If a linear DNA has 'n' number of restrictions sites for a specific enzymes the number of fragments and fargements with cohesive sites at both the ends produced respecitvely are .

Options

A. n& ( n+1)
B. (n+1) and n
C. (n+1) &(n-1)
D. (n-1) & n

Answer

To solve the problem, we need to determine the number of fragments produced when a linear DNA molecule is cut by a restriction enzyme at 'n' restriction sites. We will also find out how many of those fragments have cohesive (sticky) ends at both ends.

### Step-by-Step Solution:

1. **Understanding the Concept of Restriction Sites**:
- A restriction enzyme cuts the DNA at specific sites called restriction sites. If there are 'n' restriction sites on a linear DNA molecule, each cut will create fragments of DNA.

2. **Calculating the Total Number of Fragments**:
- When a linear DNA molecule is cut at 'n' restriction sites, it will produce a total of 'n + 1' fragments. This is because each cut creates an additional fragment. For example, if there are 3 cuts (n = 3), the DNA will be divided into 4 fragments (3 cuts + 1).

3. **Identifying Fragments with Cohesive Ends**:
- Fragments that have cohesive (sticky) ends at both ends are typically produced when the restriction enzyme cuts the DNA in a way that leaves overhanging ends.
- For 'n' restriction sites, the number of fragments with cohesive ends at both ends will be 'n - 1'. This is because only the fragments that are formed between the cuts will have cohesive ends. For example, if there are 3 cuts, there will be 2 fragments with cohesive ends (between the first and second cut, and between the second and third cut).

4. **Final Answers**:
- Therefore, the total number of fragments produced is **n + 1**.
- The number of fragments with cohesive ends at both ends is **n - 1**.

### Summary:
- Total number of fragments = n + 1
- Number of fragments with cohesive ends at both ends = n - 1