In: Biology
Using the alkaline lysis protocol, explain how the following scenarios would affect the experimental outcome:
a. 100uL of solution 1 was added to a cell pellet, but the pellet was not re-suspended
b. Solution 3 was made without potassium/sodium acetate
Answer: a. If the pellet is not resuspended in the solution 1
when plasmid isolation is performed using an alkaline lysis method
there are chances that you may see low quality plasmid, RNA
contamination and low yield of plasmid. Now, this can happen
because cells are not exposed to the active content of solution I
i.e. EDTA and RNAse. EDTA chelates the divalent ions which are
required for the activity of DNAse which degrades the plasmid and
also for the integrity of the cell membrane. EDTA inactivates DNAse
and destabilizes the cell membrane. Failure to suspend pelleted
cells in the solution I may affect the overall quality and quantity
of isolated plasmid and it can also affect the efficacy of other
two steps in plasmid isolation.
b. Potassium acetate causes protein and the SDS complex to settle
down in the mixture so the resulting solution appears somewhat
transparent after addition of solution 3. The most important
function of potassium acetate is that it renatures the plasmid but
the genomic DNA is not able of re-nature and hence it settles down
along with protein precipitate so in the supernatant only plasmid
is left. If solution three does not contain potassium acetate we
won't be able to precipitate protein and genomic DNA and plasmid
will not renature. The resulting solution will be very turbid and
white in color and when you run that into agarose get you will
hardly see any bands of the plasmid.
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