PSwk8key - Problem Set wk 8 MCB 110, Spring ‘11...

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Unformatted text preview: Problem Set wk 8 MCB 110, Spring ‘11 1. True or False __T__ High sequence conservation evolutionarily is generally an indication of functional significance __T __ Insulator proteins establish boundaries around a promoter and its enhancers so that other promoter/enhancer elements are not impacted __F__ Transcription factors with leucine zipper motifs can only form homodimers __T__ Histone tails are important for formation of tightly wrapped DNA around the histone octamer __ F__ Acetlyation of histone N ­terminal tails always leads to gene repression __ F__ Methylation of histone tails always results in gene activation __ F__ All the nucleosomes at active genes are evicted by remodeler complexes both at the promoter and along the gene sequence so that polymerase can travel along the DNA unhindered It’s false because chromatin remodelers are only shifting the nucleosomes, Pol II can still be paused even after promoter clearance. __ F__ Euchromatin is a condensed structural state of chromatin __ T __ ATP is required to slide a nucleosome __ F__ HDAC and HAT proteins perform the same functions __ T __ NELF and DSIF need to be phosphorylated for Pol II elongation __ F__ TAT is one of the host proteins that HIV virus uses 2. Chromatin immunoprecipitation is a valuable tool to determine which sequence or region of DNA a particular DNA ­binding protein binds. However it requires some prior knowledge about the sequence, either from DNaseI footpringting or gel shift. What kind of technique could you use that requires no prior knowledge and also gives you a comprehensive and unbiased look at which sequences are bound? What would your experimental (positive and negative) controls be? ChIP ­seq or ChIP ­chip: both can give you information on where a protein binds at a whole genome level. Positive controls would be Input (you save some of your total chromatin before IP) and the negative control would be no antibody (we didn’t talk about nonspecific IgG as controls in class but that’s the negative controls people usually use). These are explained more in #5 3. You’ve identified (cloned) a new protein that is localized in the nucleus. a) How can you determine if it’s a transcription factor by looking at its sequence? b) You did a ChIP ­seq with an antibody against this protein and another one with an antibody against H3 lysine 27 methylation. The sequences you obtained from the two samples overlap nearly perfectly, what does this say about your transcription factor? a) By looking in its peptide sequence whether it has a structure motif that’s a conserved DNA ­binding domain, i.e. zinc finger? Remember bHLH and b ­Zip are usually dimerization as well as DNA binding domains, the part that actually contacts DNA is usually rich in lysine and arginine b) It’s probably a transcription repressor since your protein has binding sites that colocalize with H3K27me3, which is a histone mark that indicates inactive genes. 4. Your lab is studying a human oncoprotein, X, that is a component of the mammalian transcription factor AP1 and involved in activation of proliferation genes. You know the identity and the sequence of the protein with the following amino acid sequences: mtakmettfy ddalnasflp sesgpygysn pkilkqsmtl nladpvgslk phlraknsdl ltspdmgllk laspelerli iqssnghitt tptptqflcl knvtdeqegf aegfvralae lhsqntlpsv tsaaqpvnga gmvapavasv aggsgsggfs aslhseppvy anlsnfnpga lssgggapsy gaaglafpaq pqqqqqpphh lpqqmpvqhp rlqalkeepq tvpempgetp plspidmeSQ ERIKAERKRM RNRIAASKCR KRKleriarl eekvktlkaq nselastanm lreqvaqlkq kvmnhvnsgc qlmltqqlqt f You make three deletion mutants: 1 ­the blue domain is deleted; 2 ­ the purple region is deleted, and 3 ­the green region is deleted. To test the transcriptional effects of these mutants, you did an in vitro transcription assay and found that the wildtype protein was the most active with all three mutants being non ­active. Then you did a gel shift assay and got the following results: experiments. First you did an in vitro transcription assay using a wild-type promoter with the AP1 site and all the reconstituted general transcription factors. Then you did a gel shift assay using the wild-type promoter with the AP1 binding site, and last you did a size exclusion chromatography experiment to test the difference between mutant 2 and 3. Here is your data: a) what type of transcription factor is this? (think back on the three different motifs we’ve learned Leucine zipper (it’s actually c ­Jun), look at the green domain: it has a leucine residue at every 7th position. a) What type of transcription factor is this, and what purpose do each of these domains that you deleted have, how do you know? b) What is the purpose/function of each of the three domains that you deleted in your mutants? How do you know? Blue = activation domain because it can bind DNA just fine but this mutant can no longer activate transcription Green = dimerization domain, i.e. leucine zipper Purple = DNA binding since it contains many R and K residues c) Would you hypothesize that this transcription factor works as a monomer, a dimer, a trimer, or a tetramer? Why? What type of DNA sequence motifs would you expect its binding site to have? Dimer, inverted repeats 5. While many transcription factors function by affecting initiation of transcription, some transcription regulators function by affecting elongation, partially