Do absorbances of different compounds at one wavelength sum up?

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. 





> 
> […] 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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