Part A The bacterial gene little protein (lilP) makes a small protein of 11 animo acids (AA) in length. The DNA sequence of the lilP gene is shown below. 5'-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACTAGaaatattatttaa-3′ 3'-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGATCtttataataaatt-5'
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- The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt-5’ Answer the following questions: Q1. Which is the coding strand? Which is the template strand? [10%] Top-bottom. Bottom-Top. Both can be used as either coding or template for this gene.The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt What is the length in AA’s of the Sh protein? Assume fMet is NOT CLEAVED [10%] 39 AA 13 AA 12 AA 21 AAThis 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.
- 3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ Make vertical lines between codons to make this assignment easier to do. Which strand is the template strand? The first strand is the template strand as it going 3-5 Copy the template strand in mRNA. Label the 5’ and 3’ ends. 5' AUG UUA CCC GCU GCG CGA AGC AAA GUC UAA 3” 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. Met-Leu-Pro-Ala-Ala-Arg-Ser-Lys-Val What do you notice about the mRNA strand compared to the non template strand of DNA? 2. What amino acid does the second codon code for on the DNA template strand? ______Assume the 6th amino acid in the strand is changed from a T to a C. What amino acid does it code for now? _______ What type of mutation is this? 3. Assume the 6th amino acid is changed from T to G on the DNA template strand. What type of mutation is this? What effect would…Based on sequences A,B,C. Provide an anticodon sequence that would build this protein. Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGGFigure 1 is a bacterial gene (1-180). The first base to be transcribed is the base located at position 77. 45 5' TTGGT CTTGG TCGGA TTCCA GAGGA TGAAG TGTTG ACAGC GCATT 3' 3 AACCA GAACC AGCCT AAGGT CTCCT ACTTC ACAAC TGTCG CGTAA 5' 46 5 AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5 77 90 110 135 136 5 TGATG TAGCT GATTC TGTTG AAAGG CTCCT TTTGG AGCCT TTTTT 3 3' ACTAC ATCGA CTAAG ACAAC TTTCC GAGGA AAACC TCGGA AAAAA 5 156 180 Figure 1. Illustrate how termination of transcription occurs in the gene above. (Hint: position from 156 to 180)
- what is the anticodon sequence that would build this protein? AUGUUUGUACAUUUGUGUGGGAGUCACCUGGUUGAGGCGUUGUAUUUGGUUUGUGGCGAGCGCUUUUACCAGUUAGAGAAUUACUGATranscribe the following DNA sequence into RNA, and then into amino acids 5’-GTATACTTGTGGGCCAGGGCATTAGCCACACCAGCCACCACTTTCGGATCGGCAGCC-3’ 3’-CATATGAACACCCGGTCCCGTAATCGGTGTGGTCGGTGGTGAAAGCCTAGCCGTCGG-5’5'-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 13
- What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'17) Synthesis of the mRNA starts at the boxed A/T base pair indicated by the box and proceeds left to right on the sequence below. Transcribe and translate this bacterial gene. 5'-GGACCGCGGGGCAGGATTGCTCCGGGCTGTTTCATGACTIGICAGGTGGGATGACTTGGATGGAAAAGTAGAAGGTCATG-3 3'-CCTGGCGCCCCGTCCTAACGAGGCCCGACAAAGTACTGAACAGTCCACCCTACTGAACCTACCTTTTCATCTTCCAGTAC-5′ 1 -+--at - --+-- 80What’s the resulting amino acid sequence? 3’CAG TTA AGC CTC GGT TAC CAG GAT ACG GGA 5’