In a population of 100 individuals (which follows Hardy Weinberg Principle). If the frequency of dominant allel (A) = 0.6, then calculate the number of individual belong to genotype Aa (Heterozygous individuals) :-

Options

A. 36
B. 48
C. 16
D. 96

Explanation

To solve the problem of calculating the number of heterozygous individuals (genotype Aa) in a population of 100 individuals, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Given Information**:
- Total number of individuals in the population (N) = 100
- Frequency of the dominant allele (A) = P = 0.6

2. **Calculate the Frequency of the Recessive Allele**:
- According to the Hardy-Weinberg principle, the sum of the frequencies of the dominant and recessive alleles equals 1.
- Therefore, frequency of the recessive allele (a) = Q = 1 - P
- Q = 1 - 0.6 = 0.4

3. **Calculate the Frequency of Heterozygous Individuals**:
- The frequency of heterozygous individuals (genotype Aa) is given by the formula:
\[ \text{Frequency of Aa} = 2PQ \]
- Substituting the values of P and Q:
\[ \text{Frequency of Aa} = 2 \times 0.6 \times 0.4 \]
- Calculate:
\[ \text{Frequency of Aa} = 2 \times 0.6 \times 0.4 = 0.48 \]

4. **Calculate the Number of Heterozygous Individuals**:
- To find the actual number of heterozygous individuals in the population, multiply the frequency of heterozygous individuals by the total number of individuals:
\[ \text{Number of Aa individuals} = \text{Frequency of Aa} \times N \]
- Substituting the values:
\[ \text{Number of Aa individuals} = 0.48 \times 100 = 48 \]

5. **Final Answer**:
- Therefore, the number of individuals belonging to genotype Aa (heterozygous individuals) is **48**.