Question

Shown in the answers are four E. coli partial diploids. They each were assayed for production of the Z and Y genes (ß-galactosidase and lactose permease, respectively) in the absence or presence of lactose. Which strain will not show wild-type regulation of lac permease enzyme activity encoded by Y?

A.  i^sp ⁺ o ⁺z ⁺ y ^-/ i ^-p ⁺ o ⁺z ^- y ⁺

B.  i⁺p ⁺ o ^cz ⁺ y ^–/ i ^–p ⁺ o ⁺z ^–y ⁺

C.  i⁺p ^– o ⁺z ⁺ y ⁺/ i ^–p ⁺ o ⁺z ⁺y ⁺

D. i ^– p ^– o ^cz ⁺ y ^–/ i ⁺p ⁺ o ⁺z ^–y ⁺

E. None

Answer 1

Strain (A) will not show wild-type regulation of lac permease

Permease will not be produced in strain A as there is mutation in lac repressor of chromosomal operon. This 'I​​​​​s​' mutation is the mutation on lac repressor which allolactose binding site is mutated, so this mutant repressor cannot be inactivated by presence of lactose. This mutation in lac repressor is dominant over WT lac repressor.

In otger three strains, expression of permease is normal (indusible).