With this DNA sequenec: - 5'-GCAATGGAGAGAATCTGCGCG-3'- - 3'-CGTTACCTCTGTTAGACGCGC-5' - -Identify the sequencce of RNA in which the protein product will be detrnemined
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Q: GCA UGC CGA UAC
A: 1. The tRNA anticodons for the amino acid sequence shown above is - GCA UGC CGA UAC
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Q: Given this MRNA strand: 3 - AUGAGGAAGGUA - 5"; what are the components of the polypeptide?
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Q: Given this mRNA strand: 3’ - AUGAGGAAGGUA - 5’; what are the components of the polypeptide?
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Q: COMPLEMENTARY
A: COMPLEMENTARY DNA SEQUENCE OF GACGGCTTAAGATGC is given below
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A:
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- Given the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs ( what letter changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - AGGCTCGCCTGT - 5' Mutated DNA Sequence: 3' - AGTCTCGCCTGT - 5'Given the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs (what letters changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions (with spaces), and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - CCTCGTTATGTG - 5' Mutated DNA Sequence: 3' - CCTCGTTATTTG - 5'This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.
- What is the length in AA’s of the LilP protein? Assume fMet is NOT CLEAVED. Write out the sequence of the polypeptide in AA: use the three letter notation, e.g. Met-Ser-Pro-Identify the primary sequence for the polypeptide that yields these fragments upon treatment: His-met-thr-met-ala-trp; Leu-asn-asp-phe; Val-lys obtained from chymotrypsin Leu-asn-asp-phe-his-met; Ala-trp-val-lys; Thr-met obtained from CNBRThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.
- A) Based on the mRNA sequence below, provide the corresponding DNA template (5'-3') and protein sequences (N-C terminus) using the single letter abbreviations for each 5' GCA UAU CCU UGU GAU 3' B) Identify the two unique amino acids in the protein sequence above, provide their full names and brief explanation why you chose them C) Draw the two amino acids from 3. connected with a peptide bond to each other (with free amino and carboxy termini) at physiological pH|5'-TAGCTGATCGAATATGCGGTCTCTATCTTCGTAGACGA-3' 3'-ATCGACTAGCTTATACGCCAGAGATAGAAGCATCTGCT -5' Determine the amino acids that will be encoded by this sequence Second letter First letter U C A G U UUU Phe UUC UUA UUG Leu CUU CUC CUA CUG Leu GUU GUC GUA GUG Val UCU UCC UCA UCGJ AUU AUC lle AUA ACA AUG Met ACG CCU CCC C CCA CCG ACU ACC GCU GCC GCA GCG Ser - Pro Thr Ala A UGU UACTyr Cys UGC. UAA Stop UGA Stop A UAG Stop UGG Trp G CAC His CAA Gin CAG GAUT GAC Asp GAA AAU Asn ACC Ser AGU AAG LYS AA Glu GAGJ Oa. N-Met-Arg - Ser-Leu-Ser - Ser-C Ob. N-Met-Pro-Arg - Asn-Asp - Ser-C d. N-Met-Lys - Val-Glu-Ala-C Oc. N-Asp-Pro-Lys - Ser - Val-Ile-C Oe. N- Met-Ala-Asp-Pro-Lys - Ser-C G CGU CGC CGA CGG AGA AGG. GGU GGC GGA GGG Arg SCAO Gly U UCAG UUA DUAG Arg G Third letter 13Determine the sequence of a polypeptide treated with trypsin and chimotripsine. Below are the fragments generated with each treatment. Determine the original sequence for both fragmentations (reduerde that they must be equal in the order of amino acids) Quimotripsina 1. Leu-His-Lys-Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly-Pro-Ser 1. Gln-Gln-Ala-Gln-His-Leu-Arg-Ala-Cys-Gln-Gln-Trp 2. Arg-lle-Pro-Lys-Cys-Arg-Lys-Phe Trypsin 1. Arg 2. Ala-Cys-Gln-GIn-Trp-Leu-His-Lys 3. Cys-Arg 4. Gln-Ala-Asn-Gln-Ser-Gly-Gly-Gly- Pro-Ser 5. lle-Pro-Lys 6. Light 7. Phe-Gin-Gln-Ala-Gln-His-Leu-Arg
- Draw and label the following RNA tetranucleotide: 5’phosphoryl-A-2’O-methyl-C-U-G-3’-phosphateA portion of the sequence from the DNA coding strand of the chick ovalbumin gene is shown. Determine the partial amino acid sequence of the encoded protein. CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA-(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA Enter the 3-letter abbreviation for each amino acid in sequence, separated with dashes, and no spaces (example: xxx-xxx-XXX-XXX...) The amino acid sequence is .1104bp..…........Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?