generated from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). B I U Ꭶ Paragraph Arial 10pt E Ev Av AV I %0 F Q
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- The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the MRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this difference? Explain in detail. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). A Ix BIUS Paragraph Arial 14pxGiven the DNA sequence of the restriction enzyme: gi|6329444|dbj|AB034757.1| Hynobius retardatus mRNA for larval beta-globin, complete cds GCAGAATCTGACTCAAGAAATCCCTCCTCACCCAACACCACCAGCAGCCATGGTTCACTGGACAGCAGAGGAGAAGGCAGCCATCAGCTCTGTGTGGAAGCAGGTGAACGTGGAGAGCGATGGACAGGAGGCCCTGGCCAGGTTGCTGATCGTCTACCCCTGGACCCAGAGATACTTCAGCTCTTTTGGGGACCTGTCGAGCCCAGCTGCCATTTGTGCCAACGCCAAGGTCCGTGCCCATGGCAAGAAGGTCCTGTCCGCCCTGGGAGCCGGCGCCAACCACCTGGATGACATCAAAGGCAACTTTGCTGATCTGAGCAAGCTTCACGCAGACACACTCCATGTGGACCCCAATAACTTCCTGCTCCTGGCAAACTGCCTGGTGATCGTCTTGGCCCGCAAGCTGGGAGCCGCCTTCAACCCTCAAGTCCATGCGGCCTGGGAGAAGTTCCTGGCCGTCTCCACCGCGGCTCTGTCCAGAAACTACCACTAGAGACTGGTCTTTGGGTTTAATTCTGTGAACGTCCCTGAGACAAATGATCTTTCAATGTGTAAACCTGTCATTACATCAATAAAGAGACATCTAACAAAAAAAAAAAAAAAAAAAAAAAAAA Identify two blunt-end cutters Identify two sticky-end cutters. For each, Provide the sequence of the Restriction enzyme, Highlight using a specific color where the DNA sequence where the restriction enzyme will cut the DNA Indicate the…What restriction enzyme (or enzymes) would you use to cut the following... (Gene of Interest is Bolded) 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 tatgccaatg…
- For the following sequence design the forward and reverse primer... explain and justify your answer. Gene of Interest: a 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 tatgccaatg…The enzymatic activity necessary for proofreading is: O reverse transcription O ligase O endonuclease O exonuclease O polymerase5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. Write the resulting amino acid sequence using the 3 letter code. Write the answer in a all capital letters. Leave a space between the amino acids. Do not write 5' and 3'. 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a T, then the result will be A) A nonsense mutation B) A frameshift mutation C) A silent substitution D) A missense mutation 5'GGT ACG TTG GGG CTC CAT3' This sequence is transcribed and translated. If the G in Bold changes to a A, then the result will be A) A nonsenese mutation B) A frameshift mutation C) A silent substitution D) A missense mutation
- Table 1 shows a list of restriction endonucleases with their recognition sequence and the sites of cleavage indicated by arrows. Table 1 Enzyme name Recognition sequence and position of cut 5'GIAATTC3 5'G!GATCC3' 5'GIGTACC3 5'GCIGGCCGC3' 5'IGATC3' 5'GGTACIC3' 5'ALGATCT3 EcoRI ВатHI Аcс651 Notl Sau3A Kpnl BglII (i) Which restriction enzyme(s) produce blunt ends? (ii) Are there any pair of neoschizomers in the list? Explain. (iii) Are there any pair of isocaudomers in the list? Explain.Given below is the DNA template. What are the gene products? 3’ TACCGGCCTATCTAGGGCCATGGCTTAATTCCC 5’ 5’ ATGGCCGGATAGATCCCGGTACCGAATTAAGGG3’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 TINY
- Choose one of the strands and transcribe the strand. Show the steps (with proper label) and do a post transcriptional processing Once the transcript is made, do the process of translation. Again follow the steps. Use the Wobble Table for reference 5’ CTATATTTATGTGCTATATCCAGGACTGCCCCTAGGAAATAAAAAA…AAAAAAA 3’3’ GATATAAATACACGATATAGGTCCTGACGGGGATCCTTTATTTTTT…TTTTTTT 5’Based on the following wild type DNA sequence, indicate if each of the mutations should be classified as : insertion, deletion, missense, nonsense, silent (Use the provided Genetic Code table and remember you have been given DNA sequence). Wild Type: AUGAUUCUUAAAAGU Mutant 1: AUGAUUCUUUAAAGU Mutant 2: AUGAUUCUUGAAAGU Mutant 3: AUGAUCCUUAAAAGU Mutant 4: AUGAUCCUAAAAGU Mutant 5: AUGAUCCUUAAACAGU Socond letter Key: Ala = Alanine (A) Arg Arginine (R) Asn = UUU } UAU Tyr UGU UGC Cys UGA STOP UGG Trp UCU UCC UUC Phe Ser Asparagine (N) Asp = Aspartate (D) Cys Cysteine (C) Gin = Glutamine (Q) Glu = Glutamate (E) Gly = Glycine (G) His = Histidine (H) le = Isoleucine (1) Leucine (L) Lys Lysine (K) Met = Methionine (M) Phe = Phenylalanine (F) Pro Proline (P) Ser = Serine (S) Thr Threonine (T) Trp Tryptophan (W) Tyr Tyrosine (Y) - Valine (V) UCA UCG UAA STOP UAG STOP UUA Leu UUG S CCU CC CGU CUU CUC His CGC Arg Leu Pro CAA Gin CGA CCA CCG CUA CUG CGG Leu = AGU AUU AUC } lle AUA ACU ACC ACA Ser AAC…1. You are investigating a protein that has the amino acid sequence N ... Ala – Thr - Asn – Trp – Lys - Arg - Gly – Phe – Thr ... C within its primary structure. You found that several of the mutations affecting this protein produced shortened protein molecules that terminated within this region. In one of the mutants, the Asn became the terminal (last) amino acid. (a) What DNA single-base changes(s) would cause the protein to terminate at the Asn residue? (b) What other potential sites do you see in the DNA sequence encoding this protein where mutation of a single base pair would cause premature termination of translation? >