Q.94 In a population which is in Hardy-Weinberg equilibrium, the frequency of the recessive genotype of a
certain trait is 0.09 . The percentage of individuals with heterozygous genotype is ____ %
Correct Answer: 42%
In a Hardy-Weinberg equilibrium population, a recessive genotype frequency
of 0.09 indicates that the heterozygous genotype frequency
is 42%. This calculation follows standard principles of population genetics
and is frequently tested in biology and genetics examinations.
Hardy-Weinberg Basics
The Hardy-Weinberg principle describes a population in which allele and genotype frequencies
remain constant across generations under ideal conditions such as random mating, no mutation,
no migration, no selection, and a very large population size.
The fundamental equations are:
p + q = 1 (allele frequencies)
p2 + 2pq + q2 = 1 (genotype frequencies)
where:
- p = frequency of the dominant allele
- q = frequency of the recessive allele
Step-by-Step Calculation
Given
Recessive genotype frequency:
q2 = 0.09
Step 1: Calculate allele frequency (q)
q = √0.09 = 0.3
Step 2: Calculate allele frequency (p)
p = 1 − q = 1 − 0.3 = 0.7
Step 3: Calculate heterozygous genotype frequency
2pq = 2 × 0.7 × 0.3 = 0.42
= 42%
Genotype Frequencies Breakdown
| Genotype | Frequency Equation | Calculation | Percentage |
|---|---|---|---|
| Homozygous recessive (aa) | q2 | 0.09 | 9% |
| Heterozygous (Aa) | 2pq | 2 × 0.7 × 0.3 = 0.42 | 42% |
| Homozygous dominant (AA) | p2 | 0.72 = 0.49 | 49% |
Option Analysis
Typical multiple-choice options for this problem include:
42%, 49%, 30%, and 70%.
- 42%: Correct heterozygous frequency (2pq).
- 49%: Incorrect; represents homozygous dominant (p2).
- 30%: Incorrect; equals q, not genotype frequency.
- 70%: Incorrect; equals p, not heterozygous proportion.
This problem tests precise application of the Hardy-Weinberg equations and is essential for
exam preparation in genetics, evolutionary biology, and
molecular biology.
