BamHI cut sequence: G//GATCC and each sequence is 250 nucleotides long. How many DNA segments would be created by cutting the normal gene with BamHI?
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BamHI cut sequence: G//GATCC and each sequence is 250
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- AGTGCAGCTC CAGCCGACCC CCGGGACGTA GACAAGGGCA GGCGCGCGGT 50 GAAGACTGCG GCCGGCCGCG TAGCCGCGCT GTGGTCCCGC TGGCCTTCTC 100 TTGGGCCGCG CACCCTCCGG ACCTGCGTCG GGATCCTCCC GGAGTCTCCC 150 GCTCTCTCCT CCCGGCTTGC GAACATGCGG CCCCTTAAGC CCGGCGCCCC 200 TTTGCCCGCA CTCTTCCTGC TGGCGCTGGC TTTGTCCCCG CACGGAGCCC 250 ACGGGAGGCC CCGGGGGCGC AGGGGAGCGC GCGTCACGGA TAAGGAGCCC 300 AAGCCGTTGC TTTTCCTCCC CGCGGCCGGG GCCGGCCGGA CTCCCAGCGG 350 CTCCCGGAGC GCAGAAATAT TCCCAAGAGA CTCTAACTTA AAAGACAAAT 400 TCATAAAGCA TTTCACAGGG CCGGTCACAT TTTCACCAGA ATGCAGCAAA 450 CATTTCCACC GACTCTATTA CAATACCAGG GAGTGCTCAA CGCCAGCTTA 500 TTACAAAAGA TGTGCTAGAT TGTTAACAAG ATTAGCAGTG AGTCCACTGT 550 GCTCCCAGAC CTAGCAAAAC TACCCTACAT TTCCTAAGAA TGTACATCTA 600 ATTTGAAGAA AAAGTGCCTC AAATCATGCA AAATGTAAAA AAAGATGAAA 650 TTTATATTTT TATGGATATT AAGATGAGTA AAATAAGAGA CTTCCCAGAA 700 ATAACTGGTT AGCTGTTTCC TGTCATAGAA TGGAGTCTTT CTTGCTTTAT 750 CTTTTTGTGT ATACAGTAAT TTATAATTTT GTAAAACAGA GTTTGAATCG 800 CATATTGAAA ATTAGATATT AAAAATTGTG…Sequence 1 : TACGCTACGGTAATC Sequence 2: TACGCTACTATCGTATranscribe the DNA into mRNA. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
- MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH MVHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFSDGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH MVH ltp eeksavta lwgkvnvdevggealg rllv vypwtqrffesfg dlstp davmgnpkvkahgkkvlgafsdglahl dnlkgtf atlselhcdkhvdpenfrllgnvlcvlahhfgkef tppvqaayqkvvagvanalahkyh MVH ltp eeksavta lwgkvnvdevggealg rllv vypwtqrffesfg dlstp davmgnpkvkahgkkvlgafsdglahl dnlkgtf atlselhcdkhvdpenfrllgnvlcvlahhfgkef tppvqaayqkvvagvanalahkyh…**ALL GROUPS HAVE A MUTATION MAKE SURE TO IDENTIFY THEM** Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAGConvert the following DNA sequences to RNA sequences.Sequence # 1: CCGGTTCAGGCTTCACCACAGTGTGGAACGCGGTCGTCTCCGGCGACCSequence # 2:CTAAGGTTGCTAATCTCAGCGCTCCGCTGACCCCTCAGCAAAGGGCTTGSequence # 3:GCTCAATCTCGTCCAGCCATTGACCATCGTCGAGGGGTTTGCTCTGTTACSequence # 4:CAAAACGAAATCGAGCGCCATCTGCGGGCCCCGATTACGGACATCAGASequence # 5: TCCAACTCGGGGTCCGCATCGCTCCGCCGGCGACCGACGAAGTTCCGA
