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Combining extinction coefficients?

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Combining extinction coefficients?

Postby dae » Tue Sep 18, 2007 9:25 pm

This was a problem I ran into in my Biochem lab, I solved it by combining the extinction coefficients but I've never seen that done before. Does this method work or am I way off?

A solution containing NAD+ and NADH has an optical density in a 1 cm cuvette of 0.311 at 340 nm and of 1.2 at 260 nm. Calculate the concentrations of the oxidized and reduced forms of the coenzyme in the solution. Both NAD+ and NADH absorb at 260 nm, but only NADH absorbs at 340 nm. The extinction coefficients are given below:

Table Y. Data for molar extinction coefficients of NAD and NADH at 2 wavelengths
Compound | Am ( 1 / M x cm ) 260 nm | Am ( 1 / M x cm ) 340 nm
NAD+ | 18,000 | ~ 0
NADH | 15,000 | 6,220


NOTE: A m is the molar extinction coefficient.
---------------
I found the [NADH] concentration via Beer-Lambert's law (A=ecl) at 340 nm:
c=A/(el) = 0.311/(6220 M^-1 cm^-1 * 1 cm) = 5*10^-5 M

And now my confusion,
To solve for the total concentration, I combined the extinction coefficients at 260 nm:
Etotal = (Enad^-1 + Enadh^-1)^-1 = (1/18000 + 1/15000)^-1 = 8181.81 M^-1 cm^-1

Then c(NAD+ + NADH) = 1.2/(8181.81 M^-1) = 1.466*10^-4 M
and c(NAD+) = 9.667*10^-5 M

Thanks!
dae
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Postby blcr11 » Wed Sep 19, 2007 4:20 pm

Not quite, though you're not that far off, either. The absobance at each wavelength is the sum of the absorbance (not the extinctions) of the components at each wavelength, but the extinction coefficients (more properly, molar absorptivities) don't add. You have to solve a set of simultaneous equations for the concentrations of the components. If you have two components, then you need two wavelengths at which you know the extinctions of both components; if there are three components, then you need three wavelengths...yada yada. Check out the principles of muliwavelength (or multicomponent) spectroscopy.
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Postby blcr11 » Wed Sep 19, 2007 4:47 pm

To be more specific, the system of equations for a two-component system is:

A1 = C1*E11 + C2*E21
A2 = C1*E12 + C2*E22

where A's are the absorbance at wavelenghts 1 or 2
E11 = extinction coefficient of component 1 at wavelength 1
E21 = extinction coefficient of component 2 at wavelength 1
and so on, where I'm assuming pathlengths of 1 cm and it is implied that you will multiply by any necessary dilution factor.

C1 and C2 are the unknown concentrations of the components and are the solutions to the system of equations.
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Postby dae » Wed Sep 19, 2007 4:55 pm

Thanks, that makes a ton more sense.
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Postby blcr11 » Wed Sep 19, 2007 7:05 pm

I should have read your problem a little more carefully. You have a simplification. If you let wavelenght 1 be 340 nm and wavelength 2 be 260 nm, then you can determine the concentration of NADH directly, just as you did. Then, applying the addition of absorbances along with the now known concentration of NADH, you can solve directly for the concentarion of NAD.

The first equation reduces to A(340) = C1*E11, because E21 = 0. Then, A(260) = C1*E12 + C2*E22 which can be solved directly for C2 since you know C1 from the data at 340 nm. It should give you the same result as solving a system of equations; this is just a special case of the more general case where both components absorb at both wavelengths. In either case, you can't add extinction coeffients.
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