In: Chemistry
The extinction coefficients of the four natural nucleoside-5’monophosphates pdA, pdG, pdC and pdT at 260 nm are 1.506*104 , 1.218*104 , 7.10*103 , and 8.56*103 L/mol/cm, respectively (Cavaluzzi, M.J., Nucleic Acids Res., 2004, 32, e13). Please determine:
a) The extinction coefficient of a single-stranded, 5’-phosphorylated Dickerson Dodecamer sequence d(pCGCGAATTCGCG).
Answer: ε260 = 1.244*105. HOw they got that?
b) Please calculate absorbance and percent transmittance of a 10 micromolar (μM) solution of the single-stranded Dickerson Dodecamer, as described in (a), at 260 nm when a cell with the path length of 1.0 cm was used.
What is A? %T?
I don't know the answer for that.
You have already mentioned the Lambert-Beer law.
For the solution of a single compound S in a cuvette of width d, the absorbance Eλ at a particular wavelength λ is proportional to the concentration c of S.
Eλ = (io / i)=ϵλ⋅c⋅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
−linker−S−linker−S−linker
and
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 Eλ 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 ϵλ. The higher ϵλ, the stronger the attenuation of incoming light. Note that ϵλ is
Let's tie the pieces together!
Consequently, we can sum up the combined absorbances in the following way:
Eλ=(ϵA⋅cA+ϵC⋅cC+ϵG⋅cG+ϵT⋅cT)⋅d
Eλ = at 260
Eλ = 1.244 x 10^4