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Violagirl Violagirl
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10 years ago
Relative activity of a sample?
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 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 (Neutral Face?)

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.
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Violagirl Author
wrote...
10 years ago
I found answers for both questions. Was going to see now if someone could check my work to see if it makes sense?

Question one was:

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?

U = (ΔA/Δmin)/((6220 M^-1cm^-1)(1cm))X 10^6 micrometer/M X10^-3 L

A/min = (6220 M cm (cm^-1) (.10) / 10^6 uM/M) (.001 L) = .622 A/min

Relative Activity:

U/mL= U/(volume of fraction assayed) X dilution used, if any.

(.10)/(1 mL) = .10 U/mL

Does this look right? Many thanks again.
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