Oligonucleotide synthesis

For the first three problems, the goal of synthesis is: 5í GTTAGATCCGTAAGC

With 95% coupling efficiency the products are:

    1. If the net yield of synthesis of full length product were 21% instead of 46%, what would be a likely explanation?

    2. Every foreshortened oligo below the line should be capped by acetylation. Suppose that the acetylation were inefficient, however. For each of the following contaminants, which cycle must have had a bad capping reaction?

    1. 5í TTAGATCCGAAGC
    2. 5í TTAGATCGTAAGC
    3. 5í TTAGTCCGTAAGC
    4. 5í TTGATCCGTAAGC

3. What would happen to the reaction yield and types of contaminants produced if any of the following steps were inefficient?

    1. Detritylation
    2. Activation
    3. Coupling
    4. Capping
    5. Oxidation

    4. Insulin has the following structure (in its mature form)


    It is processed from a precursor peptide that looks like this:

    MALWMRLLPLLALLALWGPDPAAAFVNQHLCGSHLVEALYLVCGERGFFYTPKTRREAEDLQVGQVELGGG

    PGAGSLQPLALEGSLQKRGIVEQCCTSICSLYQLENYCN



    The cDNA (mRNA copy) encoding this precursor looks like this:


    1 GCTGCATCAG AAGAGGCCAT CAAGCACATC ACTGTCCTTC TGCCATGGCC CTGTGGATGC

    61 GCCTCCTGCC CCTGCTGGCG CTGCTGGCCC TCTGGGGACC TGACCCAGCC GCAGCCTTTG

    121 TGAACCAACA CCTGTGCGGC TCACACCTGG TGGAAGCTCT CTACCTAGTG TGCGGGGAAC

    181 GAGGCTTCTT CTACACACCC AAGACCCGCC GGGAGGCAGA GGACCTGCAG GTGGGGCAGG

    241 TGGAGCTGGG CGGGGGCCCT GGTGCAGGCA GCCTGCAGCC CTTGGCCCTG GAGGGGTCCC

    301 TGCAGAAGCG TGGCATTGTG GAACAATGCT GTACCAGCAT CTGCTCCCTC TACCAGCTGG

    361 AGAACTACTG CAACTAGACG CAGCCCGCAG GCAGCCCCCC ACCCGCCGCC TCCTGCACCG

    421 AGAGAGATGG AATAAAGCCC TTGAACCAGC

    Design a set of oligonucleotides that can be annealed and ligated to represent the B-chain sequence in double stranded DNA, and be sure that none of the oligos is longer than 40 nt. Write out the oligonucleotides in standard format, with 5í on the left and 3í on the right. Here is the Genetic Code, if you need it:

    T C A G
    T TTT Phe (F)
    TTC "
    TTA Leu (L)
    TTG "    		 TCT Ser (S)
    TCC "
    TCA "
    TCG "    		 TAT Tyr (Y)
    TAC
    TAA Ter
    TAG Ter    		 TGT Cys (C)
    TGC
    TGA Ter
    TGG Trp (W)
    C CTT Leu (L)
    CTC "
    CTA "
    CTG "    		 CCT Pro (P)
    CCC "
    CCA "
    CCG "    		 CAT His (H)
    CAC "
    CAA Gln (Q)
    CAG "    		 CGT Arg (R)
    CGC "
    CGA "
    CGG "
    A ATT Ile (I)
    ATC "
    ATA "
    ATG Met (M)    		 ACT Thr (T)
    ACC "
    ACA "
    ACG "    		 AAT Asn (N)
    AAC "
    AAA Lys (K)
    AAG "    		 AGT Ser (S)
    AGC "
    AGA Arg (R)
    AGG "
    G GTT Val (V)
    GTC "
    GTA "
    GTG "    		 GCT Ala (A)
    GCC "
    GCA "
    GCG "    		 GAT Asp (D)
    GAC "
    GAA Glu (E)
    GAG "    		 GGT Gly (G)
    GGC "
    GGA "
    GGG "


    5. Repeat the process of designing oligonucleotides for the A-chain, but with the following restrictions: Assume you have plenty of DNA polymerase-Klenow fragment (with a 5í-3í synthetic activity and 3í-5í exonuclease activity) and plenty of dATP, dTTP, dGTP, and dCTP, but you only have a budget of $80 with which to buy oligonucleotides. Each oligonucleotide costs $1 per base to make, so you canít make the entire sequence synthetically. You will need to use an enzyme to fill in gaps!



    Polymerase Chain Reaction


    6.The insulin precursor on the previous page starts with an ATG (methionine start codon) and ends with a TAG (amber stop codon). You wish to generate a dsDNA molecule containing this coding sequence only, and by PCR. Design two oligonucleotides (of length 21 each) that can be used to generate this precise PCR product from the human genome (there are no introns in this segment of the gene). Write your answer in standard format with 5í ends on the left.