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. 0.24
B. 0.16
C. 0.48
D. 0.36

Answer

To find the frequency of heterozygotes in a random mating population at equilibrium, we can use the Hardy-Weinberg principle. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify Allele Frequencies**:
- The frequency of allele A (denoted as p) is given as 0.6.
- The frequency of allele a (denoted as q) is given as 0.4.

2. **Verify Allele Frequencies**:
- Check that the sum of allele frequencies equals 1:
\[ p + q = 0.6 + 0.4 = 1.0 \]
- This confirms that the frequencies are correctly defined.

3. **Use the Hardy-Weinberg Equation**:
- The Hardy-Weinberg equation is:
\[ p^2 + 2pq + q^2 = 1 \]
- In this equation:
- \( p^2 \) represents the frequency of homozygous dominant genotype (AA).
- \( q^2 \) represents the frequency of homozygous recessive genotype (aa).
- \( 2pq \) represents the frequency of heterozygous genotype (Aa).

4. **Calculate the Frequency of Heterozygotes**:
- To find the frequency of heterozygotes (Aa), we use the formula \( 2pq \):
\[ 2pq = 2 \times p \times q \]
- Substitute the values of p and q:
\[ 2pq = 2 \times 0.6 \times 0.4 \]
- Calculate:
\[ 2pq = 2 \times 0.6 \times 0.4 = 2 \times 0.24 = 0.48 \]

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

### Final Answer:
The frequency of heterozygotes in the population is **0.48**.