360 out of 1000 individuals in a population have a genotype of AA while 480 have Aa genotype. The rest 160 belong to aa. Frequency of allele A in this population is

The correct option is 0.6.

To find the frequency of the allele A in the given population, we can use the gene frequency calculation method. Let's break down the information given in the problem:

Total population size = 1000 individuals
Number of individuals with genotype AA (homozygous dominant) = 360
Number of individuals with genotype Aa (heterozygous) = 480
Number of individuals with genotype aa (homozygous recessive) = 160

Each individual in a diploid population carries two alleles for a specific gene. Therefore, the total number of alleles in this population of 1000 individuals is:

$1000\times 2=2000$

Now, let's calculate the total number of "A" alleles in the population:
- Each AA individual contributes 2 "A" alleles: 360 × 2 = 720 alleles
- Each Aa individual contributes 1 "A" allele: 480 × 1 = 480 alleles
- Each aa individual contributes 0 "A" alleles: 160 × 0 = 0 alleles

Total number of "A" alleles = 720 + 480 = 1200 alleles

The frequency of the allele A is the proportion of "A" alleles out of the total alleles in the population. We calculate this as follows:

$\text{Frequency of allele A}=\frac{\text{Total number of A alleles}}{\text{Total number of alleles in the population}}$

$\text{Frequency of allele A}=\frac{1200}{2000}=0.6$

Alternatively, we can use the genotype frequencies to calculate the allele frequency:
Frequency of AA genotype (p²) = 360 / 1000 = 0.36
Frequency of Aa genotype (2pq) = 480 / 1000 = 0.48

The frequency of allele A (p) is given by the formula:

$p=\text{Frequency of AA}+\frac{1}{2}\left(\text{Frequency of Aa}\right)$

$p=0.36+\frac{0.48}{2}=0.36+0.24=0.6$

Thus, the frequency of allele A in this population is 0.6.