Question

Given the following mutants and conditions, predict the expression level of the of the lac Z gene. Please indicate your answer with one of followings: no expression, basal level of expression, or activated level of expression. Explain why in the context of activities by activator and repressor.

A mutant of E.coli that has a mutation in the allolactose binding site of repressor :

(1) In the presence of glucose, presence of lactose –

(2) In the presence of glucose, absence of lactose –

(3) In the absence of glucose, absence of lactose -

(4) In absence of glucose, presence of lactose -

Answer 1

Lac operon is an inducible operon which means that the transcription of cistron takes place when the substrate is present in the medium. Lactose sugare represents to be the substrate for lac operon and thus, presence of lactose and simultaneous absence of glucose will upregulate the operon. Further, the inactivation of repressor i.e. allolactose by its binding to the repressor site causes the operon to upregulate gene expression. Thus, the answers can be found as below:

Part 1) Low level of expression: The glucose in the medium will prevent expression of the operon by masking the operator and hence downregulating gene expression. The repressor would be mildly active and thus further prevent gene expression.

Part 2) No expression: Since lactose or the activator is altogether absent from the medium, there would be no gene expression since the glucose present in the medium will completely mask the operator site and repress the gene expression.

Part 3) No expression: Although repressor is absent in this case, but the absence of activator as well evidents no expression of the operon.

Part 4) High expression: Since repressor or glucose is absent from the medium for masking the operator site and lactose or activator is present for linear mobilization of RNA polymerase over the operator, high level of gene expression would take place.