Question

Most proteins contain aromatic residues that absorb UV light at 280 nm. Estimates of the molar extinction coefficients (ε) are made based on sequence composition and the relative strengths of isolated side chain chromophores. The formula is: ε (λ = 280 nm) (M-1 cm-1) = (# Trp)(5500) + (# Tyr)(1490) + (# disulfide bonds)(125). What is the estimated ε (λ = 280) for TP53 at the convenient laboratory unit of (mg/ml)-1 cm-1? Note that there are 3 disulfide bridges in TP53.

Answer 1

Aλ = ε c L

A(280nm) = ε_{(average of Try+Tyr+disulfide bonds)} * c * L--------Beer Lamberts Law

Molar Extinction Coefficient for TP53 = (Number of Tryptophan residues X 5500) + (Number of Tyrosine residues X 1490) + (Number of disulphide bonds ×125)

Number of Tryptophan residues= 4

Number of Tyrosine residues= 9

Number of disulphide bonds = 3

Molar Extinction Coefficient = (4 X 5500) + (9 X 1490) + (3×125)

Molar Extinction Coefficient = 35785 M^-1cm^-1

Molecular weight of TP53: 43683.1

Molar Extinction Coefficient = 35785 M^-1cm^-1 / 35785 (mg/ml)^-1 cm^-1