Question

A 1 mg/ml solution of a 30,000 molecular weight protein with 6 tyrosines and no tryptophan has a higher absorption at 280 nm than a 1 mg/ml solution of a 60,000 molecular weight protein with one tryptophan and no tyrosines. Assuming the molar extinction coefficient (∈) of tryptophan and tyrosine at 280 nm are 5,600 M-1 cm-1 and 1,200 M-1 cm-1 respectively, calculate the absorbances expected for the two protein solutions listed above.

Answer 1

A= (ε x c x l)

Now, the value of ε is given for Tryptophan and tyrosine residue but in a protein the effective value of ε will depend on the number of tryptophan and/or tyrosine residues

c= concentration in M

Therefore for first protein,

Effective ε= 6x1200= 7200 M^-1cm^-1

c= 1/30,000 M

l= path length= 1cm (in most cases)

A= 7200 x (1/30000) x 1

   = 0.24

For second protein,

Effective ε= 1x5600= 5600 M^-1cm^-1

c= 1/60,000 M

l= path length= 1cm (in most cases)

A= 5600 x (1/60000) x 1

   = 0.093