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mr Goodwill
14 days ago
15

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?
Biology
1 answer:
Tresset [654]14 days ago
5 0

Answer:

1128.

Explanation:

The phenotype for dominant tasters is 1482.

The phenotype for recessive tasters = 2400 – 1482 = 918.

Recessive phenotypes = 2400 – 1482 = 918.

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

The frequency of the recessive allele q = √0.38 = 0.62.

For the dominant allele frequency, p = 1-q = 1- 0.62 = 0.38.

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

The number of heterozygous individuals within the population = 0.47 × 2400 = 1128.

Thus, the result is 1128.

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