Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 14, Problem 2ITD
Summary Introduction
To review:
The difference between the most effective schedule and other schedules of the drugs 5-fluorouracil (5-FU) and cisplatin (CDDP) added to cultured gastric cancer cells.
Introduction:
Cancer is a genetic disease in which there is an uncontrolled or abnormal growth of cells. Cancer cell grows faster than normal cells and leads to the formation of tumor. Some cancer drugs target rapidly dividing cells, and inhibit their growth or cause their death. This is called chemotherapy.
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Chapter 14 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 14.1 - Prob. 1SBCh. 14.2 - Prob. 1SBCh. 14.2 - Prob. 2SBCh. 14.2 - Prob. 3SBCh. 14.2 - Prob. 4SBCh. 14.3 - What is the importance of complementary base...Ch. 14.3 - Why is a primer needed for DNA replication? How is...Ch. 14.3 - DNA polymerase III and DNA polymerase I are used...Ch. 14.3 - Prob. 4SBCh. 14.4 - Why is a proofreading mechanism important for DNA...
Ch. 14 - Working on the Amazon River, a biologist isolated...Ch. 14 - Prob. 2TYKCh. 14 - Pyrimidines built from a single carbon ring are:...Ch. 14 - Which of the following statements about DNA...Ch. 14 - Which of the following statements about DNA is...Ch. 14 - Prob. 6TYKCh. 14 - Prob. 7TYKCh. 14 - Prob. 8TYKCh. 14 - Prob. 9TYKCh. 14 - Prob. 10TYKCh. 14 - Discuss Concepts Eukaryotic chromosomes can be...Ch. 14 - Prob. 12TYKCh. 14 - Prob. 13TYKCh. 14 - Discuss Concepts During replication, an error...Ch. 14 - Design an Experiment Design an experiment using...Ch. 14 - Prob. 16TYKCh. 14 - Prob. 1ITDCh. 14 - Prob. 2ITD
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- Human 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_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_forwardIn relation to central dogma of molecular biology answer the following questions: A- Give two reasons why the DNA replication is asymmetrical process (i.e. the DNA replication outcome is different between the leading and lagging strands)? B-The following segment of DNA is part of the transcription unit of a gene. You know already that RNA polymerase moves in a specific direction along this piece of DNA to convert one of the DNA strands into a single strand RNA transcript so that this entire region of DNA is made into RNA. 5′-GGCATGGCAATATTGTAGTA-3′ 3′-CCGTACCGTTATAACATCAT-5′ Given this information, a student claims that the RNA produced from this DNA is: 3′-GGCATGGCAATATTGTAGTA-5′ Give two reasons why this answer is incorrect. C- Imagine that the mRNA codons consisted of only two nucleotides instead of three nucleotides. Would there be a sufficient number of codons for all twenty amino acids? Explain your answer. D- The length of a particular gene in human DNA,…arrow_forward
- Eukaryotic licensing factors prevent DNA replication from being initiated at origins more than once in the cell cycle. After replication has begun at an origin, a protein called Geminin inhibits licensing factors that are required for MCM2-7 to bind to an origin and initiate replication. Thus, when Geminin is present, MCM2-7 will not bind to an origin. At the end of mitosis, Geminin is degraded, allowing MCM2-7 to bind once again to DNA and relicense the origin. Marina Melixetian and her colleagues suppressed the expression of Geminin protein in human cells by treating the cells with small interfering RNAs (siRNAs) complementary to Geminin messenger RNA . Forty-eight hours after treatment with siRNA, the Geminin-depleted cells were enlarged and contained a single giant nucleus. Analysis of DNA content showed that many of these Geminin-depleted cells were 4 n or greater. Explain these results.arrow_forwardEukaryotic licensing factors prevent DNA replication from being initiated at origins more than once in the cell cycle. After replication has begun at an origin, a protein called Geminin inhibits licensing factors that are required for MCM2-7 to bind to an origin and initiate replication. Thus, when Geminin is present, MCM2-7 will not bind to an origin. At the end of mitosis, Geminin is degraded, allowing MCM2-7 to bind once again to DNA and relicense the origin. Marina Melixetian and her colleagues suppressed the expression of Geminin protein in human cells by treating the cells with small interfering RNAs (siRNAs) complementary to Geminin messenger RNA (M. Melixetian et al. 2004. Journal of Cell Biology 165:473–482). (Small interfering RNAs form a complex with proteins and pair with complementary sequences on mRNAs; the complex then cleaves the mRNA, so there is no translation of the mRNA; . Forty-eight hours after treatment with siRNA, the Geminin-depleted cells were enlarged and…arrow_forwardIn your own words, explain how cancer cells differ from normal cells in regard to the following: Telomeres, which are products of telomerase enzymearrow_forward
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