Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 18, Problem 57SA
Summary Introduction
To explain:
The telomerase activity is higher in cancer cell than in normal somatic cells.
Introduction:
Telomerase is an enzyme, also referredas terminal transferase. It is ribonucleoprotein found in adult germ cells, somatic cells, fetal tissue, snd tumor cells. It is active in most cancer cells and less active in somatic cells. Its main function is to restore the telomeres (short stretches of DNA) that become shorter when a cell divides.
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Budding yeasts such as S. cerevisiae exhibit telomerase activity throughout their life cycles, whereas human somatic cells do not exhibit telomerase activity. Suggest why this is so.
Why do adult human cells (other than germ cells and stem cells) NOT express the enzyme telomerase? In other words what benefit does not having telomerase provide to these cells?
How is telomerases related to ageing? How Can telomerase- related ageing be addressed therapeutically?
Chapter 18 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 1RQCh. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - Prob. 4RQCh. 18 - Prob. 5RQ
Ch. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - Prob. 10RQCh. 18 - Prob. 11RQCh. 18 - Prob. 12RQCh. 18 - Prob. 13RQCh. 18 - Prob. 14RQCh. 18 - Prob. 15RQCh. 18 - Prob. 16RQCh. 18 - Prob. 17RQCh. 18 - Prob. 18RQCh. 18 - Prob. 19RQCh. 18 - Prob. 20RQCh. 18 - Prob. 21RQCh. 18 - Prob. 22RQCh. 18 - Prob. 23RQCh. 18 - Prob. 24RQCh. 18 - Prob. 25RQCh. 18 - Prob. 26RQCh. 18 - Prob. 27RQCh. 18 - Prob. 28RQCh. 18 - Prob. 29RQCh. 18 - Prob. 30RQCh. 18 - Prob. 31RQCh. 18 - Prob. 32RQCh. 18 - Prob. 33RQCh. 18 - Prob. 34RQCh. 18 - Prob. 35RQCh. 18 - Prob. 36RQCh. 18 - Prob. 37RQCh. 18 - Prob. 38RQCh. 18 - Prob. 39RQCh. 18 - Prob. 40RQCh. 18 - Prob. 41RQCh. 18 - Prob. 42RQCh. 18 - Prob. 43RQCh. 18 - Prob. 44RQCh. 18 - Prob. 45RQCh. 18 - Prob. 46RQCh. 18 - Prob. 47FBCh. 18 - Prob. 48FBCh. 18 - Prob. 49FBCh. 18 - Prob. 50FBCh. 18 - Prob. 51FBCh. 18 - Prob. 52FBCh. 18 - Prob. 53FBCh. 18 - Prob. 54FBCh. 18 - Prob. 55FBCh. 18 - Prob. 56FBCh. 18 - Prob. 57SACh. 18 - Prob. 58SACh. 18 - Prob. 59SACh. 18 - Prob. 60SACh. 18 - Prob. 61SACh. 18 - Prob. 62TQCh. 18 - Prob. 63TQCh. 18 - Prob. 64TQCh. 18 - Prob. 65TQCh. 18 - Prob. 66TQCh. 18 - Prob. 67TQCh. 18 - Prob. 68TQCh. 18 - Prob. 69TQCh. 18 - Prob. 70TQCh. 18 - Prob. 71TQCh. 18 - Prob. 72TQCh. 18 - Prob. 73TQCh. 18 - Prob. 74TQCh. 18 - Prob. 75TQ
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- How is telomerase related to aging? How can telomerase-related aging beaddressed therapeutically? What is a potential danger of telomerase activation?arrow_forwardWhy would replication come to a halt in the absence of topoisomerase II?arrow_forwardCurrently available in the marketplace are pills that supposedly increase telomerase activity. Based on your knowledge of DNA replication, explain why this would be desirable.arrow_forward
- If a mutation that inactivated telomerase occurred in acell (telomerase activity in the cell = zero), what do youexpect the outcome to be?arrow_forwardThe restriction enzymes KpnI and Acc65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from KpnI cleavage cannot be ligated directly to the sticky end formed from Acc65I cleavage. Explain why.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_forward
- The restriction enzymes Kpn I and Acc 65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from Kpn I cleavage cannot be ligated directly to the sticky end formed from Acc 65I cleavage. Explain why.arrow_forward"The molecule serving as the genetic material is expected to absorb at the wavelengths shown to be mutagenic." Explain this statement ?arrow_forwardThe topoisomerase enzyme catalyzes the interconversion of the relaxed and supercoiled forms of DNA by making transient breaks in one or both DNA strands. What is the difference between I and II?arrow_forward
- Telomeres contain a 3′ overhang region, as shown. Does telomerase require a 3′ overhang to replicate the telomere region? Explain.arrow_forwardIn producing genetically engineered human insulin in bacteria, why is it important to use the samerestriction enzyme to cut both the human DNA and the bacterial plasmid?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
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