Barrier insulators stop the spread of heterochromatin HР-1 HMȚase HAT FIGURE 12-26 In this model, barier insulators recruit enzymatic activities such as histone acetyltransferase (HAT) that promote euchromatin formation. The letter "M" stands for methylation, and the letter "A" for acetylation. Heterochromatin Euchromatin Barrier insulator

Barrier insulators stop the spread of heterochromatin HР-1 HMȚase HAT FIGURE 12-26 In this model, barier insulators recruit enzymatic activities such as histone acetyltransferase (HAT) that promote euchromatin formation. The letter "M" stands for methylation, and the letter "A" for acetylation. Heterochromatin Euchromatin Barrier insulator

Name: In Figure 12-26, provide a biochemical mechanism forwhy HP-1 can bind
Uploaded: 2024-03-20T06:42:41.000Z
Channel: Bartleby
Description: In Figure 12-26, provide a biochemical mechanism forwhy HP-1 can bind to the DNA only on the left side of thebarrier insulator. Similarly, why can HMTase bind onlyto the DNA on the left of the barrier insulator? Barrier insulators stop the spread of

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 7P

See similar textbooks

Similar questions

Compared to heterochromatin, euchromatin is comprised of densely packed nucleosomes and with higher transcription activity comprised of densely packed nucleosomes and with lower transcription activity O comprised of loosely packed nucleosomes and with lower transcription activity Ocomprised of loosely packed nucleosomes and with higher transcription activity
IPSCs are nearly identical to human embryonic stem cells in terms of gene expression, but there may be other ways in which they are not equivalent. For example, the telomeres of IPSCs often vary in length, with many IPSCs cells having telomeres shorter than those of embryonic. How might shortened telomeres affect the life-span of IPSCs or of differentiated cells derived from them?
Chromatin remodeling by the SWI /SNF complex requires hydrolysis of ATP. What purpose does this serve?
Why is transcriptionally active chromatin ∼10 times more susceptible to cleavage by DNase I than transcriptionally silent chromatin?
I. The retinoic acid receptor (RAR) is a transcription factor that is similar to steroid hormone receptors. Thesubstance (ligand) that binds to this receptor is retinoicacid. One of the genes whose transcription is activatedby retinoic acid binding to the receptor is myoD. Thediagram that follows shows a schematic view of theRAR proteins produced by genes into which one oftwo different 12-base double-stranded oligonucleotides had been inserted in the ORF. The insertion site(a–m) associated with each mutant protein is indicatedwith the appropriate letter on the polypeptide map.For constructs encoding proteins a–e, oligonucleotide 1(5′ TTAATTAATTAA 3′ read off either strand) wasinserted into the RAR gene. For constructs encoding proteins f–m, oligonucleotide 2 (5′ CCGGCCGGCCGG 3′)was inserted into the gene.NH2 f g h i j k l m COOHa b c d eThe wild-type RAR protein can both bind DNA and activate transcription weakly in the absence of retinoic acid(RA) and strongly in RA’s presence. Each…
Mating Ste12 Tec1 Budding 17. (8 points) Scientists are studying the transcription factors Ste12 and Tec1. They have found that when ste12 binds to one part of the genome alone it activates the transcription of genes important in mating pheromones. They also found that when Tec1 binds to another part of the genome alone it activates the transcription of cellular budding genes. Lastly, they found that in another part of the genome, Ste12 and Tec1 both bind to the DNA, they activate the transcription of genes that control the formation of filaments. a) These proteins bind to enhancers. Is Ste12 and Tec1 activators or mediators? How do you know? Filamentation b) Is this system an example of combinatorial control or coordinated control, both or neither? Explain your answer. sold ad bosow
in 3 5% Shown below is a schematic diagram showing a 3000 bp region of yeast genomic DNA. (Note: Transcription starts at Transcription Start Site (TSS).) TSS +1 (i) Draw the location in this diagram and name the specific regions that can be recognized by Transcription Factor IIB and Transcription Factor IID. (ii) How does Transcription Factor IIH trigger the initiation of transcription? 3-
Search about transcription elongation factor P-TEFb with its relationship with CDK9 and RNA Pol II, what is the role of this particular transcription elongation factor compared with other general transcription factors and explain the mechanism on how eukaryotic transcription is controlled by the roles of this transcription factor (be specific on reaction or modification on RNA Pol II, which domain as you learn from the research article) and integrate the concept of chromatin remodeling in terms of packaging of DNA, and accessibility of the template for the RNA polymerase https://advances.sciencemag.org/content/6/50/eabc1450/tab-pdf
Briefly describe three ways that ATP-dependent chromatin-remodeling complexes may change chromatin structure.
1. enzymes that catalyze histone acetlation are closesly associated with transription factors, which are proteins that promote transcription. why is this a good biochemical strategy? 2. sp1 is a sqeuence specific human DNA binding protein that binds to a region on the DNA called the GC box, a promoter element with the sequence GGGCGG. Binding of Sp1 to the GC box enhances RNA polymerase 2 activity 50- to a 100 fold. How would you use affinity chromatography to purify SP1?
Cytosine deaminationoccurs ~100 cytosinesper genome per day in a human genome. Eukaryotic cells also contain residues of 5’-methylcytosine, which is involved in regulating gene transcription rates. Mutation of 5’-methylcytosine by deaminationconverts it to thymine. This presents the cell with a much more severe problem than normal cytosine deaminationof cytosine to uracil. Why?
5' 3' ORF for gene X The sequence for gene X above has a hypomorphic mutation (which you name Mutation #1) that is not indicated on the diagram. Based on the experimental data below, what specific sequence element is the mutation most likely in? Wild Type Northern Gene X O -10 region O 3' of the ORF ribosome binding site O -35 region Mutant O within the ORF Wild Type Western Gene X direction of transcription Mutant 3' +1 5'