At a particular locus, frequency of allete A is 0.6 and that of allele a is 0.4. what would be the frequency of heterozygotes in a random mating population at equilibrium?

Options

A. 0.36
B. 0.16
C. 0.24
D. 0.48

Answer

To solve the problem of finding the frequency of heterozygotes in a random mating population at equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Allele Frequencies**:
- Given:
- Frequency of allele A (p) = 0.6
- Frequency of allele a (q) = 0.4

2. **Understand the Hardy-Weinberg Principle**:
- According to the Hardy-Weinberg equilibrium, the frequencies of the genotypes can be expressed as:
- \( p^2 \) = frequency of homozygous dominant (AA)
- \( 2pq \) = frequency of heterozygotes (Aa)
- \( q^2 \) = frequency of homozygous recessive (aa)
- The equation is: \( p^2 + 2pq + q^2 = 1 \)

3. **Calculate the Frequency of Heterozygotes**:
- We need to calculate \( 2pq \):
- \( 2pq = 2 \times p \times q \)
- Substitute the values of p and q:
- \( 2pq = 2 \times 0.6 \times 0.4 \)

4. **Perform the Calculation**:
- Calculate \( 2 \times 0.6 \times 0.4 \):
- \( 2 \times 0.6 = 1.2 \)
- \( 1.2 \times 0.4 = 0.48 \)

5. **Conclusion**:
- The frequency of heterozygotes (Aa) in the population at equilibrium is **0.48**.

### Final Answer:
The frequency of heterozygotes in a random mating population at equilibrium is **0.48**.