A2 · Practice questions · The Hardy-Weinberg principle

Start from q².

Six original Cambridge-style questions on Hardy-Weinberg calculations and the conditions under which the principle applies.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Square root, subtract, multiply.

q² = frequency of the recessive phenotype.
q = √q², then p = 1 − q.
Carriers = 2pq; homozygous dominant = p².
Valid only if large, random mating, no selection, mutation or migration.

Calculation

[3 marks]

Cystic fibrosis is caused by a recessive allele. In one population, 1 in 2500 babies is born with cystic fibrosis. Use the Hardy-Weinberg equations to calculate the frequency of heterozygous carriers. Show your working.

q² = 1 ÷ 2500 = 0.0004, so q = √0.0004 = 0.02. ✓
p = 1 − 0.02 = 0.98. ✓
2pq = 2 × 0.98 × 0.02 = 0.0392 (about 1 in 25 people). ✓

Calculation

[3 marks]

In a population of 1500 moths, 540 have the recessive light wing colour. Calculate the number of moths expected to be homozygous dominant.

q² = 540 ÷ 1500 = 0.36, so q = 0.6. ✓
p = 1 − 0.6 = 0.4, so p² = 0.16. ✓
Number homozygous dominant = 0.16 × 1500 = 240. ✓

Recall

[4 marks]

State four conditions that must be met for the Hardy-Weinberg principle to apply to a population.

The population is very large (so there is no genetic drift). ✓
Mating is random. ✓
There is no selection (all genotypes are equally likely to survive and reproduce). ✓
There is no mutation and no migration into or out of the population (also: diploid organisms reproducing sexually). ✓

Calculation

[3 marks]

A sample of 500 plants contains 245 with genotype RR, 210 with Rr and 45 with rr. Calculate the frequency of allele r and state whether the genotype frequencies fit the Hardy-Weinberg prediction.

Number of r alleles = 210 + (2 × 45) = 300 out of 1000, so q = 0.3 (and p = 0.7). ✓
Expected: RR = 0.49 × 500 = 245; Rr = 0.42 × 500 = 210; rr = 0.09 × 500 = 45. ✓
Observed numbers match the expected numbers exactly, so the population fits the Hardy-Weinberg prediction. ✓

Explain

[3 marks]

Explain why, when a recessive allele is rare, most copies of that allele are found in heterozygous individuals.

If q is small, q² is very much smaller than q (for example q = 0.01 gives q² = 0.0001). ✓
2pq is approximately 2q when p is close to 1, so heterozygotes (0.0198) far outnumber homozygous recessives (0.0001). ✓
So the recessive allele is mostly hidden in carriers, where it is not exposed to selection. ✓

Analysis

[3 marks]

Over 20 generations, the frequency of a recessive allele in a population of mice fell from 0.30 to 0.12. Suggest three reasons why the Hardy-Weinberg principle does not apply to this population.

Selection against the recessive phenotype, for example aa mice are more visible to predators. ✓
Genetic drift, if the population is small. ✓
Migration of mice with a higher frequency of the dominant allele into the population, or non-random mating / mutation. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.