Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 2P
The Enzymatic Activities of DNA Polymerase I (a) What are the respective roles of the
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2a) 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.
DNA ligase has the ability to relax supercoiled circular DNA in the
presence of AMP but not in its absence.
(a) What is the mechanism of this reaction, and why is it dependent
on AMP?
(b) How might one determine that supercoiled DNA had in fact
been relaxed?
DNA ligase has the ability to relax supercoiled circular DNA in the presence
of AMP but not in its absence.
(a) What is the mechanism of this reaction, and why does it depend on AMP?
(b) How could you determine that supercoiled DNA had in fact been relaxed?
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
- 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 polymerasearrow_forwardThe 3′ → 5′ exonuclease activity of Pol I excises only unpaired 3′-terminal nucleotides from DNA, whereas this enzyme’s pyrophosphorolysis activity removes only properly paired 3′-terminal nucleotides. Discuss the mechanistic signifi cance of this phenomenon in terms of the polymerase reaction.arrow_forwardUltraviolet light can cause covalent linkages between consecutive pyrimidine bases in DNA (up to 100 per second in a single cell in sunlight!). These bulky lesions (i.e. cyclobutane pyrimidine dimers and 6-4 photoproducts), which inhibit DNA and RNA polymerases, are mostly reversed by CPD photolyase when light >300 nm is available to power the reaction. In the dark, however, which DNA repair system is best able to correct these errors? a) non-homologous end-joining b) mismatch repairc) nucleotide-excision repaird) base-excision repair e) homology-directed repairarrow_forward
- DNA 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_forwardIn E. coli, all newly synthesized DNA appears to be fragmented (an observation that could be interpreted to mean that the leading strand as well as the lagging strand is synthesized discontinuously). However, in E. coli mutants that are defective in uracil–DNA glycosylase, only about half the newly synthesized DNA is fragmented. Explain.arrow_forwardA solution containing single stranded DNA with the sequence 5’ATGGTGCACCTGACTCCTGAGGAGAAGTCTNNNNN’3 undergoes DNA replication in vitro in the presence of all four nucleotides plus an amount of dideoxyadenosine triphosphate sufficient to compete for incorporation with deoxyadenosine triphosphate. How many and what DNA fragments are expected?arrow_forward
- 3aarrow_forwardAlthough DNA polymerases require both a template and a primer, the following single-stranded polynucleotide was found to serve as a substrate for DNA polymerase in the absence of any additional DNA.3′ HO-ATGGGCTCATAGCCGGAGCCCTAACCGTAGACCACGAATAGCATTAGG-p 5′What is the structure of the product of this reaction?arrow_forwardBearing in mind the different number of hydrogen bonds that form between the two different purine- pyrimidine pairs in DNA, how would you explain the fact that DNA that is rich in cytosine-guanine pairs requires heating to a slightly higher temperature in order to separate the strands than DNA that is rich in adenine-thymine pairs?arrow_forward
- In the following sequence, a cytosine was deaminated and is now a uracil (underlined). 5’-GGTAUTAAGC-3’ a. Which repair pathway(s) could restore this uracil to cytosine? b. If the uracil is not removed before a DNA replication fork passes through, what will be the sequences of the two resulting double helices? Provide the sequences of both strands of both helices. Label the old and new strands and underline the mutation(s). c. Could the mismatch repair pathway fix the mutations you’ve indicated in part b? d. If the cell undergoes mitosis, and the replicated DNAs are distributed into the two daughter cells. Will 0, 1, or 2 daughter cells have a mutation in this sequence?arrow_forwardMany of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. (a) The dnaE gene encodes the a subunit of DNA polymerase III. What effect is expected from a mutation in this gene? How could the mutant strain be maintained? (b) The dnaQ gene encodes the e subunit of DNA polymerase. What effect is expected from a mutation in this gene?arrow_forwardWhat is the role of Mg2+ in this reaction?arrow_forward
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