The figure below shows 6 genes that are differentially expressed based on their specific enhancers. The different activators (A-F) are proteins that can bind to the appropriate control element in the enhancers of these genes. Use the diagram to answer the following question: What gene(s) would be transcribed if you A, C, E and F enhancers had activators bound to them? No need to explain your answer – just provide the gene(s). ABCD Promoter Brain specific gene ABC FPromoter Liver specific gene A C E Promoter Heart specific gene A Skin specific gene Promoter AB D Promoter Gut specific gene EF Promoter Nose specific gene

The figure below shows 6 genes that are differentially expressed based on their specific enhancers. The different activators (A-F) are proteins that can bind to the appropriate control element in the enhancers of these genes. Use the diagram to answer the following question: What gene(s) would be transcribed if you A, C, E and F enhancers had activators bound to them? No need to explain your answer – just provide the gene(s). ABCD Promoter Brain specific gene ABC FPromoter Liver specific gene A C E Promoter Heart specific gene A Skin specific gene Promoter AB D Promoter Gut specific gene EF Promoter Nose specific gene

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Description: The figure below shows 6 genes that are differentially expressed based on theirspecific enhancers. The different activators (A-F) are proteins that can bind to theappropriate control element in the enhancers of these genes. Use the diagram toanswer

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 21QP: Transcriptional regulators are proteins that bind to promoters (the 5-flanking regions of genes) to...

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MyoD is a transcriptional activator that turns on theexpression of several muscle-specific genes in humancells. The Id gene product inhibits MyoD action.a. One possibility is that the Id protein directly represses the expression of these muscle-specificgenes. Explain how Id would function if it were arepressor.b. Another possibility is that Id inhibits musclespecific gene transcription indirectly, by preventingMyoD function. Explain how Id could function asan indirect repressor.c. Suppose you know the amino acid sequence ofthe Id protein. How might this information supportthe hypothesis in part (a) or in part (b)?
a. Would you expect a cell to continue or to stopdividing at a nonpermissive high temperature if itis a temperature-sensitive Ras mutant whose protein product is fixed in the GTP-bound form atnonpermissive temperature?b. What would you expect if you had a temperaturesensitive mutant in which the Ras protein staysin the GDP-bound form at high temperature?
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
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…
Discuss the following argument: “if the expression of every gene depends on a set of transcription regulators, then the expression of these regulators must also depend on the expression of other regulators, and their expression must depend on the expression of still other regulators, and so on. cells would therefore need an infinite number of genes, most of which would code for transcription regulators.” how does the cell get by without having to achieve the impossible?
Wilms tumor 1, or nephroblastoma, is caused by mutations in the WT1 gene, which encodes a transcription factor. You have identified a novel variant in WT1: Arg422Pro. You have control cells and cells that have been engineered to carry the homozygous WT1 p.Arg422Pro mutation. You want to assess effects of this mutation on a variety of endpoints. For each endpoint listed below, choose the one technique is best suited to answer the question. Choose from: array CGH, qRT-PCR, qPCR, RNA-seq, FISH, in situ hybridization, western blot, immunostaining, WT1 ChIP-seq, WT1 ChIP-PCR, ATAC-seq, 3C Endpoint Technique? WT1 protein amount (quantitative) Western blot WT1 protein binding to all enhancers, genome-wide Chip-seq WT1 mRNA amount (quantitative) WT1 protein subcellular localization Quantitative assessment of all mRNAs in these cells (genome-wide) RNAseq Chromatin interactions between a specific WT1 chromatin binding site (identified above)…
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. An interesting mutation in lacI results in repressorswith 110-fold increased binding to both operator andnonoperator DNA. These repressors display a “reverse”induction curve, allowing β-galactosidase synthesis inthe absence of an inducer (IPTG) but partly repressingβ-galactosidase expression in the presence of IPTG. Howcan you explain this? (Note that, when IPTG binds a repressor, it does not completely destroy operator affinity,but rather it reduces affinity 110-fold. Additionally, ascells divide and new operators are generated by thesynthesis of daughter strands, the repressor must findthe new operators by searching along the DNA, rapidlybinding to nonoperator sequences and dissociating fromthem.)
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ennar region of gene X, which determines the length of the tail in mice, is mutated so that transcription factors bind it at a much higher affinity compared to the wild-type sequence. What is the most likely phenotypic outcome? Tail length will not change because the enhancer is a non-coding sequence Tail length will increased due to increased activity of the gene's promoter Tail length will decreased because any mutation will cause a loss-of-function of these regulatory regions Not just the tail will be enlarged because increased activity of the enhancer will impact many genes
. One mechanism by which antisense RNAs act as negative regulators of gene expression is by base pairingwith the ribosome binding site on the sense mRNA toblock translation. In a second, alternative mechanism,the act of transcribing an antisense RNA can somehow prevent RNA polymerase from recognizing thesense promoter for the same gene. Design an experimental approach that would enable you to distinguishbetween these two modes of action at a specific gene.(Hint: What would be the outcome in each case ifhigh levels of the antisense RNA were transcribedfrom a gene on a plasmid?)
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