Question

In a population of 100 individual, which is in Hardy-Weinberg equilibrium, the number of individual of different phenotypic classes are as follows: 64AA, 32 Aa and 4 aa. The frequency of A and a in this population would be-

• A. A=0.64, a=0.36
• B. A=0.60, a=0.40
• C. A=0.80, a=0.20
• D. A=0.30, a=0.70

Hint:

Always start by calculating \( q \) from the homozygous recessive frequency (\( q^2 \)) if possible.
It is the most reliable way to find allele frequencies in a population under H-W equilibrium.
Check your work: $0.8 + 0.2$ must equal 1.0!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a population of $N = 100$ individuals with known genotype counts and asked to find the allele frequencies ($p$ for $A$ and $q$ for $a$) assuming Hardy-Weinberg equilibrium.

Step 2: Key Formula or Approach:
The Hardy-Weinberg equation is: $p^2 + 2pq + q^2 = 1$, where:
\( p^2 \) = frequency of homozygous dominant ($AA$)
\( 2pq \) = frequency of heterozygotes ($Aa$)
\( q^2 \) = frequency of homozygous recessive ($aa$)
Allele frequency can be calculated directly from the genotype counts.
Detailed Explanation:

• Direct Allele Counting Method:
• Total number of alleles in the population = $2 \times N = 2 \times 100 = 200$.

• Number of $A$ alleles = $(2 \times 64 \text{ from } AA) + (1 \times 32 \text{ from } Aa) = 128 + 32 = 160$.

• Frequency of $A$ ($p$) = $160 / 200 = 0.80$.

• Number of $a$ alleles = $(2 \times 4 \text{ from } aa) + (1 \times 32 \text{ from } Aa) = 8 + 32 = 40$.

• Frequency of $a$ ($q$) = $40 / 200 = 0.20$.

• Verification via Genotype Frequency Method:
• Frequency of $aa$ ($q^2$) = $4 / 100 = 0.04$.

• Therefore, $q = \sqrt{0.04} = 0.20$.

• Since $p + q = 1$, then $p = 1 - 0.20 = 0.80$.
Final Answer:
The frequency of the $A$ allele is 0.80 and the frequency of the $a$ allele is 0.20.