Why is every uracil replaced by 1-methyl-3'-pseudouridylyl? Explain why it is included
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Why is every uracil replaced by 1-methyl-3'-pseudouridylyl? Explain why it is included please.
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- 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 13Identify the following mutations and describe what the possible effect on the protein will be. (4) 5’GAT TTT AGC TTA GCC CAT 3’ 5’ GAT TAG CTT AGC CCA T 3’ 3’CTA AAA TCG AAT CGG GTA 5’ 3’ CTA ATC GAA TCG GGT A 5’ 5’ GAT TTT AGC TTA CCC CAT 3’ 5’ GAT TTT AGC TAA CCC CAT 3’ 3’ CTA AAA TCG AAT GGG GTA 5’ 3’ CTA AAA TCG ATT GGG GTA 5’Determine 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
- For the following sequence design the forward and reverse primer... explain and justify your answer. Full sequence would be: 1 tctagagtca tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg…Consider the following polypeptide sequence: Gly Val Tyr Ile Val Asp His Phe Thr Gly Asn Tyr Ala Leu Met Glu Asp Lys Aps Val Val Ala Tyr Glu His Ala Pro Lys Leu Asp Asp Phe Val Glu Glu Ala Ley Lys Val Glu Ala Gly Glu Val Pro Ala Ala Pro A) List the expected fragments that will result from proteolytic degradation with trypsin. B) List the expected fragments that will result from the reaction with CNBr. C) List the first three amino acids that will be identified by Edman degradation in the fragments generated in B).Draw and label the following RNA tetranucleotide: 5’phosphoryl-A-2’O-methyl-C-U-G-3’-phosphate
- Each of the following pairs of primers has a problem with it. Tell why the primers would not work well. (a) Forward primer 5'GCCTCCGGAGACCCATTGG 3' Reverse primer 5'TTCTAAGAAACTGTTAAGG 3' (b) Forward primer 5'GGGGCCCCTCACTCGGGGCCCC 3'Reverse primer 5'TCGGCGGCCGTGGCCGAGGCAG 3' (c) Forward primer 5'TCGAATTGCCAATGAAGGTCCG 3'Reverse primer 5'CGGACCTTCATTGGCAATTCGA 3'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.Consider the peptide Asp-Lys-Phe-Glu-Asn-Tyr-Gln-Val-Cys. In a single beaker, you treat this peptide with 2 proteases. One protease cleaves at the N-terminus of aromatic R groups and the other cleaves at the C-terminus of polar, non-ionizable R groups. Following the enzymatic digestion, you want to separate your peptide fragments so that you can identify them. You choose to separate the fragments using an anion exchange column. Beginning at pH=6 you apply your peptide fragments to the column and you gradually decrease the pH of the column stopping the separation when the pH of the column equals 4. Omitting chemical structures, write the amino acid sequence of the peptide fragments that are produced from this digest. Write the order that these fragments will elute from the column (if at all). (Relevant pKa values are: 2.1, 3.8, 4.3, 8.3, 9.6, 10.1, and 10.5)
- 8:52 Protein 2-10092015113649.pdf https:api.schoology.comv1attachment169963839... Name Class Date Interpreting Diagrams: Understanding the Main Ideas The Genetic Code (MRNA) Lysine Lysine Asparagine Asparagine Arginine Arginine Serine Serine Isoleucine Methionine Isoleucine Isoleucine Threonine Threonine Threonine U Threonine c Glutamic acid Glycine Glutamic acid Glycine Aspartic acid Giycine Aspartic acid Glycine Valine Valine Valine Valine Alanine Alanine Alanine Alanine "Stop" codon "Stop" codon Leucine Trytophan Cysteine Cysteine Al Gl "Stop" codon Tyrosine Тугosine Serine Serine Phenylalanine Serine Phenylalanine Serine Leucine Glutamine Giutamine CHistidine Histidine Arginine Arginine Arginine Arginine Al Leucine Leucine Leucine Loucine Proline Proline Proline Proline Icl A G Second Base in Code Word Use the information in the accompanying figure to complete the following table. The first row has been completed to help you get started. DNA codon MRNA codon IRNA Anticodon Amino…he Sequence below comes from the alpha-2 globin of the human hemoglobin gene cluster found in chromosome 16. The globin region of the hemoglobin protein itself consists of 2 alpha chains and 2 beta chains. 1 actcttctgg tccccacaga ctcagagaga acccaccatg gtgctgtctc ctgccgacaa 61 gaccaacgtc aaggccgcct ggggtaaggt cggcgcgcac gctggcgagt atggtgcgga 121 ggccctggag aggatgttcc tgtccttccc caccaccaag acctacttcc cgcacttcga 181 cctgagccac ggctctgccc aggttaaggg ccacggcaag aaggtggccg acgcgctgac 241 caacgccgtg gcgcacgtgg acgacatgcc caacgcgctg tccgccctga gcgacctgca 301 cgcgcacaag cttcgggtgg acccggtcaa cttcaagctc ctaagccact gcctgctggt 361 gaccctggcc gcccacctcc ccgccgagtt cacccctgcg gtgcacgcct ccctggacaa 421 gttcctggct tctgtgagca ccgtgctgac ctccaaatac cgttaagctg gagcctcggt 481 agccgttcct cctgcccgct gggcctccca acgggccctc ctcccctcct tgcaccggcc 541 cttcctggtc…The proximal histidine residues have been replaced by glycine residues by mutation of the cloned genes for both the α and β subunits of hemoglobin. With the tetrameric mutant hemoglobin (all subunits being mutant, α H F8 G, β H F8 G), it was found that the “proximal” coordination bonds to hemes in the mutant protein could be replaced by having the small molecule imidazole in the buffers. Oxygen binding curves for the tetrameric mutant hemoglobin were measured. A. The degree of cooperativity in oxygen binding for the mutant hemoglobin (with imidazole present) would be expected to 1) increase 2) decrease 3) not be affected) compared with the normal protein. B. Justify your answer to part A in terms of what you know about the structural basis of cooperativity in hemoglobin. C. How would the Hill coefficient for the mutant be expected to change compared with nH for normal hemoglobin, which is ~3?