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 solve the problem of determining the sequence of genes A, B, C, and D based on the given distances in map units, we will follow these steps:

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

### Step 2: Analyze the Distances
We will analyze the distances to find the relative positions of the genes. The distance between two genes in map units indicates how far apart they are on the chromosome, with smaller distances indicating closer proximity.

### Step 3: Start with the Largest Distance
The largest distance is A-B = 6 map units. This suggests that A and B are the farthest apart. We can place A and B at opposite ends of our gene map.

### Step 4: Position Gene A and Gene B
Let’s place Gene A at position 0 and Gene B at position 6 on a number line:
- A = 0
- B = 6

### Step 5: Add Gene C
Next, we know A-C = 5 map units. Since A is at 0, Gene C must be at position 5:
- C = 5

### Step 6: Add Gene D
Now, we need to place Gene D. We know:
- A-D = 3.5 map units, which means D is at position 3.5 from A (0). Therefore:
- D = 3.5

### Step 7: Check the Distances
Now, we check the distances between the genes based on our placements:
- A-D = 3.5 (correct)
- A-B = 6 (correct)
- A-C = 5 (correct)
- C-D = 1.5 (C at 5 and D at 3.5 gives us 5 - 3.5 = 1.5, correct)
- B-C = 1 (B at 6 and C at 5 gives us 6 - 5 = 1, correct)

### Step 8: Determine the Sequence
From our 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**

### Final Answer
The correct sequence of the genes is: **A - D - C - B**

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