The F' plasmid appears to "rescue" mutant strain 3 by restoring wild-type inducible expression. The F' plasmid does not change the phenotype of mutant strain 4; there is still lower than normal expression in the absence of glucose.
There are two possible mutations that could lead to low expression of the lac operon even in the absence of glucose -- a mutation in the gene for the CAP protein (known as the crp gene) or a mutation in the CAP-cAMP binding site. Recall that binding of the CAP-cAMP complex to DNA facilitates transcription of the lac operon when glucose levels are low or absent. To determine the effect of introducing a wild type F' plasmid into a bacterial cell with each of these mutations, you draw out the two possible scenarios.
Two possible partial diploids: (1) F' lac+/lac+ with crp- gene (2) F' lac+/cap- with crp+ gene For each of the partial diploids, determine whether expression of the lac operon will be inducible (wild type) or low expression in the absence of glucose (mutant). Then use this information to identify the mutations in mutant strain 3 and mutation strain 4.
Choose the two correct answers.
A. F' cap+ / crp+ cap- shows low expression in the lactose only condition. Mutation 4 is a CAP-cAMP binding site mutation.
B. F' cap+ / crp+ cap- shows high expression in the lactose only condition. Mutation 3 is a CAP-cAMP binding site mutation.
C. F' cap+ / crp- cap+ shows high expression in the lactose only condition. Mutation 3 is a mutation in the crp gene that produces CAP.
D. F' cap+ / crp- cap+ shows low expression in the lactose only condition. Mutation 4 is a mutation in the crp gene that produces CAP.