The map distance between genes A and B is 3 units, between B and C is 10 units and between C and A is 7 units. Which of the following shows the correct order of the genes in a linkage map on the above data ?

Options

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

Answer

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

### Step 1: Understand 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

### Step 2: Analyze the options
We need to check each option to see if the distances add up correctly according to the provided distances.

### Step 3: Evaluate each option

**Option A: A, B, C**
- Distance A to B = 3 units
- Distance B to C = 10 units
- Total distance A to C = 3 + 10 = 13 units (but the distance given is 7 units)
- **Conclusion:** This option is incorrect.

**Option B: A, C, B**
- Distance A to C = 7 units
- Distance C to B = 10 units
- Total distance A to B = 7 + 10 = 17 units (but the distance given is 3 units)
- **Conclusion:** This option is incorrect.

**Option C: B, C, A**
- Distance B to C = 10 units
- Distance C to A = 7 units
- Total distance B to A = 10 + 7 = 17 units (but the distance given is 3 units)
- **Conclusion:** This option is incorrect.

**Option D: C, A, B**
- Distance C to A = 7 units
- Distance A to B = 3 units
- Total distance C to B = 7 + 3 = 10 units (which matches the distance given between B and C)
- **Conclusion:** This option is correct.

### Final Answer
The correct order of the genes in a linkage map is **C, A, B**.

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