Hardy-Weinberg Principle

The Hardy-Weinberg principle is the cornerstone of theoretical population genetics. Independently derived in 1908 by the British mathematician G. H. Hardy and the German physician Wilhelm Weinberg, it provides a mathematical framework that links allele frequencies in a population's gene pool to the genotype frequencies of individuals — under a specific and well-defined set of idealised conditions. Its real power lies in being a null model: by stating precisely what allele and genotype frequencies would do if no evolution were occurring, it gives biologists a benchmark against which actual populations can be compared. Any departure from Hardy-Weinberg expectations is evidence that evolution is at work.

Spec mapping: This lesson sits in AQA 7402 Section 3.7.2 (genetic diversity and adaptation — the use of the Hardy-Weinberg principle to calculate allele, genotype and phenotype frequencies in populations). Refer to the official AQA specification document for exact wording. It builds directly on Mendelian inheritance (Section 3.7.1, course 4) and underpins the rest of this course — natural selection (lesson 2), drift and gene flow (lesson 3), and speciation (lesson 4).

Connects to: Mendelian inheritance and dihybrid crosses (Section 3.7.1, course 4 DNA, Genes and Inheritance); natural selection (Section 3.7.2, lesson 2 of this course); chi-squared goodness-of-fit testing (lesson 0); selection pressures on enzyme alleles (Section 3.1.4, Biological Molecules).

---

The Conceptual Setup — Gene Pools and Allele Frequencies

Population genetics treats a population as a collective genetic entity — a gene pool containing all the alleles at a given locus across all individuals in the population. For a diploid species with population size N, the gene pool at a single locus contains 2N alleles. The proportions of the different alleles in this pool are the allele frequencies.

Key Definitions:

• Allele frequency — the proportion of a particular allele relative to all alleles at that locus in the gene pool.
• Genotype frequency — the proportion of individuals in the population carrying a particular diploid genotype.

For a single autosomal locus with two alleles, A (dominant) and a (recessive):

• Let p = frequency of A (so the A-pool fraction of the gene pool).
• Let q = frequency of a.
• Since these are the only two alleles, every gamete must carry one or the other: p + q = 1.

This first equation is trivial bookkeeping. The genuinely informative equation links allele frequencies to genotype frequencies.

---

The Hardy-Weinberg Equation

If gametes carrying A and a unite at random — that is, every A-gamete is equally likely to fuse with another A-gamete as with an a-gamete, and the same for a — then the genotype frequencies of the next generation can be derived from the binomial expansion of (p + q)²:

p² + 2pq + q² = 1

with:

• p² = frequency of homozygous dominant genotype (AA);
• 2pq = frequency of heterozygous genotype (Aa) — the factor of 2 arises because heterozygotes can form in two ways (A from father × a from mother, or vice versa);
• q² = frequency of homozygous recessive genotype (aa).

The full Hardy-Weinberg "package" is therefore two equations:

1. p + q = 1 — allele-frequency bookkeeping.
2. p² + 2pq + q² = 1 — genotype-frequency prediction under random mating.

A-Level misconception watch: A common mistake is to confuse the two equations or to apply them interchangeably. p + q = 1 holds for alleles in the gene pool, and is true regardless of any biology. p² + 2pq + q² = 1 is the biological prediction — it requires the Hardy-Weinberg assumptions to hold. Mixing them up costs marks.

---

Conditions for Hardy-Weinberg Equilibrium

The genotype equation holds only if all five idealised conditions are met:

1. No mutation. No new alleles are being created or lost by mutation between generations.
2. No natural selection. All genotypes have equal fitness and equal survival to reproductive age.
3. No migration (no gene flow). No alleles are entering or leaving the population.
4. Random mating. Individuals mate without preference for particular genotypes (panmixia).
5. Large population size. The population is effectively infinite — large enough that random sampling effects (genetic drift, lesson 3) do not change allele frequencies appreciably from generation to generation.

If all five hold, the genotype frequencies of any generation predicted by p² + 2pq + q² will persist unchanged across successive generations — the population is in Hardy-Weinberg equilibrium. The proof (which Hardy and Weinberg independently established) is to show that the gametes produced by the parental generation have the same allele frequencies p and q as the parents, and that random union of those gametes regenerates the same genotype frequencies in the offspring.

