Question

29. Complete the problem on the slide titled 'Sample Problem". If the molecular weight of Protein G:B1 is 8 kDa, what is the concentration (in mg/mL) of a solution of this protein which has an absorbance (A) =0.2 at 2 = 280 nm? How many grams of protein are present in 1 mL of this solution? Protan G:B1 3 tyrosine & One tryptophan

Answer 1

According to Beer's law:

A = $\epsilon$cl

Where 'A' is the absorbance of the solution

'c' is the concentration of the solution

'$\epsilon$' is the molar absorption coefficient of the protein (in M^-1.cm^-1)

'l' is the sample length = 1 cm (in general)

i.e. 0.2 = $\epsilon$c * 1 cm

i.e. c = (0.2/$\epsilon$) M

i.e. c = (0.2/$\epsilon$) mol/L

i.e. c = (0.2/$\epsilon$) mol/L * 8000 g/mol (Note: kDa = 1000 g/mol)

i.e. c = (1600/$\epsilon$) g/L or (1600/$\epsilon$) mg/mL

i.e. c = (1.6/$\epsilon$) g/mL

Note: Find the '$\epsilon$' value of the protein from the internet/any reference or standard book, then substitute in the final answer then you will get the exact value in g/mL