Experimental design (30%) Please design a knockout experiment for a specific gen

Question

Experimental design (30%) Please design a knockout experiment for a specific gene that is assigned to you as shown below. Please list the following information on your word file as the answer. 1. What is the selectable marker? (2%) TRP on the vector and on the 2. What is the location of that selectable markerandits flanking sequence genome? (596) 3. Forward primer GC content and Tm(2.5%) 4. Reverse primer -Gc content and Tm(2.5%) (2.5%) 5. The location of the primers on your gene and on the vector (label them with numbers) 6. The recombinant PCR sequence (2.5%) your the 7. Draw and brief description to explain how to replace gene with a schematic diagram

Explanation / Answer

The trp1 sequence

5’ATGTCTGTTATTAATTTCACAGGTAGTTCTGGTCCATTGGTGAAAGTTTGCGGCTTGCAGAGCACAGAGGCCGCAGAATGTGCTCTAGAT TCCGATGCTGACTTGCTGGGTATTATATGTGTGCCCAATAGAAAGAGAACAATTGACCCGGTTATTGCAAGGAAAATTTCAAGTCTTGTAAAAGCATATAAAAATAGTTCAGGCACTCCGAAATACTTGGTTGGCGTGTTTCGTAATCAACCTAAGGAGGATGTTTTGGCTCTGGTCAATGATTACGGCATTGATATCGTCCAACTGCATGGAGATGAGTCGTGGCAAGAATACCAAGAGTTCCTCGGTTTGCCAGTTATTAAAAGACTCGTATTTCCAAAAGACTGCAACATACTACTCAGTGCAGCTTCACAGAAACCTCATTCGTTTATTCCCTTGTTTGATTCAGAAGCAGGTGGGACAGGTGAACTTTTGGATTGGAACTCGATTTCTGACTGGGTTGGAAGGCAAGAGAGCCCCGAAAGCTTACATTTTATGTTAGCTGGTGGACTGACGCCAGAAAATGTTGGTGATGCGCTTAGATTAAATGGCGTTATTGGTGTTGATGTAAGCGGAGGTGTGGAGACAAATGGTGTAAAAGACTCTAACAAAATAGCAAATTTCGTCAAAAATGCTAAGA AATAG3’

The genome sequence of Dgat1 or diacylglycerol O-acyltransferase 1 in Saccharomyces cerevisiae

5’ATGTCAGGAACATTCAATGATATAAGAAGAAGGAAGAAGGAAGAAGGAAGCCCTACAGCCGGTATTACCGAAAGGCATGAGAATAAGTCTTTGTCAAGCATCGATAAAAGAGAACAGACTCTCAAACCACAACTAGAGTCATGCTGTCCATTGGCGACCCCTTTTGAAAGAAGGTTACAAACTCTGGCTGTAGCATGGCACACTTCTTCATTTGTACTCTTCTCCATATTTACGTTATTTGCAATCTCGACACCAGCACTGTGGGTTCTTGCTATTCCATATATGATTTATTTTTTTTTCGATAGGTCTCCTGCAACTGGCGAAGTGGTAAATCGATACTCTCTTCGATTTCGTTCATTGCCCATTTGGAAGTGGTATTGTGATTATTTCCCTATAAGTTTGATTAAAACTGTCAATTTAAAACCAACTTTTACGCTTTCAAAAAATAAGAGAGTTAACGAAAAAAATTACAAGATTAGATTGTGGCCAACTAAGTATTCCATTAATCTCAAAAGCAACTCTACTATTGACTATCGCAACCAGGAATGTACAGGGCCAACGTACTTATTTGGTTACCATCCACACGGCATAGGAGCACTTGGTGCGTTTGGAGCGTTTGCAACAGAAGGTTGTAACTATTCCAAGATTTTCCCAGGTATTCCTATTTCTCTGATGACACTGGTCACACAATTTCATATCCCATTGTATAGAGACTACTTATTGGCGTTAGGTATTTCTTCAGTATCTCGGAAAAACGCTTTAAGGACTCTAAGCAAAAATCAGTCGATCTGCATTGTTGTTGGTGGCGCTAGGGAATCTTTATTAAGTTCAACAAATGGTACACAACTGATTTTAAACAAAAGAAAGGGTTTTATTAAACTGGCCATTCAAACGGGGAATATTAACCTAGTGCCTGTGTTTGCATTTGGAGAGGTGGACTGTTATAATGTTCTGAGCACAAAAAAAGATTCAGTCCTGGGTAAAATGCAACTATGGTTCAAAGAAAACTTTGGTTTTACCATTCCCATTTTCTACGCAAGAGGATTATTCAATTACGATTTCGGTTTGTTGCCATTTAGAGCGCCTATCAATGTTGTTGTTGGAAGGCCTATATACGTTGAAAAGAAAATAACAAATGCCAGATGATGTTGTTAATCATTTCCATGATTTGTATATTGCGGAGTTGAAAAGACTATATTACGAAAATAGAGAAAAATATGGGGTACCGGATGCAGAATTGAAGATAGTTGGGTAA 3’

Forward Primer     5’CAGAATTGAAGATAGTTGGGTAA3’

                              3’GTCTTAACTTCTATCAACCCATT5’

Tm = 4(G+C) +2(A+T) = 4(1+7)+ 2(6+9) =4x8+ 2x15=620C

GC content = [8/23] x100 =34.78%

Reverse Primer = 5’ATGTCAGGAACATTCAATGATAT3’

                              3’TACAGTCCTTGTAAGTTACTATA5’

Tm =4(G+C) +2(A+T)= 4(3+4)+ 2(7+9)= 28+32 =600C

GC content = [7/23] x100 =30.43%

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