In a population of 1000 individuals 360 belong to genototype AA ,480 to Aa and the remaining 160 to aa Based on this data ,the frequency of allele A in the population is

Options

A. 0.6
B. 0.7
C. 0.4
D. 0.5

Explanation

To find the frequency of allele A in the population, we can follow these steps:

### Step 1: Identify the Genotype Frequencies
We are given the following genotypes in a population of 1000 individuals:
- AA (homozygous dominant): 360 individuals
- Aa (heterozygous): 480 individuals
- aa (homozygous recessive): 160 individuals

### Step 2: Calculate the Frequency of Each Genotype
To find the frequency of each genotype, we divide the number of individuals with each genotype by the total population (1000).

- Frequency of AA (p²) = 360 / 1000 = 0.36
- Frequency of Aa (2pq) = 480 / 1000 = 0.48
- Frequency of aa (q²) = 160 / 1000 = 0.16

### Step 3: Use the Hardy-Weinberg Principle
According to the Hardy-Weinberg principle, the following relationship holds:
\[ p^2 + 2pq + q^2 = 1 \]
Where:
- \( p \) = frequency of allele A
- \( q \) = frequency of allele a

From the genotype frequencies calculated:
- \( p^2 = 0.36 \)
- \( 2pq = 0.48 \)
- \( q^2 = 0.16 \)

### Step 4: Calculate the Frequency of Allele a (q)
To find \( q \), we can take the square root of \( q^2 \):
\[ q = \sqrt{q^2} = \sqrt{0.16} = 0.4 \]

### Step 5: Calculate the Frequency of Allele A (p)
Using the relationship \( p + q = 1 \):
\[ p = 1 - q = 1 - 0.4 = 0.6 \]

### Step 6: Conclusion
The frequency of allele A in the population is:
\[ p = 0.6 \]

### Summary
- The frequency of allele A (p) is 0.6.