6. In a population at Hardy-Weinberg equilibrium, for gene-X only two alleles, namely A and a, are found. If frequency of allele A is 0.2 and the frequency of allele a is 0.8, the frequency of the heterozygote genotype Aa in that population will be _____ (correct to 2 decimal places).
Hardy-Weinberg Equilibrium: Calculating Heterozygote Frequency (Aa) Using Allele Frequencies
Introduction
The Hardy-Weinberg Equilibrium (HWE) is one of the most fundamental principles of population genetics. Proposed independently by G. H. Hardy and Wilhelm Weinberg in 1908, this principle explains how allele frequencies and genotype frequencies remain constant from one generation to the next in an ideal population where evolutionary forces are absent. It serves as the mathematical foundation for studying evolution, genetic variation, and inheritance within populations.
According to the Hardy-Weinberg principle, if a gene has only two alleles, represented as A and a, with allele frequencies p and q respectively, then the genotype frequencies are predicted by the equation:
p² + 2pq + q² = 1
Here, p² represents the frequency of the homozygous dominant genotype (AA), 2pq represents the frequency of the heterozygous genotype (Aa), and q² represents the frequency of the homozygous recessive genotype (aa).
Correct Answer
Correct Answer: 0.32
Detailed Explanation
For a gene with two alleles, Hardy-Weinberg equilibrium predicts genotype frequencies using the equation:
p² + 2pq + q² = 1
where:
- p = Frequency of dominant allele (A)
- q = Frequency of recessive allele (a)
- p² = Frequency of genotype AA
- 2pq = Frequency of genotype Aa
- q² = Frequency of genotype aa
Since the question asks for the frequency of the heterozygous genotype Aa, we use the formula:
Heterozygote Frequency = 2pq
Step 1: Identify the Given Data
| Parameter | Value |
|---|---|
| Frequency of allele A (p) | 0.20 |
| Frequency of allele a (q) | 0.80 |
Step 2: Apply the Hardy-Weinberg Formula
2pq = 2 × 0.20 × 0.80
2pq = 2 × 0.16
2pq = 0.32
Step-by-Step Calculation Summary
| Calculation | Result |
|---|---|
| p | 0.20 |
| q | 0.80 |
| 2 × 0.20 × 0.80 | 0.32 |
| Frequency of Aa | 0.32 |
Genotype Frequencies in This Population
Using the Hardy-Weinberg equation:
| Genotype | Formula | Frequency |
|---|---|---|
| AA | p² = (0.20)² | 0.04 |
| Aa | 2pq = 2 × 0.20 × 0.80 | 0.32 |
| aa | q² = (0.80)² | 0.64 |
Verification:
0.04 + 0.32 + 0.64 = 1.00
This confirms that the genotype frequencies satisfy the Hardy-Weinberg equation.
Hardy-Weinberg Equation
| Expression | Meaning |
|---|---|
| p + q = 1 | Total allele frequency |
| p² | Frequency of homozygous dominant (AA) |
| 2pq | Frequency of heterozygous (Aa) |
| q² | Frequency of homozygous recessive (aa) |
| p² + 2pq + q² = 1 | Total genotype frequency |
Assumptions of Hardy-Weinberg Equilibrium
| Assumption | Description |
|---|---|
| Large Population | No genetic drift |
| Random Mating | Individuals mate randomly |
| No Mutation | No new alleles arise |
| No Migration | No gene flow into or out of the population |
| No Natural Selection | All genotypes have equal fitness |
Biological Significance
The Hardy-Weinberg principle provides the null model for population genetics. It enables scientists to compare observed genotype frequencies with expected frequencies and determine whether evolutionary forces such as natural selection, mutation, migration, genetic drift, or non-random mating are acting on a population. The principle is widely applied in medical genetics, conservation biology, evolutionary biology, forensic science, and human population studies.
Final Answer
Given:
p = 0.20
q = 0.80
Frequency of heterozygous genotype:
2pq = 2 × 0.20 × 0.80 = 0.32
Correct Answer: 0.32
