Question

Solve for the genetic structure of a population with $12$homozygous recessive individuals (yy), $8$homozygous dominant individuals (YY), and $4$ heterozygous individuals (Yy).

Short answer

Frequency of homozygous recessive individual =$\frac{12}{24}=0.50$

Frequency of homozygous dominant individual = $\frac{8}{24}=0.33$

Frequency of heterozygous dominant individual = $\frac{4}{24}=0.17$

Step-by-step solution

Hardy- Weinberg principle : 

The Hardy- Weinberg equilibrium, model, theory, or rule is also known as the Hardy- Weinberg principle. In the absence of evolutionary effects, the frequencies of alleles and genotypes in a population will, on average, remain constant from generation to generation.

 Explanation : 

Hardy-Weinberg equation :

$p^2+2pq+q^2=1$

where,

$p^2$is frequency for homozygous genotype YY.

$q^2$is frequency for homozygous genotype yy.

$2pq$is frequency for heterozygous genotype Yy.

Total number of individuals = $24$

Frequency of homozygous recessive individual = $\frac{12}{24}=0.50$

Frequency of homozygous dominant individual = $\frac{8}{24}=0.33$

Frequency of heterozygous dominant individual =$\frac{4}{24}=0.17$