Question

1. Ability to taste phenylthiocarbamide (PTC) is mostly determined by the gene PTC. The letters T and t are used here to simplify analysis. TT individuals taste a strong bitter taste; Tt people experience a slightly bitter taste: tt individuals taste nothing. A fifth-grade class of 20 students tastes PTC that has been applied to small pieces of paper, rating the experience as “very yucky” (TT), “I can taste it” (Tt), and “I can’t taste it” (tt). For HW, the students test their parents, with these results:                              -Of six TT students, four have two TT parent and the remaining two have one each of TT and Tt parents                                      - Of four Tt students, who have two parents who are Tt, and two have one parent who is TT and one parent who is tt.                  -Of the 10 students who can’t taste PTC, four have two parents who are also tt, but four students have one parent who is Tt and one who is tt. The remaining two student have two Tt parents.
   1. A. Calculate the frequencies of T and t alleles in the two populations (parents and children). Is the Hary-Weinberg Equilibrium maintained?

Answer 1

Answer

Frequencies of T and t alleles in the two populations (parents and children) are as follows

Formula-

Frequency of allele T = [f(AA) + 1/2f(Aa)] / Total number of alleles

Frequency of allele t = [f(aa) + 1/2f(Aa)] / Total number of alleles

Allele frequency for parents

Total number of TT in parents = 12 (i.e Total 24 T allele)

Total number of Tt in parents = 16 (i.e total 32 allele)

Total number of tt in parents = 12 (i.e Total 24 t  allele)

Total number of allels = 80 (24 + 32+ 24)

Frequency of allele T = f(AA) + 1/2f(Aa) / Total number of allels = 24 + 16 = 40 / 80 = 0.5

Frequency of allele t = f(aa) + 1/2f(Aa) / Total number of allels = 24 + 16 = 40 /80 = 0.5

The Hardy-Weinberg equation is expressed as: p² + 2pq + q² = 1

let p be 0.5 and q be 0.5

hence p² + 2pq + q² =  0.5² + 2 (0.5 *0.5) + 0.5² = 1 ( Hardy-Weinberg equation is proved)

Allele frequency for child

Total number of TT in child = 6 (i.e Total 12 T allele)

Total number of Tt in child = 4 (i.e total 8 allele)

Total number of tt in child = 10  (i.e Total 20 t  allele)

Total number of allels = 40 (12 + 8+ 20)

Frequency of allele T = f(AA) + 1/2f(Aa) / Total number of allels = 12 + 4 = 16 / 40 = 0.4

Frequency of allele t = f(aa) + 1/2f(Aa) / Total number of allels = 20 + 4 = 24 = 0.6

The Hardy-Weinberg equation is expressed as: p² + 2pq + q² = 1

hence p² + 2pq + q² =  0.4² + 2 (0.4 *0.6) + 0.6² = 1 ( Hardy-Weinberg equation is proved)

From the above calculation proves that Hary-Weinberg Equilibrium is maintained in both populations.