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

Explanation

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

### Step-by-Step Solution:

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

2. **Understand the Hardy-Weinberg Principle**:
- In a population at Hardy-Weinberg equilibrium, the genotype frequencies can be predicted from the allele frequencies.
- The equation is:
\[ p^2 + 2pq + q^2 = 1 \]
- Here, \( p^2 \) represents the frequency of homozygous dominant (AA), \( q^2 \) represents the frequency of homozygous recessive (aa), and \( 2pq \) represents the frequency of heterozygotes (Aa).

3. **Calculate the Frequency of Heterozygotes**:
- The frequency of heterozygotes (Aa) is calculated using the formula:
\[ 2pq \]
- 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 is 0.48.

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

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