Violagirl Posted September 22, 2013 Share Posted September 22, 2013 The question: Assayed for LDH activity were 5 uL of a sample that was diluted 6 to 1. The activity in the reaction vessel, which has a volume of 1 mL, is 0.10 U. What is the A/min (absorbancy/minute) observed? What is the relative activity of the original sample? Here are the formulas:U=(ΔA/Δmin)/((6220 M^-1cm^-1)(1cm))X 10^6 micrometer/M X10^-3 LRelative Activity:U/mL= U/(volume of fraction assayed) X dilution used, if any.This is how I thought about solving the problem:A/min = 6220 M^01 cm^-1 (cm) x 10^6 uM/M x 0.001 L/ .10 U = 6.22 x 10^7 (????)For the relative activity of the original sample, since it has to be in units of U/mL of fraction, I was thinking I'd have to take .10 U/ 1 mL (since that is the volume of the fraction) and I'd get .10 U/mL as my relative activity. Is this how you'd would solve it? Help is definitely appreciated. Link to comment Share on other sites More sharing options...
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