Population genetics

Question

In a certain population, 1 person in 10,000 is albino (an absence of pigment melanin in the cells, AR). What

fraction of this population is expected to be heterozygous? Do you need to assume the Hardy-Weinberg law?

Answer 1

Yes, you would need to assume that the population is in Hardy Weinberg equilibrium,

A population's genotypic distribution in HW equilibrium can be described as A² + 2Aa + a², where A is the frequency of one of the alleles and a is the frequency of the other allele, and A² are a² theproportions of the two homozygous groups. Albinism is a recessive, so the frequency of 0.0001 is equal to a². To find the allele frequency of a, take the square root of a², or 0.01. Then since A + a = 1 (total allele frequencies must be equal to 100%), subtracxt 0.01 from 1, giving 0.99 as the allelefrequency of A. The term 2Aa gives the proportion of heterozygotes (remember that there will be two squares with heterozygotes in a Punnett square representation of this problem) So the proportion of heterozygotes is 2 x 0.99 x 0.01 = 0.0198, or 19.8%.

Hope this helps..