Question

11) The ailele for wrinkled wings (W) is dominant to the normal wing allele (w) in fruit flies. If a fruit fly population is at Hardy-Weinberg equilibrium and 91% of flies have wrinkled wings, what are the three genotype frequencies? (6 points)

Answer 1

Hardy-Weinberg equilibrium:

• Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.
• If allele frequencies are p and q, then the genotype frequencies are p2, q2, 2pq.
• A population is considered to be in Hardy-Weinberg equilibrium when: In a randomly mating population, allele frequency and genotype frequency remain constant.
• Thus, under Hardy-Weinberg equilibrium:

p+ q=1 and

p²+2pq + q² = 1.

Total = 100 (percent)

• Numbers of wrinkled winged = 91 = WW or Ww (dominant)
• Number of normal wing = 100-91 = 9 = ww (recessive)

Thus,

• Frequency of ww genotype = 9/100 = 0.09 = q²
• Frequencyof w allele = q = 0.3
• Frequency of W allele = W = 1-0.3 = 0.7

Answers:

• Frequency of normal wings of genotype ww = q² = 0.09
• Frequency of wrinkled wings with homozygous dominant genotype WW = p²= (0.7)²= 0.49
• Frequency of wrinkled wing with heterozygous dominant genotype Ww = 2pq = 2 x 0.3 x 0.7 = 0.42