growth the equation gives 0.963 = (7.079 − 3.778) ÷ (0.301 × t), where 7.079 and 3.778 are the log₁₀ of the two cell counts and 0.301 = log₁₀2. Calculate the time t, in hours.

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The method

  1. Numerator — difference of the logs: 7.079 − 3.778 = 3.301
  2. Rearrange for t and substitute: 0.963 × 0.301 × t = 3.301 0.289863 × t = 3.301
  3. Divide: t = 3.301 ÷ 0.289863 = 11.4 hours
Answer: t ≈ 11.4 hours (11.39).
The growth-rate constant uses logs of the cell counts. Take the difference of the two logs first (3.301), keep 0.301 (= log₁₀2) in the denominator, then rearrange to make t the subject and divide. Don’t take a “log of a log” — the counts are already logged.
9

Q₁₀ temperature coefficient

Unit 4 (A2), January 2026, Q2(b)Enzymes1 mark

From a graph of rate against temperature, the rate at 20 °C is about 22 a.u. and at 30 °C about 34 a.u. Using Q₁₀ = rate at (t + 10 °C) ÷ rate at t °C, calculate Q₁₀.

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The method

  1. Read both rates — numerator is the higher temperature: rate at 30 °C = 34; rate at 20 °C = 22
  2. Divide the higher-temperature rate by the lower: Q₁₀ = 34 ÷ 22 = 1.55
Answer: Q₁₀ = 1.55.
Q₁₀ is always the rate 10 °C up divided by the rate now — the numerator is the higher temperature. Dividing the wrong way round gives the reciprocal (~0.65), a classic trap. A Q₁₀ near 2 is typical for enzyme reactions; ~1.5 here shows the reaction is slowing as it nears its optimum.
10

Volume of a sphere, then a ratio

Unit 1 (AS), October 2020, Q8(c)(i)Geometry3 marks

An LDL particle has a given diameter; its volume is V = ⁴⁄₃πr³ (using π ≈ 3.14). A cholesterol molecule has a known volume. Work out the LDL volume and express the ratio of LDL volume to cholesterol volume.

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The method

  1. Volume of the LDL particle (⁴⁄₃πr³, halving the diameter for r): V ≈ 7235 nm³
  2. Form the ratio LDL : cholesterol and simplify: ≈ 14 : 1
Answer: ≈ 14 : 1 (13 : 1 if π = 3 is used).
Volume of a sphere = ⁴⁄₃πr³ — halve the diameter for the radius first, then cube it. Round the volume to a whole number before forming the ratio with the cholesterol volume. The value of π you use shifts the answer slightly (14:1 with 3.14, 13:1 with 3).

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