Question

In a population in Hardy-Weinberg equilibrium, if the frequency of the homozygous dominant genotype (AA) is 0.64, what is the frequency of the dominant allele (A)?

• A. 0.0064
• B. 0.32
• C. 0.8
• D. 0.4

Hint:

Always distinguish between allele frequencies (\(p\), \(q\)) and genotype frequencies (\(p^2\), \(2pq\), \(q^2\)).
Taking the square root of the homozygous dominant genotype frequency (\(p^2\)) directly yields the dominant allele frequency (\(p\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Hardy-Weinberg principle states that allele and genotype frequencies in a large, random-mating population remain constant from generation to generation in the absence of evolutionary influences.
The mathematical model utilizes the equation:
\[ p^2 + 2pq + q^2 = 1 \] and:
\[ p + q = 1 \] where \(p\) represents the frequency of the dominant allele and \(q\) represents the frequency of the recessive allele.

Step 2: Key Formula or Approach:
The genotype frequencies are defined as:
- Frequency of homozygous dominant (AA) = \(p^2\)
- Frequency of heterozygous (Aa) = \(2pq\)
- Frequency of homozygous recessive (aa) = \(q^2\)

Step 3: Detailed Explanation:
We are given that the frequency of the homozygous dominant genotype (AA) is \(0.64\).
Using the genotype frequency definition:
\[ p^2 = 0.64 \] To find the frequency of the dominant allele (\(p\)), we take the square root of both sides:
\[ p = \sqrt{0.64} \] \[ p = 0.8 \] Thus, the frequency of the dominant allele \(A\) is \(0.8\).
Using the allele frequency relation, we can also determine that the recessive allele frequency \(q\) is:
\[ q = 1 - p = 1 - 0.8 = 0.2 \] The calculated dominant allele frequency of \(0.8\) matches Option (C).

Step 4: Final Answer:
The frequency of the dominant allele (A) is 0.8.