Final answer:
The partial diploid E. coli strain mentioned will make ß-galactosidase in the presence of lactose due to the functional lacI+ allele and the constitutive expression from the mutant O(c) operator, allowing the lacZ+ allele to be expressed.
Step-by-step explanation:
The strain of E. coli described, lacI+ O+ lacZ+ lacY-/lacI(s) O(c) lacZ- lacY+, will produce ß-galactosidase when lactose is present, but not in the presence of glucose.
In this partial diploid strain, the lacI+ allele provides functional repressor that can bind to the O+ operator in the absence of lactose, preventing lac operon expression.
However, the presence of lactose inactivates the repressor, allowing transcription from the O+ operator since the lacI(s) allele that makes a super-repressor cannot bind to the mutant O(c) operator.
This O(c) operator is constitutive and allows for continuous transcription of the lacZ+ and lacY+ genes, but in a wild-type background, the system would be repressed by the lacI+ gene product.
Therefore, expression of ß-galactosidase depends on the presence of lactose as the inducer.
The lacY- mutation means lactose permease will not be made by the first operon, but the lacY+ allele in the second operon can still synthesize permease when the operon is de-repressed by lactose.