Concept explainers
Interpretation:
The amino acids sequence for the given mRNA base sequence code has to be predicted.
Concept Introduction:
Codon: A sequence of three ribonucleotides in the mRNA chain that codes for a specific amino acid; also a three-
Genetic code: The sequence of nucleotides, coded in triplets (codons) in mRNA that determines the sequence of amino acids in protein synthesis.
Translation: A tRNA molecule is a single polynucleotide chain held together by regions of base pairing in a partially helical structure. An amino acid is bonded to its specific tRNA by an ester linkage. Connecting specific amino acid at end of the tRNA is known as charging tRNA. Once done, tRNA is ready to be used in the protein synthesis.
At the other end of the tRNA, three anticodons are present which are complementary to the codons present in mRNA. Once the anticodons pairs off with codons, the amino acid at terminal end of the tRNA is delivered and attached to the growing protein chain.
Illustrated relationships are:
DNA informational strand : 5’ ATG CCA GTA GGC CAC TTG TCA 3’
DNA Template strand: 3’ TAC GGT CAT CCG GTG AAC AGT 5’
mRNA: 5’ AUG CCA GUA GGC CAC UUG UCA 3’
protein: Met Pro Val Gly His Leu Ser
Notice: 5’ end of the mRNA strand codes for the N-terminal amino acid, whereas the 3’ end of the mRNA strand codes for the C-terminal amino acid. Proteins are always written N-terminal to C-terminal, reading left to right.
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Check out a sample textbook solutionChapter 26 Solutions
Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
- Using the genetic code table provided below, write out the sequence of three different possible mRNA sequences that could encode the following sequence of amino acids: Met-Phe-Cys-Trp-Glu C A G U C UUU Phe UCU Ser UUC Phe UCC Ser UCA Ser UUA Leu UUG Leu UCG Ser CUU Leu CCU Pro CUC Leu CUA Leu CUG Leu CCG A CAU His CGU Arg CCC Pro CAC His CGC Arg UAU Tyr UGU Cys U UAC Tyr UGC Cys C Stop UGA Stop Stop UGG Trp UAA A UAG G CCA Pro CAA Gln Pro CAG AUU lle AUC lle AUA lle AUG Met ACG G 등등 Gln CGA Arg CGG Arg ACU Thr AAU Asn AGU Ser ACC AAC Asn AGC Ser Thr ACA Thr Thr AAA Lys AGA Arg AAG Lys AGG Arg GUU Val GCU Ala GAU Asp GGU Gly GGC Gly GUC Val GCC Ala GAC Asp GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly U C A G U C A G SUAUarrow_forwardA segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)arrow_forwardTAC/ACC/GAC/GAA/AAT/TGT/TAC/CGT/TCA/AAC asap pleasearrow_forward
- Use the chart below to find the correct amino acids for the following mRNA strand: GCUAUGUUU ala-met-stop ala-met-stop ala-met-phe phe-ala-metarrow_forwardA fragment of a polypeptide, Met-Thr-Ile-Ser-Asp-Ile is encoded by the following sequence of DNA:Strand A - TACGATGACGATAAGCGACATAGC - Strand B - ATGCTACTGCTATTCGCTGTATCG -Which is the transcribed (template) strand? Write the sequence of the resulting mRNA transcript. Add labels to the strands above to show the 3’ and 5’ ends.arrow_forwardTranslate the following mRNA nucleotide sequence into an amino acid sequence, starting at the second base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’arrow_forward
- Indicate the amino acid sequence of the protein encoded by the following mRNA molecule. Use the genetic code table and assume that the very first “AUG” the ribosome encounters will serve as the start codon and specify methionine. 5’-AAUUCAUGCCCAAAUUUGGGGCACGAAGCUUCUUAGGCUAGUCCUAAAAAA-3’arrow_forward: (2) (a) Using the codon table, show the consequences of adenine base addition to thebeginning of the following coding sequence on the subsequent translation. CGA-UCG-GAA-CCA-CGU-GAU-AAG-CAU asaparrow_forwardThe genetic code consists of a series of three-base wordsthat each code for a given amino acid.(a) Using the selections from the genetic code shown below, de-termine the amino acid sequence coded by the following seg-ment of RNA: UCCACAGCCUAUAUGGCAAACUUGAAG AUG= methionine ;CCU= proline; CAU= histidine ;UGG= tryptophan AAG= lysine ; UAU= tyrosine ;GCC= alanine ;UUG= leucine ;CGG= arginine ;UGU= cysteine ;AAC =asparagine ;ACA=threonine ;UCC= serine ;GCA=alanine ;UCA=serine(b) What is the complementary DNA sequence from which this RNA sequence was made? (c) If you were sequencing the DNA fragment in part (b), how many complementary chain pieces would you obtain in the tube containing ddATP?arrow_forward
- Translate the following mRNA: 5-A U G A A A U U U C U U U A G G U C G A A -3 NH3+- Met-Leu-Phe-Val- COO- NH3+- Met-Thr-Val-Ser- COO- NH3+- Met-Glu-Gln-Ser- COO- NH3+- Met-Asp-Ser-Pro- COO- NH3+- Met-Lys-Phe-Leu- COO-arrow_forwardDystrophin is mutated in the disease, causing a codon to change from GGA to UGA. What is the consequence of this change? (arrow_forwardIf there are 64 possible codons in the genetic code and the amino acid is specified by each, as read in the 5’ to 3’ direction from the mRNA sequence, which ones are STOP codons?arrow_forward
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