If we have a protein that has a molar extinction coefficient of 17900 M-1cm-1 at 280nm.

(a) By looking at the amino acid composition of the protein, we know it has 1 Tyr in it, but some how the professor had forgotten how many Trps where in the protein. Work out the number of Trps in the protein, given that the molar extinction coefficient at 280nm is 5500 M-1cm-1 for Trp and 1400 M-1cm-1 for Tyr

(b) If the absorbance at 280 nm of a solution of the above protein is 0.56, what is the concentration of the protein in solution, given that the pathlength of the cuvette is 1cm

a) In this protein, there are 3 Tryptophan residues

b) c = 3.128 x 10^-5 M

Explanation:

Molar extinction coefficient (ε) tells us how strongly some chemical substance can absorb light at particular wavelength. In proteins, aromatic amino acids such as Tyrosine and Tryptophan are capable of absorbing the light at 280 nm.

a) ε (protein) = 17900 M-1cm-1

We know that protein has one Tyr, which ε equals 1400 M-1cm-1

If we deduct this value from total ε of protein we wil get ε that comes from all Trp in protein.

ε (Trp) = ε (protein) - ε (Tyr)

ε (Trp) = 17900 M-1cm-1 - 1400 M-1cm-1

ε (Trp) = 16500 M-1cm-1

Then, to get the exact numbers od Trp in protein, we divide this ε with known ε of Trp which is 5500 M-1cm-1

number of Trp: 16500 / 5500 = 3

In this protein, there are 3 Tryptophan residues.

b) The equation which connects all this values is:

A = εlc

A is the amount of light absorbed by the sample for a particular wavelength

ε is the molar extinction coefficient

l is the distance that the light travels through the solution

c is the concentration of the absorbing species per unit volume

A = 0.56

ε = 17900 M-1cm-1

l = 1cm

c = ?

c = A / (εxl)

c = 0.56 / (17900 M-1cm-1 x 1cm)

c = 3.128 x 10^-5 M