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. Identify the colonies that are blue in color. These colonies have cells with functioning β-galactosidase, which indicates that the lacZ gene is active and the enzyme is cleaving X-gal into a blue product.
2. Identify the colonies that are colorless or white. These colonies likely have cells with mutations in the lacZ or lacY genes, preventing the production of functional β-galactosidase or its proper activity, resulting in the inability to cleave X-gal and produce the blue color.
3. Collect samples from the three mutant (colorless or white) colonies for genetic analysis to confirm the presence of mutations in the lacZ or lacY genes.
4. If the genetic analysis shows that the lacZ and lacY genes are wild type in the mutant colonies, consider the possibility of mutations in other regulatory regions of the lac operon, such as the promoter (P) or the operator (O) regions, which are crucial for the transcriptional regulation of the lac operon.
5. Investigate the promoter and operator regions of the lac operon in the mutant colonies for mutations or alterations that could affect the binding of regulatory proteins, thereby influencing the expression of the lacZ and lacY genes and leading to the observed mutant phenotype.