Prescott's Microbiology
10th Edition
ISBN: 9781259281594
Author: Joanne Willey, Linda Sherwood Adjunt Professor Lecturer, Christopher J. Woolverton Professor
Publisher: McGraw-Hill Education
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Chapter 15, Problem 5CHI
Summary Introduction
The gene expression in yeast offers many advantages in producing proteins as compared to other eukaryotic organisms. There are many expression systems in the severalgenera ofyeasts. For example, Saccharomyces, Pichia and Yarrowia. Expression vector is used in this process. In most of the cases, the vectors that integrate with the host chromosome is widely used.
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Searching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4. GAL4 activates the transcription of galactose genes. Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism. The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T.
In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added. These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to. The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced. Results showed that only 10 different DNA sequences had GAL4 bound. Since the…
Explain how the following mutations would affect transcription of the yeast GAL1 gene in the presence of galactose. (a) A deletion within the GAL4 gene that removes the region encoding amino acids 1 to 100. (b) A deletion of the entire GAL3 gene. (c) A mutation within the GAL80 gene that blocks the ability of Gal80 protein to interact with Gal3p. (d) A deletion of one of the four UASG elements upstream from the GAL1 gene. (e) A point mutation in the GAL1 core promoter that alters the sequence of the TATA box.
Debes 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 proteins
Chapter 15 Solutions
Prescott's Microbiology
Ch. 15.2 - Retrieve, Infer, Apply What is the advantage of...Ch. 15.2 - Prob. 2RIACh. 15.2 - Prob. 3RIACh. 15.2 - Prob. 4RIACh. 15.2 - Prob. 5RIACh. 15.3 - What functions are served by the 5 cap and the 3...Ch. 15.3 - Retrieve, Infer, Apply What elements in archaeal...Ch. 15.3 - Prob. 2RIACh. 15.3 - Prob. 3RIACh. 15.4 - Prob. 1RIA
Ch. 15.4 - Prob. 2RIACh. 15.4 - Prob. 3RIACh. 15.4 - Prob. 4RIACh. 15.5 - Retrieve, Infer, Apply List two similarities and...Ch. 15.5 - Prob. 2RIACh. 15.5 - Retrieve, Infer, Apply How are cis-encoded RNAs,...Ch. 15 - Prob. 1CHICh. 15 - All of the subunits in bacterial RNA polymerases...Ch. 15 - Would you expect that one day microbiologists...Ch. 15 - In the chapter opening story, it was stated that...Ch. 15 - Prob. 5CHI
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- The MAT locus allows yeast to switch mating type through a very complex mechanism. However, it has informed us a great deal about what aspects of gene expression typical to all organisms? Options: higher order changes in chromatin affect transcriptional efficiency that general transcription factors must first bind directly to histone tails and only then can they interact with their cognate binding sites that DNA methylation is involved in this silencing mechanism that SIR2 is required for all types of transcriptional repression that expression of Pol III genes provides a means of identifying active chromatinarrow_forwardRefer to figure, which shows the distribution of histone H2A.Z on nucleosomes near a transcription start site. What experimental technique would have been used to generate the data for this figure? Briefly describe the operation of this technique.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_forwardBased on Figure 9-19, can you predict the position of amutation that would affect the synthesis of one isoformbut not the other?arrow_forward
- There are similarities and differences during regulation of gene expression in both prokaryotes and eukaryotes. Promoters, transcription factors and RNA polymerase are essential elements in transcription but their properties and function may differ.a) Predict the outcome or consequences of mRNA transcription by RNA polymerase II in eukaryote without the presence of transcription factors (TF).arrow_forwardA newly identified protein from the cells of the Panopyra plant on Pandora was shown to inhibit translation of its target genes by binding to the 5’ UTR of the mRNA and preventing ribosome binding. A possible way this inhibition may be relieved by an sRNA would be: Group of answer choices 1.The sRNA acts as a marker, flagging the inhibitory protein for ubiquitination and allowing translation to take place. 2.The sRNA acts as a decoy, sequestering the inhibitory protein and allowing translation to take place. 3.The sRNA acts as an enhancer, suppressing the inhibitory protein and allowing translation to take place. 4.The sRNA acts as a silencer, suppressing the inhibitory protein and allowing translation to take place. Which one is the correct?arrow_forwardGenes in both prokaryotes and eukaryotes are regulated by activators and repressors.a. Compare and contrast the mechanism of functionof a prokaryotic repressor (for example, Lac repressor) with a typical eukaryotic repressor protein(a direct repressor).b. Compare and contrast the mechanism of functionof a prokaryotic activator (for example, CAP) witha typical eukaryotic activator protein.arrow_forward
- Microbiologists describe the processes of transcription and translation as “coupled” in bacteria. This term indicates that bacterial mRNA can be undergoing transcription at the same moment it is also undergoing translation. How is coupling possible in bacteria? Is coupling of transcription and translation possible in single-celled eukaryotes, such as yeast? Why or why not?arrow_forwardAn electrophoretic mobility shift assay can be used to study the binding of proteins to a segment of DNA. In the results shown here, an EMSA was used to examine the requirements for the binding of RNA polymerase |l (from eukaryotic cells) to the promoter of a protein-encoding gene. The assembly of general transcription factors and RNA polymerase Il at the core promoter is described in Week 4. In this experiment, the segment of DNA containing a promoter sequence was 1100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to an EMSA. Lane 1: No proteins added Lane 2: TFIID Lane 3: TFIIB Lane 4: RNA polymerase IIl Lane 5: TFIID + TFIIB Lane 6: TFIID + RNA 1 2 3 4 5 6. 7 polymerase II Lane 7: TFIID + TFIIB + RNA polymerase Il 1100 bp Explain the results.arrow_forwardone promoter can regulate the expression of different polypeptides (proteins) in both prokaryotes and eukaryotes but in different manners. explain how this regulation (to allow the expression of different polypepetides from only one promoter) differs between prokaryotes and eukaryotes?arrow_forward
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