NEET Class neet Answered

Since there are only two alleles, we will label them ‘A’ (PTC taster) and ‘a’ (non-PTC taster).

The frequency of the dominant allele (A) is designated by p.

The frequency of the recessive allele (a) is designated by q.

In the Hardy-Weinberg equation (p² + 2pq + q² = 1), q² is the frequency of the ‘aa’ genotype (non-PTC taster homozygous, in this case).

Once we know q, we know p, because p + q = 1

Therefore,  p = 1 – q, or p = 1 – 0.4 = 0.6

Now we know p and q (p = 0.6 and q = 0.4)

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

Therefore, frequency of heterozygotes (Aa) in the population is 2pq

Plug our values for p and q into this and you get:

Frequency of heterozygotes = 2*p*q = 2*0.6*0.4 = 0.48

To be complete, we predict all the genotype frequencies to be:

Frequency of AA = p² = (0.6)² = 0.36

Frequency of Aa = 2pq = 2*0.6*0.4 = 0.48

 Frequency of aa = q² = (0.4)² = 0.16

In a population of 4000, multiplying every term by 4000

p² + 2pq + q² = 4000

p² = 1440

q² = 640

2pq = 1920

Hence, the frequency of heterozygote tasters in a population of 4000 is 1920.

The correct answer is option 2.