GENETIC ANALYSIS: AN INTEG. APP. W/MAS
2nd Edition
ISBN: 9781323142790
Author: Sanders
Publisher: Pearson Custom Publishing
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 16, Problem 10P
In enhancer trapping experiments, a minimal promoter and a reporter gene are placed adjacent to the end of a transposon so that genomic enhancers adjacent to the insertion site can act to drive expression of the reporter gene. In a modification of this approach, a series of enhancers and a promoter can be placed at the end of a transposon so that transcription is activated from the transposon into adjacent genomic DNA. What types of mutations do you expect to be induced by such a transposon in a mutagenesis experiment?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
What is the difference between a promoter proximal element and a distal enhancer? What are the similarities?
When Laybourne and Kadonaga studied the effects of histone proteins on eukaryotic
transcription using an in vitro transcription assay explain why:
a) they used two different DNA templates that contained different promoter structures.
b) when they included both activator protein and histones, they always added the histone proteins
before adding the activator to the transcription assay mixture.
(Ctri) -
Negative supercoiling of DNA favors the transcription of genes because it facilitates unwinding. However, not all promoter sites are stimulated by negative supercoiling. The promoter site for topoisomerase II itself is a noteworthy exception. Negative supercoiling decreases the rate of transcription of this gene. Propose a possible mechanism for this effect and suggest a reason why it may occur.
Chapter 16 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
Ch. 16 - 14.1 What are the advantages and disadvantages of...Ch. 16 - Prob. 2PCh. 16 - 3. Genetic maps and physical maps are both...Ch. 16 - 14.5 What are the advantages and disadvantages of...Ch. 16 - 14.6 You have cloned the mouse ortholog (see...Ch. 16 - 14.13 The CBF genes of Arabidopsis are induced by...Ch. 16 - 14.14 When the S. cerevisiae genome was sequenced,...Ch. 16 - 14.15 Translational fusions between a protein of...Ch. 16 - In enhancer trapping experiments, a minimal...Ch. 16 - 14.19 In Genetic Analysis, we designed a screen to...
Ch. 16 - How would you design a genetic screen to find...Ch. 16 - 14.21 The eyes of Drosophila develop from imaginal...Ch. 16 - 14.22 Given your knowledge of the genetic tools...Ch. 16 - Mutations in the CFTR gene result in cystic...Ch. 16 - Prob. 16PCh. 16 - 14.25 How would you conduct a screen to identify...Ch. 16 - In land plants, there is an alternation of...Ch. 16 - 14.27 The Drosophila evenskipped (eve) gene is...Ch. 16 - Prob. 20PCh. 16 - 14.29 As shown in Figure, mutations in the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- 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_forwardHow do you think that transcription randomizes positions of nucleosomes and repression restores the ordering after transcription? How might you test to see if there was an exchange of histone subunits during transcription or if the nucleosome is truly transferred as a single unit? Would you expect the DNA band representing the distance from the restriction enzyme site to the hypersensitive site to be a single band or a smear? Defend your answer.arrow_forwardYou have isolated two different mutants (reg1 and reg2) causing constitutive expression of the emu operon (emu1 emu2). One mutant contains a defect in a DNA-binding site, and the other has a loss-of-function defect in the gene encoding a protein that binds to the site. Is the DNA-binding protein a positive or negative regulator of gene expression? Explain. To determine which mutant has a defect in the site and which one has a mutation in the binding protein, you decide to do an analysis using F′ plasmids. Assuming you can assay levels of the Emu1 and Emu2 proteins, what results do you predict for the two strains (i and ii; see descriptions below) if reg2 encodes the regulatory protein and reg1 is the regulatory site? Explain. F′ (reg1− reg2+ emu1− emu2+)/reg1+ reg2+ emu1+ emu2− F′ (reg1+ reg2− emu1− emu2+)/reg1+ reg2+ emu1+ emu2−arrow_forward
