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"#

1 Answer
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 = 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...