Question

A single gene determines the body color of the insect, Cotinis mutabilis (Figeater beetle). The
gene is located on an autosomal chromosome. Brown body color (B) is dominant to Green body
color (b). In the American Southwest, the allele frequency for the recessive allele is 0.03 (10
points)

a. What is the frequency of the Brown body color allele (B)?
b. In a population of 400 beetles:
How many normal alleles (B) would be expected in the gene pool?
How many homozygous Brown beetles would be expected?
How many homozygous Green beetles?
How many heterozygote beetles?

Answer 1

According to the Hardy Weinberg equation

p+q=1 (P is the frequency of the dominant allele or the frequency of B; q is the frequency of the recessive allele or the frequency of b)

p+0.03=1

or, p=1-0.03=0.97 or 97%

a. So, the frequency of the brown body color allele (B) is 97%

b. Each individual carries two alleles. so in a population of 400 beetles, the number of alleles will be 400 X 2 = 800.

The frequency of allele B is 97%.

97% of 800 is = 776.

So, 776 normal alleles (B) would be expected in the gene pool

Again, according to the equation

p2+2pq+q2=1

p2 is the frequency of homozygous dominant (brown) individual (BB)

so, p2 =0.972=0.94 or 94%

Hence in a population of 400 beetles, 94% beetles = 376 homozygous Brown beetles would be expected

q2  is the frequency of homozygous recessive (green) individual (bb)

So, q2 = 0.032= 0.0009=0.09%

Hence in a population of 400 beetles, 0.09% beetles = 0.36 homozygous green beetles would be expected

2pq is the frequency of heterozygous beetles (Bb)

So, 2pq= 2X0.97X0.03=0.058 or 5.8%.

Hence in a population of 400 beetles, 5.8% beetles = 23.2 homozygous green beetles would be expected