In a mutant that lacks adenylyl cyclase, the enzyme that synthesizes cAMP, predict which of the following conditions of extracellular lactose and glucose would cause regulation of the lac operon to differ from that of wild-type cells. a. no lactose, no glucose b. no lactose, abundant glucose c. abundant lactose, no glucose d. abundant lactose, abundant glucose

Verified step by step guidance

1. Understand the role of adenylyl cyclase: Adenylyl cyclase is responsible for converting ATP to cyclic AMP (cAMP). In the context of the lac operon, cAMP binds to the catabolite activator protein (CAP), which then binds to the promoter region of the lac operon to enhance transcription when glucose levels are low.
2. Analyze the effect of no adenylyl cyclase on cAMP levels: Without adenylyl cyclase, cAMP cannot be synthesized. This means that CAP cannot be activated, leading to reduced transcription of the lac operon regardless of lactose presence when glucose is low.
3. Evaluate condition a (no lactose, no glucose): In both mutant and wild-type cells, the lac operon would be off due to the absence of lactose, the inducer. The lack of adenylyl cyclase and thus cAMP does not change the regulation under this condition.
4. Evaluate condition b (no lactose, abundant glucose): Similar to condition a, the lac operon would be off in both mutant and wild-type cells due to the absence of lactose. High glucose levels would normally inhibit lac operon activity through catabolite repression, but the absence of cAMP in the mutant does not affect this since the operon is already off.
5. Evaluate condition c (abundant lactose, no glucose): In wild-type cells, low glucose would lead to high cAMP levels, activating CAP and thus enhancing lac operon transcription. In mutant cells, the absence of cAMP means CAP cannot be activated, leading to lower expression of the lac operon despite the presence of lactose. This condition shows a difference in regulation between mutant and wild-type cells.