Question

In E coli the lac operon is used to control the expression of the genes involved in the used and metabolism of lactose . describe how the genes are not expressed in the absence of lactose.

describe how the genes are expressed when lactose is present both in the presence and of glucose. what would happen if there were an l^-  mutation? what would happen if there were an O^c mutation? which one of those two mutations is recessive ?

Answer 1

Answer:

• Lac operon consists of three structural genes called Lac Z, Lac Y and Lac A and a regulatory element (promoter and operator).
• Lac I gene present outside the lac operon, act as regulatory gene and control expression of lac operon genes.
• Lac Z codes for beta galactosidase and converts lactose into allolactose, Lac Y codes for Lac permease and transport lactose into the cells and Lac Z codes for transacetylase which neutralizes toxic effects of lactose.

In absence of lactose:

• In absence of lactose, Lac I gene produces lac repressor which binds to operator region and prevent binding of RNA Polymerase thus blocks transcription.

In presence of lactose:

• In presence of lactose few molecules of lac operon enzyme catalyze conversion of lactose to allolactose.
• Allolactoseact as inducer, binds to the lac repressor and induce conformational changes causing dissociation of lac repressor from operator.
• In absence of lac repressor RNA Polymerase binds to the promoter and starts transcription of genes which catabolize lactose.

When both lactose and glucose are equally provided to bacterial cell then very low expression of lac operon genes occurs.

• At high glucose concentration the growth rate is maximum and lactose catabolism is repressed, called catabolite repression.
• If there is mutation in Lac I gene then Lac repressor becomes defective and unable to bind with Lac operator.
• Mutated I gene is represented by I- gene and if there is mutation in operator region then it prevents binding of Lac repressor.
• Mutated operator is represented by Oc .

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