Question

Assuming Hardy‐Weinberg equilibrium, the genoypte frequency of heterozygotes, if the frequency of two alleles at a locus are 0.7 and 0.3; will be

• A. 0.84
• B. 0.63
• C. 0.42
• D. 0.21

Hint:

Always double-check that your calculated genotype frequencies (\(p^2 + 2pq + q^2\)) add up to exactly 0. Here: \(0.49 + 0.42 + 0.09 = 00\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Hardy-Weinberg principle states that in a large, random-mating population free from evolutionary forces, allele and genotype frequencies remain constant from generation to generation.
Key Formula or Approach:
For a biallelic locus with alleles \(A\) and \(a\), let the frequency of \(A\) be represented by \(p\) and the frequency of \(a\) be represented by \(q\).
The sum of the allele frequencies is: \[ p + q = 1 \]
In a population at equilibrium, the expected frequencies of the genotypes are defined by the binomial expansion: \[ p^2 + 2pq + q^2 = 1 \]
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 (\(aa\)).

Step 2: Detailed Explanation:
We are given the allele frequencies: \[ p = 0.7 \] \[ q = 0.3 \]
We need to calculate the frequency of the heterozygotes (\(Aa\)), which is represented by the term \(2pq\).
Substituting the values into the formula: \[ \text{Heterozygote frequency} = 2 \times p \times q \] \[ \text{Heterozygote frequency} = 2 \times 0.7 \times 0.3 = 0.42 \]
Thus, the expected frequency of heterozygotes is \(0.42\) (or \(42\%\)).

Step 3: Final Answer:
The genotype frequency of heterozygotes under Hardy-Weinberg equilibrium is 0.4