A.C. 3.4Q1. Protein synthesis is carried out by the processes oftranscription and translation.A short length of DNA is shown:TACTCGTCGACGATGATCFirst base(a) State how many codons are present.(b) Using the table below, find the sequence of aminoacids resulting from the transcription and translation ofthe length of DNA. Show your working.UUUUU Phenyl- UCUUUC alanineFUCCUCA-Leucine LuccUUG-LeGCUUCUCCUACUGA AUA-Leucine LAUUIAUC IsoleucineMethioninestart codonAUG MMetGUUTGUCGUAGUG-Valine VCCUCCCCCACCGACUACCACAACGCGCUTGCCGCAGCGSecond base-SerineS-ProlineP-Threonine-AlanineUAUTUACAUAAUAGCAUCACCAACAGATyrosine YStop codonStop codon-Histidine H-GlutamineAAATAAG-LysineAAC-AsparagineNGAU AsparticGAC acid DGAG GlutamicGUGU-Cysteine CEUGCUGAUGGAGUAGCKAGG-ArginineCGUTCGCCGACGGGGUTGGCGGAGGGJStop codon ATryptophan-ArginineRSerine SRGlycineUCAGGSCAGSCAQSCAGThird base

Transcription is a process in which DNA is converted into various types of RNA including protein…

Q1. Protein synthesis is carried out by the processes of transcription and translation. A short length of DNA is shown: TACTCGTCGACGATGATC First base (a) State how many codons are present. (b) Using the table below, find the sequence of amino acids resulting from the transcription and translation of the length of DNA. Show your working. U C U UUU Phenyl- UUC alanine G UUA UUG CUUT CUC CUA CUG A AUA UCU UCC UCA Leucine Lucc] AUU I AUC -Isoleucine GUU GUC GUA GUG F -Leucine L AUG MMethio Methionine start codon -Valine V CCU CCC CCA CCG ACU ACC ACA ACG GCUT GCC GCA GCG Second base -Serine S -Proline P -Threonine -Alanine UAU UAC. A A -Tyrosine Y UAA Stop codon UAG Stop codon CAUT CAC TAAG-Lysine UGUT UGCJ UGA UGG CGUT CGC CGA CGG G -Histidine H CAG-Glutamine AAC-Asparagine AGU-Serine S -Cysteine C Stop codon A Tryptophan G -Arginine R GAU Aspartic GGU AUT GAC acid D GGC GAA Glutamic GGA E GGGJ KAGG Arginine R -Glycine SCAG SCAG SCAG Third base

Q1. Protein synthesis is carried out by the processes of transcription and translation. A short length of DNA is shown: TACTCGTCGACGATGATC First base (a) State how many codons are present. (b) Using the table below, find the sequence of amino acids resulting from the transcription and translation of the length of DNA. Show your working. U C U UUU Phenyl- UUC alanine G UUA UUG CUUT CUC CUA CUG A AUA UCU UCC UCA Leucine Lucc] AUU I AUC -Isoleucine GUU GUC GUA GUG F -Leucine L AUG MMethio Methionine start codon -Valine V CCU CCC CCA CCG ACU ACC ACA ACG GCUT GCC GCA GCG Second base -Serine S -Proline P -Threonine -Alanine UAU UAC. A A -Tyrosine Y UAA Stop codon UAG Stop codon CAUT CAC TAAG-Lysine UGUT UGCJ UGA UGG CGUT CGC CGA CGG G -Histidine H CAG-Glutamine AAC-Asparagine AGU-Serine S -Cysteine C Stop codon A Tryptophan G -Arginine R GAU Aspartic GGU AUT GAC acid D GGC GAA Glutamic GGA E GGGJ KAGG Arginine R -Glycine SCAG SCAG SCAG Third base

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

See similar textbooks

Similar questions

Before we start the actual translation process. If we look at our transcript from sequence I, we can predict how long the polypeptide will be. 58650545454544335 Find the Open reading frame. Then tell me how many codons are in the ORF? Provide a single integer Type your answer... 13.349545
Do question 28
Can you explain it?
1 point Before we start the actual translation process. If we look at our transcript from sequence I, we can predict how long the polypeptide will be. 544545-33454454 Find the Open reading frame. Then tell me how many amino acids are in this poly-peptide? Provide a single integer Type your answer... A8885-349953-
Template strand: 5' AATCATAACTCATTG 3' A. Write the CODING strand of this sequence, including the 5' and 3' end designations. B. Write the amino acid sequence coded by this mRNA. The start codon is the first codon in the sequence that codes for Met. First position (5'-end) U < G บบบ UUC phe UUA UUG CUU Couleu CUC Coc CUA CUG AUU AUG AUC ile GUU GUC GUA GUG Second position C UCU UCC UCA UCG val CCU co CCC CCA 000 CCG ACU ACC ACC AUA ACA thr AUA ACA AUG met ACG GCU GCC ser GCA GCG pro ala UAU ORD tyr UAC UAA Stop UAG Stop CAU CAC CAA CAG - AAU AAC www AAA 10 asn AAG 7 his GAU GAC GAA GAG gin }| lys asp glu UGU UGC cys UGA Stop UGG trp CGU C00 CGC CGA 000 CGG AGU AGC ACA AGA AGG GGU GGC GGA GGG Initiation Termination arg ser arg gly DUAL SUAU DUAU DUAG Third position (3-end)
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’