Final answer:
The strain will make lactose permease constitutively because the Oc mutation in one of the operons causes continuous expression of the lactose permease gene, as it prevents binding by the repressor protein. The correct option is B.
Step-by-step explanation:
Under what conditions will this strain make permease? The correct option is B, constitutively. Here's why:
The lac operon in E. coli controls the production of certain enzymes needed for the metabolism of lactose. In a partial diploid or merodiploid organism, we have two copies of the lac operon, which results in a combination of regulatory sequences and genes from both operons being present.
The given genotype is lacI+ O+ lacZ+ lacY- / lacIS Oc lacZ- lacY+. Here, lacI+ indicates a normal repressor that can bind to the operator to turn off the operon. But the second copy has a super-repressor mutation lacIS, which leads to a repressor that cannot be inactivated by allolactose.
The Oc mutation means the operator is not recognized by the repressor protein.
Therefore, the lacZ- gene (which codes for ß-galactosidase) is not expressed, but the lacY+ gene (which codes for lactose permease) is expressed regardless of the presence of lactose, constitutively, due to the Oc mutation on the same operon. Option B. is the correct one.