The Hardy-Weinberg equilibrium relates the frequencies of alleles in a population to Homozygote and Heterozygous frequencies.
One allele is denoted p, and the another q. And when you add up the frequencies they have to add up to 1. Because 1 is the same thing as 100% of the population.
These frequencies can be related in a similar way to Heter/Homozygote frequencies. We know that if there are two alleles then there are three possibilities for any individual in the population. The can be Homozygous for the p allele, or Hetrozygous pq, or Homozygous for q.
In a punnet square of two Heterozygotes, remember how you get a 11 ratio? It's the same principle except you're dealing with a population and so you don't have such helpful numbers.
If you can imagine that we interbreed heterozygous individuals, the chance that we'd get a homozyote (p), we'd have to have two p's come together. The chance in this example is 50%(0.5) from both parents, and because if both these things must happen, we times the likelihood of these independent events together. And so we get 0.5x0.5 =0.25, or a quarter. Which is what the the punnet square is telling us! Hardy-Weinberg just extends this reasoning to a more general application, by simply saying that if you times the frequency of an allele by itself you find out out the proportion you'd expect to be Homozygous for that Allele! You can do exactly the same thing for the other allele.
For Heterozygous you times the likelyhood of both alleles together! But remember like the punnet square you're only working out one of the two combinations that can lead to Heterozygosity (pq or qp), so you have to times the number you get by two (2pq).
And so we arrive at our equation:
HomoP + Hetro + HomoQ = 1
p2 + 2pq + q2 = 1
Try to work out some of the problems by yourself when you start to catch my explanation!
Last edited by Piprod01; 24-05-2012 at 00:58.