Question

A-B-D are three genes present in a chromosome, in that order. The map distance between A – B = 5 cm, and B – D = 10 cm. If the observed double cross-over percentage is 0.4, the coefficient of coincidence will be

• A. 0.8
• B. 0.6
• C. 0.5
• D. 0.9

Hint:

Expected DCO $= 0.05 \times 0.10 = 0.005 = 0.5\%$. Coefficient of Coincidence $C = \frac{0.4\%}{0.5\%} = 0.8$ (Interference $I = 1 - C = 0.2$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The coefficient of coincidence ($C$) is the ratio of observed double crossovers to expected double crossovers.

Step 2: Key Formula or Approach:
1. Map distance $A\text{--}B = 5\,\text{cM} = 0.05$ recombination frequency ($RF_1$).
2. Map distance $B\text{--}D = 10\,\text{cM} = 0.10$ recombination frequency ($RF_2$).
Expected double crossover frequency assuming independent crossovers:
\[\text{Expected DCO} = RF_1 \times RF_2 = 0.05 \times 0.10 = 0.005 = 0.5\%\]

Step 3: Detailed Explanation:
Observed double crossover frequency $= 0.4\% = 0.004$.
Formula for Coefficient of Coincidence ($C$):
\[C = \frac{\text{Observed DCO Frequency}}{\text{Expected DCO Frequency}} = \frac{0.4\%}{0.5\%} = \frac{0.4}{0.5} = 0.8\]

Step 4: Final Answer:
Therefore, the coefficient of coincidence is 0.8.