Question

You are studying a population of Darwin’s finches. You genotype a large group of hatchlings at the A locus and then re-genotype them as adults. You observe the following genotype frequencies:

                        A₁A₁   A₁A₂   A₂A₂

Hatchlings:      0.10     0.30     0.60

Adults:            0.20     0.40     0.40

a) Is the hatchling population at Hardy-Weinberg equilibrium (Yes/No)? Show your work to receive full credit. (The math is simplest if you use fractions.)

Answer 1

Answer.

In orden to determinate if the hatchling population is at Hardy-Weinberg Equilibrium you have to calcúlate the allelic frequency as follows:

Hatchling population:

p(A1)= A1A1 frequency + 1/2 of A1A2 frequency= 0.10 + (0.30/2) = 0.10 + 0.15= 0.25

p(A2)= 1/2 of A1A2 frequency + A2A2 frequency= (0.30/2) + 0.60= 0.15+ 0.60= 0.75

Now with this new allelic frequencies we can estimate the expected genotype frequencies:

A1A1= p² = (0.25)²= 0.0625

A1A2= 2pq= 2x(0.25x0.75)= 0.375

A2A2= q²= (0.75)²= 0.5625

Once we have all of this data we can determinate if the hatchling population is at Equilibrium by doing a Chi square test:

Chi square= ((observed₁-expected₁)/expected₁)² + (observed₂-expected₂)/expected₂)² + (observed₃-expected₃)/expected₃)²

Chi square= ((0.10-0.0625)/0.0625)² + ((0.30-0.375)/0.375)² + ((0.60-0.5625)/0.5625)²)= 0.36+0.04+0.0044= 0.4044

This value is compared with the Chi square table, with one degree of freedom and probability of 0.05 which is 3.84

Now, as 0.4044 < 3.84 , which means that there is not a significative difference, therefore the Hatchling Population is at Hardy-Weinberg Equilibrium