Thanks! While performing a protein concentration measurement, you had to dilute

Question

Thanks!

While performing a protein concentration measurement, you had to dilute protein A two times serially as follows to measure within the linear range of the spectrophotometer. a) 1 mu l of protein A with 4 mu l buffer (this dilution is designated as protein A_1), b) 2 mu l protein A_1 with 18 mu l buffer (this second dilution is designated as protein A_2). The concentration of protein A_2 is 0. 6 mg/mL. What is the dilution factor for protein A_2? What is the initial concentration (before the serial dilution) of protein A?

Explanation / Answer

Ans. Let the initial concertation of protein be X g/ mL

#A1. 1 µL of stock (initial) solution is mixed with 4.0 µL buffer

Total volume of #A1 solution = 1.0 µL + 4.0 µL = 5.0 µL

Amount of protein in the resultant solution = Concertation of stock x volume of stock taken

                                                = (X g/ mL) x 1.0 µL

                                                = (X g/ mL) x 0.001 mL                       ; [1.0 µL = 0.001 mL]

                                                = 0.001 X g

Concentration of resultant solution (final A1) = Amount of protein / total volume of A1 soln.

                                                = 0.001 X g / 5 µL

                                                = 0.001 X g / 0.005 mL                       ; [1 mL = 1000 µL]

                                                = 0.2 X g/mL

#A2. 1 Total volume of #A2 solution = 2.0 µL + 18.0 µL = 20.0 µL

Amount of protein 2.0 µL of #A1 = Concertation of #A1 x volume of #A1 taken

                                                = (0.2 X g/mL) x 2.0 µL

                                                = (0.2 X g/mL) x 0.002 mL                   ; [1.0 µL = 0.001 mL]

                                                = 0.0004 X g

Since 2.0 µL of #A1 is the single source of protein in solution #A2, the total amount of protein in the resultant solution is equal to the amount of protein in 2.0 µL of #A1.

So,

Concentration of resultant solution (final A2) = Amount of protein / total volume of A2 soln.

                                                = 0.0004 X g / 20 µL

                                                =0.0004 X g / 0.020 mL                      ; [1 mL = 1000 µL]

                                                = 0.02 X g/mL

Part 3: Final calculation.

Given, the final concentration of #A2 solution = 0.6 mg/ mL

                                                                        = 0.0006 g/ mL           ; [1 g =1000 mg]

Also, as calculated, the concentration of #A2 solution = 0.02 X g/mL                               

That is,

            0.02 X g/mL = 0.0006 g/ mL

            Or, 0.02 X = 0.0006

            Or, X = 0.0006 / 0.02

            Hence, X = 0.03

Thus, concertation of initial solution = X g/mL = 0.03 g /mL

                                                = 30 mg/mL            

                                                           

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