next up previous
Next: Rudiments of human Up: Stat 260: Statistics Previous: Stat 260: Statistics

Hardy-Weinberg Equilibrium.

How they got involved.

G.H. Hardy was a mathematician who was responding to a suggestion that in outbred populations (e.g. humans) the frequency of a dominant phenotype should rise to 3/4 under Mendel's law. He considered the equilibrium result a consequence of ``a little mathematics of the multiplication-table type'' (Hardy (1908)). W. Weinberg was interested in working-out whether the ability to bear twins was a Mendelian trait. (In fact some Australian research found a gene for twinning segregating in certain sheep).

Hardy-Weinberg equlibrium

Both Hardy & Weinberg supposed a large population, mating ``random'' w.r.t. the trait under discussion (this does not mean choose your mate randomly), equal viability of offspring, ..., with other demographic issues ignored. Suppose that the proportions of AA, Aa, and aa genotypes in such a population are P, 2Q and R, respectively, with P+2Q+R=1. Consider what happens after one round of ``mating'':

Let the population frequencies in the new generation of AA, Aa, and aa be and , respectively. Then

The condition for population stability (i.e. the genotype frequencies being the same from one generation to the next) is , and , and satisfy this equation. Thus we reach a stable set of genotype frequencies in one round of ``random mating'' at a locus, regardless of the initial genotype frequencies.

Exercise 1: Show that if , then , and , for a unique p, . What is the interpretation of p? Of these expressions?

Why do we care about Hardy-Weinberg equilibrium? Many people care about it because it creates great technical simplicity: it means that only one allele frequency is necessary to specify the genotype frequencies. i.e. if p is the frequency of the allele A, then the frequencies of the genotypes AA, Aa, and aa are and respectively. This extends to a locus with more than 2 alleles in a straightforward way. Suppose the locus has alleles , then the frequency of the genotype is and the frequency of the genotype , is . This is what is meant by ``Hardy-Weinberg equilibrium''.

The way Hardy-Weinberg equilibrium can be interpreted is by imagining that all the alleles are in a big box and to create a the genotype of a new person we just pick 2 alleles at random. Hardy-Weinberg equilibrium is assumed in many population genetic analyses. When data are examined, it is sometimes approximately true, sometimes not.

Multiple loci: Linkage equilibrium.

Consider 2 loci, with alleles A and a, and with alleles B and b. Denote the population frequency of A by p and B by q. Then under random mating and certain population assumptions, Hardy-Weinberg equilibrium will be achieved after one generation. ie the frequency of AA, Aa, and aa will be and BB, Bb, bb will be . Under these same conditions we have the following pertaining to the joint genotypes at and .

Assertion: After many of generations the joint genotypic frequencies at and will be independent, i.e. :

Proof (very short): Let denote the population frequency of the haplotype AB in the generation, and let r be the recombination fraction between and . Then for the haplotype AB we get

Using the result a number of times (n to be exact...) gives the following expression and the convergence result:

Putting these haplotype frequencies together for the different haplotypes gives the results concerning the joint genotype frequencies in the table above. Note that the independence of alleles within haplotypes is termed linkage equilibrium.

A longer and more straightforward proof can be constructed by enumerating mating pair combinations at the two loci and arguing as we did in deriving Hardy-Weinberg equilibrium.

Exercise 2: Do this.

This result generalises easily, though laboriously, to any number of loci.



next up previous
Next: Rudiments of human Up: Stat 260: Statistics Previous: Stat 260: Statistics



Simon Cawley
Mon Apr 20 20:03:22 PDT 1998