Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 15, Problem 20P
A muscle enzyme called ME
a. Does this information indicate the presence of enhancer and
b. Why does deletion D effectively eliminate transcription oflacZ?
c. Given the information available from deletion analysis, can you give a molecular explanation for the observation that ME
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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 genes
RNAP can access the lacz promoter irrespective of whether CAP-CAMP is binding the CAP O
True
O False
You are teaching a class on the regulation of eukaryotic gene expression. In order to demonstrate this complex process, you decide to draw for the class a typical eukaryotic gene/transcription unit with its major regions, such as the promoter regions, where the RNA polymerase II and transcription factors would bind
From the list given - choose all components that you think are part of a typical eukaryotic gene
From the list given - choose all the regulatory sequences that you think would control the expression of this eukaryotic gene
From the list given - choose all of the regulatory proteins that would bind the eukaryotic gene to control its expression
Chapter 15 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
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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- Gal4 is a transcription factor that activates transcription of galactose metabolism genes in yeast. These genes are ‘turned on’ when yeast cells need to metabolize galactose. To identify promoter sequences necessary for regulation of transcription of GAL1, reporter gene fusions were made and introduced into yeast cells. Deletions of GAL1 promoter were cloned upstream of LacZ gene. β-Galactosidase activity was measured in presence of galactose. Shown below is a representation of the results obtained. In the diagrams below (not to scale!): • Construct 1 contains ~ 130bp of the promoter, which is predicted to have all the predicted/putative proximal promoter elements (indicated by the solid boxes) needed to regulate transcription of GAL1.• The stippled box is the core promoter.• The arrow represents the transcriptional start site for the reporter gene Lac Z• Number of + signs represents level of transcription• Star represents a mutation in DNA sequence at that location (few nucleotides…arrow_forwardAn enhancer is surrounded by four genes (A, B, C, and D), as shown in the accompanying diagram. An insulator lies between gene C and gene D. On the basis of the positions of the genes, the enhancer, and the insulator, the transcription of which genes is most likely to be stimulated by the enhancer? Explain your reasoning.arrow_forwardIn the sea urchin, early development may occur even in the presence of actinomycin D, which inhibits RNA synthesis. However, if actinomycin D is present early in development but is removed a few hours later, all development stops. In fact, if actinomycin D is present only between the sixth and eleventh hours of development, events that normally occur at the fifteenth hour are arrested. What conclusions can be drawn concerning the role of gene transcription between hours 6 and 15?arrow_forward
- A strain of Arabidopsis thaliana possesses a mutation in the APETALA2 gene. As a result of this mutation, much of the 3′ UTR of the mRNA transcribed from the gene is deleted. What is the most likely effect of this mutation on the expression of the APETALA2 gene?arrow_forwardConsider this list (below) of steps involved in transcription. These steps are out of order. TRANSCRIPTION: 1. mRNA travels through a nuclear pore and enters the cytoplasm 2. the mRNA polymerase attaches at the start of a specific gene 3. RNA polymerase reads the gene surface4. a transcription factor bonds to a promoter site5. DNA molecule is unwound 6. a complimentary mRNA is produced What is the correct order of this transcription?arrow_forwardExamine Figure 17.7. What would be the effect on transcription if a mutation occurred in the gene that encodes GAL3, so that no functional GAL3 was produced?arrow_forward
- Progesterone is a steroid hormone (also described as a ligand) that prepares the body for pregnancy. It binds to the progesterone receptor (PR) protein in the cytoplasm of various cells. Ligand bound PR acts as a transcriptional activator, binds to the DNA in the promoter region of several genes and leads to transcriptional activation of these genes. Ligand bound PR has been shown to increase the expression of a gene, FKBP5. You are studying the activity of wild-type (WT) and mutant PR in cells by examining expression of FKBP5. Results are obtained as shown in the figure below, where the asterisk indicates when progesterone was (or was not) added to the cells. From the results, which of the following statements can be concluded? WT PR without progesterone WT PR with progesterone Time Time mutant PR without progesterone mutant PR with progesterone Time Time The wild-type PR is unable to increase FKBP5 expression in the absence of ligand The wild-type PR increases FKBP5 expression after…arrow_forwardCD3 is a signaling protein that is typically found only in the plasma membrane of immune system T lymphocytes. CD3 is composed of several different polypeptides, including a gamma chain, CD3γ. Scientists analyzed the promoter of the CD3γ chain gene for regulatory sequences that might have positive or negative effects on expression of the gene. The scientists cloned fragments of the CD3γ gene that included the first transcribed nucleotides plus up to 789 nucleotides of upstream regulatory sequences into plasmids in which the gene for the firefly enzyme luciferase immediately follows the fragments. The plasmids were then introduced into a line of T lymphocytes (Figure 1), and the cells were allowed to grow for a short while. Because the regulatory sequences of the CD3γ gene immediately precede the luciferase gene in the plasmids, the activity, either positive or negative, of the regulatory sequences affected the amount of luciferase gene expression by the T lymphocytes. Luciferase…arrow_forwardLack of phosporylation of the C-terminal domian of Pol Il will result in which of the following? O Assembly of the full set of general transcription factors plus Pol II at the promoter, but unphosphorylated Pol II cannot leave the promoter region No assembly of the general transcription factors at a promoter No binding of Pol II at a promoter The unphosphorylated Pol II can initiate and elongate but cannot terminatearrow_forward
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