Question

Assume a population in Hardy-Weinberg equilibrium with these genotypic frequencies: AA = 0.25, Aa = 0.50, and aa = 0.25. If you remove all the homozygous dominants and allow the remaining population to reproduce (again under Hardy-Weinberg conditions), what will be the frequency of homozygous dominants in the next generation?

Please explain all the steps

Answer 1

Frequency before the removal of homozygous dominant.

Genotype of dominant allele is 0.25

Frequency of dominant allele is 0.5

Genotype of recessive allele is 0.25

Frequency of recessive allele is 0.5

Frequency after removal of homozygous dominant.

Total frequency = 0.75

Frequency of AA = 0

Frequency of Aa = 0.5/0.75 = 2/3

Frequency of aa = 0.25/0.75 = 1/3

New allelic frequency of A = 1/2 *2/3

= 1/3 or 0.33

New allele frequency of a - 1/2 *2/3+ 1/3 = 2/3 = 0.67

New frequency of AA = 0.33*0.33 = 0.11

New frequency of Aa = 2*0.33*0.67 = 0.44

New frequency of aa = 0.67*0.67 = 45