Question

1.Tyrosine has an absorption peak near 280 nm with a molar extinction of 1,450. What would be the absorbance of tyrosine solution that is 1.5x10^-3 M in a 1 cm cell? in a 2 mm cell? in a 0.1 mm cell?

2. Using the molar extinction coefficient from problem 1, determine the E1%280 for tyrosine. The E1%280 is also an extinction coefficient but of a 1% solution measured at the peak near 280 nm. A 1% solution contains 1 gram of tyrosine in 100 mL of total solution. You will have to look up or figure out the molecular weight of tyrosine to solve this problem.

3.What will be the absorbance in a 1 cm cell of tyrosine solution that is 1 mg/mL in concentration? a 1 g/L solution? If these are the same why are they the same? If they are not, why not?

Answer 1

1.

Absorption of tyrosine can be calculated by Beer-Lambert law A=Ɛ*L*C

Where, Ɛ is molar extinction coefficient, L is path/cell length and C is concentration of solution which is usually in M.

Here, A = 1450 * 1 * 1.5x10^-3 M

               = 2.175

For cell length of 2 mm (2 mm = 0.2 cm)

Here, A = 1450 * 0.2 * 1.5x10^-3 M

               = 0.435

For cell length of 0.1 mm (0.1 mm = 0.01 cm)

Here, A = 1450 * 0.01 * 1.5x10^-3 M

               = 0.02175

2.

Molecular weight of tyrosine is 181.19 g.
so, 1 M solution will have 181.19 g/L or 18.119 g/100 ml
Now 18.119 g/100 mL has absorbance of 1450 as given.
Then 1 g/100 mL will have 1*1450/18.119 = 80
Therefore, E1%280 = 80.

3.

Molar concentration of tyrosine 1 mg/ml is 5.52 mM or 0.005 M

A = 1450 * 1 * 0.005 M

    = 7.25

Molar concentration of tyrosine 1 g/L is 5.52 mM or 0.005 M

A = 1450 * 1 * 0.005 M

    = 7.25

Absorbance for both are same as 1 mg/mL and 1 g/L are same ratios only the powers are changed.

Its like 1/1 and 1000/1000, answer will always be 1.