A protein has 3 tyrosine residues and 'n' tryptophan residues both of which are the only amino acids absorbing at 280nm. If the absorbance of the protein having a concentration of 10μM (in a cuvette of path length 2 cm) is 0.59, the number of tryptophan residues in the protein must be

[Given: ε₂₈₀ (Tyrosine) = 1500 M^-1cm^-1

ε₂₈₀ (Tryptophan) = 5000 M^-1 cm^-1]

Concept:

Absorption spectroscopy is a powerful analytical technique used in chemistry, physics, and various other fields to study the interaction of electromagnetic radiation with matter. It provides valuable information about the electronic structure, chemical composition, and concentration of substances in a sample. 

Given:

• ε280 (Tyrosine) =1500 M-1cm-1
• ε280(Tryptophan) = 5000 M-1cm-1
• Absorbance (A) = 0.59
• Concentration (C) = 10µM or 10-5 M
• Path Length (l) = 2 cm 

Explanation:

According to Lambert-Beer Law 

A = εlc 

A_Tyr= 3 X 1500 X 2 X 10^-5

A_Trp= n X 5000 X 2 X 10^-5

∴ A_Total = n₁A_Tyr + n₂A_Trp

0.59 = (2 x 10^-5)(4500 + 5000n₂)

29500 = 4500+5000n₂

n₂ = 5

Conclusion:

The number of tryptophan residues in the protein will be 5.