Question

a gene in a randomly - mating population has two alleles (a, a), and 1 in 400 individuals in the population is homozygous recessive (aa). calculate the frequency of heterozygotes (aa) in the population.\
\\(\bigcirc\\) 0.9025\
\\(\bigcirc\\) 0.095\
\\(\bigcirc\\) 0.05\
\\(\bigcirc\\) 0.0475\
\\(\bigcirc\\) 0.0025

Explanation:

Step1: Find recessive allele frequency

Given $q^2 = \frac{1}{400} = 0.0025$, so $q = \sqrt{0.0025} = 0.05$.

Step2: Find dominant allele frequency

Since $p + q = 1$, $p = 1 - 0.05 = 0.95$.

Step3: Calculate heterozygote frequency

Heterozygote frequency is $2pq = 2 \times 0.95 \times 0.05 = 0.095$.

Answer:

B. 0.095