Question

Wild-type E. coli can grow on minimal medium. You have isolated a mutant strain of E. coli that grows poorly in minimal medium that contains lactose and arabinose and lacks glucose. In ONPG-containing minimal medium lacking glucose the mutant strain produces very low levels of b-galactosidase activity in both the absence and presence of IPTG. The strain does not grow well when transformed with a plasmid containing the lacZ and lacY genes. The strain also does not grow well when transformed with a plasmid containing the araBADoperon.

A. Given the information above, what are 2 possible mutations that would cause the strain to grow poorly when both lactose and arabinose are present? How could you determine which gene is mutated?

B. Discuss the molecular mechanisms that explains why the mutant stain of E.coli grows poorly in minima medium lacking glucose and containing lactose and arabinose.

Answer 1

Mutant E.coli strain grows poorly in minimal media containing lactose and arabinose and lacking glucose. In ONPG-containing minimal mediun lacking glucose the mutant strain produces very low level of -galactosidase activity. ONPG mimics lactose and is hydrolyzed by -galactosidase into galactose and o-nitrophenol but is unable to act as inducer for the lac operon without the presence of another lactose analog like IPTG. Th strain does not grow well when transformed in a plasmid having lacz and lacy genes and also when transformed containing araBAD operon.

A. The possible mutations that may take place are z- and y- alongwith Is (super repressor) and CAP and adenyl cyclase. The z- mutation results in no -galactosidase formation although permease activity is restored, so lactose will not able to be metabolized; in y- mutation the lactose will not able to enter the cell as there is no permease, so -galactosidase will not function properly. The super repressor Is always switched off the lac operon; the CAP and adenyl cyclase, both weakens z,y and a expression.PCR method can be employed to identify the mutated gene.

B. Several mutations that can cause the mutant E.coli strain not to grow on the medium containing lactose and arabinose. The mutants affecting the mode of mutation are z-, y-, Is, P-, CAP and adenyl cyclase. For z- mutation, lactose wont be metabolized due to non availability of  -galactosidase, the lactose will enter the cell due to permease activity but inside the cell it cant be metabolize into glucose and galactose.; for y- mutation there will be no permease, as a result lactose remains outside the cell and  -galactosidase will not able to function. The P- mutation lead to the non-availability of RNA polymerase to bind with the promoter. So the genes z,y and a are always off. The super repressor Is tends to make the genes off irrespective of lactose presence or absence. The DNA binding site is still available but the original lactose binding site is mutated so lactose does not able to bind with it, the Is always be bound to the operator gene restricting RNA polymerase to act. The CAP and adenyl cyclase both weaken the expression of z,y and a genes, the CAP binds to the inducer cAMP which bind to the promoter. With the mutation of CAP, it cant bind with promoter and with adenyl cyclase, it will not able to make cAMP which in turn not able to bind with CAP to activate transcription of z,y and a genes which results in reduced expression of these genes.