The lac operon in E. coli regulates β-galactosidase expression through lactose presence and glucose absence. Both conditions (A) absence of glucose and (D) presence of lactose induce the gene.

Correct Answer

Options (A) and (D) induce β-galactosidase gene expression. Maximal induction requires lactose present to inactivate the repressor and glucose absent to activate via CAP-cAMP.​

Option (A): Absence of Glucose

Absence of glucose raises cAMP levels, enabling CAP-cAMP binding upstream of the promoter to enhance RNA polymerase recruitment. This positive regulation boosts transcription even if lactose inactivates the repressor. Without glucose absence, catabolite repression limits expression.​

Option (B): Absence of Lactose

Absence of lactose keeps the lac repressor bound to the operator, blocking transcription of β-galactosidase. Allolactose (from lactose) normally inactivates the repressor; without it, basal expression occurs. This prevents wasteful enzyme production.​

Option (C): Presence of Glucose

Presence of glucose triggers catabolite repression by lowering cAMP, preventing CAP activation. Even with lactose, transcription remains low as RNA polymerase binds poorly. Glucose prioritization inhibits lac operon fully.​

Option (D): Presence of Lactose

Presence of lactose produces allolactose, which binds the repressor, releasing it from the operator for derepression. This negative regulation allows transcription initiation. Full induction still needs glucose absence.​