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

Step-by-step explanation:

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

Mathematically, A =
`\epsilon * t * 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 * 1 * x + 11200 * 1 * y`

x =
`(0.66 - 11200y)/(2000)` ........... (1)

At 280 nm,

0.221 =
`5400 * 1 * x + 1500 * 1 * 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 * (0.66 - 11200y)/(2000) + 1500y`

0.221 = 1.782 - 30240y + 1500y

0.221 = 1.782 - 28740y

28740y = 1.561

y =
`54.3 * 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 =
`(0.66 - 11200 * 54.3 * 10^(-6))/(2000)`

=
`25.8 * 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`.