Hardy Weinberg principle?

Question

The principle is that sum total of all allelic frequencies is 1.Individual frequencies for example can be named p,q.In a diploid cell , p and q represent the frequency of allele A and a respectively . The frequency of AA individuals in a population is p² as the probability of an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities. Similarly for aa it will be q².Can anybody explain that how it will be 2pq for Aa?

Answer 1

There are two equations you need here. p + q=1 and p²+2pq+q²=1. Here is an example problem for you.

The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium.

To calculate the percentage of heterozygous individuals in the population you would multiply 2pq. 2*.19*.81=.31 or 31% (rounded up from .3078)

To calculate the percentage of homozygous recessive individuals in the population you would take q², which is .81*.81=.66 or 66%.

Good luck to you in your studies!