Key Point: No real population satisfies all five conditions exactly. The model is a baseline that lets us detect deviations. A statistically significant departure of observed genotype frequencies from Hardy-Weinberg predictions tells us at least one condition is being violated — i.e. evolution is occurring.

---

Worked Example 1 — Cystic Fibrosis Carrier Frequency

Cystic fibrosis (CF) is an autosomal recessive disorder caused by mutation in the CFTR gene. Approximately 1 in 2,500 babies in the UK population is born with CF.

Step 1: Affected individuals are homozygous recessive (aa), so q² = 1/2500 = 0.0004.

Step 2: q = √0.0004 = 0.02.

Step 3: p = 1 − q = 0.98.

Step 4: Carrier frequency 2pq = 2 × 0.98 × 0.02 = 0.0392.

Step 5: Homozygous-dominant frequency p² = 0.98² = 0.9604.

Genotype	Frequency	Approximate ratio
AA (homozygous normal)	0.9604	~96.0%
Aa (carrier)	0.0392	~3.9% (≈ 1 in 25)
aa (affected)	0.0004	~0.04% (≈ 1 in 2,500)

The clinically useful number is the carrier frequency: roughly 1 in 25 UK-population members carries the CF allele despite the rarity of the affected phenotype. This explains why CF is a recurring genetic-counselling concern even though only ~0.04% of births show the disease.

---

Worked Example 2 — Codominance and the MN Blood Group

The MN blood group is controlled by a single locus with two codominant alleles, M and N. In a survey of 1,000 individuals: 360 MM, 480 MN, 160 NN.

Step 1 — Allele frequencies from observed genotype counts:

Each MM individual contributes 2 M alleles; each MN contributes 1 M and 1 N; each NN contributes 2 N. Total alleles in the sample = 2 × 1,000 = 2,000.

• Number of M alleles = (2 × 360) + 480 = 1,200.
• p(M) = 1,200 / 2,000 = 0.6.
• q(N) = 1 − 0.6 = 0.4.

Step 2 — Predicted genotype counts under Hardy-Weinberg:

• Expected MM = p² × 1,000 = 0.36 × 1,000 = 360.
• Expected MN = 2pq × 1,000 = 0.48 × 1,000 = 480.
• Expected NN = q² × 1,000 = 0.16 × 1,000 = 160.

Observed and expected values match exactly. This population is consistent with Hardy-Weinberg equilibrium for the MN locus.

Step 3 — Formal test with chi-squared: Linking to lesson 0, we can apply χ² = Σ (O − E)² / E:

Class	O	E	(O − E)² / E
MM	360	360	0
MN	480	480	0
NN	160	160	0
Total			χ² = 0.00

df = 3 − 1 − 1 = 1 (subtracting an extra df for the allele frequency estimated from the data; this is a Hardy-Weinberg-specific df rule). Critical value at p = 0.05 = 3.841. We fail to reject H₀: the population is in Hardy-Weinberg equilibrium.

Note on df for Hardy-Weinberg goodness-of-fit: For a two-allele locus, df = number of genotype classes − 1 − number of parameters estimated from the data = 3 − 1 − 1 = 1. Because we estimated p from the data, we lose an extra degree of freedom. This is beyond AQA mark-scheme requirements but is the correct undergraduate treatment; A-Level mark schemes typically accept df = categories − 1 (here, df = 2).

---

Worked Example 3 — Sickle Cell and Heterozygote Advantage

Sickle cell anaemia is caused by a recessive allele HbS at the β-globin locus. In a West African population, the frequency of HbS (q) is 0.14.

Predicted genotype frequencies under Hardy-Weinberg:

• p = 1 − 0.14 = 0.86.
• p² (HbA HbA, normal) = 0.7396 (~73.96%).
• 2pq (HbA HbS, carrier — sickle cell trait) = 0.2408 (~24.08%).
• q² (HbS HbS, full sickle cell anaemia) = 0.0196 (~1.96%).

The HbS allele frequency is much higher in malaria-endemic regions than would be expected if HbS conferred only a fitness cost (sickle cell anaemia is usually fatal pre-reproduction without treatment). The reason is heterozygote advantage (overdominance): HbA HbS carriers have substantially increased resistance to Plasmodium falciparum malaria because the parasite's intracellular life cycle is disrupted in sickle-prone erythrocytes. The locus is therefore subject to balancing selection — selection that maintains both alleles in the population because the heterozygote out-competes both homozygotes.

