Biochemistry
6th Edition
ISBN: 9781337359573
Author: Reginald H. Garrett; Charles M. Grisham
Publisher: Cengage Learning US
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
Consider the structure of Cro repressor protein from bacteriophage lambda E. It is a DNA binding protein, and like many sequence-
specific DNA binding proteins, it must function as a homodimer Ex.
Notice the mutual docking of a phenylalanine residue from each subunit into a hydrophobic pocket of the partner subunit. These
hydrophobic interactions are required for dimerization.
The noncovalent interactions highlighted in yellow are also required for dimerization. These interactions represent examples of:
Osecondary structure
O tertiary structure
O quaternary structure
O secondary AND quaternary structure
Ⓒ tertiary AND quaternary structure
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
Cynt
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
G
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_forwardRifamycins have been used for the treatment of many diseases, including HIV-related Tuberculosis. Explain how Rifamycins inhibit the activities of bacterial DNA dependent RNA polymerase.arrow_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
- Ultraviolet light can cause covalent linkages between consecutive pyrimidine bases in DNA (up to 100 per second in a single cell in sunlight!). These bulky lesions (i.e. cyclobutane pyrimidine dimers and 6-4 photoproducts), which inhibit DNA and RNA polymerases, are mostly reversed by CPD photolyase when light >300 nm is available to power the reaction. In the dark, however, which DNA repair system is best able to correct these errors? a) non-homologous end-joining b) mismatch repairc) nucleotide-excision repaird) base-excision repair e) homology-directed repairarrow_forwardvvnicn the following statements are correct about the repair of a DNA duplex containing the sequence below that is grown INE coli (select all that apply)? Strand A Strand B GATCTAGCCGGCATCCGAT CTAGATCGGACGTAGGCTA Methyl ✔A. MutH cleaves Strand A O B. DNA repair will result in the bold A in strand B being replaced with a C O C. DNA repair will result in the bold G in strand A being replaced with a T ✔ D. Defect will not be properly repaired in dam(-) E coli O E. The mammalian repair system would also correct the mismatch shown based on the methylation status of the DNAarrow_forwardDraw the structure of a phosphonamidite monomer suitable for SPPS of adenosine RNA Nucleosides. Circle all protecting groups and identify their canonical deprotection conditions.arrow_forward
- www D le C 3⁰ A B Indicate True (T) or False (F) for the following statements. Only use the letter (T/F) in the space provided 1. The name of this process is best known as Rho dependent termination 2. The enzyme C called DNA polymerase incorporates ribonucleotides into B called the mRNA False 3. The DNA region A contains inverted palindrome sequences which results in formation of a stem-loops structure 4. During this process, the structure D called terminating hairpin forms and increases the enzyme affinity which terminates transcriptionarrow_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_forwardThe 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_forward
- Molecules 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_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_forwardMultiple 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_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