Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 13P
Chemical Mutagenesis of DNA Bases Show the
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Show the nucleotide sequence changes that might arise in a dsDNA (coding strand segment GCTA) upon mutagenesis with (a) HNO2 (b) bromouracil and (c) 2-aminopourine
1) Which statement below explains the trick in sanger sequencing that produces fluorescently labeled fragments at every length within a fragment?
a) When synthesizing a copy of the DNA to be sequenced, a high concentration of fluorescently labeled dideoxynucleotides (ddNTPs) are used along with a low concentration of deoxynucleotides (dNTPs) to produce the chain termination events at every location in the sequence.
b) When synthesizing a copy of the DNA to be sequenced, fluorescently labeled dideoxynucleotides (ddNTPs) are used instead of deoxynucleotides (dNTPs) to produce the chain termination events at every location in the sequence.
c) When synthesizing a copy of the DNA to be sequenced, a low concentration of fluorescently labeled dideoxynucleotides (ddNTPs) are used along with a high concentration of deoxynucleotides (dNTPs) to produce the chain termination events at every location in the sequence.
d) When synthesizing a copy of the DNA to be sequenced, fluorescently labeled…
Given the following Wild Type and Mutated DNA sequences:
1.) Identify where the base pair change occurs ( what letter changed?)
2.) For BOTH sequences, write the mRNA strands, define the codon regions and amino acid sequences.
3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein.
Wild Type DNA Sequence: 3' - AGGCTCGCCTGT - 5'
Mutated DNA Sequence: 3' - AGTCTCGCCTGT - 5'
Chapter 28 Solutions
Biochemistry
Ch. 28 - Semiconservative or Conservative DNA Replication...Ch. 28 - The Enzymatic Activities of DNA Polymerase I (a)...Ch. 28 - Multiple Replication Forks in E. coli I Assuming...Ch. 28 - Multiple Replication Forks in E. coli II On the...Ch. 28 - Molecules of DNA Polymerase III per Cell vs....Ch. 28 - Number of Okazaki Fragments in E. coli and Human...Ch. 28 - The Roles of Helicases and Gyrases How do DNA...Ch. 28 - Human Genome Replication Rate Assume DNA...Ch. 28 - Heteroduplex DNA Formation in Recombination From...Ch. 28 - Homologous Recombination, Heteroduplex DNA, and...
Ch. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Chemical Mutagenesis of DNA Bases Show the...Ch. 28 - Prob. 14PCh. 28 - Recombination in Immunoglobulin Genes If...Ch. 28 - Helicase Unwinding of the E. coli Chromosome...Ch. 28 - Prob. 17PCh. 28 - Functional Consequences of Y-Family DNA Polymerase...Ch. 28 - Figure 28.11 depicts the eukaryotic cell cycle....Ch. 28 - Figure 28.41 gives some examples of recombination...Ch. 28 - Prob. 21PCh. 28 - Prob. 22P
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- Examination of the structure of DNA polymerases bound to nucleotide analogs reveals that conserved residues come within van der Waals contact of C-2' of the bound nucleotide. What is the potential significance of this interaction?arrow_forwardWhat are the three possible outcomes of point mutations?arrow_forwardA temperature-sensitive mutation is one in which the defect is not presented functionally until the temperature is raised. In the case described below, the enzymes function normally in bacteria at 37 °C, but are non-functional at 40 °C. Predict the detailed molecular consequences of a loss of function in a temperature-sensitive mutant for each of the following enzymes: a) DNA gyrase, b) DNA polymerase III, c) DNA ligase, d) DNA polymerase I.arrow_forward
- You prepare a reaction mix containing (i) DNA polymerase III, (ii) DATP, DCTP, dGTP, Mg2+, and 2,3'dideoxy-TTP (called ddTTP*), (iii) a short primer with the sequence 5'-CCTG-3, and (iv) a source DNA fragment with the sequence, 5'-AATCGTTCACGTTAGCAGG-3. What is the product of this reaction? Note that in ddTTP, both the 2' and 3' positions on the ribose sugar lack hydroxyl groups. No reaction, because the primer is not complementary to any sequence in the source DNA. O CCTGCT O CCTGC O T'T'AGCAAGT'GCAAT CGTCC O CCTGCT'AACGT GAACGAT'Tarrow_forwardAlthough DNA polymerases require both a template and a primer, the following single-stranded polynucleotide was found to serve as a substrate for DNA polymerase in the absence of any additional DNA.3′ HO-ATGGGCTCATAGCCGGAGCCCTAACCGTAGACCACGAATAGCATTAGG-p 5′What is the structure of the product of this reaction?arrow_forwardGiven the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs (what letters changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions (with spaces), and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - CCTCGTTATGTG - 5' Mutated DNA Sequence: 3' - CCTCGTTATTTG - 5'arrow_forward
- With 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_forwarda) It is known that double stranded DNA is denatured at low pH. pKa values should allow the determination of whether this is due to perturbation of the hydrogen bonding in A-T and/or G-C base pairs. The table gives values for the pKas of different protonated groups in the nucleobases.Nucleobase Position & pKa A N1, 3.5 G N7, 1.6; N1, 9.2 C N3, 4.2 T N3, 9.7a) Draw the A-T and G-C base pairs. - Label the bases with the one-letter code. – - Number the atoms in the rings and label the atom that attaches to the sugar. - Mark the groups that interact in normal…arrow_forwardAlthough DNA polymerases require both a template and a primer, the following single-stranded polynucleotide was found to serve as a substrate for DNA polymerase in the absence of any additional DNA.arrow_forward
- Help ASAParrow_forwardThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.arrow_forwardDNA isolated from an organism can be sheared into fragments of uniform size (∼1000 bp), heated to separate the strands, then cooled to allow complementary strands to reanneal. The renaturation process can be followed over time. Explain why the renaturation of E. coli DNA is a monophasic process, whereas the renaturation of human DNA is biphasic (an initial rapid phase followed by a slower phase).arrow_forward
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