The diagram below represents a hypothetical operon in the bacterium E. coli. The operon consists of two structural genes (A and B), which code for the enzymes "Aase" and "Base", respectively, and also includes P (promoter) and O (operator) regions as shown. A В When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes "Aase" and "Base" are both synthesized at a 50-fold higher rate than in the absence of X. (X has a molecular weight of about 200.) Which of the following statements is true of the operon described above? The region of the A gene that codes for the carboxyl-terminal amino acid of “Aase" is near the left end of the A gene. The P region contains nucleotide sequences to which the RNA polymerase holoenzyme (including the o subunit) binds specifically but which the core enzyme does not recognize. The addition of X to the growth medium causes a repressor protein to bind tightly to the O region. The mRNA copied from this operon will be covalently linked to a short piece of DNA at the 5' end. The mRNA from this operon will contain a single "start" codon in a position corresponding to the right end of gene B.

The diagram below represents a hypothetical operon in the bacterium E. coli. The operon consists of two structural genes (A and B), which code for the enzymes "Aase" and "Base", respectively, and also includes P (promoter) and O (operator) regions as shown. A В When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes "Aase" and "Base" are both synthesized at a 50-fold higher rate than in the absence of X. (X has a molecular weight of about 200.) Which of the following statements is true of the operon described above? The region of the A gene that codes for the carboxyl-terminal amino acid of “Aase" is near the left end of the A gene. The P region contains nucleotide sequences to which the RNA polymerase holoenzyme (including the o subunit) binds specifically but which the core enzyme does not recognize. The addition of X to the growth medium causes a repressor protein to bind tightly to the O region. The mRNA copied from this operon will be covalently linked to a short piece of DNA at the 5' end. The mRNA from this operon will contain a single "start" codon in a position corresponding to the right end of gene B.

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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In addition to observing similarities to the lac operon, you also notice that this gene is regulated via attenuation, similar to the trp operon. Based on this similarity to this model operon, you could state that ___________. Group of answer choices If a terminator loop forms in the DNA, the expression of the structural genes is halted. The formation of the terminator hairpin followed by a series of Uracil (UUUUUUU) functions similar to Rho-Independent termination to result in the stopping of transcription. The transcription of a leader sequence affects the translation of the structural genes. Never mind – all of these statements are true! Attenuation will be the primary means of transcriptional regulation, with a repressor used as a back up option.
Two closely linked open reading frames (ORFs) have been identified on the E. coli chromosome and they are predicted to encode genes sugX sugY which is involved in the catabolism (utilization) of a non-glucose sugar. In the figure below, the top line (with the big arrow) is the coding strand, and the bottom line is the non-coding (template) strand of the operon. Since the RNA sequence would be the “same” as the coding strand, please label (show the location of), in relation to the ORFs, the promoter(s) and 4 other genetic elements in this region that may be involved in the transcription and translation of sugX and sugY as well as the regulation of their expression. (Note: genetic elements are the regions of DNA or RNA.) If you are not comfortable with the coding strand of the DNA, feel free to draw the mRNA strand underneath and label the relevant features on the RNA strand, but the RNA strand must be aligned with the DNA strands in position and scale. Leader(s)/antileader(s) and…
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?
Let’s suppose you have isolated a mutant strain of E. coli in which the lac operon is constitutively expressed. In other words, the operon is turned on in the presence or absence of lactose. One possibility is that the mutation mayblock the transcription of the lacI gene, thereby preventing the synthesis of lac repressor. A second possibility is that the mutation could alter the sequence of the lac operator in a way that prevents lac repressor from binding to the operator. How would you distinguish between these two possibilities?
The symbols a, b, and c in the table below represent theE. coli lac system genes for the repressor (I ), the operator(O) region, and the β-galactosidase (Z), although notnecessarily in that order. Furthermore, the order inwhich the symbols are written in the genotypes is notnecessarily the actual sequence in the lac operon.Activity (+) or inactivity (−) of Z geneGenotype Inducer absent Inducer presenta- b+ c+ + +a+ b+ c- + +a+ b- c- - a+ b- c+/a- b+ c- + +a+ b+ c+/a- b- c- - +a+ b+ c-/a- b- c+ - +a- b+ c+/a+ b- c- + +a. Which symbol (a, b, or c) represents each of the lacgenes I, O, and Z ?b. In the table, a superscript minus sign on a gene symbolmerely indicates a mutant, but some mutant behaviorsin this system are given special mutant designations.Using the conventional gene symbols for the lac operon,designate each genotype in the table.
There is Hyaluronic acid synthesis occuring in Group X Strep and it is controlled by an operon with 3 genes, called hasXYZ. Based on the 3-line diagram model, a. How many ribosome binding sites are there for the protein? b. How many promoters are there for the genes? c. How many start codons are there for the protein? d. How many RNA Polymerase binding locations are there for the genes? e. How many proteins will be fully functional? f. How many mRNA strands are made?
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
A strain of E. coli has the genotypes shown below at the lac operon, where I = regulator gene, P = promoter, O = operator, Z = βgalactosidase gene, and Y = permease gene. The superscript+ indicates a wild-type allele,c indicates a constitutive mutation, and− indicates a defective mutation. For each genotype, indicate whether the enzyme will be synthesized or not synthesized when lactose is present or absent by placing a + for synthesis occurring and a − for synthesis not occurring in the appropriate blank. Explain your reasoning for each answer.
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…
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−
Bacterial DNA containing an operon encoding three enzymes is introduced into chromosomal DNA in yeast (a eukaryote) in such a way that it is properly flanked by a promoter and a transcriptional terminator. The bacterial DNA is transcribed and the RNA correctly processed, but only the protein nearest the promoter is produced. Can you suggest why?