Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 28, Problem 10P
Homologous 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?
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Homologous recombination in E. coli forms heteroduplex regions of DNA containing mismatched bases. Why are these mismatches not eliminated bythe mismatch repair system?
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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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- 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_forwardIf DNA polymerase requires a perfectly pairedprimer in order to add the next nucleotide, how is it thatany mismatched nucleotides “escape” this requirementand become substrates for mismatch repair enzymes?arrow_forwardIn addition to correcting DNA mismatches, themismatch repair system functions to prevent homologousrecombination from taking place between similar but notidentical sequences. Why would recombination betweensimilar, but nonidentical sequences pose a problem forhuman cells?arrow_forward
- Richard Boyce and Paul Howard-Flanders conducted an experimentthat provided biochemical evidence that thymine dimers areremoved from DNA by a DNA repair system. In their studies, bacterialDNA was radiolabeled so the amount of radioactivity reflectedthe amount of thymine dimers. The DNA was thensubjected to UV light, causing the formation of thymine dimers. When radioactivity was found in the soluble fraction, thymine dimershad been excised from the DNA by a DNA repair system.But when the radioactivity was in the insoluble fraction, the thyminedimers had been retained within the DNA. The following tableillustrates some of the experimental results involving a normalstrain of E. coli and a mutant strain that was very sensitive to killingby UV light: Explain the results found in this table. Why is the mutant strainsensitive to UV light?arrow_forwardIn 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_forwardThe 3′-exonuclease activity of E. coli DNA polymerase I was found to show no discrimination between correctly and incorrectly base-paired nucleotides at the 3′-terminus; properly and improperly base-paired nucleotides are cleaved at equal rates there. How can this observation be reconciled with the fact that the 3′-exonuclease activity increases the accuracy with which template DNA is copied?arrow_forward
- Nonhomologous end-joining (NHEJ) of a doublestrand break almost always results in perfect resealingof the DNA lesion, without the loss or gain of nucleotide pairs. Yet CRISPR/Cas9, which produces doublestrand breaks, is a highly efficient method of makingsmall deletions or insertions at the targeted site. Howcan you resolve this apparent contradiction?arrow_forwardDNA 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…arrow_forwardLet’s say that a stretch of repeated AT issuccessfully sequenced. From what you know of the difficulties ofsequencing long repeated sequences, what other problems mightyou encounter in assembling these fragments?arrow_forward
- What is the advantage of homology-directed repair (HDR) ?arrow_forwardDNA 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_forwardSupercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?arrow_forward
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