Question

Bacteriophage \(\lambda\) vectors can accommodate foreign DNA of about

• A. \(2\text{--}5\) kb
• B. \(15\text{--}20\) kb
• C. \(50\) kb
• D. \(300\) kb

Hint:

Approximate cloning capacities: \[ \boxed{ \begin{aligned} \text{Plasmid} &\rightarrow 2\text{--}10\ \text{kb} \lambda\text{ phage} &\rightarrow 15\text{--}20\ \text{kb} \text{Cosmid} &\rightarrow 35\text{--}45\ \text{kb} \text{BAC} &\rightarrow 100\text{--}300\ \text{kb} \text{YAC} &\rightarrow 200\text{--}1000\ \text{kb} \end{aligned} } \]

Answer 1

The Correct Option is B

Step 1: Recall the cloning capacity of common vectors. Different cloning vectors accommodate different insert sizes. \[ \begin{aligned} \text{Plasmids} &\rightarrow 2\text{--}10\ \text{kb} \lambda\text{ phage vectors} &\rightarrow 15\text{--}20\ \text{kb} \text{Cosmids} &\rightarrow 35\text{--}45\ \text{kb} \text{BACs} &\rightarrow 100\text{--}300\ \text{kb} \end{aligned} \]

Step 2: Choose the correct insert size. Hence, bacteriophage \(\lambda\) vectors can carry approximately \[ \boxed{15\text{--}20\ \text{kb}} \] of foreign DNA. Therefore, \[ \boxed{(B)} \] is the correct answer.