What is the advantage of using transcriptional reporters to study the activity of a gene's promoter?No answer text provided.It allows you to easily measure the approximate, relative transcriptional activity of a single promoter of interest inthe cell in different conditions or genetic strain backgrounds.It allows you to measure exactly how many MRNA transcripts are being produced from a given promoter everysecond.It allows you to measure the activity of a promoter when only RNA Polymerase is present - transcriptionalreporters are not affected by all the other transcription factors that are present in the cell. LysR regulators like llvY repress transcription from their own promoters. This is called a negativefeedback loop. Why might negative feedback be a good way to control expression of a transcriptionalrepressor?This type of negative feedback loop ensures that ilvC transcription will always be repressed when llvY levels arelow, and ilvC will be transcribed highly when llvY levels are high.This type of negative feedback regulation ensures that ilvY is never transcribed. Since llvY is a transcriptionalrepressor, it will slow down cell growth by blocking transcription. So, llvY represses it's own transcription so thatno llvY can ever be made, and the cell can grow.This type of negative feedback loop ensures that there is always a moderate amount of llvY. When IlvY levels falltoo low, transcription from the promoter will increase since IlvY won't be able to repress it's own promoter. Thisnegative feedback also ensures that IlVY levels don't rise too high - when llvY levels are high, then transcriptionfrom the ilvY promoter is tightly repressed by all the IlvY protein, so no more is made.This type of negative feedback allows levels of IlvY to increase to very high levels when they are needed. IlvYrepresses transcription from its own promoter, so llvY levels will increase exponentially once a threshold level ofIlvY protein is made.

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Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

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Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Phorbol esters have been observed to induce the transcription of AP-1–influenced genes. Explain how this processcould occur. What are the consequences of AP-1 transcription? What role does intermittent exposure to phorbol estershave on an individual’s health?
If there were less substrate than enzyme, then Vmax could never be reached. True or False? Enzymes increase the rate of the forward and reverse reaction proportionally. True or False? The AG of a reaction is related by the Arrhenius equation to the rate constant of a reaction. True or False? Glycogen phosphorylase is phosphorylated and activated by the hormone glucagon. This is an example of transcriptional regulation. True or False? At steady state, the rate of formation of [ES] is equal to the rate of breakdown of [ES). True or False? The Lineweaver-Burk transformation linearizes kinetic data regardless of whether the enzyme follows Michaelis-Menten kinetics. True or False?
Now read this abstract from a 2013 journal article What is the authors' explanation of how Gal80p works? Note UASG from the question above is the same as UASGAL The DNA-binding transcriptional activator Gal4 and its regulators Gal80 and Gal3 constitute a galactose-responsive switch for the GAL genes of Saccharomyces cerevisiae. Gal4 binds to GAL gene UASGAL. (upstream activation sequence in GAL gene pro- moter) sites as a dimer via its N-terminal domain and activates transcription via a C-terminal transcription activation domain (AD). In the absence of galactose, a Gal80 dimer binds to a dimer of Gal4, masking the Gal4AD. Galactose triggers Gal3-Gal80 interaction to rapidly initiate Gal4-mediated transcription activation. Just how Gal3 alters Gal80 to relieve Gals0 inhibition of Gal4 has been unknown, but previous analyses of Gal80 mutants suggested a possible competition between Gal3-Gal80 and Gal80 self-association interactions. Here we assayed Gal80-Gal80 interactions and tested for…
Gene expression may change in response to external cues through the activity of multiple signal transduction cascades. What is the primary mechanism that determines if a gene is expressed or not?
You wish to find the cis-acting regulatory DNA elements responsible for the transcriptional responses of two genes, c-fos and globin. Transcription of the c-fos gene is activated in response to fibroblast growth factor (FGF), but it is inhibited by cortisol (Cort). On the other hand, transcription of the globin gene is not affected by either FGF or cortisol, but it is stimulated by the hormone erythropoietin (EP). To find the cis-acting regulatory DNA elements responsible for these transcriptional responses, you use the following clones of the c-fos and globin genes, as well as two “hybrid” combinations (fusion genes), as shown in the diagram below. The letter A represents the intact c-fos gene, D represents the intact globin gene, and B and C represent the c-fos–globin gene fusions. The c-fos and globin exons (E) and introns (I) are numbered. For example, E3(f) is the third exon of the c-fos gene and I2(g) is the second intron of the globin gene. (These labels are provided to help you…
Understanding why these receptors are present when they are not supposed to be is important in finding new treatments for Cushing's syndrome. When the sequences for the promoters for the receptor genes were analyzed there was no common sequences found. This led to the theory that the loss of methylation of cytosine in the promoters of these genes was leading to their inappropriate expression. How could the loss of cytosine methylation lead to overexpression of the receptor genes? O a. Increased histone acetyltransferase activity O b. Decreased histone deacetylase activity Oc Decreased histone acetyltransferase activity O d. Increased histone deacetylase activity
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You are researching the regulation of cell-size control and discover that a signaling pathway that acts through a particular enzyme-coupled receptor is important for the size growth (enlargement) of rat kidney cells. Receptor activation causes activation of adenylyl cyclase, which ultimately leads to the activation of PKA. PKA then phosphorylates a transcription factor called BTS on threonine 84. This phosphorylation is required to allow BTS to bind to specific regulatory DNA sequences, increasing the transcription of Zrt2, a gene encoding a protein vital for kidney cell growth. You find that kidney cells without this receptor are 25% smaller than normal cells. When you measure cells expressing a constitutively active version of PKA, you find they are 25% bigger than normal kidney cells. Based on these results, predict whether a kidney cell’s size would be bigger or smaller than normal kidney cells in each of the following circumstances. Explain why in 1-2 sentences for each. You add…
You are researching the regulation of cell-size control and discover that a signaling pathway that acts through a particular enzyme-coupled receptor is important for the size growth (enlargement) of rat kidney cells. Receptor activation causes activation of adenylyl cyclase, which ultimately leads to the activation of PKA. PKA then phosphorylates a transcription factor called BTS on threonine 84. This phosphorylation is required to allow BTS to bind to specific regulatory DNA sequences, increasing the transcription of Zrt2, a gene encoding a protein vital for kidney cell growth. You find that kidney cells without this receptor are 25% smaller than normal cells. When you measure cells expressing a constitutively active version of PKA, you find they are 25% bigger than normal kidney cells. Based on these results, predict whether a kidney cell’s size would be bigger or smaller than normal kidney cells in each of the following circumstances. Explain why in 1-2 sentences for each. You…
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Researchers have found that a certain signal transduction pathway, illustrated in the figure below, may be responsible for the development of cancer in pancreatic cells. In normal pancreatic cells, where the pathway remains inactive, a membrane-bound receptor called Ptc inhibits a downstream protein known as Smo. The inhibition of Smo blocks the activa- tion of a complex of proteins known as the HSC, which results in the cleaving of one of its proteins called Ci. The cleaving of Ci, in turn, prevents the pathway from initiating the syn- thesis of division-facilitating enzymes. A Ptc B Smo с inhibition D A model of a Ptc signal transduction pathway under normal conditions. HSC Ci cleavage A certain protein called the Hedgehog protein (Hh) is found to activate this signal transduction pathway and trigger uninterruptable cell division. cell division not initiated Which of the following describes the most plausible mechanism by which Hh triggers uninterruptable cell division? Hh binds to…

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