Hardy-Weinberg Equilibrium

The easiest starting point is the recessive phenotype, because only the genotype aa is outwardly unambiguous. Its frequency equals $q^{2}$. Taking the square root of $q^{2}$ 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: $p^{2}$ for homozygous dominant (AA), 2pq for heterozygous (Aa) and $q^{2}$ for homozygous recessive (aa). Example: if 16 % show the recessive phenotype, then $q^{2} = 0.16$, 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.

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 $p\cdot{}q$, together $p\cdot{}q + q\cdot{}p = 2pq$. 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.

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.

You use the known frequency of affected individuals, because this equals the frequency of the homozygous recessive genotype $q^{2}$. Taking the square root of $q^{2}$ 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 $q^{2} = 0.0001$, so $q = 0.01$ and $p = 0.99$. The carrier frequency is then $2pq = 2\cdot{}0.99\cdot{}0.01 = 0.0198$, 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.

If the observed genotype frequencies deviate clearly from those expected under $p^{2} + 2pq + q^{2}$, 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.