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Question on my practice exam. The answers to the questions were not released, otherwise I could work backwards to figure out how to solve it. As it stands, though, I can only guess the method to solve this question.

The extinction coefficients of the 4 natural nucleoside-5'monophosphates @ 260nm (Cavaluzzi,MJ,Nucleic Acids Res., 2004, 32,e13).

*pdA = 1.506 E4 L/mol/cm
*pdG = 1.218 E4 --
*pdC = 7.10 E3 --
*pdT = 8.56 E3 --

Find the Extinction Coefficient of a single stranded, 5'-phosphorylated Dickerson Dodecamer sequence d(pCGCGAATTCGC)

If I saw this question on a test, my first instinct would be to add up the values for pdC+pdG+pdC etc...(following the sequence of letters). Or maybe average them all.

We did learn the Beer-Lambert Law used in absorption spectrum, where e was extinction coeffecient: $$A=E(extinction)*b*c$$ but I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

Does anyone know how to solve this? Is there an equation I'm missing?

Thanks in advance.

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  • $\begingroup$ It would seem that you'd just add them up. Presumably the extinction coefficients are the molar extinction coefficients for the lone precursors. One mole of the stand would contain multiples of each precursor. // You're not asked to solve for the absorbance at all. So you don't need the Beer-Lambert Law. $\endgroup$ – MaxW Mar 15 '16 at 0:43
  • $\begingroup$ Does it make sense to add them? I don't understand the concept of extinction coefficient enough to confidently make that decision. What exactly IS an extinction coefficient? Absorbance I can visualize, path length = size of the cuvette, Molarity no problem. extinction coefficient? $\endgroup$ – Jess L Mar 15 '16 at 0:52
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[…] I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

We have all the data that we need.

You have already mentioned the Lambert-Beer law.

For the solution of a single compound S in a cuvette of width d, the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \log_{10}{\left(\frac{I_0}{I} \right)} = \epsilon_\lambda\cdot c \cdot d \]

A higher concentration of S will result in higher absorbance.

Now imagine that the light-absorbing molecules S are linked together by units that do not absorb the light at the wavelength observed.

Let's compare two equimolar solutions of $$\ce{-linker-S-linker-S-linker}$$ and $$\ce{-linker-S-linker-S-linker-S-linker-S-linker-S-linker}$$

The second will show a higher absorbance, since the concentration of S is higher!

Now let's have a look at equimolar solutions of different compounds S1 and S2 and their absorbance.

We have outlined above that the absorbance $E_\lambda$ is proportional to the concentration of the light-absorbing species.

But apparently, different compound to not all absorb the same at a particular wavelength!

Exactly this is expressed with the molar absorption coefficient $\epsilon_\lambda$. The higher $\epsilon_\lambda$, the stronger the attenuation of incoming light. Note that $\epsilon_\lambda$ is

  • always given for a particular wavelength
  • a compound-specific constant

Let's tie the pieces together!

  1. When light-absorbing units are linked together by 'blind' linkers, the effective concentration of the absorbing units counts.
  2. Different compounds can absorb at the same wavelength, but due to their specific $\epsilon_\lambda$, they will do this to different amounts.

Consequently, we can sum up the combined absorbances in the following way:

$$E_\lambda = \left(\epsilon_A \cdot c_A + \epsilon_C \cdot c_C + \epsilon_G \cdot c_G + \epsilon_T \cdot c_T \right)\cdot d$$

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  • $\begingroup$ @JessL. My pleasure :) $\endgroup$ – Klaus-Dieter Warzecha Mar 15 '16 at 6:50
  • $\begingroup$ @JessL. Welcome to Chemistry.SE! $\endgroup$ – Klaus-Dieter Warzecha Mar 15 '16 at 6:51
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    $\begingroup$ There seems to be a typo in the first equation, should it not be $\log_{10}(I_0/I)$, and decadic log as this is the normal way of reporting extinction coefficients? $\endgroup$ – porphyrin May 11 '17 at 20:08
  • $\begingroup$ @porphyrin Yes! Thanks a lot fot spotting the typo! $\endgroup$ – Klaus-Dieter Warzecha May 12 '17 at 1:21

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