Question

The molar absorptivities of tryptophan and tyrosine at 240 nm are 2.00 x 103 dm3 mol-1 cm-1 and 1.12 x 104 dm3 mol-1 cm-1. At 280 nm, they are 5.40 x 103 dm3 mol-1 cm-1 and 1.50 x 103 dm3 mol-1 cm-1. A solution of the two has absorbances of 0.660 at 240 nm and 0.221 at 280 nm in a 1.0 cm thick cell. What are the concentrations of these two amino acids in this solution?

Answer 1

Explanation:

Let us assume that the molar concentrations of tryptophan and tyrosine be x and y respectively.

Mathematically,        A = $\epsilon \times t \times C$

where,     A = absorbance

             $\epsilon$ = molar absorption coefficient

                t = thickness of the cell

                C = molar concentration

So, first calculate the molar concentration of tryptophan at 240 nm as follows.

            0.66 = $2000 \times 1 \times x + 11200 \times 1 \times y$

              x = $\frac{0.66 - 11200y}{2000}$  ........... (1)

At 280 nm,

            0.221 = $5400 \times 1 \times x + 1500 \times 1 \times y$

            0.221 = $5400x + 1500y$  ........... (2)

Now, we will substitute the value of x from equation (1) into equation (2) as follows.

         0.221 = $5400 \times \frac{0.66 - 11200y}{2000} + 1500y$

         0.221 = 1.782 - 30240y + 1500y

         0.221 = 1.782 - 28740y

         28740y = 1.561

            y = $54.3 \times 10^{-6} M$

or,            = 54.3 $\mu M$ ............ (3)

Hence, the molar concentration of tyrosine is 54.3 $\mu M$ and putting this value into equation (1) we will get the value for concentration of tryptophan as follows.

              x = $\frac{0.66 - 11200 \times 54.3 \times 10^{-6}}{2000}$

                 = $25.8 \times 10^{-6}$

or,              = 25.8 $\mu M$

Therefore, we can conclude that the concentration of tryptophan is 25.8 $\mu M$ and concentration of tyrosine is 54.3 $\mu M$.