An mRNA transcript has the following complete sequence: 5'-AUGCCUACGUUACGGACC-3'Rewrite the sequence to give it amissense mutation. Circle/highlightthe mutation. Explain how yourspecific mutation would affect theformation and structure of the2nd (middle) Base of a Codon1stUCAG3rdBaseBaseUUU PheUUC PheUUA LeuUUG LeuUCU PheUCC PheUCA LeuUCG LeuUAU TyrUAC TyrUAA STOPUAG STOPUGU CysUGC CysUGA STOPUGG Trpprotein.UCUU LeuCUC LeuCUA LeuCUG LeuCCU ProССС ProCCA ProCCG ProCAU HisCAC HisCAA GlnCAG GlnCGU ArgCGC ArgCGA ArgCGG ArgAGU SerAGC SerAGA ArgAGG ArgAAU AsnAAC AsnAUU lleAUC lleAUA lleAUG MetACU ThrACC ThrACA ThrACG ThrAAAA LysAAG LysGUU ValGUC ValGUA ValGUG ValGCU AlaGCC AlaGCA AlaGCG AlaGAU AspGAC AspGAA GluGAG GluGGU GlyGGC GlyGGA GlyGGG GlyUCAG

Name: An mRNA transcript has the following complete sequence: 5'-AUGCCUACGU
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Description: An mRNA transcript has the following complete sequence: 5'-AUGCCUACGUUACGGACC-3'Rewrite the sequence to give it amissense mutation. Circle/highlightthe mutation. Explain how yourspecific mutation would affect theformation and structure of the2nd (mi

Missense Mutation: A missense mutation is a point mutation that results in a codon that codes for a…

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An mRNA transcript has the following complete sequence: 5'-AUGCCUACGUUACGGACC-3' Rewrite the sequence to give it a missense mutation. Circle/highlight the mutation. Explain how your specific mutation would affect the formation and structure of the 2nd (middle) Base of a Codon 1st U C A G 3rd Base Base UUU Phe UUC Phe UUA Leu UUG Leu UCU Phe UCC Phe UCA Leu UCG Leu UAU Tyr UAC Tyr UAA STOP UAG STOP UGU Cys UGC Oys UGA STOP UGG Trp protein. U CCU Pro CCC Pro CCA Pro CCG Pro CGU CUU Leu CUC Leu CUA Leu CUG Leu CAU His CAC His CAA Gin CAG Gin C CGG Arg AUU lle AUC lle AUA lle AUG Met ACU Thr ACC Thr ACA Thr ACG Thr AAU Asn AAC Asn AAA Lys AAG Lys AGU Ser AGC Ser AGA Arg AGG Arg A GAU Asp GAC Asp GAA Glu GAG Glu GGU Gly GGC Gly GGA Gly GGG Gy GCU Ala GCC Ala GUU Val GUC Val GUA Val GUG Vel G GCA Ala GCG Ala POAG

An mRNA transcript has the following complete sequence: 5'-AUGCCUACGUUACGGACC-3' Rewrite the sequence to give it a missense mutation. Circle/highlight the mutation. Explain how your specific mutation would affect the formation and structure of the 2nd (middle) Base of a Codon 1st U C A G 3rd Base Base UUU Phe UUC Phe UUA Leu UUG Leu UCU Phe UCC Phe UCA Leu UCG Leu UAU Tyr UAC Tyr UAA STOP UAG STOP UGU Cys UGC Oys UGA STOP UGG Trp protein. U CCU Pro CCC Pro CCA Pro CCG Pro CGU CUU Leu CUC Leu CUA Leu CUG Leu CAU His CAC His CAA Gin CAG Gin C CGG Arg AUU lle AUC lle AUA lle AUG Met ACU Thr ACC Thr ACA Thr ACG Thr AAU Asn AAC Asn AAA Lys AAG Lys AGU Ser AGC Ser AGA Arg AGG Arg A GAU Asp GAC Asp GAA Glu GAG Glu GGU Gly GGC Gly GGA Gly GGG Gy GCU Ala GCC Ala GUU Val GUC Val GUA Val GUG Vel G GCA Ala GCG Ala POAG

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 16QP: Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also,...

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Given the following mRNA, write the double-stranded DNA segment that served as the template. Indicate both the 5 and the 3 ends of both DNA strands. Also write out the tRNA anticodons and the amino acid sequence of the protein encoded by the mRNA message. DNA: mRNA: 5-CCGCAUGUUCAGUGGGCGUAAACACUGA-3 protein: tRNA:
An unprocessed pre-mRNA has the following structure. Which of the following is not a possible size (in bp) of the mature mRNA? 205bp 180bp 150bp 100bp
What will be the overall anti-codon sequence in tRNA for this mRNA? 5’-GUAGCCUUAUCUAGCGAUCACCGUCCGUAUUACUAGUGGCCAGACUCUUUUCACCAUGUAUAGUUG-3’
An mRNA transcript is listed below and contains both start and termination codons. Assume that the initial methionine will stay on the polypeptide in this case. What amino acid sequence will be specified during translation? List the amino acids. The start codon is highlighted. 5’ – CAGCCAAGCAUGCUCGCAAAUGGACGUUGAUAUUUUGUC – 3’
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
mRNA sequence of A gene If we have the following mutations, find the type of the mutation (silent or missense or nonsense?) 17CàU 36GàA 49GàU 115AàC 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’
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?
A normal mRNA that reads 5’ – UGCCAUGGUAAUAACACAUGAGGCCUGAAC– 3’ has an insertion mutation that changes the sequence to 5’ -UGCCAUGGUUAAUAACACAUGAGGCCUGAAC– 3’. Translate the original mRNA and the mutated mRNA, and explain how insertion mutations can have dramatic effects on proteins. (Hint: Be sure to find the initiation site.)
Consider the mRNA sequence below. Assume that the following mRNA segment has been translated. 5’-UACCGAAUGUCU-3’ Note for letters a and b: Use the three-letter abbreviation of amino acid; separate amino acids with a hyphen: do not include the stop codon. Example: ala-cys-glu a. Using the table of the genetic code, determine the sequence of amino acids. b. If mutation occurs by substitution of the 12th nucleotide with cytidine-5’-monophosphate, what is the resulting amino acid sequence? c. What type of mutation occurred? Choose from same sense, missense and non-sense.
How many open reading frames are present in the following mRNA sequence? You may find the codon table helpful. 5' AAUGGGAAUUCCCGAUGAAUAUAATCGTCCTGCATUGAUGGUCCAUGCAAGU 3'OH
mRNA sequence of A gene Write the amino acid sequence of the gene A. 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCCUACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACGACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGCUGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGAGGUAGAAGCCGCUGGGGCUUGGGGCU-3’
) A normal mRNA that reads 5'- UGCCAUGGUAAUAACACAUGAAGGCCUGAAC-3' was an insertion mutation that changes the sequence to 5'- UGCCAUGGUUAAUAACACAUGAGGCGUGAAC-3'. Translate the original mRNA and the mutated mRNA and explain how insertion mutations can have dramatic effects on proteins. ( Hint; Be sure to find the initiation site).