## Combining extinction coefficients?

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### Combining extinction coefficients?

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!

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!

*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.

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.

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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