Biology · Genetics

Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium links the allele frequencies p and q in an ideal population with the genotype frequencies and stays constant from generation to generation as long as no evolutionary factors act.

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Formula

LaTeX: p^2 + 2pq + q^2 = 1
p, q, p², 2pq, q² are dimensionless frequencies (0 to 1); p + q = 1

Variables & units – Hardy-Weinberg Equilibrium

SymbolMeaningUnit
pFrequency of the dominant allele A– (0…1)
qFrequency of the recessive allele a– (0…1)
Frequency of the homozygous dominant genotypes (AA)– (0…1)
2pqFrequency of the heterozygous genotypes (Aa), the carriers– (0…1)
Frequency of the homozygous recessive genotypes (aa)– (0…1)

Derivation & background – Hardy-Weinberg Equilibrium

Godfrey H. Hardy and Wilhelm Weinberg showed independently in 1908 that allele and genotype frequencies do not change without evolutionary factors. The genotype frequencies are the expansion of (p + q)² under random mating. Assumptions: very large population, no mutation, selection, migration or genetic drift, and random mating.

Exam blueprint

Validity range

Applies to an ideal population: very large, no mutation, selection, migration or genetic drift, and with random mating; for one gene with two alleles.

Derivation steps

Under random mating the alleles combine like the expansion of (p + q)².

  1. 1The allele frequencies add up to p + q = 1.
  2. 2Squaring: (p + q)² = p² + 2pq + q² = 1 gives the three genotype frequencies.

Rearrangements

Recessive allele frequency from the phenotype

Only the homozygous recessive phenotype (aa) has directly the frequency q².

Dominant allele frequency

Follows from p + q = 1.

Heterozygote proportion (carriers)

The factor 2 counts Aa and aA.

Task variant

16 % of a population show the recessive phenotype. Determine p, q and the heterozygote proportion.

q² = 0.16 → q = 0.4; p = 1 − 0.4 = 0.6; 2pq = 2·0.6·0.4 = 0.48, i.e. 48 % carriers. Check: 0.36 + 0.48 + 0.16 = 1.

1 in 10,000 has a recessive genetic disease. What is the carrier frequency?

q² = 0.0001 → q = 0.01; p = 0.99; 2pq = 2·0.99·0.01 = 0.0198 ≈ 2 %, i.e. about 1 carrier in 50 people.

Common mistakes

Confusing q with q².

The recessive phenotype gives q²; q is obtained only by taking the square root.

Forgetting the factor 2 in 2pq.

Heterozygotes arise in two ways (Aa and aA), hence 2pq.

Counting heterozygotes into the dominant phenotype p².

The dominant phenotype comprises p² + 2pq; only the genotype AA is p².

Assuming equilibrium despite selection or a small population.

With evolutionary factors the formula does not hold; deviations point exactly to that.

Exam context

  • Typical in advanced genetics/evolution: carrier frequencies of genetic diseases and testing for equilibrium.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

If 16 % of individuals show the recessive phenotype, then q² = 0.16, so q = 0.4 and p = 1 − 0.4 = 0.6. Then p² = 0.36 (36 % AA) and 2pq = 2·0.6·0.4 = 0.48 (48 % Aa). Check: 0.36 + 0.48 + 0.16 = 1.00 ✓.

Applications

Population genetics, calculating carrier frequencies for recessive genetic diseases, human genetics, evolutionary biology, detecting selection through deviation from equilibrium

Quanta exam set

Curated exam set for "Hardy-Weinberg Equilibrium":

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Which formula describes Hardy-Weinberg Equilibrium?

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How do you rearrange p² + 2pq + q² = 1 for Recessive allele frequency from the phenotype?

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Which common mistake happens with Hardy-Weinberg Equilibrium?

