Question

Use the Beer-Lambert Law to calculate the molar extinction coefficient for GFP in Tris buffer at 397 nm (or the peak you observed closest to 397 nm).

peak is A=.023 at wavelength of 397.

GFP is 1.5 mg/ml solution, 0.1 mL

Tris buffer is 10 mM

You add GFP into 1.9mL of 10 mM of tris buffer.

Path length is 1cm

what is the molar coefficent using A=ECL

Answer 1

Concentration of GFP is

0.1 mL 1.5 mg/mL solution is diluted to 2 mL using a tris buffer.

Concentration in 2 mL = 1.5 x 0.1/2 = 0.075 mg/mL

molar mss of GFP is 27000 g/mol

One liter of 0.075 mg/mL solution contains 75 mg of GFP or 0.075 g of GFP

number of moles in 0.075 g = 0.075/27000 = 2.7 x 10^-6 moles

Since molarity is the concentration in 1000 mL

concentration in molaruty = 2.7 x 10^-6 M

By Beer Lamberts law

molar extinction coefficient = absorbance/(concentrationx pathlength) = 0.023/(2.7 x 10^-6 x 1)

= 0.00851 x  10⁶ M^-1cm^-1