D Question 5 2 pts A student determined that 10 ?L of unknown protein solution f

Question

D Question 5 2 pts A student determined that 10 ?L of unknown protein solution for the Bradford experiment was too concentrated to fall within the range of the standard curve. The student created a stock dilution of their unknown protein solution by adding 116 ?L of unknown protein solution into enough PBS buffer to have 300 ?1of solution. | The student then transferred 68 ? L of the stock dilution into a cuvette, with enough PBS buffer to make 100 ??before adding 1000 ?L of Bradford reagent. After sitting for 10 minutes, the student measured the absorb then calculated that the concentration of protein in the cuvette was 2.99 ance at 595 nm, Calculate the total concentration () of protein in the original unknown protein solution.

Explanation / Answer

Ans. Given-

#I. 116.0 uL of stock solution was diluted to 300.0 uL. Let’s call is solution 1.

#II. 68.0 uL of solution 1 was diluted to (68 uL + 100 uL + 1000 uL) = 1168 uL = 1.168 mL volume – whose OD was directly measure. Let’s call it solution 2.

#III. [Protein] of solution 2 = 2.99 ug/mL.

# Step 1: Total protein content in solution 2 = Total vol. of soln. 2 x [Protein] in it

                                    = 1.168 mL x (2.99 ug/mL)

                                    = 3.49232 ug

Since solution 2 is prepared by diluting 68 ug of solution 1, the total protein content in 68 uL of solution 1 must be equal to the total protein content of solution 2.

That is,

            68 uL of solution 2 contains 3.49232 ug protein.

So,

            [Protein] in solution 1 = 3.49232 ug/ 68 uL = 0.05135765 ug/uL

# Step 2: Solution 1 is prepared by diluting 116.0 uL of stock solution was diluted to 300.0 uL.

Now, using    C1V1 (stock solution) = C2V2 (solution 1)

            Or, C1 x 116 uL = 0.05135765 ug uL-1 x 300 uL

            Or, C1 = (0.05135765 ug uL-1 x 300 uL) / 116 uL = 0.132822 ug/ uL

# Therefore, [Protein] in original stock soln. = 0.132822 ug/ uL = 132.822 ug/ mL

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