Question

A large population of laboratory animals has been allowed to breed randomly for a number of generations. After several generations, 9% of the animals display a recessive trait (aa), the same percentage as at the beginning of the breeding program. The rest of the animals show the dominant phenotype, with heterozygotes indistinguishable from the homozygous dominants. Assuming that the locus that causes this trait is in Hardy-Weinberg equilibrium, perform the following calculations. Describe the basis of each calculation and list your calculation steps. (14 points)  

Answer 1

According to Hardy-Weinberg equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1.

So p+q =1

p2+ 2pq+q2 =1

here p = frequency of dominant allele

q is the frequency of the recessive allele.

p2 frequency of homozygous dominant

q2 frequency of homozygous recessive

2pq frequency of heterozygous dominant.

Given that there were 9% population that were homozygous recessive.

So, q2 = 9% or q = 0.09.

q = 0.3

p+q =1

p = 1-q

= 1-0.3

= 0.7

So frequency of dominant allele is 0.7 and recessive allele is 0.3.

Frequency of homozygous dominant = p2 = 0.7*0.7 = 0.49.

Frequency of heterozygous dominant = 2*0.3*0.7 = 0.42.

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