Question

The intracellular concentrations of cAMP in Escherichia coli are controlled by adenylate cyclase, which is encoded by the Cya gene. Withdrawal of the glucose from the growth medium causes a reduction of ATP and a dramatic stimulation of the cyclase activity and the production of cAMP.

Consider two E. coli strains with the following genotypes, where “+” indicates a wild-type gene with normal function and “-” indicates a loss-of-function allele:

Strain A: Cya^- CAP⁺ I⁺ P⁺ O⁺ Z^- Y⁺

Strain B: Cya⁺ CAP^- I⁺ P⁺ O⁺ Z⁺ Y⁺

When growing the two strains in media containing only lactose, predict the following

1. the transcription levels of the lac Operon in the two strains, are they higher or lower than that in the wild-type E. coli;
2. the survival of the two strains as compared to the wild-type E. coli.

Your answer:

Answer 1

The absence of glucose, cAMP-CAP is formed which binds to the CAP site. The bounded complex at CAP enabling RNA polymerase to initiate transcription.

Strain A: Cya- CAP+ I+ P+ O+ Z- Y+

cya- = No production of cAMP so no binding to CAP and no initiation of transcription.
Lac operon is lower than that in the wild-type E. coli
and survival is low compared to the wild-type E. coli.

Strain B: Cya+ CAP- I+ P+ O+ Z+ Y+

CAP- = No cap region is there to binding of CAmp, so no initiation of transcription.
Lac operon is lower than that in the wild-type E. coli
and survival is low compared to the wild-type E. coli.