supported by ChIP ­seq where you see Pol II localized to genes that are actually untranscribed. a) Without doing an RNA transcript analysis, how can you use ChIP to determine which genes have Pol II sitting at the promoter and which ones have Pol II actively transcribing. What features of the polymerase would you exploit for this purpose? You can perform ChIP ­seq with antibodies against different serine phosphorylation in the Pol II CTD. If you see some genes have enrichment primarily near the promoter region in your Ser5 ChIP sample, then those genes might not be undergoing active transcription (remember ChIP is done on many cells, so it’s an average of what’s happening at a promoter over up to millions of cells, presumably receiving the same transcription signals). If however, you see some genes enriched at the both promoter and in the actual gene for both Ser 5 and Ser 2 ChIP samples, then they might be actively transcribed by Pol II. b) What should you include as negative and positive controls? Input = positive control. It is a fraction of the total DNA you collected before pulling down (or enriching for) the DNA fragments that your protein of interest binds. No antibody = negative control. In this tube, any DNA that will nonspecifically interacts with the beads will also be enriched, so by comparing DNA in this tube to your antibody ­IP tube, you can ignore any nonspecific DNA fragments. 6. What is the TAR RNA? What is it transcribed from and how does it stimulate transcription of the HIV gene? It is the first ~70nt of HIV mRNA transcribed from the integrated proviral DNA. It is generated before host Pol II is paused, together with Tat protein, it can recruit P ­ TEFb and thus stimulate transcription elongation 7. List at least two things that P ­TEFb phosphorylates Ser2 on Pol II CTD and NELF, the negative elongation factor 8. What are the three different ways that nucleosomes or effects on nucleosomes can regulate transcription? Acetylation: neutralize the positive charge on the histone tail lysines so DNA can’t wrap as tightly Methylation: on specific residues can be recognized by different domains on transcription co ­activators, i.e. PHD fingers recognizes H3K4me3 activating mark Nucleosomes can cover the TATA box or enhancer regions, so they need to be shifted by ATPases in chromatin remodeling complexes for gene activation 9. What are the advantages for many transcription factors to form heterodimers? Heterodimer formation generates diversity in binding sites, so different combination of transcription factors that dimerize can bind different sequences and thus activate different genes 10. The HIV protein Tat is required for HIV propagation. What is the function of Tat? Tat activates transcription elongation of HIV DNA. Tat bound to HIV TAR RNA binds and recruits pTEFb kinase to the viral promoter. Recruitment of pTEFb then leads to pTEFb phosphorylation of RNAP II, NELF, and DSIF. Phosphorylation of NELF leads to NELF dissociation from Pol II and converts DSIF to a positive factor for transcription elongation. 11. Cyclic AMP serves as a second messenger for intracellular signaling. Describe how cAMP can lead to transcription activation. Start your description from binding of a hormone to a receptor. Hormone binding to a G protein coupled receptor leads to active adenylate cyclase and increased production of cAMP. cAMP binds to the regulatory subunits of Protein Kinase A (PKA) which results in release of the catalytic subunits of PKA from the regulatory subunits. The catalytic subunits of PKA then enter the nucleus where they phosphorylate CREB. The phospho ­CREB can then bind to its CRE site (CREB response element) 12. Regulators are regulated in various different manners. List all of the mechanisms you’ve learned from class. Protein synthesis Ligand binding activates your regulator Protein phosphorylation Addition of a subunit Unmasking protein via dissociation of an inhibitor Dissociation of an inhibitor protein which unshields the NLS (nuclear localization sequence) thus allowing nuclear import Cleavage of the intracellular domain of a membrane bound protein leading to free intracellular domain in the cytosol 13. What mechanism of regulation is used to regulate NF ­kB transcription factor? Dissociation of an inhibitory subunit which unshields the NLS of NF ­kB allowing nuclear import of NF ­kB into the nucleus. 14. What feature of Nuclear Receptor ligands is key for their function? Nuclear Receptor ligands are lipid ­soluble and can therefore cross the plasma membrane. 15. Imagine you are working with a Nuclear Receptor protein called Rambo. Your PI gives you three different candidate ligands for this Nuclear Receptor. Describe an experiment that would allow you to determine which of these ligands is the real ligand for Rambo. Explain what the data would show for the real ligand and for the other ligands. Use immunofluorescence to follow localization of Rambo in the presence or absence of your ligand. For this experiment you would use an antibody against Rambo that is chemically linked to a fluorescent dye. Data for the real ligand: in the absence of ligand Rambo would be cytosolic. In the presence of ligand Rambo would be localized in the nucleus. Data for the other ligands: Rambo would stay cytosolic in the presence or absence of ligand. ...
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This document was uploaded on 09/12/2011.

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