# Given the following molar absorptivities of amino acids in "0.1 M" phosphate buffer at "pH 7", find the ratio of the absorbance at "260 nm" over "280 nm", the A260:A280 ratio, for a mixture of "2 mol" of "Trp" and "1 mol" of "Tyr"?

## epsilon_("Trp", 260) ~~ "3765 L/mol"cdot"cm" epsilon_("Trp", 280) ~~ "5563 L/mol"cdot"cm" epsilon_("Tyr", 260) ~~ "585 L/mol"cdot"cm" epsilon_("Tyr", 280) ~~ "1185 L/mol"cdot"cm"

Nov 11, 2017

Given the above quantities:

epsilon_("Trp", 260) ~~ "3765 L/mol"cdot"cm"
epsilon_("Trp", 280) ~~ "5563 L/mol"cdot"cm"

epsilon_("Tyr", 260) ~~ "585 L/mol"cdot"cm"
epsilon_("Tyr", 280) ~~ "1185 L/mol"cdot"cm"

We know from Beer's law that

$A = \epsilon b c$,

where $A$ is absorbance, $b$ is the path length of the cuvette, and $c$ is the concentration of the substance...

And so, ${A}_{2} / {A}_{1} = {\epsilon}_{2} / {\epsilon}_{1}$ for a fixed number of mols of the same substance. Note that $\epsilon$ is an intensive quantity, made extensive when multiplying by the mols of substance.

Therefore, we can formulate an equation for the $A 260 : A 280$ ratio, with ${n}_{k}$ being the mols of the $k$th amino acid:

$\textcolor{b l u e}{{A}_{260} / {A}_{280}} = \frac{{\sum}_{i} {n}_{i} {\epsilon}_{260 , i}}{{\sum}_{i} {n}_{i} {\epsilon}_{280 , j}}$

$= \left(\text{2 mol Trp" cdot "3765 L/mol"cdot"cm" + "1 mol Tyr" cdot "585 L/mol"cdot"cm")/("2 mol Trp" cdot "5563 L/mol"cdot"cm" + "1 mol Tyr" cdot "1185 L/mol"cdot"cm}\right)$

$= \textcolor{b l u e}{0.659}$