At a particular locus,the frequency of the A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to the Hardy-Weinberg principle,for a population in genetic equilibrium,the genotype frequencies are given by the expansion of $(p + q)^

Vedclass · Accepted Answer

$0.48$

Vedclass · Answer

(B) According to the Hardy-Weinberg principle,for a population in genetic equilibrium,the genotype frequencies are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.Here,$p$ is the frequency of the dominant allele $(A)$ and $q$ is the frequency of the recessive allele $(a)$.Given: $p = 0.6$ and $q = 0.4$.The frequency of heterozygotes $(Aa)$ is represented by $2pq$.Substituting the values: $2 	imes 0.6 	imes 0.4 = 2 	imes 0.24 = 0.48$.Therefore,the frequency of heterozygotes in the population is $0.48$.

At a particular locus,the frequency of the $A$ allele is $0.6$ and that of the $a$ allele is $0.4$. What would be the frequency of heterozygotes in a random mating population at equilibrium?

Similar Questions

If the frequency of the '$A$' allele is $0.4$,find the frequency of the '$B$' allele and the heterozygous genotype in a random mating population at equilibrium.

The sum of the frequencies of all the alternative alleles is:

According to the $Hardy-Weinberg$ principle,what remains constant and is maintained from generation to generation?

$A$ population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele $A$ is $0.7$,the genotype frequency of $Aa$ is

There are two alleles ($A_1$ and $A_2$) in a population. If the frequency of one allele $(A_1)$ is $0$,then the frequency of the other allele $(A_2)$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module