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#### FantaRT

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##### Help with Beer's law on a specific enzyme catalysis
« on: December 02, 2016, 12:01:49 PM »

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 $$2*10^{-4} M$$ 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 $$GSSG + NADPH + H^+ -> 2GSH + NADP^+$$

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 :
$$C_0 = \frac{2*10^{-4}mol*3ml}{1000ml}= 6*10^{-7}mol/3ml$$

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

$$C_{final} = \frac{A}{ε*l} = \frac{0.25}{6220*1} = 4*10^{-5}$$

but if I'm going to use this calculator site I'm taking

$$4*10^{-3}$$

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 $$[NADPH]_{concumed}=C_0 - C_{final}$$ 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 $$[GSH]_{final}=2*[NADPH]_{concumed}$$

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

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

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.
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#### Borek

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #1 on: December 02, 2016, 09:38:45 PM »

1) The initial concentration of NADPH was diluted in a 1.5ml solution and so the new concentration became :
$$C_0 = \frac{2*10^{-4}mol*3ml}{1000ml}= 6*10^{-7}mol/3ml$$

No idea what you are doing here. Nothing looks correct, and atrocities you do to units deserves a serious punishment.

Express all concentrations as M, don't invent your own units.

Quote
but if I'm going to use this calculator site  I'm taking

$$4*10^{-3}$$

Where is my mistake ?

In use of a random web site as if it was reliable, perhaps. 4×10-3 - which is higher than the initial concentration even before dilution - can't be right.

Quote
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 $$[NADPH]_{concumed}=C_0 - C_{final}$$ 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 $$[GSH]_{final}=2*[NADPH]_{concumed}$$

Logic looks OK.

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

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

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.

What was the limiting reagent?
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#### FantaRT

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #2 on: December 03, 2016, 12:30:12 AM »

So I took it from the beginning once again. And you were right. I need some kind of punishment.

1. I calculate the dilution of the initial concentration of NADPH using C1*V1 = C2*V2:
$$[NADPH]_{initial} = \frac{2*10^{-4}M*1.5ml}{3ml} = 1*10^{-4}M$$

2. Calculate the final concentration of NADPH with Beer's Law.
$$[NADPH]_{final}=\frac{A}{ε*l}=\frac{0.25}{6220*1}=4*10^{-5}M$$

3. Calculate how much NADPH have been consumed.
$$[NADPH]_{consumed} = [NADPH]_{initial} - [NADPH]_{final} = 6*10^{-5}M$$

4. Calculate how much GSH have been produced
$$[GSH]_{final} = 2*[NADPH]_{consumed} = 2*6*10^{-5} =1.2*10^{-4}M$$

Quote
What was the limiting reagent?
Here is my thought :
If the limiting reagent was the NADPH, then we wouldn't expect to get a concentration of 4*10^-5 M remaining, but 0M.
The enzyme cannot be the limiting reagent because theoretically stays at the same concentration.
The only one left is the oxidized GSSG. If that's true and by taking in mind the reaction equation, we can conclude that there was 6*10^-5 M of GSSG at the beginning of the reaction and each one of them react with one mole of NADPH (since we have found that the consumed NADPH concentration is 6*10^-5 M).

If that's right then there is no reason to calculate step 4.

What is now your opinion ?
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#### Babcock_Hall

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #3 on: December 03, 2016, 03:25:46 AM »

If the problem asked you for the initial concentration of GSSG, then why are you concerned about the concentration of GSH?  You may be adding complexity to the problem unnecessarily.
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#### FantaRT

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #4 on: December 03, 2016, 06:33:28 AM »

If the problem asked you for the initial concentration of GSSG, then why are you concerned about the concentration of GSH?  You may be adding complexity to the problem unnecessarily.

I thought that maybe help for the calculation of the GSSG.
As I understand you find it not necessary.

The problem does not give out anything else (i.e kinetics of the reaction). And so, i don't know how else can i approach it. Do you think that the above analysis/approach i made is totally wrong ?
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#### Borek

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #5 on: December 03, 2016, 09:07:30 AM »

we can conclude that there was 6*10^-5 M of GSSG at the beginning of the reaction

find the initial concentration of oxidized GSSG
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#### FantaRT

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #6 on: December 03, 2016, 09:50:17 AM »

we can conclude that there was 6*10^-5 M of GSSG at the beginning of the reaction

find the initial concentration of oxidized GSSG

I don't know if it is. He didn't give us the answers. We are going to discuss it at classroom next Friday.
You think that the answer is the 6*10^-5 M ?

Maybe somehow i need to find the initial concentration of GSSG at 1.5 ml before adding the NADPH solution . Any idea ?
« Last Edit: December 03, 2016, 10:14:22 AM by FantaRT »
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#### Borek

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #7 on: December 03, 2016, 12:27:27 PM »

You were told to calculate initial concentration of GSSG. You calculated it. Perhaps I am missing something, but I have no idea what you are trying to do and why you are trying to calculate something else, if you have what you were told to find?

Sure, there is a chance that the result is incorrect, but it won't get better if you use the data given to find - say - the speed of light
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#### FantaRT

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #8 on: December 05, 2016, 07:22:44 AM »

Ok. I will keep that answer and we see..

Thank you very much.
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#### Babcock_Hall

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##### Re: Help with Beer's law on a specific enzyme catalysis
« Reply #9 on: December 05, 2016, 08:16:48 AM »

If the problem asked you for the initial concentration of GSSG, then why are you concerned about the concentration of GSH?  You may be adding complexity to the problem unnecessarily.

I thought that maybe help for the calculation of the GSSG.
As I understand you find it not necessary.

The problem does not give out anything else (i.e kinetics of the reaction). And so, i don't know how else can i approach it. Do you think that the above analysis/approach i made is totally wrong ?
I think that once you have the value of [GSSG], it is easy to obtain [GSH].  As you imply, one is simply double the other.
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