Question

Cytochrome C has a relative molecular weight of 11,384 g/mol. Calculate the extinction coefficient of cytochrome C at 405nm from 6.1 mg/mL.

Absorbance = 3.045

Answer 1

Ans. Beer-Lambert’s Law, A = e C L - equation 1,

where,

A = Absorbance

e = molar absorptivity at specified wavelength (M-1cm-1)

L = path length (in cm)

C = Molar concentration of the solute

Given,

            A = 3.045

            L = 1.0 cm      ; [Assumed to be 1.0 cm, the standard value of path length]

            C = 6.1 mg/ mL

# Calculate molarity of solution from C = 6.1 mg/ mL

Volume of solution = 1.0 mL = 0.001 L

Mass of cytochrome C = 6.1 mg = 0.0061 g                                ; [1 mg = 0.001 mg]

Moles of cytochrome C = Mas in grams / Molar mass

                                    = 0.0061 g / (11384 g/ mol)

                                    = 5.3584 x 10^-7 mol

Now,

            Molarity = Moles of cytochrome C / Volume of solution in liters

                                    = (5.3584 x 10^-7 mol) / 0.001 L

                                    = 5.3584 x 10^-4 mol/ L                                 ; [1 M = mol/ L]

                                    = 5.3584 x 10^-4 M

Hence, concertation of 6.1 mg/ mL cytochrome C is equal to 5.3584 x 10^-4 M

# Putting the values in equation 1-

            3.045 = e x (5.3584 x 10^-4 M) x 1 cm

            Or, e = 3.045 / (5.3584 x 10^-4 M cm)

            Or, e = 5.6827 x 10³ M^-1 cm^-1

Hence, extinction coefficient of cytochrome C at 405 nm, e₄₀₅ = 5.6827 x 10³ M^-1 cm^-1