- Sequence A uuucccucuuagaauuaauucguaauauuuaucau uuaaauuuagcucccuccccccauuaauaaauaauu cuaucccaaaaucuacacaauguucuguguacacuu cuuauguuuuuuacuucugauaaacguuuuugaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Sequence B uuucccucuuagaaugaaucucguaaugaaaaguc gaaaaaaauugugcucccuccccccauuaauaaau aauucuaucccaaaaucuacaca augu ucuguguacacuucuuauguuuuuuacuucugauaa acguuuuugaaaaaaaaaaaaaaaaaaaaaaaaaaa aaaaaaaaaa Q1a) Underline the Poly A signal in both sequences. Q1b) What is the sequence of the translational stop codon? Q1c) Which of these two mRNA isoforms is likely to be more stable? Q1d) Justify your answer to part Q1c).I just don't know how to get the answer for C and D. Expecially for C, why F+died because of leu-?Below is a short segment of DNA molecule. transcribed the DNA codon into mRNA. TACCATGAGAATTGTGGTCACCTTTTT ATGGTACTCTTAACACCAGTGGAAAAA
- ATGAAAAAACGAAAAGTGTTAATACCATTAATGGCATTGTCTACGATATTAGTTTCAAGCACAGGTAATT TAGAGGTGATTCAGGCAGAAGTTAAACAGGAGAACCGGTTATTAAATGAATCAGAATCAAGTTCCCAGGG GTTACTAGGATACTATTTTAGTGATTTGAATTTTCAAGCACCCATGGTGGTTACCTCTTCTACTACAGGG GATTTATCTATTCCTAGTTCTGATAGAAAATATTCCATCGGAAAACCAATATTTTCAATCTGCTATTTGG TCAGGATTTATCAAAGTTAAGAAGAGTGATGAATATACATTTGCTACTTCCGCTGATAATCATGTAACAA TGTGGGTAGATGACCAACAAGTGATTAATAAAGCTTCTAATTCTAACAAAATCAGATTAGAAAAAGGA AGATTATATCAAATAAAAATTCAATATCAACGAGAAAATCCTACTGAAAAAGGATTGGATTTCAAGTTGT ACTGGACCGATTCTCAAAATAAAAAAGAAGTGATTTCTAGTGATAACTTACAATTGCCAGAATTAAAACA AAAATCTTCGAACTCAAGAAAAAAGCGAAGTACAAGTGTGGACCTACGGTTCCAGACCGTGACAATGAT GGAATCCCTGATTCATTAGAGGTAGAAGGATATACGGTTGATGTCAAAAATAAAAGAACTTTTCTTTCAC CATGGATTTCTAATATTCATGAAAAGAAAGGATTAACCAAATATAAATCATCTCCTGAAAAATGGAGCAC GGCTTCTGATCCGTACAGTGATTTCGAAAAGGTTACAGGACGGATTGATAAGAATGTATCACCAGAGGCA AGACACCCCCTTGTGGCAGCTTATCCGATTGTACATGTAGATATGGAGAATATTATTCTCTCAAAAAATG AGGATCAATCCACACAGAATACTGATAGTCAAACGAGAACAATAAGTAAAAATACTTCTACAAGTAGGAC…ATGTTTGTTTTTCTTGTTTTATTGCCACTAGTCTCTAGTCAGTGTGTTAATCTTACAACCAGAACTCAAT TACCCCCTGCATACACTAATTCTTTCACACGTGGTGTTTATTACCCTGACAAAGTTTTCAGATCCTCAGT TTTACATTCAACTCAGGACTTGTTCTTACCTTTCTTTTCCAATGTTACTTGGTTCCATGCTATACATGTC TCTGGGACCAATGGTACTAAGAGGTTTG a) Assume that translation begins from the leftmost nucleotide in the above sequence. Provide the translated protein sequence for the first 5 amino acids. Please show your work.Translate the mRNA into a protein using the genetic code. (Amino acid chain.) Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTC AATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTC GCGCCGAAAAAGATATGG