Question

There is a population of Drosophila flies in a small fruit shop. Due to sudden mutation, the eye color of some of the flies changes from red to white. Red (represented by R) is the dominant gene, and white (represented by r) is a recessive gene. Due to this, the gene frequency of allele r changes from 0.38 to 0.40.
What will be the new genotype frequencies in that population?
( p + q = 1, p2 + 2pq + q2 = 1)
A. p2 = 0.36, 2pq = 0.48, q2 = 0.16
B. p2 = 0.32, 2pq = 0.23, q2 = 0.45
C. p2 = 0.25, 2pq = 0.36, q2 = 0.39

Answer 1

The answer is A. p2 = 0.36, 2pq = 0.48, q2 = 0.16.

Let's first distinguish alleles frequencies from genotypes frequencies:
p - a frequency of R allele,
q - a frequency of r allele,
p² - a frequency of RR genotype,
2pq - a frequency of Rr genotype,
q² - a frequency of rr genotype.

The gene frequency of allele r changes from 0.38 to 0.40, thus:
q = 0.40

Now, we will implement this in the following equation:
p + q = 1

If q = 0.40, then:
p + 0.40 = 1
⇒ p = 1 - 0.40
⇒ p = 0.60

Now, the frequencies of both alleles are know and it is easy to calculate the genotype frequencies:
- the frequency of RR genotype: p² = 0.60² = 0.36
- the frequency of Rr genotype: 2pq = 2 · 0.60 · 0.40 = 0.48
- the frequency of rr genotype: q² = 0.40² = 0.16

Therefore, the new genotype frequencies in that population are p2 = 0.36, 2pq = 0.48, q2 = 0.16.

Answer 2

The correct answer choice for the question above is answer choice: 

A) p2 = 0.36, 2pq = 0.48, q2 = 0.16

Explanation:

p + q = 1

You simply have to designate p as red (dominant trait= R) and q as white (recessive= r)

Since the r (recessive) changed from 0.38 to 0.40.

p + q = 1 ; remember q is r (recessive)
p + 0.4 = 1 ; so p (dominant = R) is 0.6

p² + 2pq + q² = 1 

(0.6)²+ 2pq + (0.40)² = 1
(0.36) + 2pq + (0.16) = 1 ; so 2pq is 0.48

I hope this helps, Regards.