Lesson eLearning

In a three‑gene testcross, the parental genotype is A^+B^+C^+/abc crossed with a test strain abc/abc. The observed recombination frequencies are 10 % between A and B and 20 % between B and C. Among 4000 progeny, 5 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.

hard 5 marks Class 12 · Biology · Principles of Inheritance and Variation
Answer
A‑B = 10 cM, B‑C = 20 cM; coefficient of coincidence = 0.0625, interference = 0.9375; gene order likely A‑B‑C.

The correct answer is A‑B = 10 cM, B‑C = 20 cM; coefficient of coincidence = 0.0625, interference = 0.9375; gene order likely A‑B‑C. — Since recombination between A and B is less than between B and C, and A‑C recombination would be higher if C were the outer gene, the most consistent order is A‑B‑C.

Step-by-step solution

  1. For small recombination frequencies, the map distance (in centimorgans) equals the recombination frequency in percent.
  2. Thus, map distance A‑B = 10 cM and B‑C = 20 cM.
  3. Expected double‑crossovers = (recombination fraction AB) × (recombination fraction BC) × total progeny.
  4. Recombination fractions: AB = 0.10, BC = 0.20.
  5. Expected double crossovers = 0.10 × 0.20 × 4000 = 80.
  6. Observed double crossovers = 5.
  7. Coefficient of coincidence (c.o.c.) = observed / expected = 5 ÷ 80 = 0.0625.
  8. Interference = 1 – c.o.c. = 1 – 0.0625 = 0.9375.
  9. Since recombination between A and B is less than between B and C, and A‑C recombination would be higher if C were the outer gene, the most consistent order is A‑B‑C.
Common mistake: Calculating expected double crossovers using p×q instead of (recombination fraction)^2

Concept tested: Genetic mapping and interference

Published · How this solution was prepared and checked

← All Principles of Inheritance and Variation questions