TABLE Characteristics of DNA polymerases in E. coli 3' -5' Polymerase Exonuclease Exonuclease Activity 5' - 3' 5'-3' DNA Polymerase Activity Activity Function Removes and replaces primers Yes Yes Yes II DNA repair; restarts replication after damaged DNA halts synthesis Yes Yes No III Yes Yes No Elongates DNA IV Yes No No DNA repair Yes No No DNA repair; translesion DNA synthesis

TABLE Characteristics of DNA polymerases in E. coli 3' -5' Polymerase Exonuclease Exonuclease Activity 5' - 3' 5'-3' DNA Polymerase Activity Activity Function Removes and replaces primers Yes Yes Yes II DNA repair; restarts replication after damaged DNA halts synthesis Yes Yes No III Yes Yes No Elongates DNA IV Yes No No DNA repair Yes No No DNA repair; translesion DNA synthesis

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 17P

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Do virtual restriction digests of 300ng pGLO plasmid using HinDIII, EcoRI, and XhoI separately, and a double digest of pGLO with HinDIII plus EcoRI. How many nanograms of DNA should we expect to be contained within the slowest bandfrom the HinDIII-digest of pGLO?
The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGAT Assume that RNA polymerase proceeds along this template from left to right. Q.Which end of the DNA template is 5′ and which end is 3′?
Strand directed mismatch repair corrects copying erros based on Which of the following apply to the statement above, there maybe more than one. a. the strand containg breaks in the sugar backbone of the replicated DNA b. structurral instabilities created between oncorrect base pairing and associated histone proteins c. mutations that are evolutionary beneficial to an organims and recults in these areas of the genetic code to be turned off d. methylation of adenosine in GATC sequences during bacterial DNA replication e. repair sequence motifs that attract DNA polymerase gamma
DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 5´ to 3´ exonuclease activities required for removal of the RNA primer, a central domain responsible for 3´ to 5´ exonuclease proofreading, and a C-terminal domain with polymerase activity. Pol I is thought to simultaneously remove RNA primers and fill in the gaps that result. A group of proteins known as RNaseH also have 5´ to 3´ exonuclease activity and can thus remove RNA primers. However, they lack the other two functions observed for Pol I. Predict the ability of the following mutants to replicate DNA: (1) a strain with a mutant gene encoding Pol I such that it no longer has polymerase activity (but retains both types of nuclease activities); (2) a strain without RNaseH proteins; (3) a strain with a mutant gene encoding Pol I such that it no longer has 5´ to 3´ exonuclease activities (but retains 3´ to 5´ nuclease and polymerase activities); (4) a strain with…
PolyADP-ribose polymerase (PARP) plays a keyrole in the repair of DNA single-strand breaks. In the pres-ence of the PARP inhibitor olaparib, single-strand breaksaccumulate. When a replication fork encounters a sin-gle-strand break, it converts it to a double-strand break,which in normal cells is then repaired by homologousrecombination. In cells defective for homologous recom-bination, however, inhibition of PARP triggers cell death.Patients who have only one functional copy of theBrca1 gene, which is required for homologous recombina-tion, are at much higher risk for cancer of the breast andovary. Cancers that arise in these tissues in these patientscan be treated successfully with olaparib. Explain how it isthat treatment with olaparib kills the cancer cells in thesepatients, but does not harm their normal cells.
Reiji and Tuneko Okazaki conducted a now classic experiment in1968 in which they discovered a population of short fragmentssynthesized during DNA replication. They introduced a shortpulse of 3H-thymidine into a culture of E. coli and extracted DNAfrom the cells at various intervals. In analyzing the DNA aftercentrifugation in denaturing gradients, they noticed that as theinterval between the time of 3H-thymidine introduction and thetime of centrifugation increased, the proportion of short strandsdecreased and more labeled DNA was found in larger strands.What would account for this observation?
DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 59 to 39 exonuclease activity required for removal of the RNA primer, a central domain responsible for 39 to 59 exonuclease proofreading, and a C-terminal domain with polymerase activity. Pol I is thought to simultaneously remove RNA primers and fill in the gaps that result (figure 13.14). A group of proteins known as RNaseH also have 59 to 39 exonuclease activity and can thus remove RNA primers. However, they lack the other two functions observed for Pol I. Predict the ability of the following mutants to replicate DNA: (1) a strain with a mutant gene encoding Pol I such that it no longer has polymerase activity (but retains both types of nuclease activities); (2) a strain without RNaseH proteins; (3) a strain with a mutant gene encoding Pol I such that it no longer has 59 to 39 exonuclease activity (but retains 39 to 59 nuclease and polymerase activities); (4) a strain with the…
The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGAT Assume that RNA polymerase proceeds along this template from left to right.
To determine the reproducibility of mutation fre-quency measurements, you do the following experiment.You inoculate each of 10 cultures with a single E. coli bac-terium, allow the cultures to grow until each contains 106cells, and then measure the number of cells in each culturethat carry a mutation in your gene of interest. You were sosurprised by the initial results that you repeated the experi-ment to confirm them. Both sets of results display the sameextreme variability, as shown in Table Q5–1. Assuming thatthe rate of mutation is constant, why do you suppose thereis so much variation in the frequencies of mutant cells indifferent cultures?
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