You have a protein solution. You take 5 ml of the solution, and add 45 ml of buf

Question

You have a protein solution.

You take 5 ml of the solution, and add 45 ml of buffer.

You then add 30 µl of the dilute protein solution to 70 µl of buffer and 2 ml of Coomassie solution.

The resulting absorbance at 595 nm was 0.477.

At the same time, you prepared a standard curve for the same protein using the same Coomassie assay protocol (100 µl of sample + 2 ml of coomassie reagent)

The linear fit to the Coomassie standard curve was:

Y = 0.095 + 0.035x

Where y = absorbance at 595nm and x = µg protein added to the assay.

What is the protein concentration in the original, undiluted protein solution?

Explanation / Answer

Solution 1:

5.0 mL of original protein is mixed with 45.0 mL buffer.

So, total volume of solution 1 = 5.0 mL + 45.0 mL

= 50.0 mL

II. Solution 2:

3.0 uL of solution 1 is mixed with 70.0 uL buffer and 2 mL Coomassie solution.

So, total volume of solution 2 = 30.0 uL + 70.0 uL + 2000.0 uL

= 2100 uL

= 2.100 mL

The Abs of solution 2 was taken, it’s 0.477.

Calculation:

Step1: The trendline equation (standard curve equation) is in form of y = mx + c , where m = slope , c = intercept.

In the graph, Y-axis indicates Abs and X-axis depicts amount of protein. That is, according to the trendline (linear regression or standard curve equation) equation y = 0.0.35x + 0.095 obtained from the graph, 1 Abs unit (1 Y = Y) is equal to 0.035 units on X-axis (ug) plus 0.095.

Given, Abs of unknown sample = 0.477

Putting y = 0.477 in trendline equation-

            0.477 = 0.035x + 0.095

            Or, x = (0.477 – 0.095) / 0.035

            Hence, x = 10.9143

Therefore, amount of protein in solution 2 = 10.9143 ug

Note: if the unit of “x is ug/mL”, there you would need to do a different set of calculation. For the moment, it’s assumed that the unit of “x is ug” as mentioned in the question.

Step 2: We have-

            Protein content in solution 2 = 10.9143 ug in 2.100 mL

            Solution 2 is prepared from 30.0 uL (= 0.030 mL) of solution 1.

Therefore, total protein content in 30.0 uL of solution 1 = 10.9143 ug

Now,

Total protein content of solution 1 = (10.9143 ug / 0.030 mL) x Total vol. of soln. 1

                                                            = (10.9143 ug / 0.030 mL) x 50.0 mL

                                                            = 18190.476 ug

                                                            = 18.19 mg

Step 3:

Solution 1 is prepared from 5.0 mL of original protein sample. So, total protein content in 5.0 mL of original protein sample = 18.19 mg

Now, [Protein] in original sample = Amount of protein / Vol. of original sample

                                                            = 18.19 mg / 5.0 mL

                                                            = 3.638 mg/ mL

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