Question

A researcher is studying a newly discovered gene that causes increased body weight in domesticated chickens. In a mainland population, the frequency of the A1 allele is 0.2, for this gene with two alleles. If 100 of these mainland chicken are transported on a ship to an isolated island with a population of 200 A1A1 chickens, 400 A1A2 chickens, and 400 A2A2 chickens, what would the frequency of the A1 allele in the admixed population

Answer 1

Answer:

0.382

Explanation:

In solving this, let's recall the Hardy-Weinberge's equations which are as follows: p + q = 1; $p^{2} + 2pq + q^{2} = 1$ , where p represents the dominant Allele, while q represents the recessive allele. In applying this to what we want to calculate, $p^{2} = A1A1$, 2pq = A1A2, and $q^{2} = A2A2$.

Using the Hardy-Weinberge's equations, we can determine the number of individuals having the different types of genotype inherent in the 100 mainland chicken that was transported.

Since p + q = 1, i.e. given that allele frequency of A1 is 0.2, therefore allele frequency for A2 = 0.8 (p + q = 1).

Using  $p^{2} + 2pq + q^{2} = 1$, we would have the following:

$p^{2} = A1A1 = 0.2^{2} = 0.04$

$2pq = A1A2 = 2(0.2*0.8) = 0.32$

$qx^{2} = A2A2 = 0.8^{2} = 0.64$

Therefore, number of mainland chickens with the A1A1 genotype = 0.04*100 = 4; A1A2 = 0.32*100 = 32; A2A2 =0.64*100 = 64

The genotype frequencies for the admixed population would be the sum of that of the transported mainland chicken and that of the isolated chicken, which would be:

Genotype       mainland      isolated      Total    No of A1 allele present

A1A1                      4                200           204                204

A1A2                     32              400           432                216

A2A2                    64              400           464                  0

Total                                                         1100                420

Therefore, allele frequency of the A1 allele in the admixed population = 420/1100 = 0.382