A population is in Hardy-Weinberg equilibrium | vedclass

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

Question

(B) For a gene with two alleles,$A$ and $a$,if the frequency of $A$ is $p$ and the frequency of $a$ is $q$,then the frequencies of the three possible

Vedclass · Accepted Answer

$0.42$

Vedclass · Answer

(B) For a gene with two alleles,$A$ and $a$,if the frequency of $A$ is $p$ and the frequency of $a$ is $q$,then the frequencies of the three possible genotypes ($AA, Aa,$ and $aa$) are expressed by the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$.Here,$p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.Given that $p = 0.7$,we know that $p + q = 1$,so $q = 1 - 0.7 = 0.3$.The frequency of the heterozygous genotype $Aa$ is represented by $2pq$.Substituting the values: $2pq = 2 	imes 0.7 	imes 0.3 = 0.42$.

$A$ 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

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.

$A$ population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele $A$ is $0.1$, the frequency of $AA$ is . . . . . . .

In a population of $10,000$ cattle,$6400$ have homozygous dominant genotype $(AA)$ and $400$ have homozygous recessive genotype $(aa)$. Find out the number of cattle having heterozygous genotype $(Aa)$ as per Hardy-Weinberg's principle.

If $51\%$ of the members in a population possess at least one dominant allele,find the frequency of recessive individuals in that population.

In a population of $800$ rabbits showing Hardy-Weinberg equilibrium,the frequency of recessive individuals was $0.16$. What is the frequency of heterozygous individuals?

Vedclass Test Series

Exam Paper Generator

Online Exam Module