Question

Insertion of a foreign DNA at BamHI site in an E. coli cloning vector pBR322 results in the loss of antibiotic resistance towards:

• A. Ampicillin and tetracycline
• B. Tetracycline
• C. Ampicillin
• D. Gentamycin

Hint:

Logic Tip: Memorize the restriction sites for pBR322: {Bam}HI and {Sal}I sit inside the $tet^R$ gene, while {Pst}I and {Pvu}I sit inside the $amp^R$ gene. Slicing into a gene always destroys its function!

Answer 1

The Correct Option is B

Concept:
In recombinant DNA technology, cloning vectors like pBR322 are used to carry foreign DNA into host cells. pBR322 contains two specific antibiotic resistance genes that serve as selectable markers: the ampicillin resistance gene ($amp^R$) and the tetracycline resistance gene ($tet^R$).
Step 1:
The restriction endonuclease BamHI has its specific recognition sequence located precisely within the coding region of the tetracycline resistance gene ($tet^R$) on the pBR322 plasmid.
Step 2:
When foreign DNA is ligated into the vector at the BamHI site, the physical insertion of this new DNA disrupts the continuous sequence of the $tet^R$ gene.
Step 3:
Because the $tet^R$ gene is interrupted, it can no longer produce functional proteins to confer resistance against tetracycline. This phenomenon is called insertional inactivation.
Step 4:
Since the insertion occurred only at the BamHI site (within $tet^R$), the ampicillin resistance gene ($amp^R$) remains completely intact and fully functional.
Step 5:
Therefore, the recombinant plasmid will lose resistance towards tetracycline but retain resistance to ampicillin. Option (2) is the correct answer.