Frequency of A allele is 0.6 and that of a allele is 0.4 . What would be frequency of heterozygotes in random mating population ?

Options

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

Answer

To find the frequency of heterozygotes in a random mating population using the Hardy-Weinberg principle, follow these steps:

### Step-by-Step Solution:

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

2. **Confirm Allele Frequency Sum**:
- According to the Hardy-Weinberg principle, the sum of the frequencies of the alleles must equal 1.
- Check: \( p + q = 0.6 + 0.4 = 1.0 \). This confirms that the allele frequencies are correct.

3. **Use the Hardy-Weinberg Equation**:
- The Hardy-Weinberg equation is represented as \( p^2 + 2pq + q^2 = 1 \), where:
- \( p^2 \) = frequency of homozygous dominant (AA)
- \( 2pq \) = frequency of heterozygotes (Aa)
- \( q^2 \) = frequency of homozygous recessive (aa)

4. **Calculate the Frequency of Heterozygotes**:
- The frequency of heterozygotes (Aa) is given by the term \( 2pq \).
- Substitute the values of \( p \) and \( q \):
\[ 2pq = 2 \times (0.6) \times (0.4) \]
\[ 2pq = 2 \times 0.24 = 0.48 \]

5. **Conclusion**:
- The frequency of heterozygotes (Aa) in the population is \( 0.48 \) or 48%.

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

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