Question

A researcher used PTC taste paper to estimate the number of non-tasters in a group of customers entering the UD creamery. She found there were 72 non-tasters [tt], 6 slight tasters [Tt] and 58 tasters [TT]. Using the formula for Hardy Weinberg, state the number of individuals that would be tasters, slight tasters and non-tasters if the population were to be in equilibrium.

Answer 1

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