Question

Th e molar absorption coeffi cients of tryptophan and tyrosine at 240 nm are 2.00 × 103 dm3 mol−1 cm−1 and 1.12 × 104 dm3 mol−1 cm−1 , respectively, and at 280 nm they are 5.40 × 103 dm3 mol−1 cm−1 and 1.50 × 103 dm3 mol−1 cm−1 . Th e absorbance of a sample obtained by hydrolysis of a protein was measured in a cell of thickness 1.00 cm and was found to be 0.660 at 240 nm and 0.221 at 280 nm. What are the concentrations of the two amino acids?

Answer 1

Answer:

5.43x10⁻⁵ dm³mol⁻¹= Concentration of tyrosine

2.58x10⁻⁵ dm³mol⁻¹= Concentration of tryptophan

Explanation:

Lambert-Beer law is:

A = E×C×l

Where A is measured absorbance, E is absorption coefficient, C is concentration of solution and l is optical path length.

For the result at 240nm, it is possible to write:

0.660 = 2.00x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.12x10⁴ dm³mol⁻¹cm⁻¹× Ctyr × 1cm (1)

At 280 nm:

0.221 = 5.40x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.50x10³ dm³mol⁻¹cm⁻¹× Ctyr × 1cm (2)

Thus:

-2.7× (0.660 = 2.00x10³ × Ctry  + 1.12x10⁴ × Ctyr) =

-1.782 = -5.40x10³Ctry - 3.024x10⁴Ctyr

0.221 = 5.40x10³ × Ctry + 1.50x10³  × Ctyr

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-1.561 = -28740×Ctyr

5.43x10⁻⁵ dm³mol⁻¹= Ctyr

Replacing this value in (1) or (2) it is possible to find Ctry, that is:

2.58x10⁻⁵ dm³mol⁻¹= Ctry