Question

In: Chemistry

You are trying to establish the IgG response from a recently vaccinated individual, where it will...

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?

Solutions

Expert Solution

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.


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