36. An isolated population on an island has the following genotypic frequencies:
| Genotype | AA | Aa | aa |
|---|---|---|---|
| Frequency | 0.3 | 0.4 | 0.3 |
Assuming that there are only two alleles (A and a) for the gene,
the genotypic frequency of AA in the next generation will be __________.
Introduction
Understanding how genotypic frequencies change from one generation to the next
is a fundamental concept in population genetics. Questions based on an
isolated population on an island commonly test knowledge of
allele frequency calculation and the Hardy–Weinberg principle.
This solution explains the problem step by step and analyzes all possible options.
Given Question
An isolated population on an island has the following genotypic frequencies:
| Genotype | AA | Aa | aa |
|---|---|---|---|
| Frequency | 0.3 | 0.4 | 0.3 |
Assuming that there are only two alleles (A and a) for the gene,
find the genotypic frequency of AA in the next generation.
Step-by-Step Solution
Step 1: Calculation of Allele Frequencies
Frequency of allele A (p):
p = f(AA) + 1⁄2 f(Aa)
p = 0.3 + 1⁄2(0.4) = 0.3 + 0.2 = 0.5
Frequency of allele a (q):
q = 1 − p = 1 − 0.5 = 0.5
Step 2: Hardy–Weinberg Principle
Under Hardy–Weinberg equilibrium:
AA = p2, Aa = 2pq, aa = q2
Genotypic frequency of AA in the next generation:
AA = (0.5)2 = 0.25
Correct Answer
Genotypic frequency of AA in the next generation = 0.25
Explanation of All Options
Option: 0.30
Incorrect. This represents the genotypic frequency of AA in the present generation,
not the next generation after random mating.
Option: 0.40
Incorrect. This value corresponds to the heterozygous genotype (Aa), not AA.
Option: 0.50
Incorrect. This is the allele frequency of A, not the genotypic frequency of AA.
Option: 0.25
Correct. This value is obtained using the Hardy–Weinberg equation (p2).
Conclusion
By calculating allele frequencies and applying Hardy–Weinberg equilibrium conditions,
the genotypic frequency of AA in the next generation is found to be 0.25.
