Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Question
Chapter 17, Problem 38RQ
Summary Introduction
To review:
The comparison between the size, coding capacity, and other features of the genomes of prokaryotes and eukaryotes.
Introduction:
A prokaryotic organism is unicellular, has a primitive nucleus and a small size, and thus, has a small genome. Eukaryotes are organisms that are multicellular in theirorganization and have a well-developed nucleus. They have a large size and a large genome. The size of the genome of the organisms depends on the size of the organism. The coding region is not necessarily dependent on the size of the organism.
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The genome’s functional gene products are either _____________ or _____________.
A _____________ is a multienzyme complex that synthesizes RNA primers in E. coli DNA replication.
. ___________________________ factors are proteins thatregulate or initiate RNA synthesis by binding directly orindirectly to specific DNA sequences called responseelements.
Chapter 17 Solutions
Biochemistry: The Molecular Basis of Life
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- The eukaryotic mRNA is protected from degradation by a 3′ _____________.arrow_forwardQuestion 45 When TRNAS and rRNAs have bases that H-bonds with other bases far apart from each other, the RNA molecules assume its secondary and tertiary structure. A) True B) Falsearrow_forwardTranscription is the synthesis of _________, using _________ as a template.arrow_forward
- Question 1. Enzymes, proteins and deoxyribonucleic acid (DNA) macromolecules. Enzymes are not only speed up the reaction, but also are necessary for DNA repreduction. are important biological a) Compare the process of protein synthesis between eukaryote mRNA and viral RNA b) With the aid of a diagram, draw an adapter molecule that recognizes the codons of mRNA and explain its functions in DNA translation.arrow_forwardA _____________________ is a multienzyme complex thatsynthesizes RNA primers during E. coli DNA replication.arrow_forwardQuestion 2: Part a: Complete the table describing different components of intron removal from mRNA. Nu:, X and Y refer to B-type chemistry shown on the previous page. (YELLOW table shown) Part b: Complete the table describing different components of group I self-splicing intron removal from 26S rRNA in Tetrahymena. (BLUE table shown) Part c: Draw the intron with an all atom structure for Branchpoint A after intron removal from mRNA Part d: Draw the Group I self-splicing intron with an all atom structure for the Guanosine cofactor after intron removal from 26S rRNA in Tetrahymena.arrow_forward
- Question 2. Ribosomes are cellular structures that are composed of protein and RNA; this structure is responsible for catalyzing peptide bond formation between amino acids during a process known as translation. a) Many antibiotics that kill bacteria target translation. Why might this be an effective mechanism to kill bacteria? Why don't antibiotics also kill human (eukaryotic) ribosomes? b) The antibiotic Kasugamycin (KSG) destabilizes the P-site of the ribosome. Describe what parts of translation would be altered in the presence of this antibiotic. c) How does the following graph show the efficacy of translational knockdown with KSG? Met-Methionine C % of Met incorporation 100 80 60 40 20 0 + 0 2 4 6 8 KSG concentration (mg/ml) 10arrow_forwardSuppose the codon sequence GUGCAAUUCGAGGCC has a single base pair mutation to GUGCAAUUCAAGGCC. If the old protein sequence was Val-Gln-Phe-Glu-Ala, what will be the new sequence encoded by the mutant gene? ____________________________.arrow_forwardDNA within a prokaryotic cell is found in a(n)arrow_forward
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