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Scientific sources

Common notations & search queries

p^2+2pq+q^2=1p²+2pq+q²=1p+q=1Hardy-Weinberg-GleichgewichtHardy Weinberg GleichungHardy-Weinberg-RegelAllelfrequenz berechnenGenotypfrequenzTrägerfrequenz ErbkrankheitHardy-Weinberg equilibrium

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Frequently asked questions about Hardy-Weinberg Equilibrium

How do you calculate allele frequencies with the Hardy-Weinberg equation?+

The easiest starting point is the recessive phenotype, because only the genotype aa is outwardly unambiguous. Its frequency equals q². Taking the square root of q² gives the frequency q of the recessive allele. The frequency of the dominant allele follows from p = 1 − q. Then all genotypes can be calculated: p² for homozygous dominant (AA), 2pq for heterozygous (Aa) and q² for homozygous recessive (aa). Example: if 16 % show the recessive phenotype, then q² = 0.16, so q = 0.4 and p = 0.6. Then 36 % are AA, 48 % Aa and 16 % aa. As a check, the three proportions always add up to 1.

Why does 2pq mean the frequency of carriers?+

A heterozygous individual carries one dominant and one recessive allele, that is Aa. Under random combination of alleles this pairing can arise in two ways: the dominant allele comes from the father and the recessive from the mother, or vice versa. Each way has probability p·q, together p·q + q·p = 2pq. That is why the factor 2 appears in the formula. These individuals are called carriers because they can pass on the recessive allele without showing it in their own phenotype. For a recessive genetic disease the carriers are genetically important, because two phenotypically healthy carriers can have an affected child with aa. A common mistake is to forget the factor 2.

Which conditions must be met for the Hardy-Weinberg equilibrium?+

The equilibrium requires an idealized population in which no evolutionary factors act. Specifically, five conditions must be met: first, a very large population so that random fluctuations of the frequencies, genetic drift, play no role. Second, no mutations that create new alleles. Third, no selection, meaning all genotypes are equally viable and fertile. Fourth, no immigration or emigration, that is no migration. Fifth, random mating, panmixia, without favouring particular genotypes. If these conditions hold, allele and genotype frequencies stay constant across generations. In nature this is never fully given, so the rule serves as a null model: if a population deviates measurably, it is a sign that an evolutionary factor such as selection is actually acting.

How do you estimate the carrier frequency of a recessive genetic disease?+

You use the known frequency of affected individuals, because this equals the frequency of the homozygous recessive genotype q². Taking the square root of q² gives q, the frequency of the disease allele. Then p = 1 − q and the sought carrier frequency is 2pq. Example: if a disease occurs in 1 of 10,000 people, then q² = 0.0001, so q = 0.01 and p = 0.99. The carrier frequency is then 2pq = 2·0.99·0.01 = 0.0198, i.e. about 2 percent or roughly 1 carrier in 50 people. Strikingly, there are many more healthy carriers than affected individuals, because the rare allele occurs mostly hidden in heterozygotes. This estimate is a standard tool of human genetics.

What does a deviation from the Hardy-Weinberg equilibrium indicate?+

If the observed genotype frequencies deviate clearly from those expected under p² + 2pq + q², at least one equilibrium condition is violated. An excess of homozygotes can indicate inbreeding or non-random mating. A shortage of a particular genotype often points to selection, when that genotype is disadvantaged. Migration that brings in foreign alleles, or a small population with strong genetic drift, also shift the frequencies. Therefore the rule is above all a null model: it provides the expected values for the case that no evolution takes place, and through the comparison makes visible whether and how a population changes. In practice one tests the deviation statistically, classically with a chi-square test, before concluding a specific evolutionary factor.

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How do you calculate with Hardy-Weinberg Equilibrium?

Here is how to work through a typical Hardy-Weinberg Equilibrium (p² + 2pq + q² = 1) task step by step:

  1. 1

    Task

    16 % of a population show the recessive phenotype. Determine p, q and the heterozygote proportion.

    Solution path

    q² = 0.16 → q = 0.4; p = 1 − 0.4 = 0.6; 2pq = 2·0.6·0.4 = 0.48, i.e. 48 % carriers. Check: 0.36 + 0.48 + 0.16 = 1.

  2. 2

    Task

    1 in 10,000 has a recessive genetic disease. What is the carrier frequency?

    Solution path

    q² = 0.0001 → q = 0.01; p = 0.99; 2pq = 2·0.99·0.01 = 0.0198 ≈ 2 %, i.e. about 1 carrier in 50 people.