You have isolated a protein that you determined was 27.6 mg mL^-1 after a final

Question

You have isolated a protein that you determined was 27.6 mg mL^-1 after a final dialysis step in the procedure. You are ready to visualize the protein on a gel, but you need to dilute it to 5.0 mu g mu L^-1. At least 10.0 mu L of the diluted sample is required to add to 40.0 mu L of gel loading buffer for SDS PAGE electrophoresis, such that the final sample added to the gel is 1.0 mu g mu L^-1. Explain in how are going to perform the 27.6 to 5.0 mg mL^-1 dilution using the equipment normally available to you. How would from #2 be affected if you failed to mix the diluted samples to homogeneity prior to mixing an aliquot with gel loading buffer?

Explanation / Answer

Ans. Given,

            I. Initial concentration = 27.6 mg/mL             ; [1 mL = 1000 µL]

                                                = 27.6µg/ µL                          ; [1 mg = 1000 µg]

            II. Final concentration = 5.0 µg/ µL

Now,

Using C1V1 = C2V2

C1= Concentration of initial solution 1, V1= volume of initial solution 1      ;( Initial solution)

C2= Concentration of final solution 2, V2= volume of final solution 2        ; [Final solution]

            Where, C1 = 27.6µg/ µL         ; V1 =?

                        C2 = 5.0 µg/ µL           ; V2 = 10 µL

Putting the values in above equation-

            27.6 µg/ µL x V1 = 5.0 µg/ µL x 10 µL

            Or, V1 = (5.0 µg/ µL x 10 µL) / (27.6 µg/ µL)

            Hence, V1 = 1.8 µL

Preparation: Take 1.8 µL of Initial solution in a sterile Eppendorf tube. To it add 8.2 µL distilled water to make final volume of 10 µL. The resultant solution is 10 µL in volume with protein concentration of 5.0 µg/ µL.

Note: It’s better to make to make around 50 µL of 5.0 µg/ µL solution for the sake of measuring accurate volume of 9.0 µL of initial solution. Taking 1.8 µL solution through micropipette may seem difficult many times.

#2. In case the diluted sample is not mixed to homogeneity, it will have randomly uneven distribution of protein molecules in different regions of the solution. So, if a part of it is used for further analysis using electrophoresis, the resultant protein concentration may be higher or lower or equal to the correct value. Also, repeating the same analysis may give different results every time, that the result will NOT be constant or uniform.

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