In a three‑gene testcross, the parental genotype is A⁺B⁺C⁺/a b c crossed with a test strain a b c / a b c. The observed recombination frequencies are 12 % between A and B and 18 % between B and C. Among 5000 progeny, 75 double‑crossover individuals are observed. (a) Estimate the map distances (in cM) between A‑B and B‑C. (b) Calculate the coefficient of coincidence and the interference. (c) Propose the most likely gene order.
Answer
Map distances: A‑B = 12 cM, B‑C = 18 cM; coefficient of coincidence ≈ 0.69, interference ≈ 0.31; gene order A‑B‑C.
The correct answer is Map distances: A‑B = 12 cM, B‑C = 18 cM; coefficient of coincidence ≈ 0.69, interference ≈ 0.31; gene order A‑B‑C. — Since double crossovers involve the middle gene B, the most plausible linear order is A‑B‑C.
Step-by-step solution
- Because recombination frequencies are <20 %, map distance ≈ recombination %: A‑B = 12 cM, B‑C = 18 cM.
- Expected double‑crossover frequency = (12 % × 18 %) /100 = 2.16 % (0.0216).
- Expected number of double crossovers = 0.0216 × 5000 = 108.
- Observed double crossovers = 75.
- Coefficient of coincidence = observed / expected = 75 / 108 ≈ 0.694 (≈0.69).
- Interference = 1 – coincidence = 1 – 0.694 ≈ 0.306 (≈0.31).
- Since double crossovers involve the middle gene B, the most plausible linear order is A‑B‑C.
Common mistake: Forgetting to convert percentages to proportions before multiplying for expected double crossovers
Concept tested: Genetic linkage mapping and interference
Published · How this solution was prepared and checked
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