Question Details

Plasmid pBR322 has Pstl restriction enzyme site within gene ampR that confers ampicillin resistance. If this enzyme is used for inserting a gene for β-galactoside production and the recombinant plasmid is inserted in an E.coli strain

Options

A

It will lead to lysis of host cell

B

It will be able to produce a novel protein with dual ability

C

It will not be able to confer ampicillin resistance to the host cell

D

The transformed cells will have the ability to resist ampicillin as well as produce β-galactoside

Show Answer

Correct Answer :

Option C

It will not be able to confer ampicillin resistance to the host cell

It will not be able to confer ampicillin resistance to the host cell

Solution :

Plasmid pBR322 carries the ampR gene, which provides resistance to ampicillin.

The restriction enzyme PstI cuts at the specific DNA sequence CTGCAG. In pBR322, a PstI site is located **within** the coding region of the ampR gene.

When the plasmid is digested with PstI to create an opening for insertion of a new gene (e.g., a β‑galactoside‑producing gene), the cut occurs inside the ampR gene. The subsequent ligation step joins the new gene fragment into the plasmid at this site.

Because the insertion disrupts the continuity of the ampR coding sequence, the resulting recombinant plasmid no longer contains an intact, functional ampicillin‑resistance gene.

Therefore, even though the plasmid now carries the β‑galactoside gene, the host *E. coli* cells transformed with this recombinant plasmid **cannot** express ampicillin resistance.

Consequently, the correct outcome is:

It will not be able to confer ampicillin resistance to the host cell.

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