Shown below is a schematic diagram illustrating a very short gene with 5000 bp region of anunknown Schizosaccharomyces pombe genome. (Note: Transcription starts at TranscriptionStart Site (TSS).)TSS5.3'3+1(i)Name the specific regions that can be recognized by Transcription Factor IID (TF ID)and indicate the locations in the diagram above.(ii)List the mechanistic steps that can trigger the initiation of transcription by TranscriptionFactor IIH (TF IIH).

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Description: Shown below is a schematic diagram illustrating a very short gene with 5000 bp region of anunknown Schizosaccharomyces pombe genome. (Note: Transcription starts at TranscriptionStart Site (TSS).)TSS5.3'3+1(i)Name the specific regions that can be rec

Answered: Image /qna-images/answer/14147ac2-e6b4-42ba-a973-17252144f6eb.jpg

Answered: by Transcription Factor IID (TF IID)

Science

Biology

by Transcription Factor IID (TF IID)

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 13QP

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in m 5'- 3'- Shown below is a schematic diagram illustrating a very short gene with 3000 bp region of an unknown Escherichia coli genome. (Note: Transcription starts at Transcription Start Site (TSS).) TSS -3' -5' +1 (i) Name the specific regions that can be recognized by sigma factor and indicate the locations in the diagram above. (ii) How does Sigma factor trigger the initiation of transcription?
Shown here is a theoretical viral mRNA sequence 5′-AUGCAUACCUAUGAGACCCUUGGA-3′ (a) Assuming that it could arise from overlapping genes, how many different polypeptide sequences can be produced? Using the chart in Figure 12–7, what are the sequences? (b) A base-substitution mutation that altered the sequence in part (a) eliminated the synthesis of all but one polypeptide. The altered sequence is shown below. Use Figure 12–7 to determine why it was altered. 5′-AUGCAUACCUAUGUGACCCUUGGA-3′
Searching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4. GAL4 activates the transcription of galactose genes. Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism. The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T. In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added. These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to. The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced. Results showed that only 10 different DNA sequences had GAL4 bound. Since the…
Explain how the following mutations would affect transcription of the yeast GAL1 gene in the presence of galactose. (a) A deletion within the GAL4 gene that removes the region encoding amino acids 1 to 100. (b) A deletion of the entire GAL3 gene. (c) A mutation within the GAL80 gene that blocks the ability of Gal80 protein to interact with Gal3p. (d) A deletion of one of the four UASG elements upstream from the GAL1 gene. (e) A point mutation in the GAL1 core promoter that alters the sequence of the TATA box.
The following logo plot represents the preferred cis-regulatory sequences (i.e. transcription factor binding site) of bHLH transcription factor FOSL1. C 1 2 3 4 5 6 7 8 9 10 11 position Would you expect this sequence to be recognized by a monomer, a homodimer, or a heterodimer of the protein? Explain your answer. (short phrases are sufficient; please write your answer into the template below) A- В I A -l expect FOSL1 to bind as a: (monomer, homodimer, heterodimer; please choose) B - short explanation: information content (bit) !!
Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. = exons Transcription termination site (also poly A site) = introns Promoter Start of transcription 3' 5'. TAA ATG 50 TAC 130 222 850 126 132 90 ATT 5' 3 QUESTION 3: What is the length in nucleotides of the mature, processed B-globin mRNA? A.620 B.980 C.438 D.1600
You would like to add a nuclear localization sequence (NLS) of Lys-Lys-Lys-Arg-Lys to a protein that is usually found in the cytoplasm of a yeast cell. To accomplish this, you introduce the nucleotide sequence encoding the NLS into the gene that encodes the cytoplasmic protein of interest. a. What is the size of the nucleotide insert that will encode the NLS? Briefly explain. 5' 3' b. Below is a diagram of the gene encoding the cytoplasmic protein of interest in the yeast genome. If your goal is to put the NLS at the carboxyl (C) terminus of the protein, at which location (A-E) should the NLS be inserted? Briefly explain. A TATAA ATATT promoter +1 B ATG TAC D TAA ATT stop codon E 3' 5'
(c) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to suppress syn- thesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix – (reverse) turn – helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein, a different type of repressor protein. A diagram of the trp repressor dimer is also shown. reverse turn trp helix 4 70 Trp -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly- Ile- Prl -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr- Val- Pri heilix trp helix 5 80 90 Trp Ala-Thr-Ile-Thr-Arg-Gly-Ser sgn-Ser-Leu-Lys-Ala-Ala- Prl Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- Prl helix
The following double-stranded DNA sequence is part of a hypothetical yeast genome which contains a very small gene. Transcription starts at the Transcription Start Site (TSS), proceeds in the direction of the arrow and stops at the end of the Transcription Terminator (green box). 5' 3' TSS CTATAAAAATGCCATGCATTATCTAGATAGTAGGCTCTGAGAAATTTATCTCACT | | | | | | | | | | GATATTTTTACGGTACGTAATAGATCTATCATCCGAGACTCTTTAAATAGAGTGA - 5' PROMOTER TERMINATOR 3' a) Which strand (top or bottom) is the template strand? Explain why. b) What is the sequence of the mRNA produced from this gene? Label the 5' and 3' ends. c) What is the sequence of the protein produced from the mRNA? d) If a mutation (an insertion) were found where a T/A (top/bottom) base pair were added immediately after the T/A base pair shown in red, what would be the sequence of the mRNA? What would be the sequence of the protein?
Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. Question 6: How many amino acids are present in the wild-type human B-globin protein? = exons = introns Transcription termination site (also poly A site) Promoter Start of transcription 3' 5' ATG 50 TẠC TAA 126 132 |ATT 90 130 222 850 3 5' Start codon Stop codon А. 438 В. 146 C. 620 D. 206 © 2013 John Wiley & Sons, Inc. All rights reserved.
The entire genome of the yeast Saccharomyces cerevisiaehas been sequenced. This sequencing has led to the identification of all the open reading frames (ORFs, gene-sizesequences with appropriate translational initiation andtermination signals) in the genome. Some of these ORFsare previously known genes with established functions;however, the remainder are unassigned reading frames(URFs). To deduce the possible functions of the URFs,they are being systematically, one at a time, convertedinto null alleles by in vitro knockout techniques. The results are as follows:15 percent are lethal when knocked out.25 percent show some mutant phenotype (alteredmorphology, altered nutrition, and so forth).60 percent show no detectable mutant phenotype at alland resemble wild type.Explain the possible molecular-genetic basis of thesethree mutant categories, inventing examples wherepossible.
Template strand of DNA is: 3’ TACATAACCGGGCCCATATCGGCCATTTGC5’. 2a). Following transcription, what is the total number of codons in the mRNA transcript? 2 b). Where is the start codon located in this mRNA transcript? 2c). Following translation of this mRNA transcript, how many amino acids will the proteincontain and identify the amino acids sequence of this gene from a genetic code table*.*Note= using a genetic code table