Question

Describe how to select recombinant clones if a foreign DNA is inserted into the polylinker site of pUC18 and then introduced into E. coli cells.

Answer 1

The plasmid pUC18 has a lacZ gene within it. This gene encodes beta galactosidase enzyme that breaks down X-gal (colorless) to produce a blue colored compound. X gal is 5-bromo-4-chloro-3-indolyl-β-D-galactopyranoside which is cleaved by beta galactosidase to beta D-galactose and 5-Bromo-4-chloro-3-hydroxyindole. The 5-bromo-4-chloro-3-hydroxyindole is a blue compound.

When there is no foreign insert present in the pUC18 plasmid, a functional beta galactosidase is produced as the polylinker site is intact. Polylinker region is region where there is normally insertion of foreign gene. As functional beta galactosidase is produced, X-gal is broken down to beta D-galactose and 5-Bromo-4-chloro-3-hydroxyindole, thereby producing blue colonies. When the foreign DNA inserts itself in the polylinker region, the protein-coding sequence of lacZ is disrupted. Hence, a functional beta galactosidase is not produced. This will not allow X-gal to be degraded. As X-gal is colorless, white colonies are produced. Thus, the colonies that are white after transformation are the colonies that will have the foreign insert. This type of screening is called Blue-White screening. Non-transformed colonies (lacking foreign insert) also cannot grow in presence of ampicillin. The transformed colonies that have foreign insert are ampicillin resistant and can grow on ampicillin plates.

In order to select for transformed colonies in which pUC18 has the foreign insert, cells after transformation are grown in ampicillin containing X-gal plates. White colonies that grow on these plates will be bacterial cells with pUC18 containing the foreign insert gene. Non-transformed cells give blue colonies on plate lacking ampicillin.