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Schematic of the Lac Operon

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Below is a schematic of the lac operon:

SHAPE * MERGEFORMAT

A special class of mutants of E coli unable to utilize lactose as a carbon source were found to have mutations in the coding sequence for β-galactosidase (lacZ gene). They produce no β-galactosidase protein. Surprisingly, they fail to produce the polypeptide products of the permease (lacY) and the transacetylase (lacA). In addition, no mRNA from any of these genes is detected.
Three kinds of revertants were isolated:
Type I Revertants: Regained permease (LacY) and transacetylase (LacA) activities but not β-galactosidase activity. These reverting mutations were found within the lac operon.
Type II Revertants: Regained all 3 enzymatic activities.
Type III Revertants: Regained only permease and transacetylase activities by a secondary mutation that mapped in the gene for rho factor.
Lac operon DNA was isolated from the original mutant strain. The strands of the DNA were separated from each other and mixed with single strands of DNA containing the lac operon isolated from a wild type bacterium. These "mixed" DNAs were allowed to hybridize to each other. The resulting double helix structures were isolated and visualized via electron microscopy:

SHAPE * MERGEFORMAT A. (5pnts) What is a revertant mutation (in general)?

B. (5pnts) What was the genetic event that caused the original lac- mutation (ie did not allow for the expression of lacZ, lacY, or lacA)?

C. (5pnts) Why did this event result in failure to produce permease and transacetylase?

D. (5pnts) What is the nature of the Type I Reverting mutations?

E. (5pnts) What is the nature of the Type II Reverting mutations?

F. (5pnts) What is the nature of the Type III Reverting mutations?

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Q6 (A-F). 30pnts Below is a schematic of the lac operon: SHAPE * MERGEFORMAT
A special class of mutants of E coli unable to utilize lactose as a carbon source were found to have mutations in the coding sequence for β-galactosidase (lacZ gene). They produce no β-galactosidase protein. Surprisingly, they fail to produce the polypeptide products of the permease (lacY) and the transacetylase (lacA). In addition, no mRNA from any of these genes is detected. Three kinds of revertants were isolated: Type I Revertants: Regained permease (LacY) and transacetylase (LacA) activities but not β-galactosidase activity. These reverting mutations were found within the lac operon. Type II Revertants: Regained all 3 enzymatic activities. Type III Revertants: Regained only permease and transacetylase activities by a secondary mutation that mapped in the gene for rho factor. Lac operon DNA was isolated from the original mutant strain. The strands of the DNA were separated from each other and mixed with single strands of DNA containing the lac operon isolated from a wild type bacterium. These "mixed" DNAs were allowed to hybridize to each other. The resulting double helix structures were isolated and visualized via electron microscopy:

SHAPE * MERGEFORMAT

A. (5pnts) What is a revertant mutation (in general)?

A revertant mutation is a mutant that is able to revert back to its original genotype by another suppressor mutation. A revertant mutation can also revert back to the original phenotype through this kind of suppressive interaction between mutant A and mutant B. To be a revertant mutant, the second mutant must suppress the first mutation. For example, in the mutant E. coli bacteria of the lac operon. There is a second mutation that transform the Lac operon mutated genes of LacI- and LacA- to the normal genotype (LacI and LacA) and phenotype in that it can now produce permease (Lac Y) and transacetylase (Lac A) products.

B. (5pnts) What was the genetic event that caused the original lac- mutation (ie did not allow for the expression of lacZ, lacY, or lacA)?

The Lac Operon is a gene that enable the E.coli to use the sugar lactose. Lactose ...

Solution Summary

The schematic of the Lac Operon is discussed in the solution.

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I need help on the following two questions:
1. Perform a quick review of literature on recombinant protein expression in E. coli and indicate the protein yield reported in two research articles of your choice. Make sure you specify the units for the protein yield. For each article describe the nature of the protein that was overexpressed (membrane protein, cytosolic protein, ...). Estimate how much of your recombinant T7 RNA polymerase enzyme you might expect from 50ml of liquid culture by assuming a yield comparable to the ones reported in the two articles you selected. A hard copy of each of your two articles along with your answer should be handed in to your TA when entering the lab

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a. The inoculated 50ml culture containing 200ug/ml ampicillin is to be divided in two fractions of 25ml. At time zero of the induction, the first 25ml fraction is to be induced with 1mM IPTG whereas only water is to be added to the second 25ml fraction. A 1ml aliquot is to be withdrawn from each of the two fractions at both time zero and after a 1.5 hr incubation. ( /2.5 MARKS)
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Thanks again.

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