Q. 16 - Biology

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).

Step-by-Step Solution

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 = $\frac{4}{24}=0.17$

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 = $\frac{4}{24}=0.17$