Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 8P
Human Genome Replication Rate Assume
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The human genome contains about 3 billion base pairs. During the first cell division after fertilization of a human embryo, S phase is approximately three hours long. Assuming an average DNA polymerase rate of 50 nucleotides/second over the entire S phase, what is the minimum number oforigins of replication you would expect to find in the human genome? Show your solution.
The chromosome of E. coli contains 4.6 million bp. How long will it take to replicate its DNA? Assuming that DNA polymerase III is the primary enzyme involved and that it can actively proofread during DNA synthesis, how many base pair mistakes will be made in one round of DNA replication in a bacterial population containing 1000 bacteria?
Approximately how many high-energy bonds does DNA polymerase use to replicate a bacterial chromosome (ignoring helicase and other enzymes associated with the replication fork)? compared with its own dry weight of 10–12 g, how much glucose does a single bacterium need to provide enough energy to copy its DNA once?
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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- Number 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_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
- Semiconservative 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_forwardWhich statements are true? Explain why or why not.1 The different cells in your body rarely havegenomes with the identical nucleotide sequence.2 In E. coli, where the replication fork travels at 500nucleotide pairs per second, the DNA ahead of the fork—in the absence of topoisomerase—would have to rotate atnearly 3000 revolutions per minute.3 In a replication bubble, the same parental DNAstrand serves as the template strand for leading-strandsynthesis in one replication fork and as the template forlagging-strand synthesis in the other fork.4 When bidirectional replication forks from adja-cent origins meet, a leading strand always runs into a lag-ging strand.5 DNA repair mechanisms all depend on the exis-tence of two copies of the genetic information, one in eachof the two homologous chromosomesarrow_forward
- 3a) In eukaryotic cells that lack telomerase, the telomeres at the ends of the chromosomes gradually get shorter with each round of DNA replication. Describe or explain why the "normal" DNA replication machinery, excluding telomerase, can't completely and accurately replicate all the DNA at the ends of linear chromosomes. Please note that the question does NOT ask you to describe what telomerase does - it asks you to explain why cells without telomerase have this problem.arrow_forwardDNA Replication occurs on both prokaryotes and eukaryotes. Although they have a similar genetic flow, there are small differences in between. What are the differences of DNA replication in prokaryotes and eukaryotes? What is/are the major difference/s?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
- 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_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_forwardHelicase 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_forward
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