When researchers analyzed the gad operon they found the following: Expression of the gad operon at acidic and neutral pH pH 2.0 рH 7.0 Wild type +++ Mutant 1 +++ +++ Mutant 2 Mutant 3 +++ +++ Mutant 4 What is a valid conclusion from this experiment? O a. Mutant 1 is a mutation in a repressor O b. Mutant 1 is a mutation in an operator O c. Mutant 2 is a mutation in an activator O d. Mutant 2 is a mutation in an enhancer O e. All of the above are valid conclusions O f. None of the above are valid conclusions

When researchers analyzed the gad operon they found the following: Expression of the gad operon at acidic and neutral pH pH 2.0 рH 7.0 Wild type +++ Mutant 1 +++ +++ Mutant 2 Mutant 3 +++ +++ Mutant 4 What is a valid conclusion from this experiment? O a. Mutant 1 is a mutation in a repressor O b. Mutant 1 is a mutation in an operator O c. Mutant 2 is a mutation in an activator O d. Mutant 2 is a mutation in an enhancer O e. All of the above are valid conclusions O f. None of the above are valid conclusions

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter14: Gene Regulation

Section: Chapter Questions

Problem 13TYU: INTERPRET DATA Develop a simple hypothesis that would explain the behavior of each of the following...

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You have isolated two different mutants (reg1 andreg2) causing constitutive expression of the emu operon (emu1 emu2). One mutant contains a defect in aDNA-binding site, and the other has a loss-of-functiondefect in the gene encoding a protein that binds tothe site.a. Is the DNA-binding protein a positive or negativeregulator of gene expression?b. To determine which mutant has a defect in the siteand which one has a mutation in the binding protein, you decide to do an analysis using F′ plasmids. Assuming you can assay levels of the Emu1and Emu2 proteins, what results do you predict forthe two strains (i and ii; see descriptions below) ifreg2 encodes the regulatory protein and reg1 is theregulatory site?i. F′ (reg1−reg2+emu1−emu2+)/reg1+reg2+emu1+emu2−ii. F′ (reg1+reg2−emu1−emu2+)/reg1+reg2+emu1+emu2−c. What results do you predict for the two strains(i and ii) if reg1 encodes the regulatory proteinand reg2 is the regulatory site?
a. The original constitutive operator mutations in thelac operon were all base changes in o1. Why doyou think mutations in o2 or o3 were not isolated inthese screens?b. Explain how a mutagen that causes small insertions could produce an ocmutation.c. Would a strain with one of the ocmutations described in part (b) and also a lacIS mutation be ableto make β-galactosidase either in the presence orabsence of inducer? Explain
You have isolated different mutants (reg1 and reg2) causing constitutive expression of the emu operon (which has genes emu1 and emu2). One mutant contains a defect in a DNA-binding site, and the other has a loss-of-function defect in the gene encoding a protein that binds to the site Say you don’t know which mutant has a defect in the site and which one has a mutation in the binding protein. To figure it out, you construct the two partial diploid strains (i and ii below), and you then assay the levels of the Emu1 and Emu2 proteins in these two strains. F’ (reg1- reg2+ emu1- emu2+) / reg1+ reg2+ emu1+ emu2- F’ (reg1+ reg2- emu1- emu2+) / reg1+ reg2+ emu1+ emu2- What proteins do you predict will be expressed for strains i and ii if reg2 encodes the regulatory protein and reg1 is the regulatory site?
You then make a screen to identify potential mutants (shown as * in the diagram) that are able to constitutively activate Up Late operon in the absence of Red Bull and those that are not able to facilitate E. Coli growth even when fed Red Bull. You find that each class of mutations localize separately to two separate regions. For those mutations that prevent growth even when fed Red Bull are all clustered upstream of the core promoter around -50 bp. For those mutations that are able to constitutively activate the operon in the absence of Red Bull are all located between the coding region of sleep and wings. Further analysis of each DNA sequence shows that the sequence upstream of the promoter binds the protein wings and the region between the coding sequence of sleep and wings binds the protein sleep. When the DNA sequence of each is mutated, the ability to bind DNA is lost. Propose a final method of gene regulation of the Up Lateoperon using an updated drawn figure of the Up Late…
Strain ROFL4 has a premature stop mutation in the lacZ gene, resulting in a nonfunctional b-galactosidase. Otherwise all other parts of the operon are functional. 1. Can strain ROFL4 metabolize lactose if it is present? 2. If lactose is absent, will strain ROFL4 transcribe its lac operon? 3. If lactose is present, will strain ROFL4 transcribe its lac operon?
