Answer:
In a population at H-W equilibrium:
- Tasters [TT] ~ 28 individuals
- Slight tasters [Tt] ~ 67 individuals
- Non-tasters [tt] ~ 41 individuals
Explanation:
To determine if the population is at Hardy-Weinberg equilibrium, we need to use the equation p² + 2pq + q² = 1
where p² represents the frequency of the homo-zygous genotype TT (taster phenotype), q² represents the frequency of the homo-zygous genotype tt (non-taster phenotype), and 2pq represents the frequency of the heterozygous genotype Tt (slight taster phenotype).
In addition, the sum of the allele frequencies for the allele T and allele t must be 1, so p + q = 1.
In this example:
Frecuency of the T allele: [(58 x 2) + 6] / [(58 x 2) + (6 x 2) + (72 x 2)] = 122/272 = 0.449
Frecuency of the t allele: [(72 x 2) + 6] / [(58 x 2) + (6 x 2) + (72 x 2)] = 150/272 = 0.551 >>>
p + q = 0.449 + 0.551 = 1
In consequence,
p² (TT) = 0.449² = 0.2016 (i.e., 20.16% of the total population) >> 27.4 individuals
2pq (Tt) = 2 x 0.449 x 0.551 = 0.495 (i.e., 49.5% of the total population) >> 67.3 individuals
q² (tt) = 0.551² = 0.3036 (i.e., 30.36% of the total population) >> 41.3 individuals
