Question

A 10ml sample containing the reduced form of nicotinamide adenine dinucleotide (NADH) had a fluorescence of 26.0 relative to a reagent blank. When 1.00umol of NADH was added to the sample solution (without voume change), the fluorescene increased to 78.3 relative to the blank. Calculate the concentration of NADH in the sample in units of umol/ml.

Answer 1

Ans. Step 1: Calculate [NADH] in Final aliquot after spiking

Given – 1.00 umol NADH is added to 1.0 mL solution without change in volume.

So,

Increase in [NADH] due to spiking = Amount of NADH added / Volume of soln.                                                    = 1.00 umol / (10.0 mL)

                                    = 0.10 umol/ mL                  

Hence, Increase in [NADH] due to spiking = 0.10 umol/ mL

# Step 2. Calculate [NADH] in original sample:

Given,

            Fluorescence of original sample = 26.0

            Fluorescence of sample plus addition = 78.3

Now,

Increase in fluorescence =

Fluorescence of sample plus addition – Fluorescence of original sample

            = 78.3 – 26.0

            = 52.3

# Note that the increase in fluorescence of the (sample plus addition) with respected to original sample is due to increase in [NADH] concentration by 0.1 umol/ mL.

So, a fluorescence of 52.3 is equivalent to 0.10 umol mL^-1 NADH

Or, a fluorescence of 1.0     is         -           (0.10 / 52.3) umol mL^-1 NADH

Or, a fluorescence of 26.0 is           -          (0.10 / 52.3) x 26.0 umol mL^-1 NADH

= 0.0497 umol/ mL

Hence, [NADH] of original sample = 0.0497 umol/ mL = 0.05 umol/ mL