Two genes, A and B, are located on the same chromosome. In a testcross, the observed recombination frequency between them is 15 %. (a) Estimate the map distance between A and B in centimorgans. (b) If the double‑crossover frequency observed in a three‑gene testcross (genes A‑B‑C) is 2 % while the expected double‑crossover frequency (based on the individual recombination frequencies of 15 % for A‑B and 20 % for B‑C) is 3 %, calculate (i) the coefficient of coincidence and (ii) the interference. (c) Briefly explain the biological significance of a positive interference value.

Step-by-step solution

1. Recall that recombination frequency (in %) approximates map distance in centimorgans (cM).
2. Therefore, a 15 % recombination frequency corresponds to a map distance of 15 cM.
3. For the three‑gene situation, convert percentages to decimals: 0.15 for A‑B and 0.20 for B‑C.
4. Calculate the expected double‑crossover frequency = 0.15 × 0.20 = 0.03 = 3 %.
5. Observe the actual double‑crossover frequency = 2 % = 0.02.
6. Coefficient of coincidence (c.o.c.) = observed / expected = 0.02 / 0.03 = 0.67.
7. Interference = 1 – c.o.c. = 1 – 0.67 = 0.33 (or 33 %).
8. Explain significance: a positive interference (value >0) indicates that the occurrence of one crossover reduces the probability of another nearby crossover, which limits excessive chromosomal breakage and promotes accurate segregation.

Concept tested: Genetic linkage, map distance and interference

Published 27 September 2026 · How this solution was prepared and checked