Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 18, Problem 61SA
Summary Introduction
To explain:
The detection and formation of a stop codon in the mRNA (messenger Ribonucleic Acid).
Introduction:
Several genetic codons in the triplet form make the universal genetic code. The genetic code comprises a total of 64 codons. More than one codon codes for a single amino acid. Out of these 64 codons, three codons do not code for any amino acid and are known as stop codons. These are also called nonsense codons or termination codons. These codons are named as UAA (ochre), UAG (amber) and UGA (opal).
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Genetically engineered mRNAs that code for a stretch of basic residues, such as poly(Lys), induce translation termination and destruction of the nascent polypeptide. Explain how this response would protect cells from the effect of faulty transcription that produces mRNAs with mutated Stop codons.
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Question 27 options:
Guanylytransferase
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Chapter 18 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 1RQCh. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - Prob. 4RQCh. 18 - Prob. 5RQ
Ch. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - Prob. 10RQCh. 18 - Prob. 11RQCh. 18 - Prob. 12RQCh. 18 - Prob. 13RQCh. 18 - Prob. 14RQCh. 18 - Prob. 15RQCh. 18 - Prob. 16RQCh. 18 - Prob. 17RQCh. 18 - Prob. 18RQCh. 18 - Prob. 19RQCh. 18 - Prob. 20RQCh. 18 - Prob. 21RQCh. 18 - Prob. 22RQCh. 18 - Prob. 23RQCh. 18 - Prob. 24RQCh. 18 - Prob. 25RQCh. 18 - Prob. 26RQCh. 18 - Prob. 27RQCh. 18 - Prob. 28RQCh. 18 - Prob. 29RQCh. 18 - Prob. 30RQCh. 18 - Prob. 31RQCh. 18 - Prob. 32RQCh. 18 - Prob. 33RQCh. 18 - Prob. 34RQCh. 18 - Prob. 35RQCh. 18 - Prob. 36RQCh. 18 - Prob. 37RQCh. 18 - Prob. 38RQCh. 18 - Prob. 39RQCh. 18 - Prob. 40RQCh. 18 - Prob. 41RQCh. 18 - Prob. 42RQCh. 18 - Prob. 43RQCh. 18 - Prob. 44RQCh. 18 - Prob. 45RQCh. 18 - Prob. 46RQCh. 18 - Prob. 47FBCh. 18 - Prob. 48FBCh. 18 - Prob. 49FBCh. 18 - Prob. 50FBCh. 18 - Prob. 51FBCh. 18 - Prob. 52FBCh. 18 - Prob. 53FBCh. 18 - Prob. 54FBCh. 18 - Prob. 55FBCh. 18 - Prob. 56FBCh. 18 - Prob. 57SACh. 18 - Prob. 58SACh. 18 - Prob. 59SACh. 18 - Prob. 60SACh. 18 - Prob. 61SACh. 18 - Prob. 62TQCh. 18 - Prob. 63TQCh. 18 - Prob. 64TQCh. 18 - Prob. 65TQCh. 18 - Prob. 66TQCh. 18 - Prob. 67TQCh. 18 - Prob. 68TQCh. 18 - Prob. 69TQCh. 18 - Prob. 70TQCh. 18 - Prob. 71TQCh. 18 - Prob. 72TQCh. 18 - Prob. 73TQCh. 18 - Prob. 74TQCh. 18 - Prob. 75TQ
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- In prokaryotic protein synthesis, formylmethionine (fmet) is the first amino acid incorporated, whereas (normal) methionine is incorporated in eukaryotes. The same codon (AUG) serves both. What prevents methionine from being inserted into the beginning and formylmethionine in the interior?arrow_forwardHow would a null mutation in the guanylyl transferase gene affect overall mRNA stability in eukaryotic cells?arrow_forwardAs we focused on the genetic code and the transcription of genetic information stored in DNA into complementary RNA molecules. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions: Question: How were the experimentally derived triplet codon assignments verified in studies using bacteriophage MS2?arrow_forward
- The following enzyme has an exact integer number of turns. What is the length and amount of amino acid? And based on the image, what post-translational modification it has undergone?arrow_forwardHuman wildtype and mutant alleles are identical in sequence except for a single base-pair substitution that changes one nucleotide towards the end of intron 2. The wildtype and mutant sequences of the affected portion of the mRNA are listed in the following table. Explain how a single base substitution could alter the reading frame, which could result in a physiological disorder?arrow_forwardA eukaryotic cell carrying out transcription and RNA processing is incubated with 32P-labeled ATP. Where will the radioactive isotope appear in mature mRNA if the ATP is labeled at the (a) α position, (b) β position, and (c) γ position?arrow_forward
- For each of the following initiation factors, how would eukaryotic initiation of translation be affected if it were missing? A. eIF 2 B. eIF4 C. eIF5arrow_forwardExplain why the translation of a given mRNA can be inhibited by a segment of its complementary sequence, a so-called antisense RNA.arrow_forwardMost of the mutations that Yanofsky recovered were missense mutations. However, Yanofsky also recovered a nonsense mutation that changed amino acid number 15 into a stop codon. This codon normally encodes Lysine. Does the recovery of this mutation support the hypothesis that this Lysine residue is critical in the function of the tryptophan synthetase protein? Why or why not?arrow_forward
- How does a eukaryotic ribosome select its start codon? Describe the sequences in eukaryotic mRNAs that provide an optimal context for a start codon.arrow_forwardSelect the three post-transcriptional modifications often seen in the processing of mRNA in eukaryotes: Group of answer choices heteroduplex formation; base modification; capping 3'-capping; 5'-poly(A) tail addition; splicing 5'-poly(A) tail addition; insertion of introns; capping removal of exons; insertion of introns; capping 5'-capping; 3'-poly(A) tail addition; splicingarrow_forwardLactose permease, a protein of E. coli, is composed of a single polypeptide that is 417 amino acids in length. By convention, the amino acids within a polypeptide are numbered from the aminoterminus to the carboxyl-terminus. Are the following questions about lactose permease true or false? A. Because the 64th amino acid is glycine and the 68th amino acid is aspartic acid, the codon for glycine, 64, is closer to the 3′ end of the mRNA than the codon for aspartic acid, 68. B. The mRNA that encodes lactose permease must be greater than 1241 nucleotides in length.arrow_forward
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