Question

A double-stranded genomic DNA has 40% (A+T) content. What is the average size (base pairs) of the fragments generated by digestion with the restriction enzyme HaeIII (GGCC)?

• A. 81
• B. 123
• C. 256
• D. 4096

Hint:

Average size = $1 / (\text{product of individual base probabilities})$.

Answer 1

The Correct Option is B

Step 1: Concept
The frequency of a restriction site depends on the probability of finding each specific base in the sequence.

Step 2: Meaning
If $A+T = 40%$, then $A=T=20%$ and $G=C=30%$.

Step 3: Analysis
The site is GGCC. The probability $P$ of this sequence occurring is: $P = P(G) \times P(G) \times P(C) \times P(C)$ $P = 0.3 \times 0.3 \times 0.3 \times 0.3 = 0.0081$ The average fragment size is $1/P$: $Size = 1 / 0.0081 \approx 123.45$ bp.

Step 4: Conclusion
The calculated average fragment size is approximately 123 bp. Final Answer: (B)