Calculating Gene Frequencies from Genotypic Data
If genotypic frequency of dominant homozygote is $P=0.05$ and genotypic frequency recessive homozygote is $Q=0.65$ then find out the gene frequencies p & q-
- p=0.20 & q= 0.80 — Correct Answer
- $p=0.80$ & $q=0.20$
- $p=0.40~\&~q=0.60$
- p=0.30 & $q=0.50$
Explanation:
Correct Answer: p = 0.20 & q = 0.80
Using Hardy-Weinberg formulas, gene frequencies (p and q) can be calculated from known genotypic frequencies using the relationships between allele frequencies and homozygote proportions.
Step-by-Step Calculation
- Given: P (dominant homozygote AA) = 0.05; Q (recessive homozygote aa) = 0.65
- Step 1 — Find H: P + H + Q = 1 → H = 1 − 0.05 − 0.65 = 0.30
- Step 2 — Find p: p = P + ½H = 0.05 + ½(0.30) = 0.05 + 0.15 = 0.20
- Step 3 — Find q: q = Q + ½H = 0.65 + 0.15 = 0.80 (or simply q = 1 − p = 0.80)
- Verification: p + q = 0.20 + 0.80 = 1.00 ✔
Hardy-Weinberg Key Formulas
| Parameter | Formula |
|---|---|
| Gene frequency p from genotypes | p = P + ½H |
| Gene frequency q from genotypes | q = Q + ½H |
| Genotypic frequency P | P = p² |
| Genotypic frequency H | H = 2pq |
| Genotypic frequency Q | Q = q² |
Why Other Options Are Wrong
- p=0.80 & q=0.20 → values reversed; q must be larger because Q (recessive homozygote) = 0.65 is dominant
- p=0.40 & q=0.60 → does not satisfy p = P + ½H = 0.05 + 0.15 = 0.20
- p=0.30 & q=0.50 → p + q ≠ 1 (0.30 + 0.50 = 0.80); violates fundamental H-W rule
📚 About this Topic — Animal Genetics & Breeding
This multiple choice question is from Animal Genetics & Breeding, Animal Refresher. It has 4 options with a detailed explanation of the correct answer. Practice more MCQs from Animal Genetics & Breeding to strengthen your preparation.
