Question

A mutant E. coli strain, grown under conditions that normally induce the lac operon, produces high amounts of ß-galactosidase. What is a possible genotype of the cells? lacI– lacP+ lacO+ lacZ– lacY+ lacA+

lacI+ lacP– lacO+ lacZ+ lacY+ lacA+

lacI+ lacP+ lacO+ lacZ– lacY+ lacA+

lacI+ lacP+ lacOc lacZ+ lacY+ lacA

( if you are google searching for answer=the one in the quizlet is wrong)

Answer 1

The answer is lacI+ lacP+ lacOc lacZ+ lacY+ lacA

For explanation find the below

lacI– lacP+ lacO+ lacZ– lacY+ lacA+
As the lacZ gene here is a mutant one, it cannot produce functional ß-galactosidase.

lacI+ lacP– lacO+ lacZ+ lacY+ lacA+
As the promoter is mutant here, the RNA polymease cannot bind over here, so the lacZ gene will be silenced.

lacI+ lacP+ lacO+ lacZ– lacY+ lacA+
As the lacZ gene here is a mutant one, it cannot produce functional ß-galactosidase.

lacI+ lacP+ lacOc lacZ+ lacY+ lacA
As the operator here is constitutive, the ß-galactosidase will be produced in high amount becuase the repressor cannot bind to operator.