Question

Make a table listing the volume of milk power sample solution and the volume of BSA solution you will spike the sample with and the volume of water you will need to dilute to the final volume. The protein content of the dry milk powder is about 30%.

The available stock solutions are 1 mg/ml milk power solution, and 2 mg/ml standard BSA solution.

0, 20, 50, 100, 200, 500μg/mL standard BSA solutions are prepared in clean test tubes, by making dilutions to the original BSA solution (2 mg/mL) with water to a final volume of 600μL.

How do you do the calculations, if there are any. I want to know how. Thank you very much.

all of the concentrations should be withinthe linear range (0.01-1.0 mg/mL for protein analysis) and for accurate determination of the slope and intercept, the maximum amountof solute added from the standard should not be too high or too low (100% to 300% of the solute in the sample would be appropriate)

Answer 1

Ans. Calculation I: Preparing dilutions of standard BSA.

Given [BSA] = 2 mg/ mL = 2000 ug/ mL                                      ; [1 mg = 10³ uL]

Use the following equation-                       C1V1 = C2V2

            Where, C1 = concentration; V1 = Volume           - of standard BSA soln

                        C2 = Concertation;   V2 = Volume           - of diluted aliquots.

Suppose, you want to make 1.0 mL of 20 ug/ mL aliquot (tube 2).

So,       C1 = 2000 ug/ mL                ; V1 = ?

            C2 = 20 ug/ mL                    ; V2 = 1 mL

Putting the values in above equation-

            2000 ug mL^-1 x V1 = 20 ug mL^-1 x 1.0 mL

            Or, V1 = (20 ug mL^-1 x 1.0 mL) / 2000 ug mL^-1 = 0.01 mL

Preparation: Take 0.01 mL (= 10 uL) of standard BSA stock in a 1.0 mL Eppendorf tube. Make the final volume upto 1.000 mL by adding required amount of distilled water. The resultant solution os 20ug/ mL aliquot.

Create the Graph: You write “with water to a final volume of 600 uL”. But nowhere mentioned the procedure for “standard addition”.

I guess a possible procedure for standard addition method-

            Step I: 500 uL of each standard aliquot (0, 20, 50, 100, 200 and 500 ug/ mL) was taken in different tubes and labelled.

            Step 2: 100 uL of milk powder solution (1 mg/ mL) was added to all tubes.

Calculation 2: calculate final BSA concentration in the final volume of each aliquot using C1V1 = V2V2.

            For example, you use 500 uL (V1) of 20 ug/mL (C1) BSA aliquot. Final volume was made upto 600 uL (V2). So, C2 = (500 uL x 20 ug mL^-1) / 600 uL = 16.67 ug/ mL. Similarly calculate final [BSA] in all finally diluted aliquots.

            Note: Or, the volume of standard BSA and milk powder can be reversed. That is, use 500 uL of milk powder solution and 100 uL of each standard BSA aliquots. Both the methods follow standard addition method, the later one being more suitable.

# Plot absorbance vs final [BSA] to get a linear graph. The linear equation is in form of y = mx + c.

From the linear graph equation-

            [Protein] in un-spiked aliquot (aliquot 1 in which 0.00 ug/ mL BSA aliquot was added) is = (m /c ) unit of concertation used.

For example,

            If you get y = 0.020x + 0.005

            Then, [Protein] in un-spiked aliquot = m / c = 0.020 / 0.050 = 4 ug/ mL

# Calculating [Protein] in original milk powder:

You used 100 uL of milk powder solution in un-spiked aliquot (as well as all other aliquots). In this aliquot no standard BSA solution was added. So, protein content in this aliquot is solely due to milk powder solution in it.

Final volume was made upto 600 uL in each case.

Use C1V1 = C2V2 to get original [Protein] in milk powder solution.

            C1 = (4 ug mL^-1 x 600 uL) / 100 uL = 24 ug/ mL

Thus, [Protein] in milk powder solution = 24 ug/ mL

## Given the milk powder is prepared by dissolving 1 mg of milk protein in 1.0 mL water.

So, total protein content in 1.0 mL of (1 mg milk protein/ mL) is equal to protein content in 1.000 mg of milk powder (from which milk powder solution is prepared).

            Protein content in milk powder solution = 24 ug/ mL

So, 1 mL milk powder solution = 1 mg milk powder contain 24 ug protein.

Thus,

            [protein] in original milk powder = 24 ug protein/ mg milk protein