Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 29, Problem 17P
Interpretation Introduction
Interpretation:
To determine the reason for the position of a-amanitin to be consistent with its inhibition modes.
Concept introduction:
DNA or Deoxyribonucleic acid is a molecule made of two chains which coil around one another. These form a double helix which carries instructions genetical in nature like related to reproduction, growth, development, functioning of the living organisms.
a-amanitin inhibits RNA polymerase II. This toxin which slows the polymerase translocation along DNA for NTP substrates has no effect on enzyme’s affinity.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
please provide handwritten explanation
1 it is written tranlation and trancription occurs early in bacteria does that means in bacteria central dogma rule dosen't follow
I am sending you two images
Fill in the box blank with the correct answer
Different sensitivities to the mushroom toxin a-amanitin distinguish the three RNA polymerases from one another. Which of the following properties listed below also distinguish RNA Polymerase II from Pol I and Pol III?
Options:
Only RNA Pol II possesses a large subunit
RNA Polymerase I and RNA Polymerase III do not require TBP for optimal transcription efficiency
only RNA Polymerase II requires an ATP-dependent helicase to melt the DNA around the transcription start site
Only RNA Polymerase II resembles the prokaryotic RNA Polymerase
RNA Pol II has an extended N terminal region that becomes phosphorylated during intiation
Chapter 29 Solutions
Biochemistry
Ch. 29 - Prob. 1PCh. 29 - The Events in Transcription Initiation Describe...Ch. 29 - Substrate Binding by RNA Polymerase RNA polymerase...Ch. 29 - Comparison of Prokaryotic and Eukaryotic...Ch. 29 - Prob. 5PCh. 29 - Prob. 6PCh. 29 - Prob. 7PCh. 29 - Alternative Splicing Possibilities Suppose exon 17...Ch. 29 - Prob. 9PCh. 29 - Prob. 10P
Ch. 29 - Post-transcriptional Modification of Eukaryotic...Ch. 29 - Prob. 12PCh. 29 - Prob. 13PCh. 29 - The Lariat Intermediate in RNA Splicing Draw the...Ch. 29 - Prob. 15PCh. 29 - Prob. 16PCh. 29 - Prob. 17PCh. 29 - Prob. 18PCh. 29 - Figure 29.15 highlights in red the DNA phosphate...Ch. 29 - Chromatin decompaction is a preliminary step in...
Knowledge Booster
Learn more about
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
- Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.arrow_forwardCynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X Garrow_forwardThe RNA polymerase from bacteriophage T7 diff ers structurally from prokaryotic and eukaryotic RNAPs and is extremely specifi c for its own promoter. Why do these properties make T7 RNAP useful in experiments with recombinant DNA?arrow_forward
- Hi please can I have the solarrow_forward6arrow_forwardFunctional Consequences of Y-Family DNA Polymerase Structure The eukaryotic translesion DNA polymerases fall into the Y family of DNA polymerases. Structural studies reveal that their fingers and thumb domains are small and stubby (see Figure 28.10). In addition, Y-family polymerase active sites are more open and less constrained where base pairing leads to selection of a dNTP substrate for the polymerase reaction. Discuss the relevance of these structural differences. Would you expect Y-family polymerases to have 3-exonuclease activity? Explain your answer.arrow_forward
- The Enzymatic Activities of DNA Polymerase I (a) What are the respective roles of the 5 -exonudease and 3 -exonuclease activities of DNA polymerase I? (b) What might be a feature of an E. coli strain that lacked DNA polymerase I 3 -exonuclease activity?arrow_forwardHeteroduplex 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_forwardMolecules of DNA Polymerase III per Cell vs. Growth Rate It is estimated that there are 40 molecules of DNA polymerase III per E. coli cell, is it likely that the growth rate of E. coli is limited by DNA polymerase III availability?arrow_forward
- Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram illustrating replication of the circular E. coli chromosome (a) at an early stage, (b) when one-third completed, (c) when two-thirds completed, and (d) when almost finished, assuming the initiation of replication at oriC has occurred only once. Then, draw a diagram showing the E. coli chromosome in problem 3 where the E. coli cell is dividing every 20 minutes.arrow_forwardSubstrate Binding by RNA Polymerase RNA polymerase has t binding sites for ribonucleoside triphosphates: the initiation site and the elongation site. The initiation site has a greater Km for NTPs than the elongation site. Suggest what possible significance this fact might have for the control of transcription in cells.arrow_forwardChemical Mutagenesis of DNA Bases Show the nucleotide sequence changes that might arise in a dsDNA (coding strand segment GCTA) upon mutagenesis with (a) HNO2, (b) bromouracil, and (C) 2-aminopurine.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning
Biochemistry
Biochemistry
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Cengage Learning
DNA vs RNA (Updated); Author: Amoeba Sisters;https://www.youtube.com/watch?v=JQByjprj_mA;License: Standard youtube license