Question

If the gene frequency for PTC nontasters is 0.4, then what will be the number of heterozygotes tasters in a population of 4000 ?

• A. 1920
• B. 1080
• C. 2520
• D. 1440

Answer 1

The correct option is A 1920

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 $(p^2+2pq+q^2=1),q^2$ 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: $p^2+2pq+q^2=1$
Therefore, frequency of heterozygotes (Aa) in the population is 2pq
Put the values of p and q into this and you get:
Frequency of heterozygotes $=2\ast p\ast q=2\ast 0.6\ast 0.4=0.48$
To be complete, we predict all the genotype frequencies to be:
Frequency of $AA=p^2=\left(0.6\right)2=0.36$
Frequency of $Aa=2pq=2\ast 0.6\ast 0.4=0.48$
Frequency of $aa=q^2=\left(0.4\right)2=0.16$
In a population of 4000, multiplying every term by 4000
$p^2+2pq+q^2=4000$
$p^2=1440$
$q^2=640$
$2pq=1920$
Hence, the frequency of heterozygote tasters in a population of 4000 is 1920.
The correct answer is option B.