Question

On average, how many fragments would a restriction enzyme, which recognizes a specific 5 base sequence in the DNA, would produce from a double-stranded bacteriophage genome of size 5252bp?

• A. About 3
• B. About 5
• C. About 7
• D. About 9

Hint:

Standard recognition sequence lengths and frequencies:
4-base cutter: $4^4 = 256$ bp.
6-base cutter: $4^6 = 4096$ bp.
8-base cutter: $4^8 = 65,536$ bp.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question requires calculating the probability of a specific DNA sequence occurring in a genome of a given size to estimate the number of restriction sites and resulting fragments.

Step 2: Key Formula or Approach:
Assuming a random distribution of the four nucleotides (A, T, G, C) in equal proportions (25% each), the probability of finding a specific sequence of length $n$ is:
\[ P = \left( \frac{1}{4} \right)^n \]
The average distance between two sites (cut frequency) is $4^n$.
The expected number of sites is: \[ \text{Number of Sites} = \frac{\text{Genome Size}}{\text{Average Distance}} \]
Detailed Explanation:

• Sequence Length ($n$): The enzyme recognizes a 5-base sequence ($n = 5$).

• Average Cutting Distance: $4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 1024$ bp.

• This means, statistically, there is one site every 1024 base pairs.

• Expected Number of Sites: \[ \frac{5252 \text{ bp}}{1024 \text{ bp/site}} \approx 5.12 \text{ sites} \]

• Number of Fragments: In a circular genome (common in some phages) the number of fragments equals the number of sites. In a linear genome, the number of fragments is Sites + 1.

• Given the value 5.12, the closest estimate provided in the options is "About 5".
Final Answer:
Based on the probability calculation for a 5-base cutter ($1/1024$), a 5252bp genome would contain approximately 5.12 restriction sites, leading to about 5 fragments.