Hi. I was searching on internet about Beer's Law, and enzymes and i found a post that helped me on this forum. But because I want to read your idea, I became a member and that's my first post.

At University (undergraduate) our teacher gave as the following exercise:

From a solution with [tex]2*10^{-4} M[/tex] NADPH we take 1.5ml and we added to another 1.5ml which contain unknown concentration of oxidized GSSG and glutathione reductase enzyme.

The final absorbance we take for NADPH at 340nm is 0.25 and we used a vessel with 1cm width.

Absorbance coefficient for NADPH at 340nm is 6220 M-1 cm-1. The final question is to find the initial concentration of oxidized GSSG given that the reaction is [tex]GSSG + NADPH + H^+ ->

2GSH + NADP^+[/tex]

So here is my approach till now.

1) The initial concentration of NADPH was diluted in a 1.5ml solution and so the new concentration became :

[tex]C_0 = \frac{2*10^{-4}mol*3ml}{1000ml}= 6*10^{-7}mol/3ml[/tex]

2) Then i have to calculate the final concentration by using Beer's Law and the data from the spectrometer.

Here i take two different numbers.

If I'm going to calculate it be hand

[tex]C_{final} = \frac{A}{ε*l} = \frac{0.25}{6220*1} = 4*10^{-5}[/tex]

but if I'm going to use

this calculator site I'm taking

[tex]4*10^{-3}[/tex]

Where is my mistake ?

My thoughts for the next steps are:

1st to convert the final concentration from the spectrometer from M to mol/3ml.

2nd find how many moles of NADPH have been consumed by subtracting [tex][NADPH]_{concumed}=C_0 - C_{final}[/tex] and then i can find how many moles of GSH have been created because 1mole of NADPH is consuming to create 2 moles of GSH [tex][GSH]_{final}=2*[NADPH]_{concumed}[/tex]

But how can i actually find the initial concentration of oxidized GSH? I thought to use the kinetics equation

[tex]\frac{d[GSH]}{dt} = k*[GSSG]*[NADPH][/tex]

and solve for [GSSG], but since teacher didn't gave us any rate constant probably is not right to do it.

Am i approach it with the right way or not? Any hint or idea is welcomed.

Thank you.