Forward: 5'- CTG AAC CCC ATG TGG AAC GA-3' Tm: GC %: Reverse: 5'- GGC ATC CAT CAC CTA GCT ACA-3' Tm: GC %: Calculate the Tm and GC % for each one. Is the Tm for the two primers similar? Why it is important to be similar?
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- Which forward are reverse primers will amplify the following sequence by PCR? 5' GACCTCGCCGACGCCCTCGACCAGCTCCTGCGCCGCACCCGCCACCTCGCCGAGACC GAGCAGAAAACCCGCCGTCGGAGAGCCACCCGCTCCGGTACGGCAGCACAGTTGCCCGA TCTTCCAGGCCAGCGGCGCCCATTCGACGCAGAGACCGCACCGGCCCCGGCACCGGACT GGAGCGAGAGCCTGGACGACCTCATCAGCGTCGACACGGCGGCCCAGACCGGCACGAGC GAGATGGAGGGCGCGAGCGTGCCGCCGGCCGAGGCAGGCGGGTACGGGCTGTGGGACGC CGAAGCGGAAGCCGAGCAATGGTGAACGCCTCACCGGGCACGAGCGATACGCCGGG 3'For the following sequence please design an 18 base pair REVERSE primer. ATG TCA AAA GCT GTC GGT ATT GAT TTA GGT ACA ACA TAC TCG TGT GTT GCT CAC TTT GCT TAABelow are 9 possible primer pairs. ● Determine which primer pair is the best choice by considering the following: 1. primers should be 18-24 bases in length; 2. base composition should be 45-55% (G+C); 3. primers should end (3') in a G or C, or CG or GC: this prevents "breathing" of ends and increases efficiency of priming; 4. Tms tween 55-70°℃ are preferred (Tas, annealing temperatures, are approximately 5°C lower than the Tm); 5. the Tm for your primer pair should be within 2 degrees of each other, though ideally the same; 6. runs of three or more Cs or Gs at the 3'-ends of primers may promote mispriming at G or C-rich sequences (because of stability of annealing), and should be avoided; 7. 3'-ends of primers should not be complementary (i.e. base pair), as otherwise the formation of primer dimers will result; 8. primer self-complementary (ability to form secondary structures such as hairpins) should be avoided. • Explain why the other primers are not good choices. ● Underline or…
- 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?1b) When doing automated sequencing, all 4 dideoxynucleotides are added to the same sequencing reaction, and run together in a single capillary gel. What's the difference - why can an automated sequencing reaction be done with all 4 dideoxynucleotides mixed together, but not a conventional sequencing reaction?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…
- Determine the isoelectric point of the peptide product of the mutated sequence: 5' - AUG UCC AUG AUU CUG GAA AUU ACC UCC AUC AUG AAG CGC UGA CCC AUU AUU AA - 3'5'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 mutationif you have the following sequence of DNA 5' ATTGCGGAGCCTCGAT 3' do the following:
- AU Py U AGGC C UGGC G GG C Jc What modified nucleoside base is indicated by the arrow? dihydrouracil pseudouracil -ACC.Each of the following pairs of primers has a problem with it. Tell why the primers would not work well. (a) Forward primer 5'GCCTCCGGAGACCCATTGG 3' Reverse primer 5'TTCTAAGAAACTGTTAAGG 3' (b) Forward primer 5'GGGGCCCCTCACTCGGGGCCCC 3'Reverse primer 5'TCGGCGGCCGTGGCCGAGGCAG 3' (c) Forward primer 5'TCGAATTGCCAATGAAGGTCCG 3'Reverse primer 5'CGGACCTTCATTGGCAATTCGA 3'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…