In a population of 1000 individuals, 360 belo | vedclass

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(C) The total number of individuals is $1000$.Each individual has $2$ alleles, so the total number of alleles in the population is $1000 	imes 2 = 20

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) The total number of individuals is $1000$.Each individual has $2$ alleles, so the total number of alleles in the population is $1000 	imes 2 = 2000$.The genotype $AA$ has $360$ individuals, contributing $360 	imes 2 = 720$ alleles of $A$.The genotype $Aa$ has $480$ individuals, contributing $480$ alleles of $A$.The genotype $aa$ has $160$ individuals, contributing $0$ alleles of $A$.Total number of $A$ alleles = $720 + 480 + 0 = 1200$.Frequency of allele $A$ $(p)$ = $\frac{	ext{Total number of } A 	ext{ alleles}}{	ext{Total number of alleles}} = \frac{1200}{2000} = 0.6$.

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

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