Question

You studied a small, isolated human population and identified a new gene A. You found these numbers of individuals with these genotypes (capital letter represents dominance): 250 individuals of AA genotype, 500 individuals of Aa phenotypes and only 1 aa genotype, who is a 3 years old boy. Is gene A in Hardy-Weinberg equilibrium? Describe the possible reasons why gene A is (or isn’t) at HardWeinberg equilibrium. What can you conclude from the collected data?

Answer 1

AA- 250

Aa- 500

aa-1

Each genotype has two alleles either AA, Aa or aa. Individual with AA genotye has 2 A alleles. Therefore for calculating number of A alleles in AA individuals we have to multiply number of AA genotype individual with 2. Same for allele a. Total number of alleles in this whole population is 751 x 2 = 1502

frequency of allele A = p=( (2 x 250) + (500) ) / 2 x 751 = 0.66

frequency of allele a = q = ( (2 x 1) +(500) )/ 2 x 751 = 0.33

Now we will calculate expected genotype frequencies to check if population is in hardy weinbergy equilibrium by comparing with observed values

Genotype observed expected

AA 250 327.1 (p² x 751)

Aa 500 327.1 (2pq x 751)

aa 1    81.8 (q² x 751)

The expected values are considerably different from observed values. There is an increase in number of heterozygotes at expense of homozygotes.

Heterozygote advantage and negative assortative mating may be the reason for deviation of the population from hardy weinberg equilibrium.

Heterozygote advantage is observed when the survival fitness of heterozygotes is higher than homozygotes.

Negative assortative mating is the condition where people prefer to mate with individuals of different genotype.

Both of these situations increase the number of heterozygotes in a population. The expected genotype frequency are considerably different from observed genotype suggesting that population is not following hardy weinberg equilibrium.