Hardy-Weinberg principle

Biology· Unit 18: Selection and Evolution· 12 min read

1. Core Assumptions of Hardy-Weinberg Equilibrium★★☆☆☆⏱ 3 min

The Hardy-Weinberg principle describes a theoretical non-evolving population where allele frequencies remain constant across successive generations. CIE exam mark schemes require you to state all 5 core assumptions to earn full marks for any related definition question.

📘 Definition

Hardy-Weinberg Equilibrium

A stable population state where no evolutionary forces act, so allele and genotype frequencies do not change from one generation to the next.

No random mutations occur at the target gene locus
Mating between individuals in the population is completely random
There is no significant gene flow (immigration or emigration) between populations
No natural selection acts to favour or disfavour any specific genotype
The population is very large, eliminating random genetic drift effects

✓ Quick check

Test your understanding of the assumptions:

Which of the following violates Hardy-Weinberg equilibrium?
- A large isolated population
- Non-random mating for a preferred trait
- No new mutations
- Equal survival of all genotypes
Reveal answer
Non-random mating for a preferred trait —
Non-random mating directly shifts genotype frequencies away from expected equilibrium values.

Exam tip:

Never list fewer than 5 assumptions in exam answers, as CIE awards 1 mark per stated condition.

2. Hardy-Weinberg Equations and Standard Calculation Workflow★★★☆☆⏱ 4 min

The two core equations form the basis of all Hardy-Weinberg calculations. They only apply to diploid, autosomal genes with two possible alleles, and cannot be used for X-linked traits unless explicitly stated in the question.

p + q = 1

p^2 + 2pq + q^2 = 1

📐 Worked Example

If 1 in 2500 people in a population have a recessive autosomal disorder, calculate the frequency of the recessive allele q.

1
First, identify that the proportion of affected individuals equals the homozygous recessive genotype frequency q²
2
$q^2 = \frac{1}{2500} = 0.0004$
3
Take the square root of q² to get q
4
$q = \sqrt{0.0004} = 0.02$
5
Use p + q = 1 to find the dominant allele frequency p if required
6
$p = 1 - 0.02 = 0.98$

✓ Quick check

Quick calculation check:

If q = 0.1, what is the heterozygous carrier frequency?
- 0.01
- 0.18
- 0.81
- 0.9
Reveal answer
0.18 —
2pq = 2 * 0.9 * 0.1 = 0.18

Exam tip:

Always start calculations from the recessive phenotype value, as you cannot distinguish homozygous dominant and heterozygous individuals from external observation alone.

3. Interpreting Deviations from Hardy-Weinberg Equilibrium★★★☆☆⏱ 3 min

If observed genotype frequencies do not match the values predicted by the Hardy-Weinberg equations, at least one of the 5 core assumptions has been violated. This deviation is direct evidence that the population is undergoing evolutionary change.

Higher than expected q² frequency: No selection against the recessive homozygote, or new mutation increasing recessive allele count
Lower than expected q² frequency: Strong selection against the recessive homozygote, e.g. lethal recessive disorder
Higher than expected heterozygote 2pq frequency: Heterozygote advantage, e.g. sickle cell trait malaria resistance
Lower than expected heterozygote 2pq frequency: Inbreeding, which increases homozygote frequency across the genome

📐 Worked Example

Observed heterozygote frequency for the sickle cell gene in a malaria region is 0.32, but expected Hardy-Weinberg value is 0.18. Explain this deviation.

1
The deviation shows more heterozygotes are present than predicted
2
This is caused by heterozygote advantage: individuals carrying one sickle cell allele have higher resistance to malaria
3
This violates the no selection assumption of Hardy-Weinberg equilibrium

4. Exam Phrasing and Mark Scheme Best Practices★★☆☆☆⏱ 2 min

5. Common Pitfalls

Wrong move:

Using the total dominant phenotype frequency directly as p

Why:

Dominant phenotypes include both p² and 2pq genotypes, so you cannot extract p from this combined value

Correct move:

Always start calculations from the recessive homozygote q² value to derive q first

Wrong move:

Omitting one or more of the 5 core Hardy-Weinberg assumptions in written answers

Why:

CIE mark schemes award 1 mark per stated assumption, so partial lists lose easy marks

Correct move:

Memorise and write all 5 assumptions every time you are asked for Hardy-Weinberg conditions

Wrong move:

Calculating heterozygote frequency as pq instead of 2p*q

Why:

This ignores the two possible allele combinations from parental gametes (dominant from mother + recessive from father, and vice versa)

Correct move:

Double check the heterozygote formula multiplies the product of p and q by 2

Wrong move:

Rounding intermediate calculation values to 2 significant figures

Why:

Small rounding errors accumulate and produce final answers outside the CIE accepted tolerance range

Correct move:

Keep 3+ significant figures for all intermediate values, only round the final answer to 2 or 3 s.f.

Wrong move:

Applying Hardy-Weinberg equations to X-linked traits without adjusting the formula

Why:

The standard p² + 2pq + q² =1 formula only applies to diploid autosomal loci

Correct move:

Only use the standard equations for autosomal genes, and follow any special instructions given for X-linked cases

6. Quick Reference Cheatsheet

Quantity	Formula	Exam Notes
Dominant allele frequency	$p$	Calculated as $1 - q$
Recessive allele frequency	$q$	Square root of recessive phenotype proportion
Homozygous dominant frequency	$p^{2}$	Cannot be counted from standard phenotypes
Heterozygous carrier frequency	$2 pq$	Most commonly asked calculation
Homozygous recessive frequency	$q^{2}$	Equals proportion of affected individuals

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 4
Hardy-Weinberg calculation for cystic fibrosis
2023 · Paper 4
State Hardy-Weinberg assumptions
2022 · Paper 2
Calculate recessive allele frequency

What's Next

Mastery of the Hardy-Weinberg principle unlocks 6-8 mark calculation questions that appear in almost every CIE A Level Biology Paper 4, and forms the quantitative foundation for understanding how natural selection shifts allele frequencies over generations. You can now connect these theoretical calculations to real-world evolutionary scenarios, from antibiotic resistance in bacterial populations to directional selection for camouflage traits in wild animal groups. This skill will also help you interpret population genetic data in practical exam assessments.