Question

The inability in humans to taste capsaicin resides in a single gene difference between two alleles P and p. The allele P for tasting is dominant over the nontasting allele. In a population of 400 individuals in Hardy-Weinberg equilibrium, 64 are nontasters. How many individuals are heterozygous for the gene?

• A. 64
• B. 128
• C. 144
• D. 192

Answer 1

The Correct Option is D

To solve the problem of finding how many individuals are heterozygous for the capsaicin tasting gene, we need to utilize the Hardy-Weinberg principle. This principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. The equation is given by:

\(p^2 + 2pq + q^2 = 1\) 

Where:

• \(p^2\) = frequency of the homozygous dominant genotype (tasters, PP)
• \(2pq\) = frequency of the heterozygous genotype (tasters, Pp)
• \(q^2\) = frequency of the homozygous recessive genotype (nontasters, pp)

The question states that there are 64 nontasters among 400 individuals. Thus, \(q^2 = \frac{64}{400} = 0.16\).

To find \(q\), we take the square root of \(q^2\):

\(q = \sqrt{0.16} = 0.4\)

Since \(p + q = 1\), we find \(p = 1 - 0.4 = 0.6\).

Now, using the allele frequencies, we can calculate the heterozygous genotype frequency, \(2pq\):

\(2pq = 2 \times 0.6 \times 0.4 = 0.48\)

Therefore, the number of heterozygous individuals in the population is:

\(0.48 \times 400 = 192\)

Thus, there are 192 individuals who are heterozygous for the capsaicin tasting gene, making the correct answer:

192