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    Step 0: Prelab questions
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    Step 0: Prelab questions

    Friday, 2/14/2020image.pngimage.pngimage.pngimage.pngimage.pngimage.png
    Friday, 2/14/2020
    Restriction Digest
    ○
    Download and review the genbank sequence files for pPSU1 and pPSU2.
    ○
    Use Benchling to upload these plasmid sequences.
    ○
    Use Benchling to simulate PstI and EcoRV restriction digests of the two plasmids. See this Benchling tutorial for instructions.
    pPSU1 digest
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    pPSU2 digest
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    ●
    Where do PstI and EcoRV cut within their recognition sequences?
    PstI: CTGCA|G G|ACGTC
    EcoRV: GAT|ATC CTA|TAG
    ●
    What gene is encoded into the plasmids? AmpR (Ampicilin resistance)
    ●
    How many copies of the plasmid would you roughly expect in each cell? What information did you use to determine this?
    The ORI determines the copy number, ie. the expected number of plasmids per cell. Both pPSU1 and pPSU2 are plasmids with the ORI pBR322, which has a copy number of 15-20
    2. DNA Libraries
    ●
    How big is the E. Coli genome? 4,639,221 bp and 4,000 genes
    .
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    The spectinomycin gene cassette of interest is 1.2 kb. What is the probability of finding that gene in a sheared E. coli gene fragment of 2,000 bases?
    The probability of finding a 1,200bp fragment P(x>1,200) = 1-P(<=1,200) = 1-1,200/2,000 = 40%
    However, we need to find out fragments where the gene is complete. and the probability of finding the first basepair of the gene is 1,200/4,600,000 = 0.026%. Therefore, the probability of finding the gene in the sheared fragment of 2,000 bases is 0.026%x40% = 0.0104%
    ●
    How many E. coli transformants would you have to screen through to have a 99% probability of finding the spectinomycin gene?
    Assuming we select fragments with ~2,000bp in our gel extraction, as we saw in the previous question, the probability of finding that gene in a
    N=ln (1-P)/ln(1-f) = ln (1-P)/ln (1-2,000/4,639,221); N= ln(1-99%)/ln(1-2,000/4,639,221) = 10,680
    P=probability of finding a clonse

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