I want to prepare reduced form of NADPH. I am using 100mM of Tris (pH8.0) to make the buffer but the absorbance is showing to be 1.6 at 340nm which should supposed to be 0.6.
can someone please help me to resolve this?
The extinction coefficient of NADPH is 6.22 mM-1cm-1 at 340 nm. If you are using a 1-cm pathlength cuvette to make the measurement of absorbance in a spectrophotometer, a 0.1 mM solution should give you an absorbance of 0.622 at 340 nm, assuming you have first subtracted the solvent blank.
Hi there,
The best thing to do is to perform a scan between 250nm and 400nm so that you will get both specific absorbance peaks of NADPH at 260 and 340 nm. Provided you run a blank first, the absorbance at 340nm will allow you to deduce the concentration of the reduced form and the ratio 260/340 will be indicative of the possible presence of the oxidized form... The ratio is 2.6 if 100% reduced, it tends to increase if oxidization occurs...
NB: both peak values have to be within the linear range of absorbance of the spectrophotometer...
The absorbance or optical density (A) of a solution is described by this equation
A =ecL
where e = the extinction coefficient of the compound (this value is 6.22 x 106 cm2mol-1 for NADPH at 340 nm)
c is the concentration of the substance, and
L is the path length of the light in cm in the cuvette.
In a standard cuvette this value = 1.
Therefore A =ec
Thus a 100 micromolar solution (10-4 molar) of NADPH would have an optical density (A) of 0.622 in a standard 1 cm path length cuvette at 340nm.
Assuming you have no other substances that have any significant absorbance at this wavelength in your solution, your value of A = 1.6 corresponds to a concentration of 267 micromolar so you need to dilute your solution 2.67 fold (for each mL of your NADPH solution add 1.67 mL of buffer). The molecular weight of NADPH often causes errors when making up solutions because many people forget to include the mass of the sodium or potassium or whatever counter ion is in there bottled NADPH.