[IamDeem]

The absorbance of UV light at 280 nm by proteins is mostly due to the aromatic amino acids tyrosine and tryptophan. Lactate DH is a tetramer with each subunit having 332 a.a. (36,507 Da) and containing 5 residues of tryptophan (e = 5.6 x 103 M-1cm-1), 6 residues of tyrosine (e = 1.4 x 103 M-1cm-1) and 8 residues of phenylalanine (e = 0.2 x 103 M-1cm-1).

a) Estimate the molar extinction coefficient for this protein at 280 nm. (e = __________)

b) Estimate the E1% for this protein at 280 nm. (E1% = ________________)

c) Calculate the absorbance and percent transmission for a solution of this protein at a
concentration of 0.50 mg/mL from a cell with a path length of 0.50 cm measured at 280 nm. (A = ______ ; %T = _______)

[IamDeem]

So i know,

Atotal = (ecl)1  + (ecl)2 + (ecl)3
Atotal = (5.6e3 *5) +(1.4e3*6) + (8*200) = 3.8e4

then E= A/cl

im not sure what to put for c.  Does being a tetramer having anything to do with the outcome?

[aHerraez]

a)
The molar extintion coeff. measures absorbance per mole, so you must take int account that the protein has 4 subunits: each mole of protein has 20 moles Trp, 24 Tyr, 32 Phe. So E must be 4 times higher than your calculation.

b) c is 1%, i.e. 1 gram per 100 ml. Convert it to moles/L and you are on the way to solution.

c) is trivial once you have calculated E