Question

In a sample of 2,400 people, 1,482 were found to have the dominant (taster) phenotype. Assuming that the population is in Hardy-Weinberg equilibrium, approximately how many individuals in the sample are expected to be heterozygous for TAS2R38?

Answer 1

Answer:

1128.

Explanation:

The dominant taster phenotype is 1482.

The recessive taster phenotype = 2400 – 1482 = 918 .

Recessive phenotypes = 2400 – 1482 = 918

The recessive phenotype frequency = q² = 918 / 2400 = 0.38.

the recessive allele frequency q = √0.38 = 0.62.

The dominant allele frequency, p = 1-q = 1- 0.62 = 0.38

The heterozygous frequency =  2pq = 2 ×0.38 × 0.62 = 0.47 .

The heterozygous individual in the population = 0.47 × 2400 = 1128.

Thus, the answer is 1128.