Q.25 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
Introduction
Hardy Weinberg equilibrium capsaicin tasting nontasters problems test CSIR NET aspirants’ grasp of population genetics. In this classic query—400 individuals, 64 nontasters—find heterozygous individuals using allele frequencies p and q. Master Hardy Weinberg equilibrium capsaicin tasting nontasters calculations for exams.
Understanding the Genetics
Capsaicin tasting ability is dominant (P allele); nontasting is recessive (pp genotype). Hardy-Weinberg principle assumes no evolution: p² + 2pq + q² = 1. Nontasters give q² = 0.16, so q = 0.4, p = 0.6, and heterozygotes 2pq = 0.48 or 192 individuals.
Step-by-Step Solution
The inability to taste capsaicin follows a simple dominant-recessive inheritance pattern controlled by alleles P (dominant, tasting) and p (recessive, nontasting), with the population in Hardy-Weinberg equilibrium. Nontasters represent homozygous recessive individuals (pp genotype), allowing calculation of allele frequencies and heterozygote (Pp) numbers from the given data.
- Nontasters (pp) number 64 out of 400 individuals, so the frequency of the pp genotype is q² = 64/400 = 0.16.
- The recessive allele frequency is q = √0.16 = 0.4, and the dominant allele frequency is p = 1 – q = 0.6.
- Heterozygote frequency is 2pq = 2 × 0.6 × 0.4 = 0.48, so the number of heterozygotes is 0.48 × 400 = 192.
Detailed Calculation Breakdown
- q² = 64/400 = 0.16.
- q = √0.16 = 0.4.
- p = 0.6.
- 2pq = 0.48 × 400 = 192.
Genotype distribution: PP = 144 (36%), Pp = 192 (48%), pp = 64 (16%).
Option Analysis
(A) 64: Matches nontaster count but represents homozygous recessive (pp), not heterozygotes.
(B) 128: Equals 2 × 64, a common error assuming heterozygotes double recessives without allele frequencies.
(C) 144: Might arise from miscalculating q = 0.3 (36/400) or other approximation errors, yielding 2 × 0.7 × 0.3 × 400 = 168 (close but incorrect).
(D) 192: Correct, as derived from precise 2pq calculation.
Why Options Fail (Exam Tips)
Common pitfalls in Hardy Weinberg equilibrium capsaicin tasting nontasters: confusing q² with 2pq or skipping square root. Option A is pp; B assumes direct doubling; C from frequency errors.
CSIR NET Relevance
This mirrors CSIR NET Life Sciences Unit 11 questions on evolution and genetics. Practice similar PTC tasting or Rh factor problems for mastery.
