Solution
Allele Frequency from Genotype Counts
Show the problem again
In a population in Hardy-Weinberg equilibrium, with allele A having frequency θ, the genotypes AA, Aa and aa have probabilities θ², 2θ(1−θ) and (1−θ)². A random sample of 100 individuals contains 30 of type AA, 50 of type Aa and 20 of type aa. What is the maximum likelihood estimate of θ? Give an exact fraction.
Worked solution
The log-likelihood is, up to a constant, 30·2 ln θ + 50·(ln θ + ln(1−θ)) + 20·2 ln(1−θ) = 110 ln θ + 90 ln(1−θ). The exponents count alleles: each AA carries two A alleles, each Aa one, and so on, for 110 copies of A out of 200 alleles. The maximizer is 110/200 = 11/20. This equals the plain allele count, even though the three genotype probabilities are not binomial in the usual sense.
Source: Classic Hardy-Weinberg maximum-likelihood example from population genetics texts. Statement written for AxiomIQ.