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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...
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…
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