Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 16P
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137
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2.1 Given the following eukaryotic DNA strand, transcribe and translate the DNA into apolypeptide using the 3’ – 5’ strand as the template. You may use drawings, diagrams,colours and annotations to describe how the DNA strand will be synthesized into afunctional protein.
5’ - TATAAAAASSMSBMDATGSBDCCMBDBAATBSMDSTGTGTCCTMSBAG – 3’ (KEY: The letters SBMD are “made up” nucleic acids that depict non-coding regions in theDNA, hypothetically S pairs with B and M pairs with D).
(i) (.:, From the diagram to the right of the trp repressor in its
approximate binding relationship to a double-stranded DNA mole-
cule, what is the relative orientation of the macrodipoles of helices 4
and 5 to each other? Indicate the N-terminal ends of helices 4 and
DNA
5
5? Which residue in the sequence above is the N-terminal residue
of each helix?
N
N
(ii) :. Comparing the two protein sequences above, identify all
amino acid pairs that differ in electrostatic charge due to proton dis-
sociable groups (assume pH 7). Indicate the charge of both residues
for each pair.
C.
2.1 Given the following eukaryotic DNA strand, transcribe and translate the DNA into apolypeptide using the 3’ – 5’ strand as the template. You may use drawings, diagrams,colours and annotations to describe how the DNA strand will be synthesized into afunctional protein.
5’ - TATAAAAASSMSBMDATGSBDCCMBDBAATBSMDSTGTGTCCTMSBAG – 3’ (KEY: The letters SBMD are “made up” nucleic acids that depict non-coding regions in theDNA, hypothetically S pairs with B and M pairs with D).
2.2. Describe what are missense mutations and its effects on structure and function usinghaemoglobin as an example .
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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- Heteroduplex DNA Formation in Recombination From the information in Figures 28.17 and 28.18, diagram the recombinational event leading to the formation of a heteroduplex DNA region within a bacteriophage chromosome.arrow_forwardpppApCpCpUpApGpApU-OH(a) Using the straight-chain sugar convention, write the structure of the DNA strand that encoded this short stretch of RNA.(b) Using the simplest convention for representing the DNA base sequence, write the structure of the nontemplate DNA strand.arrow_forwardA mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. Incorporation of radioactivity, cpm Excess (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?arrow_forward
- Consider the structure of Cro repressor protein from bacteriophage lambda E. It is a DNA binding protein, and like many sequence- specific DNA binding proteins, it must function as a homodimer Ex. Notice the mutual docking of a phenylalanine residue from each subunit into a hydrophobic pocket of the partner subunit. These hydrophobic interactions are required for dimerization. The noncovalent interactions highlighted in yellow are also required for dimerization. These interactions represent examples of: Osecondary structure O tertiary structure O quaternary structure O secondary AND quaternary structure Ⓒ tertiary AND quaternary structurearrow_forwardA mixture of four a-[*p]-labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular- weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribo- nucleoside triphosphates was added, with the results shown in the graph.arrow_forwardA mixture of four a-[32P]–labeled ribonucleoside triphosphates was added to permeabilized bacterial cells undergoing DNA replication in the presence of an RNA polymerase inhibitor, and incorporation into high-molecular-weight material was followed over time, as shown in the accompanying graph. After 10 minutes of incubation, a 1000-fold excess of unlabeled ribonucleoside triphosphates was added, with the results shown in the graph. (a) Why was the excess of unlabeled rNTPs added?(b) How could you tell that radioactivity is being incorporated as ribonucleotides rather than as an alternative such as reduction to deoxyribonucleotides, followed by incorporation?(c) What does this experiment tell you about the process of DNA replication?arrow_forward
- 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_forwardUnlike linker DNA and deproteinized DNA, DNA segments wrapped around histone cores are relatively resistant to the hydrolytic actions of nucleases. Explain.arrow_forwardDNA polymerase I, DNA ligase, and topoisomerase I catalyze the formation of phosphodiester bonds. What is the activated intermediate in the linkage reaction catalyzed by each of these enzymes? What is the leaving group?arrow_forward
- Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram illustrating replication of the circular E. coli chromosome (a) at an early stage, (b) when one-third completed, (c) when two-thirds completed, and (d) when almost finished, assuming the initiation of replication at oriC has occurred only once. Then, draw a diagram showing the E. coli chromosome in problem 3 where the E. coli cell is dividing every 20 minutes.arrow_forwardNumber of Okazaki Fragments in E. coli and Human DNA Replication Approximately how many Okazaki fragments are synthesized in the course of replicating an E. coli chromosome? How many in replicating an “average� human chromosome?arrow_forwardMultiple Replication Forks in E. coli I Assuming DNA replication proceeds at a rate of 750 base pairs per second, calculate how long it will take to replicate the entire E. coli genome. Under optimal conditions, E. coli cells divide every 20 minutes. What is the minimal number of replication forks per E. coli chromosome in order to sustain such a rate of cell division?arrow_forward
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