Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 22, Problem 2NST
Summary Introduction
To determine: The sequence of the probe that will hybridize the ASO sequence.
Introduction: Allele-specific oligonucleotide (ASO) is a short synthetic DNA which is complementary to the sequence of target DNA. ASO analysis is done under conditions that facilitate only complementary
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The most common MCAD mutation is shown below. The coding strand is shown for both the WT and mutant. The TATA box and kozak sequences are in parenthesis. What type of mutation is present?
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Chapter 22 Solutions
Concepts of Genetics (12th Edition)
Ch. 22 - In order to vaccinate people against diseases by...Ch. 22 - Prob. 2NSTCh. 22 - Prob. 1CSCh. 22 - Prob. 2CSCh. 22 - Prob. 3CSCh. 22 - HOW DO WE KNOW? In this chapter, we focused on a...Ch. 22 - Prob. 2PDQCh. 22 - Why are most recombinant human proteins produced...Ch. 22 - One of the major causes of sickness, death, and...Ch. 22 - Sequencing the human genome, the development of...
Ch. 22 - Prob. 6PDQCh. 22 - As genetic testing becomes widespread, medical...Ch. 22 - Prob. 8PDQCh. 22 - Prob. 9PDQCh. 22 - Does genetic analysis by ASO testing allow for...Ch. 22 - Maternal blood tests for three pregnant women...Ch. 22 - What is the main purpose of genome-wide...Ch. 22 - Describe how the team from the J. Craig Venter...Ch. 22 - Prob. 14PDQCh. 22 - Prob. 15PDQCh. 22 - Dominant mutations can be categorized according to...Ch. 22 - In 2013 the actress Angelina Jolie elected to have...Ch. 22 - Prob. 18PDQCh. 22 - Should the FDA regulate direct-to-consumer genetic...Ch. 22 - Prob. 20ESPCh. 22 - Following the tragic shooting of 20 children at a...Ch. 22 - Private companies are offering personal DNA...Ch. 22 - Prob. 23ESPCh. 22 - Prob. 24ESPCh. 22 - Prob. 25ESPCh. 22 - Craig Venter and others have constructed synthetic...
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- . Explain why DNA is stable in the presence of alkali (0.3 M KOH), while RNA is quantitatively degraded to 2'- and 3'-nucleoside monophosphates under these conditions.arrow_forwardBased on the following wild type DNA sequence, indicate if each of the mutations should be classified as : insertion, deletion, missense, nonsense, silent (Use the provided Genetic Code table and remember you have been given DNA sequence). Wild Type: 5’ ATG GCT AGA GTC GAG TTG 3’ Mutant 1: 5’ ATG GCA GAG TCG AGT TG 3’ Mutant 2: 5’ ATG GCT TGA GTC GAG TTG 3’ Mutant 3: 5’ ATG GCT AGA GTT GAG TTG 3’ Mutant 4: 5’ ATG GCT AGA AGT CGA GTT G 3’ Mutant 5: 5’ ATG GCT AGA ATC GAG GTT 3’arrow_forwardhe Sequence below comes from the alpha-2 globin of the human hemoglobin gene cluster found in chromosome 16. The globin region of the hemoglobin protein itself consists of 2 alpha chains and 2 beta chains. 1 actcttctgg tccccacaga ctcagagaga acccaccatg gtgctgtctc ctgccgacaa 61 gaccaacgtc aaggccgcct ggggtaaggt cggcgcgcac gctggcgagt atggtgcgga 121 ggccctggag aggatgttcc tgtccttccc caccaccaag acctacttcc cgcacttcga 181 cctgagccac ggctctgccc aggttaaggg ccacggcaag aaggtggccg acgcgctgac 241 caacgccgtg gcgcacgtgg acgacatgcc caacgcgctg tccgccctga gcgacctgca 301 cgcgcacaag cttcgggtgg acccggtcaa cttcaagctc ctaagccact gcctgctggt 361 gaccctggcc gcccacctcc ccgccgagtt cacccctgcg gtgcacgcct ccctggacaa 421 gttcctggct tctgtgagca ccgtgctgac ctccaaatac cgttaagctg gagcctcggt 481 agccgttcct cctgcccgct gggcctccca acgggccctc ctcccctcct tgcaccggcc 541 cttcctggtc…arrow_forward
- Sickle cell hemoglobin DNA CACGTAGACTGAGG ACAC.. Sickle cell hemoglobin MRNA Sickle cell hemoglobin AA sequence 4. What type of mutation is this? Please explain why.arrow_forward#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:arrow_forwardShown is part of the coding strand of the HBB gene involved in sickle cell disease. What will be the effect on the resulting peptide of a deletion of the two nucleotides shown in bold/underline? 5' CTG ACT CCT GAG GAC 3' a) A change from glutamic acid to valine will occur b) The deletion will not result in a change to the polypeptide c) A nonsense mutation will occur d) The polypeptide will be missing one amino acid, but the reading frame will be preservedarrow_forward
- The proximal histidine residues have been replaced by glycine residues by mutation of the cloned genes for both the α and β subunits of hemoglobin. With the tetrameric mutant hemoglobin (all subunits being mutant, α H F8 G, β H F8 G), it was found that the “proximal” coordination bonds to hemes in the mutant protein could be replaced by having the small molecule imidazole in the buffers. Oxygen binding curves for the tetrameric mutant hemoglobin were measured. A. The degree of cooperativity in oxygen binding for the mutant hemoglobin (with imidazole present) would be expected to 1) increase 2) decrease 3) not be affected) compared with the normal protein. B. Justify your answer to part A in terms of what you know about the structural basis of cooperativity in hemoglobin. C. How would the Hill coefficient for the mutant be expected to change compared with nH for normal hemoglobin, which is ~3?arrow_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_forwardWhat type of mutation is shown below: CGG GGG GAG TCT TTT TAT GCA GCT changes to: CGG GGG GAG CTT TTTATG CAG CT A frame shift mutatuion which resulted from a deletion. O Point mutation in which one nucleotide was changed. O Point mutation in which a length of nucleotides were changed. O A frame shift mutatuion which resulted from an insertion.arrow_forward
- The beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'...AAGCTCGAGAGCAGCAGCTСТАТGCGCTАСТАТААTGACСАТТАТАССССТАСGTGATAG...5" promoter a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3'arrow_forwardWith this DNA sequenec: - 5'-GCAATGGAGAGAATCTGCGCG-3'- - 3'-CGTTACCTCTGTTAGACGCGC-5' - -Identify the sequencce of RNA in which the protein product will be detrnemined. -What will be the protein product sqeuence? -What will be the pl of the protein product?arrow_forwardThe beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3’ 3'...AAGCTCGAGAGCAGCAGCTСТАТGCGCTАСТАТААTGACCATTATАССССТАСGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?arrow_forward
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