In a linear chromosome, map distance between four loci are as follows : a - b - 10 , b - c - 4, a - d - 3, a - c - 6 . The expected cross - over frequency between c and d is

Options

A. `3%`
B. `9%`
C. Either 3% or 9%
D. 4 % to 12%

Answer

To find the expected crossover frequency between loci C and D based on the given distances, we can follow these steps:

### Step 1: Understand the distances between loci
We have the following distances:
- Distance between A and B = 10 units
- Distance between B and C = 4 units
- Distance between A and C = 6 units
- Distance between A and D = 3 units

### Step 2: Determine the possible positions of the loci
From the distances given:
- Since A and B are 10 units apart, we can place A at position 0 and B at position 10.
- The distance between B and C is 4 units. Therefore, C can be positioned either at 14 (if C is to the right of B) or at 6 (if C is to the left of B).
- However, we also know that A and C are 6 units apart. If A is at 0, then C must be at position 6 (since 0 + 6 = 6).
- Now, we place D. The distance between A and D is 3 units, so D can be at position 3.

### Step 3: Calculate the distance between C and D
Now we have:
- A at position 0
- D at position 3
- C at position 6

To find the distance between C and D:
- Distance between C (6) and D (3) = |6 - 3| = 3 units.

### Step 4: Determine crossover frequency
According to the genetic mapping:
- 1 unit of distance corresponds to a 1% crossover frequency.
- Therefore, a distance of 3 units corresponds to a crossover frequency of 3%.

### Conclusion
The expected crossover frequency between loci C and D is **3%**.

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