Question

Choose the correct statement in respect to sacB gene?

(a) The cloning site for PAC lies on the side of sacB gene

(b) It is responsible for producing an enzyme which is responsible for catalysing sucrose into glucose and fructose

(c) Expression of sacB in the presence of sucrose is beneficial

(d) Levan is non-toxic for E. coli

This question was posed to me during an online interview.

Origin of the question is Bacteriophage-Mu and Bacteriophage-Pi in chapter Other Vector Systems for E of Genetic Engineering

Answer 1

Right answer is (b) It is responsible for producing an enzyme which is responsible for catalysing sucrose into glucose and fructose

To elaborate: sacB gene along with E. coli promoter flanks the cloning site in PAC vector. It is responsible for producing an enzyme which is responsible for catalysing sucrose into glucose and fructose. The fructose is ploymerized to levan and is toxic to E. coli. Thus expression of sacB in the presence of sucrose is lethal.