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

Question

4). You have a protein solution. You take 5 ml of the solution, and add 45 ml of buffer. Youthen add 30 1ofthedil teproteinsolu ont 70ofbufferand2ml ofCon ass esolution The resulting absorbance at 595 nm was 0.477 At the same time, you prepared a standard curve for the same protein using the sdme Coomassie assay protocol (100 ul 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 ug protein added to the assay What is the protein concentration in the original, undiluted protein solution? (5 points)

Explanation / Answer

Ans. The original protein sample is diluted as follow-

I. 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

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.