Translate the first 10 amino acids into RNA code Explain where this process occurs in the cell Transcribe this RNA code into DNA Explain where this process would occur in the cell My protein is human insulin which the sequence of amino acids are: malwmrllpl lallalwgpd paaafvnqhl cgshlvealy Ivcgergffy tpktrreaed lqvgqvelgg gpgagslqpl alegslqkrg iveqcctsic slyqlenycn
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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:A segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)Using the genetic code below, decipher the following mRNA sequence. 5'CCCGGGAUGCGGUGUUGGUUUUAACCCGGG3' (show all work)
- The following segment of DNA codes a protein. The uppercase letters represent Exons, the lowercase letters introns. Draw the pre- mRNA, the mature mRNA and translate the codons using the genetic code to form the protein. Identify the 5’UTR and 3UTR 5’- AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3’The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post transcriptional processing which is essentially way too different from the original code in the DNA template. Given: Cytosine; a Protein with known amino acid sequence (amino acid sequence given below) MATIVNTKLGEHRGKKRVWLEGQKLLREGYYPGMKYDLELKDSQVVLRVKEEGKFTISKRERNGRVSPII DLTVQELATVFDGVEMLRVFIRNGAIVISAHHQQERVIERVNRLISKLENGESLSVCSLFHGGGVLDKAI HAGFHKAGIASAISVAVEMEGKYLDSSLANNPELWNEDSIVIESPIQAVNLSKRPPQVDVLMGGIPCTGA SKSGRSKNKLEFAESHEAAGAMFFNFLQFVEALNPAVVLIENVPEYQNTASMEVIRSVLSSLGYSLQERI LDGNEFGVIERRKRLCVVALSHGIDGFELEKVQPVRTKESRIQDILEPVPLDSERWKSFDYLAEKELRDK AAGKGFSRQLLTGDDEFCGTIGKDYAKCRSTEPFIVHPEQPELSRIFTPTEHCRVKGIPEELIQGLSDTI AHQILGQSVVFPAFEALALALGNSLWSWVGMMPIMVEVVDESQPVIGGEDFHWATALVDAKGTLKLSPAA KKQGMPFNIMDGQLAVYSPNGTKKSCGHEPCEYLPVMMSGDAIMVTSSLVH Requirement: Original DNA code (itemize the steps you would take to get to know the original DNA code of Cytosine in focus)3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ Copy the template strand in mRNA. Label the 5’ and 3’ ends. Write out the amino acid (you can use the short form) that this protein would be made of (keep in mind proteins are usually a minimum of 50 proteins.
- 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?Give the protein synthesized of the given mRNA sequence. No need to explain. Just give the answer. ACUGAAAUGACCAGAUUUAUGGCCUGAAUGACCGiven the following DNA sequence of the template strand for a given gene: 5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3' Part A ) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends). Part B ) Use the genetic code to write the peptide sequence translated in a cell from the mRNA in part A. Please use the 3 letter abbreviation for each amino acid. Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?
- The amino acid sequence of human vasopressin is Cys-Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly. Write an mRNA sequence that would encode this oligopeptide (tiny protein). Label the 5' and 3' ends of the mRNA.Below is the CODING sequence of a gene. An A is inserted at position 131. Insert the variant and write the mature mRNA sequence. Then write the amino acid sequence of the resulting transcript. AGAGCTACCAGCTTTAGCGGGATATAAACACACTTATCCTACTATCCCTA 51 TCTTCCCCCAAAGAGGGATCATGCCGTCGACGGCCAAACCGATCGAGCGG 101 AGGTTAAGGGCCATTGGAAAAGGACACTCTGTACGACTAAGAAAATAGC 151 AGCGACTGAGCTGAACGCTACGGGGGGGGGGGTATATACCCCCCCCCCCC 201 CCAGCTACGAGCGACTTTTTATTTTTTTTTTTTATTTTTTTTCTTTTTTA 251 TTTTTTTTGTTTTTTTTTTTCGCGACTACAGAGACTTCTAGCGATCGACG 301 AGCTAGCTAGCGCGGCATCGGACGCGATCGCGATCCTAGACTCTACGATA portion of an unknown enzyme has the amino acid sequence leucine – alanine – lysine – tyrosine.The DNA code for the amino acid sequence of this unknown enzyme is a. GAC CGA TTC TTA b. GAT CGC TTT ATA c. GAA CGG TAC ATC d. GAG CGU UUU AUG