Concept explainers
a.
To determine:
The reason behind the appearance of white color of luxR/lacZ colonies in the absence of autoinducer.
Introduction:
In an experiment to understand the molecular mechanism of quorum sensing, the use of two transcriptional fusion reporter genes was involved each having the 9 kb fragment of Vibrio fischeri DNA.
b.
To determine:
The reason behind the appearance of white color of luxICDABE/lacZ colonies in the absence of autoinducer.
Introduction:
In a reporter known as luxICDABE/lacZ, the luxICDABE operon regulatory sequences drive lacZ expression. It means that the structural genes of the operon are replaced by the lacZ coding sequences.
c.
To determine:
The reason behind the appearance of white color of luxR/lacZ colonies in the presence of autoinducer.
Introduction:
In the reporter known as luxR/lacZ, the luxR regulatory region drives the transcription of lacZ. It means that the luxR coding sequences are replaced by the lacZ.
d.
To determine:
The reason behind the appearance of blue colored luxICDABE/lacZ colonies in the presence of autoinducer and the time-dependency of the reaction.
Introduction:
The E.coli that contains either of the reporter (luxR/lacZ or luxICDABE/lacZ) has white colonies. When the purified autoinducer is added to the media, the luxR/lacZ colonies remain white but the luxICDABE/lacZ colonies turn blue over time.
e.
To determine:
The inference that can be made out about the transcription of the luxR gene from the given results.
Introduction:
The regulatory region of luxR drives the transcription of the lacZ gene. It indicates that the coding sequence of luxR is replaced by the coding sequence of lacZ.
Want to see the full answer?
Check out a sample textbook solutionChapter 16 Solutions
Genetics: From Genes to Genomes
- Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.arrow_forwardMutations in bacterial promoters may increase or decrease the rate of gene transcription. Promoter mutations that increase the transcription rate are termed up-promoter mutations, and those that decrease the transcription rate are termed down-promoter mutations. As shown , the sequence of the −10 site of the promoter for the lac operon is TATGTT. Would you expect the following mutations to be up-promoter or down-promoter mutations? A. TATGTT to TATATT B. TATGTT to TTTGTT C. TATGTT to TATGATarrow_forwardDebes et al recently described how aging in humans, mice, nematodes, and other eukaryotes is associated with an increase in the average speed of transcriptional elongation by RNA polymerase II. Overexpression of some proteins that decreased PolII elongation speed extended lifespan in the fly Drosophila. For each of the following proteins, predict whether overexpression of that protein (assuming all other cellular components are normal) would likely reduce transcriptional speed, and briefly justify your answer. a) Mediator proteinsb) Histone proteinsc) Insulator binding proteinsarrow_forward
- A bacterial cell with trp and lac operons in a medium with high concentration of lactose, tryptophan, and glucose. Yes or no to the following :- is attenuator sequence inhibit transcription of trp? yes/no Will LacI inhibit transcription of lac? yes/no Is cAMP bound to CAP? yes/no Is TrpR active? yes/no Is CAP bound to the CAP site? yes/no If less glucose was present, would lac expression change? yes/no If less glucose was present, would trp expression change? yes/no Please answer in short and ASAP, please answer all parts yes or no thanksarrow_forwardMany eukaryotic promoter regions contain CAAT boxes with consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?arrow_forwardComparing the -10 regions of two E. coli promoters which have identical -35 regions revealed the sequence TATAAT for the first and GATACT for the second one. Why does the first promoter cause a higher transcription rate than the second one? a. The transcription rate from the first promoter will be higher, because RNA polymerase will bind TATAAT with a higher affinity than GATACT. b. It will be higher, because formation of the open promoter complex is more easily achieved with TATAAT than with GATACT. c. It will be higher, because TATAAT of the -10 region is transcribed into UAUAAU, which forms fewer hydrogen bonds with the template strand than GAUACU. d. a and b, but not c e. a, b, and carrow_forward
- How long would it take for the E. coli RNA polymerase to synthesize the primary transcript for the E. coli genes encoding the enzymes for lactose metabolism, the 5,300 bp5,300 bp lac operon? Assume an average elongation rate of 7070 nucleotides per second. a)How far along the DNA would the transcription "bubble" formed by RNA polymerase move in 10 seconds10 seconds? b)Assuming that human Pol II transcribes at a similar rate, how long does it take to transcribe the 2,000,000 bp2,000,000 bp dystrophin gene?arrow_forwardAccording to the scenario shown, how many segments of DNA (one, two, or three) are removed during site-specific recombination within the gene that encodes the κ (kappa) light chain for IgG proteins? How many segments are spliced out of the pre-mRNA?arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forward
- The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forwardThe genes that encode the enzymes for arginine biosynthesis are located at several positions around the genome of E. coli, and they are regulated coordinately by a transcription regulator encoded by the ArgR gene. The activity of the Argr protein is modulated by arginine. upon binding arginine, Argr alters its conformation, dramatically changing its affinity for the DNA sequences in the promoters of the genes for the arginine biosynthetic enzymes. given that ArgR is a repressor protein, would you expect that ArgR would bind more tightly or less tightly to the DNA sequences when arginine is abundant? if ArgR functioned instead as an activator protein, would you expect the binding of arginine to increase or to decrease its affinity for its regulatory DNA sequences? explain your answers.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning