Question

If a single strand of DNA contain seven restriction sites for a particular restriction endonuclease and none of the restriction site is present either at 3' or 5' end then this enzyme will form how many DNA fragment reacting on it

• A. Six fragments
• B. Seven fragments
• C. Eight fragments
• D. Nine fragments

Hint:

Always identify if the DNA is linear or circular:
- Linear DNA: \(N\) sites \(\rightarrow N+1\) fragments.
- Circular DNA (Plasmid): \(N\) sites \(\rightarrow N\) fragments.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The enzymatic cleavage of a linear DNA molecule by a restriction enzyme divides the molecule into distinct fragments. The number of fragments is determined by the molecule's shape (linear or circular) and the number of cutting sites.
Key Formula or Approach:
For a linear DNA molecule: \[ \text{Number of fragments} = \text{Number of restriction sites } (N) + 1 \]

Step 2: Detailed Explanation:
1. The question states that we have a linear DNA molecule with ends at 3' and 5' (since "none of the restriction site is present either at 3' or 5' end").
2. There are \(N = 7\) internal restriction cleavage sites.
3. Each cut introduces a break in the linear strand: - 1 cut \(\rightarrow\) 2 fragments
- 2 cuts \(\rightarrow\) 3 fragments
- 7 cuts \(\rightarrow 7 + 1 = 8\) fragments.

Step 3: Final Answer:
The enzyme will produce 8 DNA fragments.