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- The BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINYShown below is an E. coli's DNA sequence coding for XXR protein. The nucleotides are numbered 1 to 330. Transcription starts at the Transcription Start Site (TSS) that is the base located at position 57. -55 5' AATAAСTTGAGATTTTGATTGACАТССТТСТСАCAGAGCCTATAATACCТАТТТС 3' 3'TТАТTGAACTСТААААСТААСТGTAGCAACAGTGTCТСGGATATTATGGATAAAG 5' 56- 5' ТACGTATAGAСАСТСAGAGGAAAGACAGAGAGAGAGTTAGCATTGTACTАТСТСТ 3' 3' АTGCATATСТсTGAGTCTCCTTтстстстстстсТСААТСGTAACATGATAGAGA 5' 65- -105 --110 -140- 5' СТTTTAGATATATCTCТАТСТСТСТСАСТССАТСТТТСТCGTGTTAACACAAСА 3' 3' GAAAATCTATАTAGAGATAGAGAGAAGTGAGGTAGAAAGAGCACAAТТСTСТTGT 5' 111- -120- -130- -150- -160----165 166- --175- --185- -195 -205- -215-----220 ------- 5' GTCACAGACTCACAGATCTTTGTCGGTGATCGGAGATGGAGTTCCGGGAGAAGCT 3' 3' CAGTGTCTGAGTGTCТAGAAACAGCCACTАGССТСТАССТСААGGCCCTСТТCGA 5' 221- -230- 240 -250 -260 -270-----275 5' TTATAAGTTCAAGTTGCAATAGGTGTTTGCCTTTGTTTTATCTCTCCTCACCGTA 3' 3'ААТАТТСААGTTCAACGTTATCCАCAAACGGAAACAAAАТAGAGAGGAGTGGCAT 5' 276- -285- -295- -305 -315…Consider the following mature mRNA from a human cell: 5' UAAUGUCGCAAUAACC 3¹ What is the sequence of amino acids in the translated protein? Second letter A First letter С U d A G บบบ UUC P U UUA UUG Leu CUU CUC CUA CUG GUU GUC GUA GUG -Phe Met-Ser-Gln Met-Arg-Lys-Ser Leu Stop-Cys-Arg-Asn-Asn C AUU AUC lle AUA ACA AUG Met ACG Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC GCU GCC GCA GCGJ Ser Pro Thr Ala UAU UAC Tyr UGU UGC. UAA Stop UGA Stop UAG Stop UGG Trp CAC CAGGin CAU His Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. AAU Asn AAA Lys AAG. There is no start codon, resulting in no translated protein. GAU ASP GAC Glu GAA GAG G CGU CGC CGA CGG Cys AGU AGC AGA AGG Arg GGU GGC GGA GGGJ Arg ser Gly MCAG U UCAG с А SCAG U SCAG U Third letter
- For the messenger RNA sequence below, find the beginning of the amino acid coding sequence and translate the sequence using the genetic code provided below. 5' - AAUUAUGGGCAAUAUGCCGGGCcGGUUAAGCG - 3' Second Letter A UGU cys u UGC Phe UCU UU U UUC UUA UAU Tyr Ser UAC UAA UAG Leu UCA Stop UGA Stop UUG UCG Stop UGG Trp CUU CU CAU His CGU c cuc Leu ccc ССА CCG Pro CÁC CAA CAG CGC CGA CGG Arg CUA CUG Gin 1st 3rd letter Ser u letter AUU ACU AAU AAC AAA AAG Asn A AUC AUA AUG lle ACC ACA AGU AGC AGA AGG Thr Lys Arg Met ACG GUU G GUC GUA GUG GCU GAU GAC GAA Asp GGU Val GCC Ala GGC Gly GCA GGA Glu GCG GAG GGG GBelow is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'c) A gene in a bacteria has the following DNA sequences (the promoter sequence is positioned to the left but is not shown): 5'-CAATCATGGAATGCCATGCTTCATATGAATAGTTGACAT-3' 3'-GTTAGTACCT TACGGTACGAAGTATACTTATCAACTGTA-5' i) By referring to the codon table below, write the corresponding mRNA transcript and polypeptide translated from this DNA strand. 2 Second letter с A UUUPhe UAU Tyr UAC. UGU UGCJ UCU) UCC UCA UUG Leu UCG Cys UUC UUA Ser UAA Stop UGA Stop A UAG Stop UGG Trp G CUU CÚC CCU ССС CAU CGU His САC Pro CC CỦA Leu ССА CAA Arg CGA CUG J CCG) CAG Gin CGG AUU ACU AAU Asn AGU Ser AUC le АСC АCА AAC AAA AGC. Thr JArg AUA AGA AUG Met ACG AAG Lys AGG. GAU Asp GUU) GCU GCC GCA GCG GGU" GGC GGA GGG GUC Val GUA GAC Ala Gly GAA Glu GAGJ GUG ii) If the nucleotide indicated by the highlighted bold letter undergoes a mutation that resulted in deletion of the C:G base pair, what will be the resulting amino acid sequence following transcription and translation? Third letter DUAG DUAG DUAG A. First…
- What is the length in AA’s of the LilP protein? Assume fMet is NOT CLEAVED. provide calculationWhat will be the overall anti-codon sequence in tRNA for this mRNA? 5’-GUAGCCUUAUCUAGCGAUCACCGUCCGUAUUACUAGUGGCCAGACUCUUUUCACCAUGUAUAGUUG-3’There are four codons that encode threonine. Consider the leader sequence in Figure 31.22A. What codons are used and with what frequency?
- What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'The following bacterial DNA sequence uses the top strand as the coding strand and the bottom strand as the template strand. Write what the messenger RNA sequence for this gene would be after transcription occurs.What is the lagging strand sequence if the leading strand of DNA is TTA CCG ATC GAA? How many amino acids can be identified if the coded mRNA sequence is GCAUUACAUGGCGGA