Question

When grown in the presence of glucose and lactose together, E. coli does not utilize lactose first because

• A. lactose permease is not present on the bacterial membrane
• B. glucose inhibits the synthesis of cyclic AMP
• C. glucose activates the synthesis of cyclic AMP
• D. glucose stimulates the efflux of cyclic AMP out of the cell

Hint:

In E. coli, glucose inhibits the synthesis of cyclic AMP, which reduces the activation of the lac operon and prevents lactose utilization.

Answer 1

The Correct Option is B, D

Step 1: Understanding the mechanism of lactose utilization in E. coli.
In E. coli, the lac operon regulates the uptake and metabolism of lactose. The presence of glucose inhibits the use of lactose by reducing the synthesis of cyclic AMP (cAMP). Low cAMP levels prevent the activation of the lac operon, so lactose utilization is suppressed when glucose is available.

Step 2: Analyzing the options.
(A) lactose permease is not present on the bacterial membrane: This is incorrect. Lactose permease is present in E. coli cells, but its activity is regulated by cAMP levels.
(B) glucose inhibits the synthesis of cyclic AMP: Correct — The presence of glucose inhibits adenylate cyclase, leading to reduced cAMP levels, which prevents activation of the lac operon.
(C) glucose activates the synthesis of cyclic AMP: This is incorrect. Glucose inhibits, not activates, the synthesis of cAMP in E. coli.
(D) glucose stimulates the efflux of cyclic AMP out of the cell: This is incorrect. Glucose affects cAMP levels by inhibiting its synthesis, not by stimulating its efflux.

Step 3: Conclusion.
The correct answer is (B) glucose inhibits the synthesis of cyclic AMP, which explains why E. coli utilizes glucose before lactose.