Assayed for LDH activity were 5 microliters of a sample that was diluted 6 to 1. The activity in the reaction vessel, which has a volume of 3 mililiters, is 0.30 U. What is the ΔA/min 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 X 3X10^-3 L
relative activity:
U/mL= U/(volume of fraction assayed) X dilution used, if any.
I have an answer for the first part. I got 0.622. The second part, i don't know how to put the dilution into the equation.
I
To calculate the ΔA/min observed, you will need to use the formula provided:
ΔA/min = (U/mL) * ((6220 M^-1cm^-1)(1cm)) * 10^(-6) * 5 microliters * 3X10^(-3) L
Since you already have the value of U as 0.30 U and the volume of the fraction assayed is 5 microliters, you can plug in these values into the formula:
ΔA/min = (0.30 U) * ((6220 M^-1cm^-1)(1cm)) * 10^(-6) * 5 microliters * 3X10^(-3) L
Simplifying the units:
ΔA/min = (0.30 U) * (6220/10^6) * 5 * 3X10^(-9) mol * (1/1000) L
Now you can calculate the ΔA/min observed.
As for the relative activity, you need to take into account the dilution used for the sample. The formula for relative activity is:
U/mL = U / (volume of fraction assayed) * dilution used
In this case, the dilution is given as 6 to 1. So the relative activity can be calculated as:
U/mL = 0.30 U / (5 microliters * (6/1))
Simplifying:
U/mL = 0.30 U / 30 microliters
Now you can calculate the relative activity.