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 with a 10-fold dilution has an A₂₈₀ reading of 0.202 (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? (this question is asking how much ending IgG was there as compared to how much starting IgG)

Answer 1

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

where,

                                                A = Absorbance

                                                e = molar extinction coefficient (M^-1cm^-1)

                                                L = path length (in cm)

                                                C = concentration

#1. Pre-vaccination aliquot.

Putting the values in equation 1-

            0.85 = (1.35 g^-1 L cm^-1) x C x 1.0 cm

            Or, C = 0.85 / (1.35 g^-1 L)

            Hence, C = 0.6296 g/L

Hence, [IgG] in 2.0 mL extract = 0.6296 g/L

# Total amount of IgG in 2.0 mL extract = [IgG] x Volume of aliquot in liters

                                                = (0.6296 g/ L) x 0.002 L

                                                = 0.0012592 g                                  ; [1 g = 10⁶ ug]

                                                = 1259.2 ug

Since the 2.0 mL aliquot/ extract is prepared from 10.0 mL blood, the total amount of IgG in 2.0 mL extract must be equal to that of 10.0 mL blood.

Hence, total IgG in 10.0 mL of pre-vaccinated blood = 1259.2 ug

#2. Post-vaccination aliquot.

Putting the values for 10-fold diluted extract in equation 1-

            0.202 = (1.35 g^-1 L cm^-1) x C x 1.0 cm

            Or, C = 0.202 / (1.35 g^-1 L)

            Hence, C = 0.1496 g/L

Hence, [IgG] in 10-fold diluted extract = 0.1496 g/L

# Original [IgG] in 2.0 mL undiluted extract = [IgG] of diluted aliquot x dilution factor

                                                                        = (0.1496 g/L) x 10

                                                                        = 1.496 g/ L

Hence, [IgG] in original 2.0 mL extract = 1.496 g/ L

# Total amount of IgG in 2.0 mL extract = [IgG] x Volume of aliquot in liters

                                                = (1.496 g/ L) x 0.002 L

                                                = 0.002992 g                         ; [1 g = 10⁶ ug]

                                                = 2992.0 ug

Since the 2.0 mL aliquot/ extract is prepared from 10.0 mL blood, the total amount of IgG in 2.0 mL extract must be equal to that of 10.0 mL blood.

Hence, total IgG in 10.0 mL of post-vaccinated blood = 2992.0 ug

# Part B: Increase in [IgG] after vaccination =

IgG in post-vaccination blood / IgG in pre-vaccination blood

= 2992.0 ug/ 1259.2 ug

= 2.376

Therefore, vaccination increase IgG content in blood by around 2.4 times.