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 15, Problem 16P
The restriction enzymes Xho
Xho
Sal
Can the sticky ends created byXho
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#16) The restriction enzymes Xhol and SalI cut their specific sequences as shown below:
XhoI 5' C | TCGAG 3' SalI | 5' GTCGAC 3'
3' GAGC | TC 5' 3' G | AGCTG 5'
Can the sticky ends created by XhoI and SalI sites be ligated? If yes, can the resulting sequences be cleaved by either XhoI or SalI?
You are studying a protein that contains the peptide sequence
RDGSWKLVI. The part of the DNA encoding this peptide is included
in the sequence shown below.
5'-CGTGACGGCTCGTGGAAGCTAGTCATC-3'
3'-GCACTGCCGAGCACCTTCGATCAGTAG-5'
This sequence does not contain any BamHI restriction enzyme sites.
The target sequence for the BamHI restriction nuclease is GGATCC.
Your goal is to create a BamHI site on this plasmid by manipulating the
DNA sequence, without changing the coding sequence of the protein. How
would you do this, ie what would the new sequence be?
b
Chapter 15 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 15 - 15.1 What purpose do the bla and lacZ genes serve...Ch. 15 - The human genome is 3109 bp in length. How many...Ch. 15 - 15.3 Ligase catalyzes a reaction between the...Ch. 15 - You have constructed four different libraries: a...Ch. 15 - Using the genomic libraries in Problem 4, you wish...Ch. 15 - The human genome is 3109bp. You wish to design a...Ch. 15 - 15.7 Using animal models of human diseases can...Ch. 15 - 15.8 Compare methods for constructing homologous...Ch. 15 - 15.9 Chimeric genefusion products can be used for...Ch. 15 - 15.10 Why are diseases of the blood simpler...
Ch. 15 - Injection of double-stranded RNA can lead to gene...Ch. 15 - Compare and contrast methods for making transgenic...Ch. 15 - 15.13 It is often desirable to insert cDNAs into a...Ch. 15 - 15.14 A major advance in the s was the development...Ch. 15 - 15.15 The bacteriophage lambda genome can exist in...Ch. 15 - 15.16 The restriction enzymes Xho and Sal cut...Ch. 15 - 15.17 The bacteriophage has a single-stranded DNA...Ch. 15 - 15.18 To further analyze the CRABS CLAW gene (see...Ch. 15 - You have isolated a genomic clone with an EcoR I...Ch. 15 - 15.20 You have identified a cDNA clone that...Ch. 15 - 15.21 You have isolated another cDNA clone of the...Ch. 15 - 15.22 You have identified five genes in S....Ch. 15 - You have generated three transgenic lines of maize...Ch. 15 - 15.24 Bacterial Pseudomonas species often possess...Ch. 15 - 15.25 Two complaints about some transgenic plants...Ch. 15 - 15.26 In Drosophila, lossoffunction Ultrabithorax...Ch. 15 - Prob. 27PCh. 15 - The highlighted sequence shown below is the one...Ch. 15 - Vitamin E is the name for a set of chemically...Ch. 15 - The RAS gene encodes a signaling protein that...Ch. 15 - 15.31 You have cloned a gene for an enzyme that...Ch. 15 - 15.32 About of occurrences of nonautoimmune type...Ch. 15 - Describe how having the Cas 9 gene at a genomic...Ch. 15 - 15.34 Would a gene drive system spread rapidly...
