Question

In humans, the widow’s peak allele is dominant over the straight hairline allele. In a population of humans, the frequency of the allele for a widow’s peak is 0.61 and the frequency of the allele for a straight hairline is 0.39. What would the genotypic frequencies be if the population is in equilibrium? _____ = Frequency of homozygous dominant individuals _____ = Frequency of heterozygous individuals _____ = Frequency of homozygous recessive individuals How many individuals would you expect of each genotype in a population of 75 humans? _____ = Number of homozygous dominant individuals _____ = Number of heterozygous individuals _____ = Number of homozygous recessive individuals

Answer 1

Let, the widow's peak allele be A and straight hairline allele be a and their frequencies be p and q respectively.

The frequency of AA individuals in a population is p​​² and that of aa individuals is q².

Thus, genotypic frequency will be:

p² + 2pq + q​​​​² =1

(p+q)² = 1

0.61 + 0.39 = 1

Frequency of homozygous dominant individuals = p² = (0.61)² = 0.3721= 37%

Frequency of heterozygous individuals = 2pq = 2 x 0.61 x 0.39 = 0.4758 = 48%

Frequency of homozygous recessive individuals = q​²​ = (0.39)² = 0.1521 =15%

In a population of 75 humans,

Number of homozygous dominant individuals = 37% of 75 = 28

Number of heterozygous individuals = 48% of 75 = 36

Number of homozygous recessive individuals = 15% of 75 = 11