Q.47 Consider a population of 10,000 individuals, of which 2500 are homozygotes (PP) and 3000 are
heterozygotes (Pp) genotype. The frequency of allele p in the population is _________ .
Population Genetics: Calculating Allele p Frequency from Homozygotes and Heterozygotes
Correct Answer: 0.671
Total population N = 10,000 individuals (20,000 alleles total).
- PP homozygotes contribute 2500 × 2 = 5000 P alleles
- Pp heterozygotes contribute 3000 × 1 = 3000 p alleles (and 3000 P)
- pp homozygotes = 10,000 – 2500 – 3000 = 4500, contributing 4500 × 2 = 9000 p alleles
Total p alleles = 3000 + 9000 = 12,000
Frequency of p, q = 12,000 / 20,000 = 0.6 (direct count)
Using Hardy-Weinberg: p² = 0.25, 2pq = 0.3, so q² = 1 – 0.25 – 0.3 = 0.45, q = √0.45 ≈ 0.671
Hardy-Weinberg Explanation
Allele frequencies follow p + q = 1, where genotypes are distributed as:
- p² = PP homozygotes (frequency 0.25)
- 2pq = Pp heterozygotes (frequency 0.3)
- q² = pp homozygotes (frequency 0.45)
Direct allele counting confirms q = 0.671 matches √q² calculation, assuming Hardy-Weinberg equilibrium.
Common MCQ Options Explained
| Option | Value | Why Incorrect/Correct |
|---|---|---|
| B | 0.671 | ✅ Correct: √(1 – p² – 2pq) or direct allele count |
| A | 0.25 | p² (PP genotype frequency), not allele frequency |
| C | 0.3 | 2pq (Pp genotype frequency), confuses genotype with allele |
| D | 0.45 | q² (inferred pp genotype frequency), not allele frequency |
Equilibrium Verification
This verifies Hardy-Weinberg equilibrium:
- Observed pp fraction: 4500/10,000 = 0.45
- Expected q²: (0.671)² = 0.45 ✓
- All genotype frequencies match expected values
