Question

The forward mutation rate from allele Z^A to allele Z^B is 0.00001 and the reverse mutation rate is 0.0000001.

Which of the following may be true for a population in which the frequency of allele Z^B is 67%?

Select one:

(i) Mutation equilibrium between Z^A and Z^B has not been reached

(ii) The frequency of Z^A is higher than expected at mutation equilibrium

(iii) The frequency of Z^B is higher than expected at mutation equilibrium

Both (i) and (iii)

Both (i) and (ii)

Answer 1

Correct answer is Both i) and iii).

Explanation with Calculation:

Let the forward mutation rate of Z^A to Z^B be X = 0.00001 or 1 x 10^-5

Let the backward mutation rate of Z^B to Z^A be Y = 0.0000001 or 1 x 10^-7

The frequency of allele Z^B (p) initially is 0.67 % or 0.67

According to Hardy-Weinberg equilibrium we know that the sum of allelic frequency of dominant and recessive alleles is always 1.

So allelic frequency of Z^A (q) = 1 - p = 1 - 0.67 = 0.33

Case - I (Mutation equilibrium between Z^A and Z^B has not been reached) :

At equilibrium , the change in allelic frequency of both alleles should be zero. Then only we can conclude that the alleles have a reached an equilibrium. If the change in allelic Frequency of any allele is observed then the alleles are not in equilibrium.

Let us take the Z^A allele.

The change in allelic frequency of Z^A is given by the formula :

▲q= Y x p - X x q

= 1 x 10^-7 x 0.67 - 1 x 10^-5 x 0.33

= 0.67 x 10^-7 - 0.33 x 10^-5

= -0.667 x 10^-5 , which is not equal to 0.

We can observe that the change in Z^A is in negative i.e.. allele frequency of  Z^A is decreasing and it is not zero. Thus the alleles have not reached equilibrium

Thus option i) is correct.

Case - II (The frequency of  Z^A is higher than expected at equilibrium) :

The rate of forward mutation of Z^A to Z^B = 1 x 10^-5 is higher than the backward mutation rate of Z^B to Z^A = 0.0000001 or 1 x 10^-7 .

Thus we expect the the allelic frequency of Z^B to be more than that of Z^A .

Let us calculate the allelic frequency of Z^A at equilibrium. This can be determined solely by using the backward and forward mutation rates.

The formula is given as

q' = Y / X + Y = 1 x 10^-7 / 1 x 10^-5 + 1 x 10^-7 =  1 x 10^-7 / 101 x 10^-7 = 1 / 101 = 0.009

Allelic frequency of Z^B (p') = 1 - q' = 1 - 0.009 = 0.991

From this we can conclude that the allelic frequency of Z^Bis more than allelic frequency of Z^A at equilibrium.

Thus option ii) is incorrect and option iii) is correct .

PS - Hardy-Weinberg equilibrium is postulated by Hardy-Weinberg principle which states that the allelic frequencies do not change from generation to generation in the absence of evolutionary factors such as natural selection, etc.