Question

Predict the level of expression for the lac z gene given the following genotype and conditions: i o z no lactose and no glucose present.

A) High expression
B) Low expression
C) No expression
D) Moderate expression

Answer 1

With both lactose and glucose absent in the 'i o z' genotype, the expression of the lac z gene would be low, because lactose is not present to serve as an inducer, and the CAP-cAMP complex has no action due to the mutation in the operator sequence.

Step-by-step explanation:

The lac operon is a well-studied model for gene expression regulation in bacteria, particularly E. coli. When analyzing the expression of the lac z gene under the given conditions, it is crucial to consider the roles of both lactose and glucose.

For the lac operon to be highly expressed, two conditions must be satisfied: there must be an absence of glucose, and lactose must be present. If glucose is absent, the binding of the catabolite gene activator protein (CAP) enhances transcription. Conversely, when lactose is present, it is converted into allolactose, the inducer molecule that binds to and inactivates the lac repressor, leading to transcription of the lac operon genes including lacZ.

In the given scenario, the presence of the genotype 'i o z' indicates that the operator sequence 'o' is mutated and non-functional, allowing for transcription regardless of the lac repressor. However, both lactose and glucose are absent, meaning there is no inducer to inactivate the repressor and no glucose to suppress the CAP activation. Therefore, the expression of the lac z gene would be expected to be low, due to lack of lactose, despite the defective operator.