Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 30, Problem 13P
Interpretation Introduction
Interpretation:
The reason for the dinucleotides are the major products of the RNA polymerase proofreading rather than the mononucleotides should be determined.
Concept introduction:
RNA synthesis (transcription) is the synthesis of an RNA molecule from the
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RNA sequence.
ate 3' and 5' ends on BOTH strands
ate which strand served as the TEMPLATE strand and which
ING strand
Complements. The sequence of part of an mRNA is
5'-AUGGGGAACAGCAAGAGUGGGGCCCUGUCCAAGGAG-3'
5'-AUGGGGAACAGCAAGAGUGGGGCCCUGUCCAAGGAG-3'
What is the sequence of the DNA coding strand? Of the DNA template
strand?
Transcription. Using strand 1 of the DNA molecule as a template, transcribe a messenger RNA molecule (a.k.a. mRNA transcript).
Strand 1
3’ End
TTG
CTT
CAC
CTT
GCG
CGC
CCG
CGC
TAA
TTG
5’ end
mRNA
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- An extra piece. In one type of mutation leading to a form of thalassemia, the mutation of a single base (G to A) generates a new 3' 3' splice site (blue in the illustration below) akin to the normal one (yellow) but farther upstream. Normal 3' end of intron 5' CCTATTGGTCTATTITCCACCCITAGGCTGCTG 3' 5' CCTATTAGTCTAIIIICCACCCTTAGGCTGCTG 3' What is the amino acid sequence of the extra segment of protein synthesized in a thalassemic patient having a mutation leading to aberrant splicing? The reading frame after the splice site begins with TCT.arrow_forwardTranslation. Write the anti-codon sequence of the MRNA transcript. Translate the MRNA transcript into peptide sequence using both the 3 letter abbreviation and 1 letter abbreviation. ANTI-CODON 3' 5' SEQUENCE AMINO ACID N- C- SEQUENCE (3 letter terminus Abbreviation) Terminus AMINO ACID N- C- SEQUENCE (1 letter terminus Abbreviation) Terminusarrow_forwardTrue or False. Explain. A) At no time during protein synthesis does an amino acid make direct contact with the mRNA being translated. B) Because the two strands of DNA are complementary, the mRNA of a gene can be synthesized using either strand as a template.arrow_forward
- the correct reading frame. next thing to do is to write out the mRNA sequence using this sense strand reading frame. The last thing to do is to translate the sequence. The reading frame DNA sequence is: The mRNA sequence is: The polypeptide sequence is:arrow_forwardPolymerase inhibition. Cordycepin inhibits poly(A) synthesis at low concentrations and RNA synthesis at higher concentrations. NH2 H. он Cordycepin (3'-deoxyadenosine) a. What is the basis of inhibition by cordycepin? b. Why is poly(A) synthesis more sensitive than the synthesis of other RNAS to the presence of cordycepin? c. Does cordycepin need to be modified to exert its effect?arrow_forwardtransformation and CRISPR. In your own words, briefly describe two differences between these technologies. (For example, these can be differences between their outcomes, procedures, reagents, or something else.)arrow_forward
- tRNA enzyme. Any given aminoacyl-tRNA synthetase: a. Attaches the amino acid to the 5′-end '5end of the tRNA b. Always recognizes only one specific tRNA c. Recognizes all tRNA molecules d. Forms an ester linkage between the amino acid and the tRNAarrow_forwardYes or no? No explanation. rna polymerase are recruiting to start transcription but promoters are DNA sequence. Taq polymerase is enzyme and can synthesize dna at 72 degrees. dna reads 5 to 3 and polymerase reads template 3 to 5.arrow_forwardplease help me with thi question. What advantages do CRISPR‑Cas systems have over restriction enzymes and engineered nucleases for editing DNA? The options are attached. Multiple answers can be chosenarrow_forward
- RNA is transcribed. Label the 5′ and 3′ ends of each strand. 17. The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGAT Assume that RNA polymerase proceeds along this template from left to right. a. Which end of the DNA template is 5′ and which end is 3′? b. Give the sequence and identify the 5′ and 3′ ends of the RNA transcribed from this template.arrow_forwardBe sure to answer all parts. Write a possible mRNA sequence that codes for each peptide. a. His-Cys-Tyr-Val-Ser 5¹- b. Phe-Val-Thr-Tyr-Glu 5'- 5'- c. Trp-Phe-Asn-Gln -3' U -3' с Table 26.2 The Genetic Code-Triplets in Messenger RNA First Base (5' end) -3' U UUU UUC UUA UUG CUU CUC CUA CUG AUL Phe Phe Leu Leu Leu Leu Leu Leu la C UCU UCC UCA UCG CCU CCC CCA CCG Second Base A UAU UAC UAA UAG CAU CAC CAA CAG Ser Ser Ser Ser Pro Pro Pro Pro Tyr 55 Tyr Stop Stop His His Gin Gin G UGU UGC UGA UGG CGU CGC CGA CGG Cys Cys Stop Trp Arg Arg Arg Arg Third Base (3¹ ond) DUAC DU AG с А Аarrow_forwardGenome. Wide Association Studies: use chip-based array to correlate disease symptoms to SNPs use genomic sequencing technology to correlate disease symptoms to SNPs use small fragment DNA sequencing technology (e.g., Sanger Sequencing) to correlate disease symptoms to SNPs sequence exomes to identify SNPs involved in disease symptoms monitor protein translation to diagnose diseasearrow_forward
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