A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele 'A' is 0.7, genotype frequency of 'Aa' is

Options

A. 0.21
B. 0.42
C. 0.36
D. 0.7

Answer

To solve the problem regarding the Hardy-Weinberg equilibrium and the genotype frequency of heterozygous individuals (Aa), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Hardy-Weinberg Principle**:
The Hardy-Weinberg equilibrium states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. The equations used are:
- Allele frequencies: \( p + q = 1 \)
- Genotype frequencies: \( p^2 + 2pq + q^2 = 1 \)

2. **Identify Allele Frequencies**:
Given that the frequency of allele 'A' (denoted as \( p \)) is 0.7, we can denote the frequency of allele 'a' (denoted as \( q \)) as:
\[ q = 1 - p = 1 - 0.7 = 0.3 \]

3. **Calculate the Heterozygous Genotype Frequency**:
The frequency of the heterozygous genotype (Aa) is given by the formula \( 2pq \). We can substitute the values of \( p \) and \( q \):
\[ 2pq = 2 \times (0.7) \times (0.3) \]

4. **Perform the Calculation**:
\[ 2pq = 2 \times 0.7 \times 0.3 = 2 \times 0.21 = 0.42 \]

5. **Conclusion**:
The genotype frequency of heterozygous individuals (Aa) is \( 0.42 \).

### Final Answer:
The genotype frequency of 'Aa' is **0.42**.

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