PLEASE MAKE THE DR BRUJIN GRAPH From these k-mers construct a de Bruijn graph and determine the sequence of the contig. AGCG ATCT ATGA ATGG ATTC CCCT CCTG CTCT CTGA CTGC CTTT GAAG GATT GCGT GCTC GTTC TATG TCAT TCTA TCTT TGAA TGAT TGGA TGTT TTCA TTCC TTTC
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PLEASE MAKE THE DR BRUJIN GRAPH
From these k-mers construct a de Bruijn graph and determine the sequence of the contig.
AGCG ATCT ATGA ATGG ATTC CCCT CCTG CTCT CTGA CTGC CTTT GAAG GATT GCGT GCTC GTTC TATG TCAT TCTA TCTT TGAA TGAT TGGA TGTT TTCA TTCC TTTC
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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…The first thing you'll need to do is annotate the DCR1 gene, as follows: 1 agagtctcct cagacgccga gatgctggtc atggcgcccc gaaccgtcct cctgctgctc 61 tcggcggccc tggccctgac cgagacctgg gccggtgagt gcgggtcggg agggaaatgg 121 cctctgccgg gaggagcgag gggaccqсag gс¶¶¶¶gcgc atgacctcag gagccgcgcc 181 gggaggaggg tcgggcgggt ctcagcccct cctcaccccc aggctcccac tccatgtggt 241 atttctacac ctccgtgtcc cggcccqgcc gcggggagcc ccgcttcatc tcagtgggct 301 acgtggacga cacccagttc gtgaggttcg acagcgacgc cgcgagtccg agagaggagc 361 cgcgggcgcc gtggatagag caggaggggc cggagtattg ggaccggaac acacagatct 421 acaaggccca ggcacagact gaccgagaga gcctgcggaa cctgcgcttc tactacaacc 481 agagcgaggc cgttgcqtga ccccqgcccg gggcgcaggt cacgactccc catcccccac 541 gtacqgcccg ggtcgccccg agtctccggg tccgagatcc gcctccctga ggccqcggga promoter: nucleotide 1 - 108; highlight in yellow exon #1: nucleotide 109 - 285; highlight in green intron (there is only one intron in this gene); highlight in light grey exon #2: nucleotide 455 - 567; highlight in blue start…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…
- Give typing answer with explanation and conclusion The following gene codes for a liver enzyme. Changing the A/T base pair to which of the following would result in a missense mutation? 5’ GCCTAATATGCCCGTATGC(AGC)nCGAATAAAATAGTACGGTCGTCGC 3’ 3’ CGGATTATACGGGCATACG(TCG)nGCTTATTTT ATCATGCCAGCAGCG5’ A. C/G B. T/A C. G/C D. Any of the above E. 2 of the above F. None of the aboveFor 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…Table I CACGT A GA CTGAGG ACTC CACGTAGACTGAG G ACAC Wild-type beta-globin gene fragment Sickle-cell beta-globin gene fragment > Circle the mutation in DNA of the sickle-cell beta-globin gene fragment Compare fragments of DNA the wild-type and mutant beta-globin genes in the Table I above, what are the similarities and differences you observe?
- GAG Glu GGG Use the genetic decoder provided to assist in answering this question.\ Second letter UGUT UGC Cys Stop UUU UTo UCU UAU Tyr ULIC Phe UACJ Ser UCC UAA Stop UGA A UAG Stop UGG Trp G UUA nan onn UCG OLeu UCA CAU CGU) CACHS Pro CAA CUU CCU His C CUC Leu C CUA LEUCCA ) Gin CAGJ CGA Arg CUG CGG ACU AAULAso AGUser AUU AUC le AUA AUG Met ACG ACC ACA The AAA AGA AAGJ AGG JArg 6 FLys GAUAn GU GUU GUC GCU GCC Asp Val GUA GAC Ala GAA Glu GAG GGC A Gly GGA GGG GCAA GUG GCG What do you predict would happen if you created a tRNA with an anticodon of 5'-AGG-3' that is charged with methionine, and added this modified tRNA to a cell-free translation system that has all the normal components required for translating RNAS? We did one like this in class... we have to think'about the directionality (5' -> 3' or 3' -> 5'). The easiest way is to draw out your anticodon and codon first, and then consider are the strands going to line up parallel or anti-parallel. O a. methionine would be…A portion of the sequence from the DNA coding strand of the chick ovalbumin gene is shown. Determine the partial amino acid sequence of the encoded protein. CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA-(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA Enter the 3-letter abbreviation for each amino acid in sequence, separated with dashes, and no spaces (example: xxx-xxx-XXX-XXX...) The amino acid sequence is .1104bp..…........Convert the following DNA sequences to RNA sequences.Sequence # 1: CCGGTTCAGGCTTCACCACAGTGTGGAACGCGGTCGTCTCCGGCGACCSequence # 2:CTAAGGTTGCTAATCTCAGCGCTCCGCTGACCCCTCAGCAAAGGGCTTGSequence # 3:GCTCAATCTCGTCCAGCCATTGACCATCGTCGAGGGGTTTGCTCTGTTACSequence # 4:CAAAACGAAATCGAGCGCCATCTGCGGGCCCCGATTACGGACATCAGASequence # 5: TCCAACTCGGGGTCCGCATCGCTCCGCCGGCGACCGACGAAGTTCCGA
- Word AaBbCcI AaBb CcD AaB AaBbCсI AaBbCcL 4 1 Normal Subtitle Title 1 No Spaci... Subtle Em..... Create and Share Adobe PDF Styles Adobe Acro P 1 1 2 4 A 3. A AL T L 4 E 5 E Question 6: The largest chromosome in Mischievous gremlinus has a length of 9 x 107 base pairs. The rate of DNA synthesis is 3,000 bp per minute at each replication fork. a. How long does it take to replicate the entire chromosome from a single origin of replication located exactly in the middle of the chromosome? Assume no pauses. I b. Experiments performed by Professor Trink reveal that it takes only 20 minutes for a growing cell to replicate this chromosome. What is the minimum number of replicons present on the chromosome? Florian Fischer Find v Replace Select Editing 153 23Figure 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)Figure 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 77 90 5' AATTG ACCTT GCTGT ATTAT AGCCA AGGAC AGATC TACGA GCATG 3' 3 TTAAC TGGAA CGACA TAATA TCGGT TCCTG TCTAG ATGCT CGTAC 5' 91 110 5 TGCGA ACCGC AAGCA TTCGT TCTCC TAGGC TACTC GATCC CGTAA 3' 3 ACGCT TGGCG TTCGT AACCA AGAGG ATCCG ATGAG CTAGG GCATT 5' 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. (i) Identify both the hexameric sequences of the promoter region in the coding strand above.