E. coli can metabolize either glucose or lactose. However, when both are present, E. coli preferentially metabolize glucose. Which of the following is not part of the biochemical mechanisms involved in that metabolic preference.

1. The presence of glucose reduces the level of cAMP.

2. Catabolite Activator Protein (CAP) is active only in the absence of cAMP.

3. CAP binding assists in the formation of a transcription complex by RNA polymerase at the lac promoter

4. CAP is a positive control factor whose presence is necessary to initiate transcription at dependent promoters.

5. None of the above

Answer: 2. Catabolite Activator Protein (CAP) is active only in the absence of cAMP

Explanation:

E. coli's switch from glucose to lactose depends on the formation of complex by cAMP and catabolite activator protein (CAP). (cGAP-CAP) The complex binds to a specific DNA sequence at the upstream of lac operon operator and promoter site receptacle to certain genes.

This binding gives increase RNA polymerase binding and gene expression activation. The CAP binds to a promoter site in order to stimulate the promoter-associated genes.; for the process. it alters its conformation, to become a genetic activator.

The option 2 in the question is wrong. In absence of cCAMP, cCAMP-CAP complex cannot be formed, no binding will take place for gene activation, and therefore gene expression from the lac operon cannot take place.

Catabolites produced from the glucose break down inhibits the formation of signal molecule cCAMP. And therefore prevent activation of enzyme needed for lactose metabolism, even in presence of lactose Therefore E. coli will prefers glucose for energy source in presence of lactose, unless when glucose is not available