In a population of 10,000 sheep,4900 have a h | vedclass

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

Question

(A) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$ and the genotype frequencies are given by $p^2 + 2pq + q^2

Vedclass · Accepted Answer

$0.7$

Vedclass · Answer

(A) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$ and the genotype frequencies are given by $p^2 + 2pq + q^2 = 1$.Here,$p$ represents the frequency of the dominant allele $(A)$ and $q$ represents the frequency of the recessive allele $(a)$.Total population = $10,000$.Number of homozygous dominant individuals $(AA)$ = $4900$.Frequency of genotype $AA$ $(p^2)$ = $\frac{4900}{10000} = 0.49$.To find the frequency of the dominant allele $(p)$,we take the square root of the frequency of the homozygous dominant genotype:$p = \sqrt{p^2} = \sqrt{0.49} = 0.7$.Thus,the frequency of the homozygous dominant allele is $0.7$.

In a population of $10,000$ sheep,$4900$ have a homozygous dominant genotype $(AA)$ and $4200$ are heterozygous $(Aa)$ sheep. The remaining sheep have a homozygous recessive genotype $(aa)$. Find the frequency of the homozygous dominant allele $(p)$ as per the Hardy-Weinberg principle.

Similar Questions

In the Hardy-Weinberg equation $p^{2} + 2pq + q^{2} = 1$,which term represents the frequency of the homozygous recessive genotype?

In a population at Hardy-Weinberg equilibrium,the frequency of allele '$A$' is $0.6$ and the frequency of allele '$a$' is $0.4$. What will be the frequency of heterozygotes in the population?

Which one of the following factors will not affect the Hardy-Weinberg equilibrium?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module