Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 29, Problem 15P
Interpretation Introduction
Interpretation:
The length of
Concept introduction:
In biology, deoxyribonucleic acid (
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pppApCpCpUpApGpApU-OH(a) Using the straight-chain sugar convention, write the structure of the DNA strand that encoded this short stretch of RNA.(b) Using the simplest convention for representing the DNA base sequence, write the structure of the nontemplate DNA strand.
Compare the DNA binding modes of minor groove binding, intercalating, and crosslinking agents and explain how these interactions give rise to specific pharmaceutical applications. Use the molecular structures of at least one DNA
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34) What amino acid sequence is coded for by the following DNA coding strand?
(Recall: the DNA template strand runs 5' to 3' and the mRNA strand runs
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5'-TTATGCGACCAGACCAGTTT-3' Coding strand
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...
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- Please input Shine-Dalgarno sequence alsoarrow_forward29.) equalities now called Chargaff's rule. Biochemist Erwin Chargaff was the first to note that, in DNA, [A]=[T] and [G]=[C], A) Using this rule, determine the percentages of all the bases in DNA that is 20% thymine. [A] = [C] = [G] [T] = 20% %3D - B) If a single strand of RNA is 20% uracil, what can you predict about the percentages of the remaining bases and why?arrow_forwardCodon-Anticodon Recognition: Base-Pairing Possibilities (Integrates with Chapter 11.) Draw base-pair structures for (a) a G:C base pair. (b) a C:G base pair. (C) a G:U base pair, and (d) a U:G base pair. Note how these various base pairs differ in the potential hydrogen-bonding patterns they present within the major groove and minor groove of a double-helical nucleic acid.arrow_forward
- 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_forwardNumber of Okazaki Fragments in E. coli and Human DNA Replication Approximately how many Okazaki fragments are synthesized in the course of replicating an E. coli chromosome? How many in replicating an “average� human chromosome?arrow_forward1arrow_forward
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- Part II: Information Transfer Background Information - Key Points The background information provided for this lab has given you a general overview of some of 24 the key terms and definitions necessary to understand the transfer of information from gene to protein. The information included below will help you work through the specific problems included in your Tutorial 4 Assignment. When working on the problems remember the base mu to pairing rules (Table 3). Table 3: Rules for nucleotide base pairing. cytosine (C) - guanine (G) adenine (A)- thymine (T) DNA RNA For Transcription: ● ● ● ● cytosine (C) - guanine (G) adenine (A)- uracil (U) Initiation is determined by the recognition of the promoter sequence in the DNA by the RNA polymerase. Stef The transcription start site is downstream of the promoter and is designated as the +1 site. Aspartic acid Alanina Valine Arginine Serine Lysine Asparagine Glutamic TEOPO|0C|AGUCAG|UC|AG/DCAG/3G/ CAGUC UGU A C A Threonine G Methionine Isoleucine…arrow_forwardComposition as a mole fraction of one of a double-stranded DNA strand T = 0.22 and C = 0.30. In the light of this information, the following values are Calculate as a fraction. If the given information is used to calculate the desired value, If it is not sufficient, indicate the result as X.arrow_forwardThe polyamines spermine and spermidine have numerous effects on both prokaryotic and eukaryotic cells. Examples include chromatin condensation, transcription, translation, and apoptosis. They are best known for their role in promoting DNA stability. Explain how polyamines enhance DNA stability and promote supercoilingarrow_forward
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