This is the classical illustration that Hardy-Weinberg's "no selection" assumption is being violated in a particular, named way: the population is not in simple Hardy-Weinberg equilibrium, and the deviation has a precise biological explanation.

---

How Hardy-Weinberg Breaks — Mapping Conditions to Forces

Condition violated	Force of evolution	Effect on allele/genotype frequencies
No mutation	Mutation	New alleles appear; allele frequencies shift very slowly (typical rates 10⁻⁸ to 10⁻⁵ per generation per locus)
No selection	Natural selection	Fitter genotypes increase, less fit decrease; can be directional, stabilising or disruptive (lesson 2)
No migration	Gene flow	Allele frequencies shift toward the immigrating population's frequencies; homogenises subpopulations (lesson 3)
Random mating	Non-random mating (assortative mating, inbreeding)	Allele frequencies unchanged but genotype frequencies skewed — excess homozygotes under inbreeding
Large population	Genetic drift	Random sampling changes allele frequencies; strong in small populations; can fix or lose alleles (lesson 3)

The five forces of microevolution map directly onto the five Hardy-Weinberg assumptions. This is not coincidence: the principle is constructed precisely so that violations of each assumption pick out one of the five known evolutionary mechanisms.

---

Using Hardy-Weinberg in Exam Questions

The standard question genres at A-Level are:

1. q² → carrier frequency. Given the incidence of an autosomal recessive condition, compute carrier frequency 2pq. Always proceed q² → q → p → 2pq, showing every step.
2. Allele frequencies → genotype frequencies. Given p and q, compute p², 2pq, q² for expected counts.
3. Genotype counts → allele frequencies. Count alleles from observed genotype data, then either compute expected counts or test goodness-of-fit with χ².
4. Equilibrium testing. Given observed and expected genotype counts, run a chi-squared test (cross-link to lesson 0) and conclude whether the population is in equilibrium.
5. Interpreting departures. Given a documented deviation from Hardy-Weinberg, identify which assumption is most likely violated and justify with biological reasoning.

Exam Tip: The single most common Hardy-Weinberg calculation at A-Level is q² → carrier frequency. Marks are awarded for stating q² = (incidence), taking the square root, computing p = 1 − q, and finally 2pq. Each line of working secures one mark.

---

flowchart TD
  A["Allele freqs p, q with p + q = 1"] --> B["Random mating (panmixia)"]
  B --> C["Gametes pair at random"]
  C --> D["Genotype freqs: p^2 + 2pq + q^2 = 1"]
  D --> E["Assumptions intact?"]
  E -- Yes --> F["Hardy-Weinberg equilibrium maintained across generations"]
  E -- No --> G["Evolution: selection, drift, gene flow, mutation, non-random mating"]

---

A-Level-Depth Misconceptions

• Confusing p² + 2pq + q² = 1 with p + q = 1. Students mix these up routinely. The first is the genotype prediction; the second is allele-frequency bookkeeping. Marks are lost when students write 2pq + q = 1 or similar hybrids.
• Treating Hardy-Weinberg as a description of reality. It is a null model — populations only approximate it. Saying "the population must be in Hardy-Weinberg" is an error.
• Forgetting the random mating clause when the question involves sex-linkage or population structure. Hardy-Weinberg applies straightforwardly only to autosomal loci with random mating. X-linked loci have a different equilibrium (allele frequencies in male hemizygotes vs female diploids equilibrate over several generations) — beyond AQA but a known A* discriminator.
• Misapplying when carrier-detection has eliminated affected individuals from the breeding pool. If genetic counselling allows carriers to choose not to have affected children, q² is being directly suppressed — the population is not in equilibrium because selection (in this case via human choice) is acting.
• Reporting carrier frequency as 2pq² instead of 2pq. A surprisingly common arithmetic slip.

---

Required Practical Cross-Reference

There is no dedicated AQA 7402 required practical for Hardy-Weinberg directly. The statistical apparatus (chi-squared testing) is covered in lesson 0 of this course and finds practical use in Required Practical 11 on species distribution (anchored in ecosystems, course 9). Hardy-Weinberg calculations are a routine paper-and-pencil exercise rather than a wet practical.

---

Specimen Question — Modelled on the AQA paper format

Question (9 marks): Phenylketonuria (PKU) is an autosomal recessive disorder. In a European population, approximately 1 in 10,000 newborns is affected.