Question

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"`

Answer 1

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 = epsilonbc`,

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 `A260:A280` ratio, with `n_k` being the mols of the `k`th amino acid:

`color(blue)(A_(260)/A_(280)) = (sum_i n_i epsilon_(260,i))/(sum_i n_i epsilon_(280,j))`

`= ("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")`

`= color(blue)(0.659)`

Oh look, this article has the same equation...