Question

2. Suppose you have six strains of E. coli. One is wild type, and each of the other five has a single one of the following mutations: lacZ, lacY, laď·0; and lach. For each of these six strains, describe the phenotype you would observe using the following assays. Explain your answers. [Notes: (1) IPTG is a colorless synthetic molecule that acts as an inducer of lac operon expression but cannot serve as a carbon source for bacterial growth because it cannot be cleaved by ß-galactosidase; (2) X-gal is a substrate of ß-galactosidase, which produces a blue color change upon being cleaved by P-galactosidase (see Figure 16.3 in your textbook), but it cannot serve as a carbon source for growth; (3) E. coli requires active lactose permease (the product of lacY) to allow lactose, X-gal, or IPTG into the cells.] Some tips to the logic of this problem. Colonies will be blue if β-galactosidase is made (meaning that a lacZ gene is present and transcribed) and if X-gal is present and can enter the cell (the cells are lacY). Transcription of the lac operon will occur if an inducer (lactose or IPTG) is present and can enter the cell (the cells are lacY'), or if mutations in the bacterial chromosome render the lac operon constitutive. High levels of glucose prevent positive control of lac operon transcription by CRP-cAMP and lead to low level expression of the lac operon in the presence of an inducer Growth on medium in which the only carbon source was lactose a. b. Colony color in medium containing glycerol as the only carbon source, X-gal, and IPTG Colony color in medium containing glycerol as the only carbon source and X-gal, but no IPTG c. d. X-gal, and IPTG Colony color in medium containing high levels of glucose as the only carbon source and X-gal, but no IPTG e.

Answer 1

a. If the only source of growth on the medium is lactose then the wild type would give blue colonies because they have an active enzyme for every process required. Lac Z^- will not grow because even if the lactose can enter the cell, the cell will be unable to break down lactose. if Lac Y^- is present then also the lactose will not be able to enter the cell and the colonies will not grow. if the colonies are Lac I^- the colonies will not be able to grow because for the working of the Lac Y it is very important that the inducer is present. If O^c is present then the cells will be able to grow because the source available is only lactose and no high levels of glucose are present. When I^s is present the cells will be able to grow as both the inducer and the Lac genes are working.

b. The E. coli cells are able to grow on glycerol. Since in glycerol metabolism, there is no need of Lac operon at all. The color of the colonies will be white because beta-galactosidase is the substrate of X gal so no color production will occur.

c. The E. coli cells are able to grow on glycerol. Since in glycerol metabolism, there is no need of Lac operon at all. The color of the colonies will be white because beta-galactosidase is the substrate of X gal so no color production will occur.

d. They are not a carbon source, therefore, no growth will be seen in all the cases.

e. The E. coli cells are able to grow on glucose. Since in glucose metabolism, there is no need of Lac operon at all. The color of the colonies will be white because beta-galactosidase is the substrate of X gal so no color production will occur.