The map distance between genes A and B is 3 units, between B and C 10 units and between C and 7 units. The order of the genes in the linkage map constructed in the above data would perhapes be

Options

A. A,B,C
B. A,C,B
C. B,C,A
D. B,A,C

Explanation

To determine the order of genes A, B, and C based on the provided map distances, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Distances**: We have the following distances:
- Distance between A and B = 3 units
- Distance between B and C = 10 units
- Distance between C and A = 7 units

2. **Determine the Largest Distance**: The largest distance is between B and C (10 units). This suggests that B and C are the farthest apart on the linkage map.

3. **Place B and C**: We can place gene B and gene C on the linkage map:
- Let's position B at one end and C at the other end, with a distance of 10 units between them.

4. **Add Gene A**: Now, we need to position gene A based on the distances to B and C:
- The distance between A and B is 3 units, so A will be placed 3 units away from B.
- The distance between A and C is 7 units. Since we have already placed C 10 units away from B, we need to check if placing A 3 units from B will satisfy the distance to C.

5. **Calculate the Position of A**:
- If A is 3 units from B, it will be positioned closer to B than C.
- The distance from A to C should be 7 units. Since B to C is 10 units, and A is 3 units from B, the distance from A to C can be calculated as:
- Distance from A to C = Distance from B to C - Distance from A to B
- Distance from A to C = 10 units - 3 units = 7 units
- This confirms that the placement is correct.

6. **Final Order of Genes**: Based on the distances and placements, the order of the genes on the linkage map is:
- B (10 units) - A (3 units) - C (7 units)

### Conclusion:
The correct order of the genes is B, A, C.