please help me answer this question. pele Questions: 1. For an LDH activity assa
ID: 150979 • Letter: P
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please help me answer this question.
pele Questions: 1. For an LDH activity assay, suppose your activity was so high that you could not achieve initial velocity so you diluted the sample (1 part sample and 5 parts water) or 1 in total 6, or 1/6 which some use convention 1:6. Then you measured 5ul of that diluted sample. The activity in the reaction vesse!, which has a volume of 1 ml, is 0.10 U. What is the A/min observed? What is the relative activity of the original sample? E.g. how do you have to adjust your answer to consider the dilution? Hint: this is saying you diluted 1:6 and then took 5ul of that dilution to test so use Formula 2, part 2 to account for your dilution (1 point).Explanation / Answer
1 unit or 1U is defined as that amount of LDH which generates 1 micromol of NADH per minute by conversion of lactate to pyruvate under 8.8 pH and 37 degree Celsius temperature.
Given activity = 0.10 U i.e. 0.1 micromoles of NADH generated per minute
but this has to be for 1 ml while the sample volume was 5 microlitre or 0.005 ml
So, for activity per ml, it becomes 0.1/ 0.005 i.e. 20 U or 20 micromoles of NADH generated per minute (observed activity)
So, activity of original solution is by considering the dilution factor
Dilution factor = final solution in parts/ actual sample in parts
Dilution factor (present case) = [5 (parts of water) +1(part of sample)] / 1(part of sample) = 6/1 or 6
So, activity of original sample = observed activity x dilution factor
= 20 x 6 i.e. 120
activity of original sample is 120 U or 120 Units/ml. or 120 micromole of NADH per minute per ml of sample. (activity of original sample)
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