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

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

$ε_{Trp, 260} \approx 3765 L/mol \cdot cm$ $epsilon_("Trp", 260) "3765 L/mol"cdot"cm"$
$ε_{Trp, 280} \approx 5563 L/mol \cdot cm$ $epsilon_("Trp", 280) "5563 L/mol"cdot"cm"$

$ε_{Tyr, 260} \approx 585 L/mol \cdot cm$ $epsilon_("Tyr", 260) "585 L/mol"cdot"cm"$
$ε_{Tyr, 280} \approx 1185 L/mol \cdot cm$ $epsilon_("Tyr", 280) "1185 L/mol"cdot"cm"$

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Answer 1 · 2017-11-11T17:23:27

Given the above quantities:

$ε_{Trp, 260} \approx 3765 L/mol \cdot cm$ $epsilon_("Trp", 260) ~~ "3765 L/mol"cdot"cm"$
$ε_{Trp, 280} \approx 5563 L/mol \cdot cm$ $epsilon_("Trp", 280) ~~ "5563 L/mol"cdot"cm"$

$ε_{Tyr, 260} \approx 585 L/mol \cdot cm$ $epsilon_("Tyr", 260) ~~ "585 L/mol"cdot"cm"$
$ε_{Tyr, 280} \approx 1185 L/mol \cdot cm$ $epsilon_("Tyr", 280) ~~ "1185 L/mol"cdot"cm"$

We know from Beer's law that

$A = ε b c$ $A = epsilonbc$ ,

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

And so, $A_{2} / A_{1} = ε_{2} / ε_{1}$ $A_2//A_1 = epsilon_2//epsilon_1$ for a fixed number of mols of the same substance. Note that $ε$ $epsilon$ is an intensive quantity, made extensive when multiplying by the mols of substance.

Therefore, we can formulate an equation for the $A 260 : A 280$ $A260:A280$ ratio, with $n_{k}$ $n_k$ being the mols of the $k$ $k$ th amino acid:

$\frac{A_{260}}{A_{280}} = \frac{\sum_{i} n_{i} ε_{260, i}}{\sum_{i} n_{i} ε_{280, j}}$ $color(blue)(A_(260)/A_(280)) = (sum_i n_i epsilon_(260,i))/(sum_i n_i epsilon_(280,j))$

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

$= 0.659$ $= color(blue)(0.659)$

[Oh look, this article has the same equation...](https://www.biotek.com/resources/application-notes/nucleic-acid-purity-assessment-using-a260/a280-ratios/)

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