Question

A population of 100 rats was analyzed for their genotype at the isocitrate dehydrogenase locus. There are two alleles (R and S) present in the population. The following results were obtained: RR 40, RS 30 and SS 30. The allelic frequencies of this population is p = f(A) = 0.575 and q = f (B) = 0.425. Is THIS population in Hardy-Weinburg equilibrium? Why?

Answer 1

Answer:

Yes, this population lies in Hardy-Weinberg equilibrium as it follows both the principles of Hardy-Weinberg.

Explanation:

Here ,

$f(A)= p=0.575$

$f(B)= q=0.425$

As per the first principle of hardy Weinberg, the sum of all the alleles at the locus must be equal to 1.

Thus,

$p+q=1\\0.575+0.425 = 1\\1=1$

Also, as per the second equation of Hardy Weinberg's equation-

$p^{2} + q^{2} +2pq =1$

$(0.575)^2+2(0.575)(0.425)+(0.425)^2=1\\0.3306+ 0.48875+ 0.180625=1\\1=1$

Hence, this population lies in Hardy-Weinberg equilibrium as it follows both the principles of Hardy-Weinberg