Question

The sequence of a polypeptide X is as following:
THCARKEDSGYPEWANVRDECW

1. The Optical Density (O.D.) at 280 nm of a solution of the above polypeptide X is 0.51 at the standard experimental conditions (i.e. optical length L = 1.0 cm) . What is the concentration of the polypeptide X in this solution (in mM)? (The molar absorptivity (e) for tyrosine and tryptophan residues at 280 nm are 1,400 M^(-1)cm^(-1) and 5,500 M^(-1)cm^(-1), respectively).

2. Estimate the molecular weight of this peptide.

3. Draw side-chain structures and peptide bonds for first five amino acid residues of the polypeptide X.

Answer 1

The aromatic rings of Tyrosine (Y) and Tryptophan (W) are responsible for the absorbance at 280nm. Phenylalanine gives absorbance at 260nm. Molar absorptivity of tyrosine and tryptophan are given in the question. It means that '1Molar solution of tryptophan in a cuvette of path length 1centimeter will give an absorbance of 5500.

Calculation of Molar Extinction Coefficient is given below. It refers to the absorbance of 1M solution measured in 1cm path length cuvette-

Molar extinction coefficient at 1M is very dense solution. Thus, it is relevant to calculate the absorbance of a protein at 1mg/ml solution, which can be calculated by dividing the Molar extinction coefficient by molecular weight.

Answer 1:

Beer Lambert's Law defines a linear relationship which says that A=cl; where

A= optical density, =molar absorptin coefficient, c=concentration of solution, l=path length

Answer 2

Calculating molecular weight- It is well known that one amino acid residue weighs 110 Daltons

Therefore, the molar extinction coefficient of 1mg/ml solution -

Answer 3