30. A population is in Hardy-Weinberg equilibrium for a gene with only two alleles (“A” and “a”). If the gene frequency of the allele “A” is 0.7, genotype frequency of
heterozygous “Aa” is ______.
Hardy-Weinberg Equilibrium: Calculating the Frequency of Heterozygous (Aa) Genotype
Introduction
The Hardy-Weinberg equilibrium is one of the most important principles in population genetics because it describes how allele frequencies and genotype frequencies remain constant from one generation to the next in an ideal population. Proposed independently by G. H. Hardy and Wilhelm Weinberg in 1908, this principle serves as the mathematical foundation for studying evolution, genetic variation, disease inheritance, and population structure.
According to the Hardy-Weinberg principle, if a population is infinitely large and experiences no mutation, migration, natural selection, genetic drift, or non-random mating, the frequencies of alleles and genotypes remain unchanged over successive generations. This equilibrium allows scientists to predict genotype frequencies directly from allele frequencies using simple mathematical equations.
Correct Answer
Correct Answer: 0.42 (42%)
Detailed Explanation
For a gene with two alleles, Hardy-Weinberg equilibrium states that the sum of allele frequencies is equal to one.
p + q = 1
where:
- p = frequency of allele A
- q = frequency of allele a
The question states that the frequency of allele A is:
p = 0.7
Therefore, the frequency of allele a is:
q = 1 − 0.7 = 0.3
Under Hardy-Weinberg equilibrium, genotype frequencies are calculated using the equation:
p² + 2pq + q² = 1
where:
- p² = frequency of genotype AA
- 2pq = frequency of genotype Aa
- q² = frequency of genotype aa
Since the question asks for the heterozygous genotype (Aa), we calculate:
2pq = 2 × 0.7 × 0.3
= 0.42
Thus, the expected frequency of heterozygous individuals is 0.42, which is equivalent to 42%.
Step-by-Step Calculation
Step 1: Identify Allele Frequencies
Frequency of allele A (p) = 0.7
Frequency of allele a (q) = 1 − 0.7 = 0.3
Step 2: Apply the Hardy-Weinberg Equation
2pq = 2 × 0.7 × 0.3
= 0.42
Step 3: Express the Result
Genotype frequency of Aa = 0.42
Percentage = 42%
Calculation Summary
| Parameter | Value |
|---|---|
| Frequency of A (p) | 0.7 |
| Frequency of a (q) | 0.3 |
| Frequency of AA (p²) | 0.49 |
| Frequency of Aa (2pq) | 0.42 |
| Frequency of aa (q²) | 0.09 |
Verification of Hardy-Weinberg Equilibrium
The genotype frequencies should add up to one.
AA + Aa + aa
= 0.49 + 0.42 + 0.09
= 1.00
This confirms that the calculated frequencies satisfy the Hardy-Weinberg equilibrium.
Hardy-Weinberg Equations
| Equation | 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 |
Why Is the Heterozygous Frequency Calculated as 2pq?
A heterozygous individual (Aa) can be formed in two equally probable ways. One possibility is that the A allele comes from one parent and the a allele comes from the other parent. The second possibility is the reverse combination, where the a allele comes from the first parent and the A allele comes from the second parent. Since both combinations produce the same genotype, their probabilities are added together, giving the expression 2pq.
Biological Significance
The Hardy-Weinberg equilibrium provides the null model for population genetics. It enables researchers to estimate genotype frequencies from allele frequencies and to detect evolutionary changes when observed frequencies deviate from expected values. The principle is widely applied in evolutionary biology, conservation genetics, genetic counseling, medical genetics, forensic science, and studies of inherited diseases.
Final Answer
Frequency of allele A (p) = 0.7
Frequency of allele a (q) = 0.3
Frequency of heterozygous genotype (Aa) = 2pq
= 2 × 0.7 × 0.3
= 0.42
= 42%
Correct Answer: 0.42 (42%)
