Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 3P
Multiple Replication Forks in E. coli I Assuming
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Which 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 chromosomes
All known DNA polymerases catalyze synthesis only in the 5' → 3' direction. Nevertheless, during semiconservative DNA replication in the cell, they are able to catalyze the synthesis of both daughter chains, which would appear to require synthesis in the 3' → 5' direction on one strand. Explain the process that occurs in the cell that allows for synthesis of both daughter chains by DNA polymerase
Suppose that 22% of the nucleotides of a DNA molecule are deoxyadenosine and during replication the relative amounts of available deoxynucleoside triphosphates are 22% dATP, 22% dCTP, 28% dGTP, and 28% dTTP. What deoxynucleoside triphosphate is limiting to the replication? Explain.
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
- 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_forwardHuman 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_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
- to 4 minutes) The schematic diagram below shows organization of the DNA replication fork. Match parts of the diagram (labeled A-F) with the corresponding term from the answer list (designated 31 parental duplex 5' 3' fork progression v A 1Lagging strand 2. An Okazaki tragment 3.Site of action of DNA topoisomerase 4 Leading strand 5. Site of action of DNA helicCase 6.Site of action of DNA ligasearrow_forward2a) There are two different DNA polymerase enzymes, DNA Polymerase I and DNA Polymerase III, that are active during prokaryotic DNA replication. Suppose you generated a mutant E. coli strain in which DNA Polymerase III was inactivated (all its enzymatic activities were non-functional) - assuming that all the other enzymes involved in replication remained fully functional, how would DNA replication in these mutant cells without DNA Pol III differ from DNA replication in normal E. coli? Briefly explain why you would expect to see that change/those changes in DNA replication in the mutant cells.arrow_forwardEukaryotic Genetic Sequence: 5'-TAC CAT GAT CCC TAT - 3' 1. What would be the newly synthesized DNA strand and explain how the strand will be replicated. Where in the cell would this occur? 2. What would be the synthesized mRNA strand, and how is it transcribed from the original DNA strand, and then converted from a pre-mRNA strand to a mature mRNA? Where in the cell does this occur? 3. What would be the anti-codons for the tRNA. What are the amino acids generated based on the RNA. How are these amino acids translated into protein and where in the cell does this happen?arrow_forward
- Matching Type Choose the directionality of the given process. (4 points) What is the directionality of the given process? * 4 points 3'-5' 5'-3' Exonuclease activity Complementary strand of the continuous strand Addition of nucleotides going to the replication fork Addition of nucleotides away from the replication forkarrow_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_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
- The E. coli chromosome is 1.28 mm long. Under optimal conditions, thechromosome is replicated in 40 minutes.(a) What is the distance traversed by one replication fork in 1 minute?(b) If replicating DNA is in the B form (10.4 base pairs per turn), how manynucleotides are incorporated in 1 minute in one replication fork?(c) If cultured human cells (such as HeLa cells) replicate 1.2 m of DNAduring a five-hour S phase and at a rate of fork movement one-tenthof that seen in E. coli, how many origins of replication must the cellscontain?(d) What is the average distance, in kilobase pairs, between these origins?arrow_forwardFigure 6-7 CONDITION (A) bacteria grown in light medium (B) bacteria grown in heavy medium TRANSFER TO LIGHT MEDIUM (C) bacteria grown an additional 1 hour in light medium RESULT centrifugal force centrifugal force centrifugal force INTERPRETATION AVIV AVVY AVVI light DNA molecules AMM heavy DNA molecules OR WW DNA molecules of intermediate weight In the original Meselson-Stahl experiment shown above bacteria were cultured in 15N-containing "heavy" medium and thenarrow_forwardPlease help with 2a) 2a) There are two different DNA polymerase enzymes, DNA Polymerase I and DNA PolymeraseIII, that are active during prokaryotic DNA replication. Suppose you generated a mutant E. colistrain in which DNA Polymerase III was inactivated (all its enzymatic activities were non-functional) - assuming that all the other enzymes involved in replication remained fullyfunctional, how would DNA replication in these mutant cells without DNA Pol III differ fromDNA replication in normal E. coli? Briefly explain why you would expect to see thatchange/those changes in DNA replication in the mutant cells.arrow_forward
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