Question

You are trying to establish the IgG response from a recently vaccinated individual, where it will be important to know how much IgG they have both before and after the vaccination. You start with two samples of 8 mL of blood, one from before vaccination and one after. You use an affinity chromatography technique to remove just the IgG (ε = 1.35 g-1L cm-1). You end with a 2 mL sample of each. The pre-vaccinated sample has an A280 reading of 0.85, and the post-vaccinated sample has an A280 reading of 0.202 with a 10-fold dilution (both with a 1 cm path length).

a) How many micrograms of IgG were in each of the original blood samples?

b) What fold change in IgG amounts was seen after vaccination?

Answer 1

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

                                                A = Absorbance

                                                e = molar absorptivity at specified wavelength

                                                l = path length (in cm)

                                                C = Molar concentration of the solute

Given,

            e = 1.35 g^-1L cm^-1 = 1350 mg^-1L cm^-1      ; [1 g = 1000 mg]

            A (before vaccination) = 0.85

            A (After vaccination, 10-times diluted sample) = 0.202

Part A: 1. [IgG] Before vaccination,

Putting the values in equation 1-

            0.85 = (1350 mg^-1L cm^-1) x [IgG] x 1 cm               ; [C = [IgG]]

            Or, [IgG] = 0.85 / (1350 mg^-1L) = 6.296 x 10^-4 mg/L

So, [IgG] before vaccination = 6.296 x 10^‑4 mg/L

Amount of IgG in 8.0 mL (= 0.008 L) blood sample = [IgG] x volume of blood in L

                                                                        = (6.296 x 10^‑4 mg/L) x (0.008 L)

                                                                        = 5.036 x 10^-6 mg

                                                                        = 5.037 ng                   ; [1 mg = 10⁶ ng]

Part A: 2. [IgG] After vaccination,

Putting the values in equation 1-

            0.202 = (1350 mg^-1L cm^-1) x [IgG] x 1 cm             ; [C = [IgG]]

            Or, [IgG] = 0.202 / (1350 mg^-1L) = 1.496 x 10^-4 mg/L

Since, this solution is 10 times diluted (1:10 dilution), total [IgG] = (1.496 x 10^-4 mg/L) / dilution factor

                                                                        = (1.496 x 10^-4 mg/L) / (1:10)

                                                                        = 1.496 x 10^-3 mg/L

So, [IgG] after vaccination = 1.496 x 10^-3 mg/L

Amount of IgG in 8.0 mL (= 0.008 L) blood sample = [IgG] x volume of blood

                                                                        = (1.496 x 10^-3 mg/L) x (0.008 L)

                                                                        =1.197 x 10^-5 mg

                                                                        = 11.97 ng

Part B: Number of times increase in [IgG] after vaccination =

[IgG] after vaccination / [IgG] before vaccination

= 11.97 ng / 5.037 ng

= 2.37 times

Thus, vaccination results a 2.37 fold increase in IgG amount in the given sample of blood.