GENETIC ANALYSIS: AN INTEG. APP. W/MAS
2nd Edition
ISBN: 9781323142790
Author: Sanders
Publisher: Pearson Custom Publishing
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Chapter 15, Problem 18P
The consequences of four deletions from the region upstream of the yeast gene DBM
Which mutations(s) affect an enhancer sequence? Explain your reasoning.
Which mutation(s) affect a silencer sequence? Explainyour reasoning.
Which mutation(s) affect the promoter? Explain yourreasoning.
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4.1 Name and discuss two transcription regulatory elements that can be found in the figure. (6)4.2. During the activation of eukaryotic transcription the promoter region needs to be accessible for the binding of transcription factors. Describe in detail one of the mechanisms involved in this process.
Many eukaryotic promoter regions contain CAAT boxes with consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?
"Upstream"
"Downstream"
Exons
Start of
transcription
Termination codon
5
3'
Promoter
initiator
codon
Introns
Polyadenylation
signal
(intervening sequences)
5' untranslated
region
3' untranslated
region
Direction of transcription
Please study the diagram above on eukaryotic gene expression. In order to
provide instructions for gene expression, a eukaryotic gene should have the
following sequences except for
O A. Promoter
B. Start codon also known as initiator codon
C. Splicing signals (dinucleotide sequence in the intron)
O D. 5' CAP sequence
Chapter 15 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
Ch. 15 - 13.1 Devoting a few sentences to each, describes...Ch. 15 - 13.2 Describe and give an example (real or...Ch. 15 - What is meant by the term chromatin remodeling?...Ch. 15 - 13.4 What general role does acetylation of histone...Ch. 15 - 13.5 Describe the roles of writers, readers, and...Ch. 15 - Outline the roles of RNA in eukaryotic gene...Ch. 15 - 13.7 What are the roles of the Polycomb and...Ch. 15 - Most biologists argue that the regulation of gene...Ch. 15 - Compare and contrast the transcriptional...Ch. 15 - The term heterochromatin refers to heavily...
Ch. 15 - 13.11 Compare and contrast promoters and enhancers...Ch. 15 - 13.12 What are the different chromatin...Ch. 15 - 13.13 Define epigenetics, and provide examples...Ch. 15 - What is one proposed role for lncRNAs?Ch. 15 - 13.17 A hereditary disease is inherited as an...Ch. 15 - Prob. 16PCh. 15 - A gene expressed in long muscle of the mouse is...Ch. 15 - The consequences of four deletions from the region...Ch. 15 - Provide a description of the mechanistic roles of...Ch. 15 - 13.20 A muscle enzyme called ME is produced by...Ch. 15 - 21. A muscle protein in mouse is produced through...
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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 genesarrow_forwardMutations 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 TATGATarrow_forwarda. How do bacteria increase the efficiency of gene expression? Is this possible in eukaryotes? b. A mutation in the promoter of Gene K disrupts an enzyme binding site and results in the loss of Gene K expression. Is this change in gene expression likely happening at the transcriptional or the translational level? Explain. c. Propose three different mutations to prevent initiation, elongation, and termination of bacterial transcription, respectively. Explain how/why each mutation would prevent its respective step. (Hint: mutations can be in genes that encode proteins or regulatory DNA sequences)arrow_forward
- Many promoter regions contain CAAT boxes containing consensus sequences CAAT or CCAAT approximately 70 to 80 bases upstream from the transcription start site. How might one determine the influence of CAAT boxes on the transcription rate of a given gene?arrow_forwardThere are similarities and differences during regulation of gene expression in both prokaryotes and eukaryotes. Promoters, transcription factors and RNA polymerase are essential elements in transcription but their properties and function may differ.a) Predict the outcome or consequences of mRNA transcription by RNA polymerase II in eukaryote without the presence of transcription factors (TF).arrow_forwardRefer to figure, which shows the distribution of histone H2A.Z on nucleosomes near a transcription start site. What experimental technique would have been used to generate the data for this figure? Briefly describe the operation of this technique.arrow_forward
- The IMD2 promoter contains three upstream transcription start sites (TSS) that are utilized under high GTP conditions and a single downstream TSS (-106) that is normally only utilized under low GTP conditions. In a wild type cell, expression of IMD2 mRNA only occurs if transcription initiates from the -106 TSS. In 300 words or less, describe: 1.) The normal function of Ssl2, and 2.) why a mutation in Ssl2, that increases its catalytic rate, would allow expression of the IMD2 ORF under high GTP conditions. (Conditions under which the IMD2 ORF is NOT expressed in the wild type.)arrow_forwardone promoter can regulate the expression of different polypeptides (proteins) in both prokaryotes and eukaryotes but in different manners. explain how this regulation (to allow the expression of different polypepetides from only one promoter) differs between prokaryotes and eukaryotes?arrow_forward6) The C-terminal domain (CTD) of yeast RNA polymerase II (RPB) subunit RPB1 contains the repeating amino acid sequence "-PTSPSYS-". Explain how this sequence plays a role in the regulation of the transcriptional activity of the yeast RPB holoenzyme.arrow_forward
- Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.arrow_forwardIn addition to Tc1, the C. elegans genome contains otherfamilies of DNA transposons such as Tc2, Tc3, Tc4, andTc5. Like Tc1, their transposition is repressed in thegerm line but not in somatic cells. Predict the behaviorof these elements in the mutant strains where Tc1 is nolonger repressed due to mutations in the RNAi pathway.Justify your answer.arrow_forward6.6arrow_forward
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