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- The map of plasmid pUC19 is shown below. The restriction site coordinate is the position of the 5’base on the top strand of each site sequence. The restriction enzyme sites are in bold type if there is only one site in pUC19. Please list the fragments in order of size, largest to smallest, which will result from a complete digestion by the restriction enzyme PvuII. Please list the fragments in order of size, largest to smallest, which will result from a complete digestion by the restriction enzyme DrdI.arrow_forwardRefer to Figure , which describes the base modifications of bacteriophage T4 DNA, and briefly describe some issues that must be dealt with in preparing a restriction map of T4 DNA.arrow_forwardUsing the plasmid map of pBCH2.0 provided above, predict how many DNA fragments would be formed if this plasmid was digested with restriction enzyme BamHI.arrow_forward
- Given the following double-stranded fragment of DNA: 5'- ACTTGGCAGGCCTTCGATCC-3' 3'- TGAАССGTCСGGAAGCTAGG-5' A hypothetical restriction endonuclease recognizes a 6bp sequence with two-fold symmetry (typical for restriction enzymes) found in this fragment and catalyzes cleavage of this DNA on both strands between GG nucleotides within the recognition sequence. This nuclease exhibits b-type cleavage (atypical for restriction enzymes). Draw the double-stranded sequence of each fragment after cleavage showing any phosphates left on the ends.arrow_forwardDecide on two restriction sites that you can use to clone this into pL4440’s MCS. Identify their sequence and explain why. Tip: The plasmid map is showed below, details of restriction site sequences can be found at https://enzymefinder.neb.com/#!#nebheaderarrow_forwardThe following is a section of the gene coding for bovine rhodopsin along with several restriction endonucleases, their recognition sequences, and their hydrolysis sites. Which endonucleases will catalyze cleavage of this section of DNA? 5-GCCGTCTACAАСССGGTCATCTAАСТАТСАТGATCААСАAGCAGTTCCGGAACT-3' Recognition Sequence Recognition Sequence Enzyme Enzyme AG | CT TGG | CCA CG I CG GG | CC CI CGG | GATC GC | GGCCGC GAGCT | C Alul Hpall Ball Mbol FnuDII NotI HealII Sadarrow_forward
- #4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:arrow_forwardDecide on two restriction sites that you can use to clone this into pL4440’s MCS. Identify their sequence. Tip: The plasmid map is in Figure 3, details of restriction site sequences can be found at https://enzymefinder.neb.com/#!#nebheaderarrow_forward1) The DNA fragment shown in Figure 1 is cleaved by the restriction enzyme EcoRI as indicated. The number in parenthesis shows the position of the cleavage site. The total length of the DNA fragment is 4000 bp. Small parts of the DNA sequence is known as shown. 5' ACCCTAGGTGTGACCGCGATCCGGCAGCATAAT 3' EcoRI (400) EcoRI (1300) EcoRI (3800) 3' CGCGAAATGCTTTAAGCGCTCTACGGGAGGG5' 3'AGCGTTAGAGTAGCCGGTAAAGGGTACGCGCCTTAA 5' Figure 1: DNA fragment with a total length of 4000 bp. The figure below depicts a gel on which marker DNA of known size has been run. Sketch the location of the bands that will appear, if the DNA fragment shown in Figure 1 is cleaved by EcoRI and afterwards run on the gel along with the marker DNA. 6000 5000 4000 3000 2000 1000 800 600 500 400 200arrow_forward
- Below is a portion of an exon from a gene that encodes protein Y in the genome of the plant Brassica. Wildtype DNA3’ CTT AAT GCT CCG AAT CCA 5’ template strand5’ GAA TTA CGA GGC TTA GGT 3’ non-template strand A new strain (Strain X) of Brassica is identified with the same region of the gene coding for protein Y:3’ CTT AAT GCT GCG AAT CCA 5’ template strand5’ GAA TTA CGA CGC TTA GGT 3’ non-template strand Compare the sequence of Wildtype with Strain X DNA, and note the following: Whether there is a mutation. If there is a mutation, what is the type of mutation (be as specific as possible) and explain the rationale for your decision. Assuming this is the only difference between the Wildtype and Strain X, describe the potential impact of the mutation on the structure and function of the protein.arrow_forwardFor the DNA sequence shown, indicate the products of its cleavage with the following restriction endonucleases (AKA restriction enzymes):5′-ACAGCTGATTCGAATTCACGTT-3′3′-TGTCGACTAAGCTTAAGTGCAA-5′a) EcoRI (the recognition sequence and cleavage site is G↓AATTC);b) AluI (the recognition sequence and cleavage site is AG↓CT).arrow_forwardBelow is a portion of an exon from a gene that encodes protein X in the genome of the plant Arabidopsis. Wildtype DNA3’ TTC AAT GCT CCG AAT ACC 5’ template strand5’ AAG TTA CGA GGC TTA TGG 3’ non-template strand A new strain (Strain B) of Arabidopsis is identified with the same region of the gene coding for protein X: 3’ TTC AAT GCT CCC AAT ACC 5’ template strand5’ AAG TTA CGA GGG TTA TGG 3’ non-template strand Compare the two DNA sequences and look for any differences. Based on what you find a. There is no mutation in Strain B compared to Strain A. b. After the point of the mutation, all the amino acids encoded by the Strain B template will be different than the Strain A protein X. c. Protein X made from the Strain B template will be much shorter than protein X made from the Strain A template d. Protein X from Strain B will have one amino acid difference that would not affect protein function. e. There is a mutation but there will not be any difference in the…arrow_forward
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