You have a stock solution of enzyme (125 kD) at a concentration of 1mg/ml. You want to achieve a final enzyme concentration of 36nM in a 1ml reaction volume. What volume of the enzyme stock solution will you add to the reaction?
M1V1 = M2V2
with,
M1 = 1mg x 1000000/125000 x 0.001 = 8000 nM
V1 = ?
M2 = 36 nM
V1 = 1 ml
we get,
Volume of stock solution required V1 = 36 x 1/8000 = 0.0045 ml
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