Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Textbook Question
Chapter 28, Problem 6P
Number of Okazaki Fragments in E. coli and Human
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4a in context to taking genomic DNA from eukaryotic cells and randomly shearing it into pieces of a constant size, why do some of the genomic DNA fragments re-nature so much more quickly than other fragments
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N), heavy parental DNA, and all nucleotides with 14N (another isotope of N) as "nucleotide
supply source" for DNA replication in a tube. Use the figure provided below to illustrate the
arrangement of light and heavy isotopes of nitrogen in DNA molecules formed in Generation
one, two and three. (Please use the shade or different color to represent the daughter strands of
DNA with light N-isotope, so as to be different from the parental DNA strands.)
Make sure to label what each color represents clearly. Otherwise, the answer is not considered
correct.
semiconservative replication
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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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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- Human Genome Replication Rate Assume DNA replication proceeds at a rate of 100 base pairs per second in human cells and origins of replication occur every 300 kbp. Assume also that human DNA polymerases are highly processive and only two molecules of DNA polymerase arc needed per replication fork. How long would it take to replicate the entire diploid human genome? How many molecules of DNA polymerase does each cell need to carry out this task?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_forwardMultiple 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_forward
- 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 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.arrow_forwardSemiconservative or Conservative DNA Replication If 15N-Iabeled E. coli DNA has a density of 1.724 g/mL, 14N-labeled DNA has a density of 1.710 g/mL, and E. coli cells grown for many generations on 14NH4+as a nitrogen source are transferred to media containing 15NH4+as the sole N-source, (a) What will be the density of the DNA after one generation, assuming replication is semiconservative? (b) Suppose replication took place by a conservative mechanism in which the parental strands remained together and the two progeny strands were paired. Design an experiment that could distinguish between semiconservative and conservative modes of replication.arrow_forwardReplication involves a period of time during which DNA is particularly susceptible to the introduction of mutations. If nucleotides can be incorporated into DNA at a rate of 20 nucleotides/second and the human genome contains 3 billion nucleotides, how long will replication take? How is this time reduced so that replication can take place in a few hours?arrow_forward
- 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_forwardFunctional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.arrow_forwardHomologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesn’t the mismatch repair system of E. coli eliminate these mismatches?arrow_forward
- When Chargaffwas performing his experiments, the tetranucleotide hypothesis, which stated that DNA was composed of GACT nucleotide repeats, was the most widely accepted view of DNA’s composition. How did Chargaff disprove this hypothesis?arrow_forwardFigure 9.10 You isolate a cell strain in which the joining together of Okazaki fragments is impaired and suspect that a mutation has occurred in an enzyme found at the replication fork. Which enzyme is most likely to be mutated?arrow_forwardMolecules of DNA Polymerase III per Cell vs. Growth Rate It is estimated that there are 40 molecules of DNA polymerase III per E. coli cell, is it likely that the growth rate of E. coli is limited by DNA polymerase III availability?arrow_forward
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