Lac Operon

Short Report: Lac Operon.

The lac operon is a set of structural genes that consists of one regulatory gene, Lac I, and three structural genes: Lac Z, Lac Y and Lac A [1,2]. Lac I codes for the repressor protein of the lac operon. Lac Z codes for b-galactosidase, which hydrolyze lactose into glucose and galatose [1,2]. Lac Y codes for permease, which increases permeability of the cell to b-galactosides [1,2]. Lac A codes for transacetylase, which detoxifies certain byproducts of lactose metabolism [1,2]. In addition, there is a Promoter region, the binding site for RNA polymerase, and an Operator region, the binding site for repressor protein.

In the present of lactose, it binds to the repressor protein and induces a conformational change that prevents the repressor protein from binding to the Operator. As a result, RNA polymerase is able to proceed, and encodes Lac Z, Lac Y and Lac A into β-galactosidase, permease, and transacetylase, respectively, Figure 1 [3]. Consequently, lactose is metabolized. On the other hand, in the absence of lactose and in the present of glucose, the repressor protein that is coded from Lac I is in its original conformation and able to bind to the Operator. This inhibits RNA polymerase from carry out its function because RNA polymerase cannot proceed past the Operator. As a result, there is no RNA is made and by extension, no enzyme is made, Figure 2 [3].

Figure 1: Lac operon in the present of lactose. [3]

Figure 2: Lac operon in the absent of lactose. [3]

The purposes of this lab were to observe different phenotypes and how the bacteria are behaving in different types of media. Four different genotypes of bacteria were used: I+P+Z+, I+P-Z+, I-P+Z+, and I+P+Z-. Genotype I+P+Z+ is the wild type, in which everything gene is normal. Genotype I+P-Z+ has mutated gene P, which do not allow promoter to bind very well; therefore, the level of β-galactosidase is very low. Genotype I-P+Z+ has mutated gene I, which make the repressor protein unable to bind to the Operator even in the present of lactose. As a result, the production of β-galactosidase can be proceeding. Finally, genotype I+P+Z- has mutated gene Z, which result in a mutant enzyme, β-galactosidase. Consequently, lactose cannot be fermented.

These four strains of bacteria were labeled A, B, C, and D. The purpose was to identify which strain is A, which strain is B, so forth and so on… Each of these strains was streaked onto four different types of media: Glu, Lac, McConkey, and Glu + X Gal. Glu media contains only salts and glucose. Lac media is rich in lactose. McConkey media contains all nutrients, lactose, and an indicator dye. If the bacteria ferment lactose, the colony will turn deep red, otherwise the colonies will be white. Glu + X Gal media contains glucose and substrate for β-galactosidase. A total of sixteen plates were streaked. Then the plates were incubated and the phenotypes were observed in order to determine which strain of bacteria is corresponded to which letter.

The glucose plates were the control. All strains of bacteria are expected to grow in these plates. In lactose plates, genotype I+P+Z+ and genotype I-P+Z+ are expected to grow also. On the other hand, genotype I+P-Z+ is expected to show little growth and genotype I+P+Z- is expected to show no growth at all. In McConkey plates, genotype I+P+Z+ and genotype I-P+Z+ are expected to have red media and red colonies. Genotype I+P-Z+ is expected to have orange media and red colonies while genotype I+P+Z- will have white media and orange colonies. In Glu + X Gal plates, genotype I+P+Z+ is expected to have blue color. Genotype I+P-Z+ will have light blue. Genotype I-P+Z+ will have dark blue whereas genotype I+P+Z- will have white color. Please refer to Table 1.

Table 1: Expected Results.

Strain Glu (Control) Lac McConkey Glu + X Gal

I+P+Z+ Growth Growth Red Media + Colonies Blue

I+P-Z+ Growth Little Growth Orange Media + Red colonies Light Blue