Question

A colorimetric LDH assay provides a method to measure the enzyme activity of LDH. NADH (MW 633.4 g/mol) absorbs light at 340 nm while NAD+ does not show any absorption at that wavelength. The Beer-Lambert Law relates the absorbance of a solute to its concentration: A = ε*c*L A is absorbance, ε is the molar extinction coefficient, c is concentration, and L is the path length of the solution in the spectrophotometer (usually in cm). In the lab you have available NADH. You dissolve 4.8 mg in 50 ml of buffer and measure an absorbance of 0.2363 at 340 nm in a 0.5 cm cuvette. Subsequently in a 1 cm cuvette, you mix 25 µl of 20 mM NADH, 25 µl pyruvate 0.2 M , 400 µl of TRIS buffer (pH=7.8), and 50 µl of a solution containing lactate dehydrogenase at a concentration of 2 mg/l. You place the cuvette in a spectrophotometer and follow the kinetics of absorbance at 340 nm. You measure a rate of decrease of absorbance (ΔA) of 0.50 min-1 .

a) Determine the absolute amount of NADH consumed after 20 seconds in the reaction (in nmol)

b) Determine the specific activity of LDH as NAD+ produced per LDH in the unit of time (in µmol/mg/min)

Answer 1

A=εcl

c=4.8 mg in 50ml or 9.6mg in 100ml or 96 mg in 1000ml/1L

MW of NADH is 633.4g/mol; 633g in 1L is 1 mol so 96g in 1 L is 0.1516 mole so 96mg in 1L is 0.15mmol

so c=0.15mmol; l=0.5cm; A=0.2363 so ε=A/cl=0.2363/(0.15x0.5)=3.15/mmol/cm

25 µl of 20 mM NADH+ 25 µl pyruvate 0.2 M+ 400 µl of TRIS buffer+ 50 µl lactate dehydrogenase=500 µl total volume. so in 500 µl it contains 1mM of NADH.

ε=3.15, l=1cm, c=1mM so A=εcl=3.15x1x1=3.15.

ΔA=0.5/min or 0.5/60sec=0.0083/sec so after 20sec A will decresased by 0.166 from actual

A=0.1666 l=1cm ε=3.15 so c=A/εl=0.166/(3.15x1)=0.0529mM=52900nM

Specific activity of LDH=Rate of reaction x volume of reaction / mass of total LDH

mass of LDH=2mg/L so in 50µL it is 0.1µg

volume 500µL

Rate of reaction=moles of substance reacted/reaction time in sec=0.0529mM/20sec=52.9µmole/liter/20sec= 2.64µmol/l/sec

Specific activity of LDH=2.64 x 500x10^-6 / 0.1µg = 0.0132µmol/µg/sec x 60sec=0.792µmol/µg/min x1000=792µmol/mg/min

a) Determine the absolute amount of NADH consumed after 20 seconds in the reaction (in nmol) =52900nM

b) Determine the specific activity of LDH as NAD+ produced per LDH in the unit of time (in µmol/mg/min)= 792µmol/mg/min