Distance between the genes a,b,c and d in map units is a-d =3.5 , b-c=1,a-b=6, c-d=1.5 and a-c=5 .Find out the sequecne of the genes

Options

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

Answer

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

### Step 1: List the distances
We have the following distances between the genes:
- A-D = 3.5
- B-C = 1
- A-B = 6
- C-D = 1.5
- A-C = 5

### Step 2: Analyze the distances
From the distances, we can start to visualize the arrangement of the genes. The distance between A and B is 6 map units, which is the largest distance given, suggesting that A and B are relatively far apart compared to the other genes.

### Step 3: Place the genes on a line
1. Start with gene A.
2. Since A-B = 6, we can place B at a distance of 6 units from A.
3. Next, we know A-C = 5, which means C must be placed 5 units from A. Since A-B is 6, C must be between A and B.
4. Now, we need to find the position of D. We know A-D = 3.5 and C-D = 1.5.

### Step 4: Positioning D
- If we place A at position 0, then B will be at position 6.
- C, being 5 units from A, will be at position 5.
- Now, since A-D = 3.5, D will be at position 3.5.
- Finally, we check C-D = 1.5. Since C is at 5, D at 3.5 means the distance is indeed 1.5 (5 - 3.5 = 1.5).

### Step 5: Final arrangement
Based on the above placements:
- A is at 0
- D is at 3.5
- C is at 5
- B is at 6

Thus, the sequence of the genes from left to right is:
**A - D - C - B**

### Conclusion
The correct sequence of the genes is A, D, C, B.

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