Question

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

• A. \(0.99\)
• B. \(0.01\)
• C. \(0.02\)
• D. \(0.10\)

Hint:

• Always pay close attention to whether the question asks for an allele frequency (\(p\) or \(q\)) or a genotype/phenotype frequency (\(p^2\), \(2pq\), or \(q^2\)).

• To find the other allele's frequency: \(q = 1 - p = 1 - 0.1 = 0.9\).

• The heterozygous frequency (\(Aa\)) would be \(2pq = 2 \times 0.1 \times 0.9 = 0.18\).

Answer 1

The Correct Option is B

Concept:

• The Hardy-Weinberg principle states that allele and genotype frequencies in a population remain constant from generation to generation in the absence of evolutionary influences.

• For a gene locus with two alleles, \(A\) (dominant) and \(a\) (recessive): itemize

• Let \(p\) be the frequency of allele \(A\).

• Let \(q\) be the frequency of allele \(a\).

• Thus, \(p + q = 1\).
The expected genotype frequencies in the population are given by: \[ p^2 + 2pq + q^2 = 1 \] where:

• \(p^2\) is the frequency of homozygous dominant individuals (\(AA\)).

• \(2pq\) is the frequency of heterozygous individuals (\(Aa\)).

• \(q^2\) is the frequency of homozygous recessive individuals (\(aa\)).
itemize

Step 1: Identify the given values from the problem statement
The problem states that the population is in Hardy-Weinberg equilibrium, and the frequency of the allele \(A\) (represented as \(p\)) is: \[ p = 0.1 \]

Step 2: Calculate the frequency of genotype AA
The frequency of the homozygous dominant genotype \(AA\) is represented mathematically by \(p^2\): \[ \text{Frequency of } AA = p^2 \] \[ \text{Frequency of } AA = (0.1)^2 = 0.01 \] Thus, the expected frequency of the genotype \(AA\) in the population is \(0.01\), which corresponds to option (B).