Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Textbook Question
Chapter 8, Problem 26P
DNA footprint protection (described in Research Technique
a. Explain why this gel provides evidence that the cloned DNA may act as a promoter sequence.
b. Approximately what length is the DNA region protected by RNA pol II and TFIIs?
c. What additional genetic experiments would you suggest to verify that this region of cloned DNA contains a functional promoter?
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d.) Enhancers may function in either orientation (forward or backward).
When a region of DNA that contains the genetic information for a protein is isolated from a bacterial cell and inserted into a eukaryotic cell in a proper position between a promoter and a terminator, the resulting cell usually produces the correct protein. But when the experiment is done in the reverse direction (eukaryotic DNA into a bacterial cell), the correct protein is often not produced. Can you suggest an explanation?
Suppose you have a 1-kb segment of cloned DNA that is suspected to contain a eukaryotic promoter including a TATA box, a CAT box, and an upstream GC-rich sequence. The clone also contains a gene whose transcript is readily detectable. Your laboratory supervisor asks you to outline an experiment. (1) determine if eukaryotic transcription factors (TF) bind to the fragment and, if so, (2) identify where on the fragment the transcription factors bind. All necessary reagents, equipment, and experimental know-how are available in the laboratory.
Complete the outline of an experiment, which determines if eukaryotic transcription factors (TF) bind to the fragment. Assume all necessary reagents, equipment, and experimental how are available in the laboratory.
Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.
For the band shift assay, two samples of the DNA fragment are analyzed by agarose gel electrophoresis: one sample is…
Chapter 8 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 8 - Prob. 1PCh. 8 - 8.2 In one to two sentences each, describe the...Ch. 8 - 8.3 Answer these questions concerning...Ch. 8 - 8.4 The diagram below shows a DNA duplex. The...Ch. 8 - The following is a portion of an mRNA sequence:...Ch. 8 - Compare and contrast the properties of DNA...Ch. 8 - The DNA sequences shown below are from the...Ch. 8 - Bacterial and eukaryotic gene transcripts can...Ch. 8 - Describe the two types of transcription...Ch. 8 - What is the role of enhancer sequences in...
Ch. 8 - Prob. 11PCh. 8 - Draw a bacterial promoter and label its consensus...Ch. 8 - For a eukaryotic gene whose transcription require...Ch. 8 - Three genes identified in the diagram as A, B and...Ch. 8 - Prob. 15PCh. 8 - 8.16 The segment of the bacterial gene involved in...Ch. 8 - Prob. 17PCh. 8 - Prob. 18PCh. 8 - 8.19 A DNA fragment from the end of the mouse...Ch. 8 - 8.20 Wild-type E. coli grow best at but can grow...Ch. 8 - A mutant strain of Salmonella bacteria carries a...Ch. 8 - 8.22 The human wild-type allele and a certain...Ch. 8 - Prob. 23PCh. 8 - A full-length eukaryotic gene is inserted into a...Ch. 8 - The accompanying illustration shows a portion of a...Ch. 8 - DNA footprint protection (described in Research...Ch. 8 - Suppose you have a 1-kb segment of cloned DNA that...Ch. 8 - Assume that a mutation affects the gene for each...Ch. 8 - 8.29 The DNA sequence below gives the first base...Ch. 8 - 8.30 Genomic DNA from a mouse is isolated,...Ch. 8 - 8.31 A portion of a human gene is isolated from...
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- The following is a portion of an mRNA sequence: 3’ –AUCGUCAUGCAGA-5’ a)During transcription, was the adenine at the left-hand side of the sequence the first or the last nucleotide used to build the portion of the mRNA shown? Explain how you know. b)Write out the sequence and polarity of the DNA duplex that encodes this mRNA segment. Label the template and coding DNA strands. c)Identify the direction in which the promoter region for this gene will be located.arrow_forwardThe human hexokinase enzyme has the same function as the bacterial hexokinase enzyme but is somewhat different in its amino acid sequence. You have obtained a mutant bacterial strain in which the gene for hexokinase is missing. If you introduce into your mutant strain a DNA plasmid engineered to contain the DNA coding sequence of the human hexokinase gene, what must you also include? a)The human hexokinase promoter b)The bacterial hexokinase promoter c)Both the human and bacterial promoters d)You cannot engineer a bacteria to produce a human enzymearrow_forwardShown below is a portion of a wild-type DNA sequence that encodes the last amino acids of a protein that is 270 amino acids long. The first three bolded base pairs indicate the frame and include the coding region. 5^ ...GCTAAGTATTGCTCAAGATTAGGATGATAAATAACTGG 3^ 3^.. CGATTCATAACGAGTTCTAATCCTACTATTTATTGACC 5^ Which strand is the template strand for transcription of this gene? Briefly explain how you know. An insertion of one base pair causes the protein to decrease in length by seven amino acids. With respect to the sequence given above, where does this insertion occur? A change of one base pair leads to the protein increasing in the length by one amino acid. With respect to the sequence given above, which base pair would you change, and what would you change this base pair for the protein to increase in the length by one amino acid?arrow_forward
- You want to clone the Human Gene A into a plasmid for producing the Protein A in bacteria. GeneA encodes an mRNA of 100 nucleotides long, but the entire gene spans more than 4000 nucleotides. There are three exons and two introns. If we were to clone this gene directly from the nuclear DNA, bacteria would not be able to express the insulin protein. Explain why this is true.arrow_forwardConsider the following sequence of DNA: 3'-TTA CGG-5'What dipeptide is formed from this DNA after transcription and translation? b. If a mutation converts CGG to CGT in DNA, what dipeptide is formed? c. If a mutation converts CGG to CCG in DNA, what dipeptide is formed? d. If a mutation converts CGG to AGG in DNA, what dipeptide is formed?arrow_forwardConsider 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_forward
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