5. A particular DNA coding segment is ACGTTAGCCCCAGCT.Write the sequence of nucleotides in the corresponding MRNA.• Determine the amino acid sequence formed from the MRNA during translation.What amino acid sequence results from each of the following mutations?replacement of the underlined guanine by adenineinsertion of thymine immediately after the underlined guaninedeletion of the underlined guaninea.b.C.

Disclaimer: " It is assumed that the given coding strand is from 5' to 3' direction. The central…

Answered: A particular DNA coding segment is…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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If DNA segments changes from GCATAG to GCATGG, this is a: MRNA Codon/Amino Acid Chart First Base Second Base Third Base A G UUU1 Phenylalanine UCUT UAUT FTyrosine (Tyr) UAC UGU, U FCysteine (Cys) UGCJ UUCJ (Phe) UCC Serine (Ser) UCA U UGA - Stop A UUA1 FLeucine (Leu) UUG- UAA1 FStop UAGJ UCG- UGG - Tryptophan (Trp) G CUU CAU1 CCU1 CGUT Histidine (His) CAC- CUC CCC Proline (Pro) CCA CGC FLeucine (Leu) CUA FArginine (Arg) CGA CAA1 Glutamine A CAGJ (Glu) CGG- CUG- CCG- ACU AAUJ Asparagine AUUT AGUT FSerine (Ser) AGC- U AUC FIsoleucine (lle) ACC AACJ(Asn) Threonine A AUA- (Thr) AAA1 FLysine (Lys) AAG- АСА AGA, FArginine (Arg) AGG- A Start Methionine AUG- (Met) ACG- G GUU GCU, GAU1 Aspartic Acid GGU U GAÇJ(Asp) GUC Fvaline (Val) GUA GCC GGC G Alanine (Ala) GCA Glycine (Gly) A GAAJ Glutamic Acid GGA GAG- (Glu) GUG- GCG- GGG- G
Use this mRNA coding sequence as your starting point. This sequence begins with a start codon and ends with a stop codon, so it is only looking at the region of DNA that directly encodes a protein sequence. 5’-AUGCACAAAUUAGAGUACCCCCCAGGAAGGUAG-3’ Make the following mutation in this sequence by changing/adding/removing only one nucleotide. Make the mutation easy to see (a different color, circled, something like that) A frameshift mutation
A segment of mRNA produced by the normal order of DNA nucleotides and the corresponding amino acid chain are given below: mRNA segment: GCC UAC AAU GCG Amino acid chain: Ala-Tyr-Asn-Ala Knowing that insertion mutations shift the triplets by one base, if an insertion mutation adds a U to the beginning of that mRNA segment, what will be the new triplet/codon grouping and the new amino acid chain? Group of answer choices a. U GCC UAC AAU GCG; Ala-Tyr-Asn-Ala b. UCC UAC AAU GCG; Ser-Leu-Gln-Cys c. UGC CUA CAA UGC G; Cys-Leu-Gln-Cys d. UGC CUA CAA UGC G; Ala-Tyr-Asn-Ala e. UGCC UAC AAU GCG; Cys-Tyr-Asn-Ala
An RNA polymer is made by using the enzyme polynucleotide phosphorylase with equal quantities of CTP and GTP. When this RNA is used in an in vitro translation system, all of the following amino acids could be incorporated into a newly made polypeptide, except: Codon Table Second position C UUU UCU UAU UGU phe tyr сys UUC UCC UAC UGC ser UAA Stop UGA Stop UAG Stop UGG trp UUA UCA UUG UCG CUU CCU CAU CGU leu his ССС pro ССА CỤC САС CGC arg CỦA САА CGA gln CUG CCG CAG CGG AUU ACU AAU AGU asn ser AUC ile ACC thr АCА AAC AGC AUA AAA lys AAG AGA arg AUG met ACG AGG GUU GCU GAU GGU asp GUC GCC ala GCA GẠC GGC val gly GUA GAA GGA glu GUG GCG GAG GGG glycine (Gly) histidine (His) proline (pro) alanine (Ala) arginine (Arg) Third position (3'-end) AGUCAG First position (5'-end)
A segment of template DNA is known to contain the following base sequence: 3' GATACCTTTGTGTAGTCATCTT 5' a) Write the mRNA that would be transcribed from this DNA fragment. b) Highlight the starter and the stopper in your mRNA sequence. Write the sequence of amino acids which would be encoded in translation. asap please
The following is as segment of mRNA: 5'-UCGGAAUGUGGUGGCAUACAGGCUUACAGAACUAAGUCUGAGAAU-3' A. How many amino acids long will be the protein translated from the only reading frame available in this segment? B. If a mutation changes the third letter of the stop codon in the only reading frame available in this segment, how many amino acids long will be the protein translated?
1. Given the DNA sequence below: 5’-ACATGTGTACAGGCTTTGTCTGAATGGCTT-3’ 3’-TGTACACATGTCCGAAACAGACTTACCGAA-5’ Translate the mRNA. (Using the 3-letter code for amino acids, write the primary structure of the peptide that will be produced.) 2. Given the DNA sequence below: 5’-ACATGTGTACAGGCTTTGTCTGAATGGCTT-3’ 3’-TGTACACATGTCCGAAACAGACTTACCGAA-5’ Translate the mRNA. (Using the 3-letter code for amino acids, write the primary structure of the peptide that will be produced.)
PLEASE HELPP
UCAAUGGGGUUUAUAGCG… (there are bases after the last G but we don’t know what they are so there may be some ambiguity for the last codon with regards to the amino acid possibilities) if the translation frame had UCA as the first codon: if the translation frame had CAA as the first codon: if the translation frame had AAT as the first codon:
Match the term with its definition comparing genomic DNA, mRNA, and proteins. You may only use each option ONCE. where transcription starts. where translation ends a chemical group that indicates the first nucleotide that was added to the mRNA a chemical group that indicates the first amino acid that was added to the polypeptide a DNA sequence that is neither transcribed nor translated a non-protein coding region upstream of the start codon in the mRNA promoter complimentary base-pairs with the codon amino-terminus a DNA sequence that is transcribed, but not intron translated [Choose ] stop codon +1 site tRNA 5-prime UTR start codon exon 5-prime triphosphate promoter amino-terminus 5-prime triphosphate stop codon
summarize these results using concise language in a neat table; Control : 5’ ATGTACGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ This is the coding strand of DNA and hence this DNA sequence is similar to mRNA sequence. So the mRNA sequence is : 5’ AUGUACGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ Mutant 1: 5’ ATGTACGAGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence 5’ AUGUACGAGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ The bold Adenine is the mutated base which is substituted in place of Cytosine. So the codon change from GCG to GAG. GCG codes for Alanine but GAG codes for Glutamic acid. So the amino acid sequence changes. Hence this mutation is missense mutation where a base substitution results in change in amino acid sequence. Mutant 2: 5’ ATGTATGCGCGATCACCATACATCATGGCACCCGCTAGCTATTAACATGTTTTTT 3’ mRNA sequence: 5’ AUGUAUGCGCGAUCACCAUACAUCAUGGCACCCGCUAGCUAUUAACAUGUUUUUU 3’ In this mutation, Cytosine is replace by Thymine and hence the codon…
A mRNA has the following sequence and direction: 5’-AUGUCAACCUAA-3’ What would be the anticodon sequences formed from this mRNA during the translation process?

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