Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 30, Problem 10P
Interpretation Introduction
Interpretation:
The transcription start site in the given DNA sequence needs to be identified.
Concept introduction:
To copy a particular segment of DNA into RNA of DNA based genes, transcription is used. In the beginning, RNA polymerase binds to a promoter sequence near the gene beginning either directly or in the presence of helper proteins.
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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…
Based on these sequences. Remove codons 24 to 66, inclusive.
Sequence ATCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGTSequence BTCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCAATCTCTTAATGACTSequence CTACAAACATGTAAACACACCCTCAGTGGACCAACTCCGCAACATAAACCAAACACCGCTCGCGCCGAAAAAGATATGG
What 2 aspects can you observe in this animal DNA that indicates that it is the regulatory part of a eukaryotic transcription unit? (The spaces are added to make the sequence easier to read).
AGAGGGCGGT CCGTATCGGC CAATCTGCTC ACAGGGCGGA TTCACACGTT GTTATATAAA TGACTGGGCG TACCCCAGGG TTCGAGTATT CTATCGTATG GTGCACCTGA CT..................
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- b) Shown below is a short gene of an unknown bacteria genome (Figure 2). (Note: Transcription starts at Transcription Start Site (TSS).) 5'TATTATAACGCATGACGAGCCATGCATTATCGGTATATGCACTGACCCGGAAAGGCTCCTTTTGGAGCCTTTTTT-3' 3'-ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTTCCGAGGAAAACCTCGGAAAAA-5' Promoter (i) (ii) TSS (iii) Terminator Figure 2 Which DNA strand (the top or the bottom) is used by polymerase as a DNA template? List the mechanistic steps that can trigger the initiation of transcription by the Sigma Factor. What are the amino acids translated from the resulting mRNA? Indicate the amino (NH₂*) and carboxyl (COO) termini of the polypeptide chain.arrow_forwardBelow is a sequence of DNA.5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many "reading frames" can be identified for this sequence? How many "open reading frames" can be identified for this sequence? What is the frame of the longest ORF? How many codons are in the longest ORF? What is the frame of the shortest ORF? How many AA are in the shortest ORF? Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortestORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longestORF (from N to C-terminal end)?arrow_forwardmRNA sequence of A gene If we have the following mutations, find the type of the mutation (silent or missense or nonsense?) 17CàU 36GàA 49GàU 115AàC 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’arrow_forward
- 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.arrow_forwardConsider 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?arrow_forwardFor 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…arrow_forward
- 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 13arrow_forwardFind 5’ UTR and 3’UTR of Mrna 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’arrow_forwardSequence: CCACCTGTACCCGGACACACCCTGGTGTCC 1. Identify the gene from which the querysequence originates (Name of gene) 2. Provide the FULLprotein sequence encoded by the gene. 3. Are different splice variants known for this gene? 4. What human disease has been connected to this gene? 5. Calculate molecular weight (kiloDalton, kD) and calculated pI (the pH where the protein carries no net electrical charge) of the protein.arrow_forward
- b) Shown below is a short gene of an unknown bacteria genome (Figure 2). (Note: Transcription starts at Transcription Start Site (TSS).) TSS 5'-TATTATAACGCATGAGGAGCCATGCATTATCGGTATATGCACTGACCCGGAAAGGCTCCTTTTGGAGCCTTTTTT-3' 3-ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTTCCGAGGAAAACCTCGGAAAAA-5 Promoter ********* Terminator Figure 2 Based on the DNA sequence of terminator, draw the structure of the hairpin loop that will be formed during the end of transcription.arrow_forwardb) Shown below is a very short gene of an unknown bacteria genome (Figure 2). Transcription starts at Transcription Start Site (TSS) and terminates at the Terminator site. TSS 5'TATTATTAACGCATGACGAGCCATGCATTATCGGTATATGCACTGACCCGGAAGGCTCCTTTTGGAGCCTTTTI 3' ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTCCGAGGAAAACCTCGGAAAAA-5 Promoter Terminator Figure 2 (i) Which strand of DNA (the top or the bottom) is used by RNA polymerase as a template? (ii) What are the amino acids translated from the resulting mRNA? Indicate the amino (NH3') and carboxyl (COO"') termini of the protein.arrow_forwardChoose 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’arrow_forward
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