The molar absorptivity of a protein in water at 280 nm can be estimated within ~

Question

The molar absorptivity of a protein in water at 280 nm can be estimated within ~5-10% from its content of the amino acids tyrosine and tryptophan and from the number of disulfide linkages (R-S-S-R) between cysteine residues. epsilon_280 nm (M^-1 middot cm^-1) 5500. n_Trp + 1490. n_Tyr + 125 n_S - S where n_Trp is the number of tryptophans, n_Tyr is the number of tyrosines, and n_S - S is the number of disulfides. This information is obtained from the amino acid sequence and from additional studies such as X-ray crystallography and nuclear magnetic resonance to locate disulfide linkages. A certain serum protein has 684 amino acids including 7 tryptophans, 22 tyrosines, 17 disulfide linkages, and carbohydrate chains of variable length. The molecular mass with the dominant carbohydrates is 82220 Da. Predict the molar absorptivity of the protein. Predict the absorbance of a 1.72 wt% protein solution in a 1.000-cm-pathlength cuvet. (Assume the density of the solution is 1.00 g/mL.) Estimate wt% and mg/mL of a protein solution with an absorbance of 1.28 at 280 nm in a 1.000-cm-pathlength cuvet. (Assume the density of the solution is 1.00 g/mL.).

Explanation / Answer

a. Molar absorptivity = €280= 5500×7+1490×22+125×17

                                                      =73405M-1Cm1

b. Absorbance of a 1% solution=Molar absorptivity ÷Molecular wt of a protein

                               =73405÷82220

                               =0.892

Absorbance of a 1.73% solution=0.892×1.72=6.87

c. % Wt= Absorbance÷€%

€%=(€280×10) ÷Mol. Wt.= 8.9

% Wt= 1.28÷8.9=0.143

mg/ml=% Wt×10=1.43 mg/ml

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