Question

There is a certain dominant/recessive gene in a population, where the dominant allele is referenced as "A", and the recessive variant as "a". The population has 1000 individuals. If the population is known to have 160 individuals that show the recessive trait, how many of the individuals who show the dominant trait are heterozygotes? What are the values of "p" and "q"?

Answer 1

Number of heterozygotes = 480

p = 0.6

q = 0.4

Step-by-step explanation:

Assuming the population is in Hardy-Weinberg Equilibrium,

p + q = 1

p² + 2pq + q² = 1

where,

p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of dominant homozygous population

2pq = frequency of heterozygous population

q² = frequency of recessive homozygous population

Here,

frequency of aa = q² = 160/1000 = 0.16

q = √0.16 = 0.4

p = 1-q = 1-0.4 = 0.6

frequency of Aa = 2pq = 2 * 0.6 * 0.4 = 0.48

Number of Aa heterozygotes = 1000 * 0.48 = 480