X-gal is a colorless, lactose-like molecule that can be split into two fragments by ββ-galactosidase. One of these product molecules creates a blue color. The photograph here shows E. coli colonies growing in a medium that contains X-gal. Find three colonies whose cells have functioning copies of ββ-galactosidase. Find three colonies whose cells might have mutations in the lacZ or the lacY genes. Suppose you analyze the protein-coding sequence of the lacZ and lacY genes of cells from the three mutant colonies and find that these sequences are wild type (normal). What other region of the lac operon might be altered to account for the mutant phenotype of these colonies?

Verified step by step guidance

1. Step 1: To identify the colonies with functioning copies of ββ-galactosidase, look for the colonies that have turned blue. This is because the enzyme ββ-galactosidase breaks down X-gal into a product that turns blue.
2. Step 2: The colonies that have not turned blue might have mutations in the lacZ or lacY genes. These genes code for proteins that are part of the lactose metabolism pathway, including ββ-galactosidase. If these genes are mutated, the enzyme may not be produced or may not function correctly, and X-gal will not be broken down.
3. Step 3: If the protein-coding sequences of the lacZ and lacY genes are normal, then the mutation causing the lack of color change could be in the regulatory region of the lac operon. This region controls the expression of the lacZ and lacY genes.
4. Step 4: The regulatory region includes the promoter, where RNA polymerase binds to start transcription, and the operator, where a repressor protein can bind to prevent transcription. If there is a mutation in these areas, it could prevent the lacZ and lacY genes from being transcribed into mRNA and translated into protein, even if the genes themselves are normal.
5. Step 5: To confirm this, you could sequence the regulatory region of the lac operon in the mutant colonies and compare it to the sequence in the wild type colonies. If there are differences, this could explain the lack of ββ-galactosidase activity and the lack of color change in the presence of X-gal.