Question

You have a protein of unknown concentration (23,503 Daltons), and you want to determine its concentration using the Bradford assay.

You prepare a standard curve by making dilutions of a standard protein solution at 10 mg/mL, using the volumes noted below as “μL stock”, and then adding water to a final volume of 1 mL for each of the dilutions (this is also what you did in Lab 1). You add 10 μL of each dilution to 990 μL of Bradford reagent and get the following readings:

μL stock	Diluted concentration?	Milligrams in reaction?	A280
5			0.06
25			0.12
50			0.22
100			0.34
250			0.68
500			1.23
750			1.8
1000			2.4

You should graph the standard curve by first calculating the concentration of your standard protein in each sample, then the milligrams added, and finally plotting that versus your absorbance reading.

You are unsure of your unknown protein concentration, so you add 10 μL to 990 μL of the Bradford reagent and obtain an absorbance measurement of 0.92. What is the concentration of your unknown in mg/mL?

Answer 1

Ans. Step 1: Calculation of aliquot concentrations.

Calculation for aliquot 1:

The concentration of protein in aliquots is calculated using C1V1 = C2V2 as follow-

C1= Concentration, and V1= volume of initial solution 1         ; standard solution,

C2= Concentration, and V2 = Volume of final solution 2         ; Aliquots solution

Putting the values in above equation-

            (10 mg/ mL) x 5 uL = C2 x 100 uL

            Or, C2 = (10 mg/ mL) x 5 uL / 100 uL = 0.5 mg/ mL

Hence, [Protein] in aliquot 1 = 0.5 mg/ mL

# Step 2. Calculating mg of protein

            Mass of protein in aliquot = Final [Protein] x Final volume of aliquot

                                                = (0.2 mg/ mL) x 1.0 mL

                                                = 0.5 mg

Since, the final volume is 1.0 mL is all cases, the amount of mg of respective aliquots is equal to the concentration value.

The calculation for rest aliquots are done in excel.

# Step 3: Standard Curve and Unknown Protein concentration.

Plot mg pf protein on X-axis vs respective absorbance of Y-axis. Generate trendline equation for the linear graph.

The trendline equation “y = 0.2311x + 0.0821” is in form of “y = mx + c”.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 0.2311x + 0.0821 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 0.2311 units on X-axis plus 0.0821.

Given, Abs of diluted unknown protein = 0.92

Now,

            Putting the value y = 0.92 in trendline equation-

            0.92 = 0.2311x + 0.0821

            Or, 0.2311x = 0.92 – 0.0821 = 0.8379

            Or, x = 0.8379 / 0.2311 = 3.6257

Hence, mass of protein in diluted unknown solution = 3.6257 mg

# Step 4: Calculation of [Protein] in original unknown

10.0 uL of original unknown sample is diluted upto 1000 uL with Bradford’s reagent to make the “diluted” unknown sample. So, the total protein content of the diluted unknown aliquot is solely due to protein content of 10 uL original unknown sample.

So,

            10 uL of original unknown sample contains 3.626 mg protein.

Or,

            [Protein] in original unknown = 3.6257 mg/ 10 uL                     ; [1 mL = 10³ ug]

                                                            = 3.6257 mg / (0.010 mL)   

                                                            = 362.57 mg/ mL