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Answer:
0.382
Explanation:
In solving this, let's recall the Hardy-Weinberge's equations which are as follows: p + q = 1; , where p represents the dominant Allele, while q represents the recessive allele. In applying this to what we want to calculate,
, 2pq = A1A2, and
.
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 , we would have the following:
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