You have isolated two different mutants (reg1 and reg2) causing constitutive expression of the emu operon (emu1 emu2). One mutant contains a defect in a DNA-binding site, and the other has a loss-of-function defect in the gene encoding a protein that binds to the site. Is the DNA-binding protein a positive or negative regulator of gene expression? Explain. To determine which mutant has a defect in the site and which one has a mutation in the binding protein, you decide to do an analysis using F′ plasmids. Assuming you can assay levels of the Emu1 and Emu2 proteins, what results do you predict for the two strains (i and ii; see descriptions below) if reg2 encodes the regulatory protein and reg1 is the regulatory site? Explain. F′ (reg1− reg2+ emu1− emu2+)/reg1+ reg2+ emu1+ emu2− F′ (reg1+ reg2− emu1− emu2+)/reg1+ reg2+ emu1+ emu2−
Strain P77 has a mutation in the lacO that prevents it from being bound by the lacI protein. Otherwise all other parts of the operon are functional. 1. Can strain P77 metabolize lactose if it is present? 2. If lactose is absent, will strain P77 transcribe its lac operon? 3. If lactose is present, will strain P77 transcribe its lac operon?
On its chromosome, an E. coli cell has a genotype of lacI− lacZ+ lacY+ lacA+. It has an F′ factor with a genotype of lacI+ lacZ+ lacY+ lacA+. What is the expected level of expression of the lac operon genes (lacZ+ lacY+ lacA+) in the absence of lactose?a. Both lac operons will be expressed.b. Neither lac operon will be expressed.c. Only the chromosomal lac operon will be expressed.d. Only the lac operon on the F′ factor will be expressed.
Suppose you have six strains of E. coli. One is wildtype, and each of the other five has a single one of thefollowing mutations: lacZ−, lacY−, lacI−, oc, andlacIS. For each of these six strains, describe thephenotype you would observe using the following assays. [Notes: (1) IPTG is a colorless synthetic molecule that acts as an inducer of lac operon expressionbut cannot serve as a carbon source for bacterialgrowth because it cannot be cleaved byβ-galactosidase; (2) X-gal cannot serve as a carbonsource for growth; (3) E. coli requires active lactosepermease (the product of lacY) to allow lactose,X-gal, or IPTG into the cells.] Colony color in medium containing glycerol as theonly carbon source and X-gal, but no IPTG.d. Colony color in medium containing high levels ofglucose as the only carbon source, X-gal, andIPTG.e. Colony color in medium containing high levels ofglucose as the only carbon source and X-gal, butno IPTG
. Suppose you have six strains of E. coli. One is wildtype, and each of the other five has a single one of thefollowing mutations: lacZ−, lacY−, lacI−, oc, andlacIS. For each of these six strains, describe thephenotype you would observe using the following assays. [Notes: (1) IPTG is a colorless synthetic molecule that acts as an inducer of lac operon expressionbut cannot serve as a carbon source for bacterialgrowth because it cannot be cleaved byβ-galactosidase; (2) X-gal cannot serve as a carbonsource for growth; (3) E. coli requires active lactosepermease (the product of lacY) to allow lactose,X-gal, or IPTG into the cells.]a. Growth on medium in which the only carbonsource was lactose.b. Colony color in medium containing glycerol as theonly carbon source, X-gal, and IPTG
. In an effort to determine the location of an operator sitefor a negatively regulated gene, you have made a seriesof deletions within the regulatory region. The extent ofeach deletion is shown by the line underneath the sequence, and the resulting expression from the operon(i = inducible; c = constitutive; − = no expression) isalso indicated.... GGAT C T T AGCCGGCTAACATGATAAATATAA......C C T AGAATCGGCCGA TTGTA C T A TTT ATAT T ...1 i2 –3 c4 –5 ca. What can you conclude from these data about thelocation of the operator site?b. Why do you think deletions 2 and 4 show no expression of the gene?
Consider the trp operon found in E. coli. You have discovered a mutant strain of E.coli that contains a mutation in its trpR gene. This mutation prevents the trpR protein from binding to its small effector molecule. What would happen if both the wild type and mutant bacteria were put in an environment that contained lots of tryptophan? The normal (wild type) bacteria would have (i)________expression of the trp operon while the mutant bacteria would have (ii) _________ expression of the trp operon. A) (i) high (ii) high B) (i) low (ii) low C) (i) high (ii) low D) (i) low (ii) high