Question

A student studying a population of butterflies wants to determine the gene frequencies for wing color. She knows that in this particular butterfly , wing color is exhibited as either yellow or black and that black is dominant. Her survey data from this year’s population shows that, as for the past five years, 40% of the butterflies have yellow wings, and 60% have black wings.

a. Use the Hardy - Weinberg equation to calculate the apparent frequencies of the yellow (b) and black (B) alleles. Show your work.

b. Does it appear that this population is in Hardy - Weinberg Equilibrium? Explain.

Answer 1

Given:

In the population,

Wing color in butterfly is under control of two alleles : black(B) and yellow(b) . Out of the two alleles, allele for black is dominant over yellow.

40% butterflies have yellow wings.

60% butterflies have black wings.

To find:

(a) Allelic frequencies

In the population, BB and Bb individuals will have black wings and bb types will have yellow wings.

bb=40%

b²=40/100

b²=0.4

b=√0.4

b=0.63

Frequency of B = 1- frequency of b (0.63)

=1-0.63

=0.37

Thus,

Frequency of black allele , p = 0.37

Frequency of yellow allele, q = 0.63

(b) Let us calculate the genotypic frequencies,

	p = 0.37	q = 0.63
p = 0.37	0.1369 (p2, homozygous black)	0.2331 (pq, heterozygous black)
q = 0.63	0.2331 (pq, heterozygous black)	0.3969 (q2, homozygous yellow)

Now, for the population to be in Hardy-Weinberg equilibrium, sum of allelic and genotypic frequencies is 1. thus,

p+q=1, and

p²+2pq+q²=1

(i) p+q=0.37+0.63=1 , thus sum of allelic frequencies is 1.

(ii) 0.1369+0.4662+0.3969 =1, thus sum of genotypic frequencies is 1.

Thus, population is in Hardy-Weinberg equilibrium because sum of allelic as well as genotypic frequencies is 1.