Question

In a population, the frequency of a recessive allele (\(q\)) is \(0.4\). Calculate the frequency of homozygous dominant individuals?

• A. \(0.16\)
• B. \(0.36\)
• C. \(0.48\)
• D. \(0.64\)

Hint:

Hardy--Weinberg equilibrium: \[ \boxed{ p+q=1 } \] and \[ \boxed{ p^2+2pq+q^2=1, } \] where \[ \boxed{ p^2=\text{Homozygous dominant},\; 2pq=\text{Heterozygous},\; q^2=\text{Homozygous recessive}. } \]

Answer 1

The Correct Option is B

Step 1: Use the Hardy--Weinberg equation. For a population in Hardy--Weinberg equilibrium, \[ \boxed{ p+q=1, } \] where
• \(p\) = frequency of dominant allele,
• \(q\) = frequency of recessive allele.

Step 2: Calculate the dominant allele frequency. Given, \[ q=0.4. \] Therefore, \[ p=1-0.4=0.6. \]

Step 3: Calculate the frequency of homozygous dominant individuals. The frequency of homozygous dominant genotype is \[ \boxed{p^2.} \] Hence, \[ p^2=(0.6)^2=0.36. \] Therefore, \[ \boxed{0.36} \] is the correct answer. Thus, \[ \boxed{(B)} \] is the correct answer.