Question

o carry out 'blue-white screening', the E.coli plasmid pUC18 carries:

	A.	The N-terminal portion of the lacI gene
	B.	The N-terminal portion of the lacZ gene
	C.	The C-terminal portion of the lacI gene
	D.	The C-terminal portion of the lacZ gene
	E.	The entire lacI gene

Answer 1

Blue white screening is based on the activity of beta-galactosidase, coded by lacZ, which cleaves lactose into glucose and galactose. The multiple cloning site is present in lacZ sequence. When the foreign gene will get inserted the fuctional beta-galactosidase will not be produced. If the gene will not be inserted mans the lacZ sequence will be intact and will produce the enzyme.

So, if X-gal will be added to the agar plate the recombinant colonies will not hydrolyze it to produce blue color because of absence of funtional enzymes.

So, in this techniques lacZ is necessary for screening.

Functional β-galactosidase is a homo-tetramer. The N-terminal domain is responsible for formation of the homo-tetramer. But in recombinant cells there is deletion of few amino acids from N-terminal domain resulting in the formation of non-functional dimer of beta-galactosidase. This deletion mutation is known as LacZ∆M15 which produce truncated beta-galactosidase which cannot metabolize lactose.

So, the pUC18 can carry C-terminal portion of lacZ becauase there is deletion in N-terminal.