- An electrophoretic mobility shift assay can be used to study the binding of proteins to a segment of DNA. In the results shown here, an EMSA was used to examine the requirements for the binding of RNA polymerase |l (from eukaryotic cells) to the promoter of a protein-encoding gene. The assembly of general transcription factors and RNA polymerase Il at the core promoter is described in Week 4. In this experiment, the segment of DNA containing a promoter sequence was 1100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to an EMSA. Lane 1: No proteins added Lane 2: TFIID Lane 3: TFIIB Lane 4: RNA polymerase IIl Lane 5: TFIID + TFIIB Lane 6: TFIID + RNA 1 2 3 4 5 6. 7 polymerase II Lane 7: TFIID + TFIIB + RNA polymerase Il 1100 bp Explain the results.arrow_forwardYou have isolated different mutants (reg1 and reg2) causing constitutive expression of the emu operon (which has genes emu1 and emu2). One mutant contains a defect in a DNA-binding site, and the other has a loss-of-function defect in the gene encoding a protein that binds to the site. Is the DNA-binding protein a positive or negative regulator of gene expression?arrow_forwardYou are studying how the expression of Gene F is regulated. Initially, you identified three potential regulatory elements in the regulatory region upstream of Gene F's transcription start site. To study these potential regulatory elements, you created three different mutants with a different potential regulatory element mutated in each. The data below shows qPCR data for expression of Gene F in Wild-type (Control) samples, each of the three different mutants (Mutant 1-3), and a negative control (No DNA Control). Were you able to identify a regulatory element that binds a transcriptional activator? If so, explain how you determined that and which mutant it is. Make sure to reference specific data from the image and your knowledge of the molecular biology involved in regulating transcription. The dotted line represents our measurement threshold. Fluorescence N Wild-Type Control Mutant 1 Mutant 2 Mutant 3 No DNA Control 10 10 20 30 Cyclearrow_forward
- One can synthesize RNA from any gene in a test tube (in vitro). You want to synthesize mRNA from a prokaryotic gene in a test tube. You added the gene, which contains its promoter, coding region, and rho-dependent transcription termination site, in a test tube. Which of the following do you need to add to this test tube so that you can synthesize RNA from this gene and terminate transcription accurately? 1) RNA polymerase core enzyme, 2) RNA polymerase holoenzyme,3) RNA polymerase II, 4) ATP, CTP, GTP, and TTP mixture 5) ATP, GTP, CTP, and UTP mixture, 6) dATP, dGTP, dCTP, and dUTP mixture 7) Rho factor, 8) primer. 3,4,7 1,6,7 2.5.7 2,4,6 2,5,7,8arrow_forwardIn the laboratory, you want to study protein that is normally toxic to E. coli cells. You wish to grow this protein in E. coli and purify it from E. coli. Your advisor suggests cloning the gene into an expression vector that uses the araBAD promoter. Explain why is it ideal to use the araBAD promoter for expression of your gene of interest in E. coli cells?arrow_forward3′-->5′ Exonuclease activity allows DNA polymerase III (Pol III) to back-up and fix a mismatched base pair that was just incorporated into a growing strand of new DNA. True Or False In one of the four ways to regulate gene expression, positive control with repression indicates that transcription is activated in the presence of a co-repressor. True Or Falsearrow_forward
- The tac promoter, an artificial promoter made from portions of the trp and lacUV5 promoters, has been introduced into a plasmid. It is a hybrid of the lac and trp (tryptophan) promoters, containing the −35 region of one and the −10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?arrow_forwardThe tac promoter, an artificial promoter made from portions of the frp and lacUV5 promoters, has been introduced into a plasmid. It is a hybrid of the lac and trp (tryptophan) promoters, containing the –35 region of one and the -10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?arrow_forwardGal4 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Bacterial Genomics and Metagenomics; Author: Quadram Institute;https://www.youtube.com/watch?v=_6IdVTAFXoU;License: Standard youtube license