1)A mutant E. coli was constructed containing a nonpolar null mutation in the lacZ gene. The...

1)A mutant E. coli was constructed containing a nonpolar null mutation in the lacZ gene. The null mutation resulted in a nonfunctional LacZ product. In the presence of lactose, would this mutant E. coli be induced to make Lac permease (LacY)? Why or why not?

2)What is the difference between a repressor and aporepressor?

Expert Solution

1.Ans. No, this mutant will not be induced to make Lac permease (LacY). LacZ is the gene which produces beta-galactosidase enzyme. Beta-galactosidase enzyme is responsible for breaking down of lactose into its monomers i.e. glucose and galactose. Glucose is the prefferred source of energy for E.coli. Hence, when glucose is not available, lactose available in the medium is taken in to use as source of energy by breaking down into glucose and galactose. Lac permease (LacZ) is the enzyme responsible for taking of lactose inside the E.coli cell. If an mutant E.coli produces nonfunctional LacZ product, then this mutant will not be able to break down lactose. Hence, there is no need for the E.coli to produce Lac permease enzyme. Hence, this mutant E.coli will not be induced to make Lac permease enzyme.

2.Ans. Repressor is the functional protein molecule upon binding of which at the operator inhibits the expression of structural genes (LacZ, LacY and LacA genes).

Aporepressor is inactive repressor which binds to a corepressor and binding of a corepressor to a repressor leads to binding of the repressor to bind the operator tightly.

gladiator answered 6 months ago

The E. coli LacZ will be used as a reporter gene and the expression of the...

The E. coli LacZ will be used as a reporter gene and the expression of the LacZ gene product will be detected by the enzymatic cleavage of the compound X-gal producing blue staining within the embryo. Why was an E. coli LacZ gene used as a reporter gene? Why is there a Drosophila weak basal promoter located on the P[lacZ] enhancer-trap element? The E. coli LacZ will be used as a reporter gene and the expression of the LacZ gene...

What would be the predicted phenotype of an E. coli mutant that contains a mutation in...

What would be the predicted phenotype of an E. coli mutant that contains a mutation in the cheA gene that abolishes its autophosphorylation activity? State the predicted phenotype. What kind of motility would it display?

Consider a colony of E. coli cells that have a mutation in the lac I gene,...

Consider a colony of E. coli cells that have a mutation in the lac I gene, resulting in a LacR protein that cannot bind to allolactose. Describe the effect that this mutation will have on the lac operon in terms of βgal protein levels when the cells are grown in media containing lactose but no glucose. Explain your reasoning.

Using the rII mutant bacteriophage and E. coli strains B and K, semour benzer studied gene...

Using the rII mutant bacteriophage and E. coli strains B and K, semour benzer studied gene structure by both recombination and complementation. Wild type phages form plaques on strains B or K; rII forms plaques on B only. A. In one set of experiments, Benzer infected E. coli strain B simultaneously with two different rII mutants. Progeny phages were collected and used to infect strain K, and, occasionally, plaques formed. B. In a second set of experiments, Benzer infected strain...

Predict what will happen to the expression of β-galactosidase (lacZ gene) in a mutant strain lacking...

Predict what will happen to the expression of β-galactosidase (lacZ gene) in a mutant strain lacking lacI-gene (cells are not able to produce LacI-protein): A. lacZ gene will not be expressed. B. lacZ gene will be expressed only when both lactose and glucose are present. C. lacZ gene will be expressed only in the presence of lactose; maximum expression will be observed in the absence of glucose. D.lacZ gene will not be repressed; maximum expression will be observed in the...

3. Describe a method to isolate E. coli mutants that do not transcribe lacZ?

Cystic fibrosis is caused by a mutation in a single gene. Homozygotes for this mutant recessive...

Cystic fibrosis is caused by a mutation in a single gene. Homozygotes for this mutant recessive allele (ff) have cystic fibrosis. Within a population, 2% of people have cystic fibrosis. Assuming Hardy-Weinberg equilibrium, the frequency of heterozygotes in the population is 0.24. If inbreeding was common in this population, what would happen to the percentage of people who have cystic fibrosis over time?

A strain of E. coli carries a mutation that completely inactivates the enzyme encoded in the...

A strain of E. coli carries a mutation that completely inactivates the enzyme encoded in the gene. Several revertants, mutants of the mutant with partly or fully restored activity, were selected, and the amino acid sequence of the enzyme was determined. The only differences found were at position 10 in the polypeptide chain. Revertant 1 had Thr. Revertant 2 had Glu. Revertant 3 had Met. Revertant 4 had Arg. Assume that the initial mutation itself, as well as each revertant,...

An E. Coli strain has the following lac genotype: Lacl+ lacP+ lacOc lacZ- / lacl- lacP+...

An E. Coli strain has the following lac genotype: Lacl+ lacP+ lacOc lacZ- / lacl- lacP+ lacO + lacZ + If lactose is PRESENT, what would the level of Beta-galactose be? 1- High 2- Low

An E. coli mutant isolated is simultaneously unable to utilize a large number of sugars, e.g....

An E. coli mutant isolated is simultaneously unable to utilize a large number of sugars, e.g. lactose, arabinose, maltose, as carbon source. However, genetic analysis shows that each of the operons responsible for metabolism of each sugar is free of mutation. Name a specific gene the mutation of which may account for this outcome? Explain.

Question