10. You have developed a transcription assay that recapitulates the in vivo regulation of the promoter for the RNA Pol II-transcribed Ratchet gene. By fractionating this extract, you have purified and cloned two specific transcription factors that, in addition to the general RNA Pol II transcription factors, are required for correct regulation of the promoter. As a first step to understand the function of these factors you make a series of deletion mutants of the Ratchet promoter and test their effect on transcription in the presence and absence of the Click and Clack factors. You find the following results. 3 5 7 Transcription +Click -Click -Clack +Clack + ++++ +++ +++ + +++ +++ +++ Based on this data, what would you predict the function of elements 3, 6, and 7 are?

10. You have developed a transcription assay that recapitulates the in vivo regulation of the promoter for the RNA Pol II-transcribed Ratchet gene. By fractionating this extract, you have purified and cloned two specific transcription factors that, in addition to the general RNA Pol II transcription factors, are required for correct regulation of the promoter. As a first step to understand the function of these factors you make a series of deletion mutants of the Ratchet promoter and test their effect on transcription in the presence and absence of the Click and Clack factors. You find the following results. 3 5 7 Transcription +Click -Click -Clack +Clack + ++++ +++ +++ + +++ +++ +++ Based on this data, what would you predict the function of elements 3, 6, and 7 are?

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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a. Very few if any eukaryotic genes contain tractswith more than 25 As or Ts in a row, yet almost alleukaryotic mRNAs have a tract with more than100 As in a row. How is this possible?b. Scientists know the nucleotide sequences that directthe termination of bacterial gene transcription, butthey generally have little idea about the nature ofthe nucleotide sequences that direct transcriptiontermination in eukaryotic cells. Explain the basisof this statement.
Mutations in bacterial promoters may increase or decrease therate of gene transcription. Promoter mutations that increasetranscription are termed up-promoter mutations, and those thatdecrease transcription are termed down-promoter mutations.The sequence of the −10 site of the promoter for the lac operonis TATGTT (see Figure 14.5). Would you expect each of thefollowing mutations to be an up-promoter or down-promotermutation?A. TATGTT to TATATTB. TATGTT to TTTGTTC. TATGTT to TATGAT
. 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?)
How does the 4 feature of transcription factors namely the structural motifs of DNA binding protein, activation domains, multiple transcription factors and enhancers help in the design of a building block tool. U can use the SrY gene as ur building block tool. Pls explain in details using those features of the transcription factors. In 400 words
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A. Identify the mutation(s) that lead to the most loss in transcriptional activity, and discusswhether those match expectations based on the consensus sequence for the Initiator.B. Hypothesize a molecular mechanism to explain how the mutations identified in Part Aresult in loss of transcription activity.C. Identify whether any mutations cause transcription to initiate from a different positionthan wild-type, and provide a brief explanation as to why that occurs for those mutantcore promoters. THIS IS A PRIMER EXTENSION ASSAY
III.ii-Eukaryotic Gene Regulation Esiankiki: My mom told me that in the case of lactase persistence, it turns out that the persistence is caused by continued expression of the lactase RNA. Xiao-Ma: That's so interesting! Here are some more notes from my biology class. I wonder how the lactase RNA can be expressed in some adults but not in others? Activators bind to genes at enhancer sequences and increase transcription. ACTIVATORS ACTIVATOR COACTIVATORS ENHANCER ACTIVATOR ATA BINDING TATA BOX coactivators function as adaptors and integrate signals from activators and repressors and relay the information to the basal transcription factors. Repressors bind to genes at silencer sequences and decrease transcription. CORE PROMOTER- REPRESSOR POLYMERASE BASAL FACTORS CODING REGION Basal transcription fac- tors and RNA polymerase bind to the core promoter of genes and are required for transcription. Figure 5. Transcription in eukaryotes. - The expression of RNA in eukaryotes is initiated by…
A bacterial species has a hypothetical sigma promoter that has the following sequence: TTGGCA - 18 bases - TATAAT What change in the level of transcription would there be if the sequence was mutated to: TTCGCA -18 bases -TATAAT Group of answer choices 1.The mutation would inhibit the promoter thereby inhibiting transcription 2.No change the consensus TATAAT sequence in the same. 3.The mutation would move the promoter away from consensus and reduce the level of transcription 4.The mutation would bind the promoter to the consensus and produce normal levels of transcription
5. As you know, the lac promoter is a poor intrinsic promoter: it is transcribed efficiently only when CAP-CAMP is bound nearby. Imagine that a culture of E. coli cells is mutagenized and spread on an agar plate containing IPTG, glucose, and X-Gal. Most of the resulting colonies are white but a few are blue. [IPTG is a lactose analog that binds to and inactivates the lac repressor; unlike lactose, IPTG does not require Lac Permease (LacY) to enter the cell. X-gal is a substrate of B-galactosidase that yields a blue product when cleaved by the enzyme; X-gal does not bind to the lac repressor. For the purpose of this problem, assume that anything less than maximal induction of B-galactosidase results in a white colony. (i) Why was it expected that most of the colonies would be white? (ii) Describe two different mutations mapping within protein-coding DNA that can account for the blue colonies. The two mutations should map in different genes and need not map to the lac operon. Name the…
The enormous complexity in initiating eukaryotic gene transcription seems at first glance to be wasteful. What is the most plausible reason for requiring so many different proteins to assemble on a large array of DNA sequences in order to initiate transcription? O The use of large numbers of proteins and regulatory sequences in DNA opens many possibilities for sophisticated combinatorial gene regulation O The need to open chromatin for transcription demands a much larger number of proteins and regulatory sequences in DNA O The use of large numbers of proteins and regulatory sequences in DNA is essential for alternative splicing of hnRNA in eukaryotes O The larger number of eukaryotic genes demands a correspondingly larger transcription apparatus than that of prokaryotes The larger size of eukaryotic cells demands a correspondingly larger transcription apparatus than that of prokaryotes
. In an effort to determine the location of an operator sitefor a negatively regulated gene, you have made a seriesof deletions within the regulatory region. The extent ofeach deletion is shown by the line underneath the sequence, and the resulting expression from the operon(i = inducible; c = constitutive; − = no expression) isalso indicated.... GGAT C T T AGCCGGCTAACATGATAAATATAA......C C T AGAATCGGCCGA TTGTA C T A TTT ATAT T ...1 i2 –3 c4 –5 ca. What can you conclude from these data about thelocation of the operator site?b. Why do you think deletions 2 and 4 show no expression of the gene?
A bacterial species has a hypothetical sigma promoter that has the following sequence: TTGGCA - 18 bases - TATAAT What change in the level of transcription would there be if the sequence was mutated to: TTCGCA -18 bases -TATAAT 1.The mutation would move the promoter away from consensus and reduce the level of transcription 2.No change the consensus TATAAT sequence in the same. 3.The mutation would bind the promoter to the consensus and produce normal levels of transcription 4.The mutation would inhibit the promoter thereby inhibiting transcription