Reference no: EM1393431
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=(?A/?min)/((6220 M^-1cm^-1)(1cm))X 10^6 micrometer/M X10^-3 L
Relative 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.
Question 2: Of the original "undiluted" sample from Question 1, 10 uL of a 5-to-1 dilution were used to measure protein concentration. With the use of a standard curve, this is found to be 140 ug. What is the specific activity of the original sample?
This one I'm not sure on at all. If someone could let me know if I'm on the right track for the first one, it'd be appreciated. Otherwise work showing and explaining how it to solve for both would be great too.
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