The discovery of the lac operon helped build the foundation upon which modern day genetics is built. Encoded within the genomes of E. coli, the lac operon is required for lactose metabolism. In turn, lactose is required to activate the lac operon to full levels of expression. Mutations in the various components of this regulatory system can be thus discriminated by culturing mutant strains on different growth media. This report examines how lac mutant strains can be phenotyped and genotyped.
Lac Operon Genetics
Practical 2. Analysis of wild type and lac operon mutant strains of the bacterium Escherichia coli
Complete the results tables below using the data you obtained in the practical.
Describe the size, colour and eosin sheen of the colonies on the EMB plates in Table 1 below.
Strain
Size
Colour
Eosin sheen
WT
Large colonies purple
Strong eosin sheen
H
Large colonies purple
Weak eosin sheen
J
Small colonies pink
No eosin sheen
K
Small colonies pink
No eosin sheen
Fill in the fluorescence results for NA+glu and NA+lac in Table 2 below.
Strain
NA+glu within 1 minute of MUG overlay
NA+glu
minutes after MUG overlay
NA+glu
minutes after MUG overlay
NA+lac within 1 minute of MUG overlay
NA+lac
minutes after MUG overlay
NA+lac
minutes after MUG overlay
WT
H
J
K
* Record the degree of fluorescence as (-) or (+) or (++) or (+++). Where (-) is no fluorescence and (+++) is the greatest fluorescence.
5. Based on your experimental observations of each strain on EMB and the NA+lac and NA+glu plates decide on the genotype/phenotype of each of the strains and fill in the Table 3 below.
Strain
phenotype
(lac+ or lac-)
Genotype*
WT
lac+
lacZ+, lacI+, lacY+
H
lac+
lacZ+, lacI-, lacY+
J
lac-
lacZ+, lacI+, lacY-
K
lac-
lacZ-, lacI+, lacY+
* lacY- or lacY+, lacZ- or lacZ+, lacI- or lacI+ (put in genotype of all three genes for each strain) (2 marks)
6. In your own words, explain how the results you present in Tables 1 and 2 support the conclusions you present in Table 3
The WT strain is the reference strain. Contemporary EMB growth medium contains both lactose and sucrose and gram-negative bacteria will preferentially metabolize sucrose in a wild-type strain. However, we are screening specifically for lac operon mutants and lactose is the only carbohydrate in the growth medium. Lactose hydrolysis produces hydrogen ions, which lowers the pH of the growth media immediately surrounding the location of colonies. In response to the increasingly acidic conditions, eosin will grow darker. This explains the dark purple color and the strong eosin sheen surrounding the wild-type colonies. The phenotype of the WT strain is lac+ and the genotype lacZ+, lacY+, and lacI+.
The H. strain is able to metabolize lactose and is therefore able to produce a functional beta-galactosidase enzyme (lacZ+). Lactose hydrolysis in this strain is capable of acidifying the surrounding growth medium, but to a lesser extent than the WT strain. This suggests that the lac operon is not fully upregulated. For full activation of the lac operon to occur, lactose is converted to allolactose by beta-glactosidase. This small-molecule inducer binds to the lac repressor (lacI) and prevents it from binding to the operator element, thereby upregulating transcriptional activity. The starvation signal, cAMP, would also need to bind to the CAP site near the promoters for full transcriptional activity to be achieved. In the absence of a starvation signal or the lac repressor, the lac operon would be expressed only at moderate levels. The H. strain is therefore phenotypically lac+ and genotypically lacZ+, lacY+, and lacI?. Other mutations would explain this phenotype, but are not among the choices in this experiment.
The J. And K. strains both produced pink colonies on the EMB plates. These strains are therefore phenotypically lac?. Genotypically, these strains could be lacZ? Or lacY?, since either mutation would result in an inability to metabolize lactose. Other mutations or combinations could also explain this phenotype, but are not being explored here. A lacZ? strain would not be able to produce a viable beta-galactosidase enzyme and therefore could not metabolize lactose and reduce the pH of the surrounding growth medium. The lacY gene produces a promiscuous permease transporter required for lactose entry into the bacterium, and in its absence beta-galactosidase has no substrate to hydrolize. Therefore, the genotype of the J. And K. strains are most likely [lacZ? Or lacY?], and lacI+.
The beta-galactosidase activity of the WT strain on nutrient agar plates with glucose or lactose, as assayed using MUG overlays, revealed the suppression of lac operon expression in the presence of glucose and its upregulation in the presence of lactose. This is consistent with glucose acting as a suppressor of lac operon expression.
In contrast, the lac operon is constitutively expressed in the H. strain in the presence of either glucose or lactose. This suggests that the mechanism responsible for suppression of the lac operon by glucose has been compromised. This is consistent with the H. strain having a genotype of lacZ+, lacY+, and lacI?.
The beta-glactosidase activity of the J. And K. strains is the same whether in the presence of glucose or lactose and are therefore indifferent to the two conditions. The J. strain is capable of low-levels of beta-galactosidase activity in the presence of glucose and lactose, but the K. strain is not. Therefore the J. strain is lacZ+, which in turn suggests the sugars are unable to efficiently enter the bacterium. This suggests the genotype of the J. strain is lacZ+, lacY?, and lacI+ and the genotype of the K. strain is lacZ?, lacY+, and lacI+.
7. In your own words, explain the mechanism by which regulation of E. coli trp operon gene expression responds to levels of tryptophan in growth medium. What are the similarities and differences compared to lac operon regulation? (3 marks)
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