The distance between the genes a,b,c and d in mapping units are a-d=3.5 , b-c=1,a-b=6,c-d=1.5,a-c=5 Find out the sequence of arrangement of these genes:

Options

A. acdb
B. abcd
C. acbd
D. adcb

Answer

To determine the sequence of arrangement of the genes A, B, C, and D based on the given distances in mapping units, we can follow these steps:

### Step-by-Step Solution:

1. **List the Distances**:
- a-d = 3.5
- b-c = 1
- a-b = 6
- c-d = 1.5
- a-c = 5

2. **Start with Gene A**:
- Assume gene A is at position 0 on a linear chromosome.

3. **Position Gene B**:
- Since the distance between A and B is 6, place gene B at position 6 (A at 0, B at 6).

4. **Position Gene C**:
- The distance between A and C is 5, so place gene C at position 5 (A at 0, C at 5).

5. **Position Gene D**:
- The distance between A and D is 3.5, so place gene D at position 3.5 (A at 0, D at 3.5).

6. **Check Distances**:
- Now we need to check the distances between the other genes:
- Distance between B and C: B (6) - C (5) = 1 (correct)
- Distance between C and D: C (5) - D (3.5) = 1.5 (correct)
- Distance between A and D: A (0) - D (3.5) = 3.5 (correct)
- Distance between A and B: A (0) - B (6) = 6 (correct)
- Distance between A and C: A (0) - C (5) = 5 (correct)

7. **Final Arrangement**:
- The arrangement based on the positions is: A - D - C - B.

### Conclusion:
The sequence of arrangement of the genes is **A, D, C, B**.