Question

On average, how many fragments would a restriction enzyme, which recognizes a specific 5-base sequence in DNA, be expected to cleave a double-stranded bacteriophage with a genome size of 6066 bp into?

• A. About 1212
• B. About 120
• C. About 60
• D. About 6

Hint:

A 4-base cutter cuts every \(4^4 = 256\) bp, a 5-base cutter cuts every \(4^5 = 1024\) bp, and a 6-base cutter cuts every \(4^6 = 4096\) bp.
Dividing 6066 bp by 1024 bp gives approximately 6, which quickly points to the correct answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Restriction enzymes cleave double-stranded DNA at specific nucleotide recognition sequences.
The average distance between recognition sites (and therefore the frequency of cleavage) depends on the length of the recognition sequence and the base composition of the genome.
Key Formula or Approach:
For a restriction enzyme with a recognition sequence of length \(n\) bases, assuming a random distribution of equal base frequencies (25% each of A, T, G, C), the probability of finding the sequence at any given position is: \[ P = \left(\frac{1}{4}\right)^n \] The expected average distance (in base pairs) between successive recognition sites is: \[ \text{Average distance} = 4^n \] The expected number of cuts in a genome of size \(G\) is: \[ \text{Number of cuts} = \frac{G}{4^n} \]

Step 2: Detailed Explanation:
The restriction enzyme recognizes a specific 5-base sequence (\(n = 5\)).
We calculate the expected average distance between recognition sites: \[ \text{Average distance} = 4^5 = 1024 \text{ bp} \] The genome size of the bacteriophage is given as \(G = 6066 \text{ bp}\).
The expected number of cleavage sites (cuts) is calculated by dividing the genome size by the average distance: \[ \text{Expected cuts} = \frac{6066 \text{ bp}}{1024 \text{ bp}} \approx 5.92 \] For a circular bacteriophage genome, 6 cuts will yield 6 fragments.
For a linear bacteriophage genome, 6 cuts will yield 7 fragments.
In either case, the number of fragments is approximately 6.

Step 3: Final Answer:
Thus, the expected number of fragments is about 6, corresponding to option (D).