A population is in Hardt-Weinberg equiblibrium 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 step by step, we will use the Hardy-Weinberg principle, which states that for a population in equilibrium, the frequencies of genotypes can be calculated using the allele frequencies.

### Step-by-Step Solution:

1. **Identify Allele Frequencies**:
- Given the frequency of allele 'A' (denoted as p) is 0.7.
- Since there are only two alleles (A and a), the frequency of allele 'a' (denoted as q) can be calculated as:
\[ q = 1 - p = 1 - 0.7 = 0.3 \]

2. **Use Hardy-Weinberg Equation**:
- The Hardy-Weinberg principle provides the equation for genotype frequencies:
\[ p^2 + 2pq + q^2 = 1 \]
- Here, \( p^2 \) represents the frequency of homozygous dominant genotype (AA), \( 2pq \) represents the frequency of heterozygous genotype (Aa), and \( q^2 \) represents the frequency of homozygous recessive genotype (aa).

3. **Calculate Genotype Frequencies**:
- We need to find the frequency of the heterozygous genotype (Aa), which is given by \( 2pq \):
\[ 2pq = 2 \times p \times q = 2 \times 0.7 \times 0.3 \]
- Calculate \( 2pq \):
\[ 2pq = 2 \times 0.7 \times 0.3 = 2 \times 0.21 = 0.42 \]

4. **Final Result**:
- The genotype frequency of 'Aa' is \( 0.42 \).

### Summary:
- The frequency of the heterozygous genotype 'Aa' in the population is **0.42**.