ESSENTIALS OF GENETICS MCC BUNDLE >BI<
9th Edition
ISBN: 9781323915370
Author: KLUG
Publisher: PEARSON
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Chapter 10, Problem 21PDQ
Summary Introduction
To review:
The following:
(a) The effect of mutation in dnaE gene in E.coli and the maintenance of mutant strain.
(b) The effect of mutation in dnaQ gene in E.coli.
Introduction:
DNA (deoxyribonucleic acid) synthesis is a process of incorporating
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Chapter 10 Solutions
ESSENTIALS OF GENETICS MCC BUNDLE >BI<
Ch. 10 -
CASE STUDY | At loose ends
A researcher was...Ch. 10 - Prob. 2CSCh. 10 - Prob. 3CSCh. 10 - Prob. 4CSCh. 10 - HOW DO WE KNOW? In this chapter, we focused on how...Ch. 10 - Review the Chapter Concepts list on p. 180. These...Ch. 10 - Compare conservative, semiconservative, and...Ch. 10 - Prob. 4PDQCh. 10 - Predict the results of the experiment by Taylor,...Ch. 10 - Prob. 6PDQ
Ch. 10 - Prob. 7PDQCh. 10 - Prob. 8PDQCh. 10 - Prob. 9PDQCh. 10 - Prob. 10PDQCh. 10 - Prob. 11PDQCh. 10 - Prob. 12PDQCh. 10 - Prob. 13PDQCh. 10 -
14. Distinguish between (a) unidirectional and...Ch. 10 - Prob. 15PDQCh. 10 - Define and indicate the significance of (a)...Ch. 10 - Outline the current model for DNA synthesis.Ch. 10 - Why is DNA synthesis expected to be more complex...Ch. 10 - Prob. 19PDQCh. 10 - Several temperature-sensitive mutant strains of E....Ch. 10 - Prob. 21PDQCh. 10 - Prob. 22PDQ
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- What is an Okazaki fragment? In which strand of replicating DNA are Okazaki fragments found? Based on the properties of DNA polymerase, why is it necessary to make these fragments?arrow_forwardConsider the ends of the DNA fragments shown below. They have been produced by digestion of a single sequence of DNA using a number of restriction endonucleases. 1. 5'A 3' 3'TTCGA5' 2. 5'G 3' 3'CAGCT5' 3. 5'AATTC3' 3' G5 4. 5'TCGAC3' 3' G5' 5. 5'GGG 3' 3'CCC 5' Which of these ends are capable of annealing and being joined by DNA ligase?arrow_forwardConsider the following sequence of DNA: 3'-TTA CGG-5'What dipeptide is formed from this DNA after transcription and translation? b. If a mutation converts CGG to CGT in DNA, what dipeptide is formed? c. If a mutation converts CGG to CCG in DNA, what dipeptide is formed? d. If a mutation converts CGG to AGG in DNA, what dipeptide is formed?arrow_forward
- What is meant by non-classical DNA-dependent DNA polymerases? What roles do they play in our cells?arrow_forwardThree common ways to repair changes in DNA structure are nucleotideexcision repair, mismatch repair, and homologous recombination repair. Which of these three mechanisms would be used to fix the following types of DNA changes?A. A change in the structure of a base caused by a mutagen in anondividing eukaryotic cellB. A change in DNA sequence caused by a mistake made by DNApolymeraseC. A thymine dimer in the DNA of an actively dividing bacterial cellarrow_forwarda) "Out of three E.coli DNA polymerases, DNA polymerases 3 has a high processivity and rate of polymerization and therefore better suited for replication of the genome" What is meant by processivity? how does the DNA polymerase 3 maintain high processivity? b) What is a replication fork ?. Give the protein/enzymes of a replication fork and describe their function?arrow_forward
- the one above: Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT Compare this mutated sense sequence given below to the original one given above and identify and classify all mutations that can be found in this new DNA sequence? TGAGCATGAAACTCACACCGGGGGCAGTTTCGCACTTAGGATTCTTGTACAGGACCTAGTATAACAAGTT 2. Using this mutated DNA strand, express it as a polypeptide by using the correct reading frame. When you get to the stop codon – you may write an “*” to denote the stop codon. 3. How many amino acids were changed in the mutated polypeptide?arrow_forwardDNA repair enzymes preferentially repair mis- matched bases on the newly synthesized DNA strand, using the old DNA strand as a template. If mismatches were instead repaired without regard for which strand served as template, would mismatch repair reduce repli- cation errors? Would such a mismatch repair system result in fewer mutations, more mutations, or the same number of mutations as there would have been without any repair at all? Explain your answers.arrow_forwardA temperature-sensitive mutation is one in which the defect is not presented functionally until the temperature is raised. In the case described below, the enzymes function normally in bacteria at 37 °C, but are non-functional at 40 °C. Predict the detailed molecular consequences of a loss of function in a temperature-sensitive mutant for each of the following enzymes: a) DNA gyrase, b) DNA polymerase III, c) DNA ligase, d) DNA polymerase I.arrow_forward
- Considering prokaryotes, what is the enzyme that helps hold DNA polymerase III in place when nucleotides are being added?arrow_forwardHow do we know that DNA synthesis is discontinuous on one of the two template strands?arrow_forwardHuman Fbh1 helicase is important in the process of DNA replication. When a mutation occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? A - Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triangle because helicase is needed to separate the strands of DNA. B - Topoisomerase would unwind the DNA, but then the process would stop at the blue triangle because helicase, the RNA primer, would not be able to attach to the DNA molecule and initiate replication. C - The process would begin at the blue triangle when topoisomerase unwinds the DNA and an RNA primer attaches to the DNA molecule and initiates replication. DNA polymerase would begin the synthesis…arrow_forward
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