A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency o…

For a population in Hardy–Weinberg equilibrium:

Where:

• \(p\) = frequency of allele A
• \(q\) = frequency of allele a
• \({p}^{2}\) = frequency of genotype AA
• \(2pq\) = frequency of genotype Aa
• \({q}^{2}\) = frequency of genotype aa

Given: \(p=0.1\)

Frequency of AA genotype: \(AA={p}^{2}\)\(=(0.1{)}^{2}\)\(=0.01\)

Therefore, the frequency of AA = 0.01.

Correct Answer: Option A (0.01)