Question

Aliquots of a 0.5 - mg/mL standard of BSA are used to construct a standard curve for the Bradford protein assay. The tubes contain the following amounts of BSA solution: 0, 20, 40, 60, 80, and 100 uL. The corresponding absorbencies after adding Bradford re agent are 0, 0.05, 0.09, 0.14, 0.19, and 0.22. If you take 20 uL of an unknown and add 80 uL of water, mix , take 10 uL of the mixture and add to Bradford reagent, and see an absorbance of 0.08, what is the protein concentration of the undiluted unknown?

Answer 1

Step-by-step explanation:

The general equation would be:

A = Ebc

where E is the molar absorbity,

c is concentration; b is the cubet in cm, and finally A is absorbance.

When we do the plot, of absorbance vs concentration, the slope would be E*b so, first, we need to know the concentration used, compute the plot and then, use the values given at the end to determine the concentration of the unknown:

Calculate first the concentration of the BSA solution, As you are not telling the final volume that you are actually adding into the tubes, I'll assume the final volume of 100 microliters.

C1 = 0.5 mg/mL * (0/100) = 0 ------> A1 = 0

C2 = 0.5 * (20/100) = 0.1 mg/mL ----> A2 = 0.05

C3 = 0.5 * (40/100) = 0.2 mg/mL -----> A3 = 0.09

C4 = 0.5 * (60/100) = 0.3 mg/mL -----> A4 = 0.14

C5 = 0.5 * (80/100) = 0.4 mg/mL -----> A5 = 0.19

C6 = 0.5 (100/100) = 0.5 mg/mL -----> A6 = 0.22

Now, computing A vs C we have the following results:

r2 = 0.996120428

B = slope = 0.44857

A = 2.86x10-3

With these data, let's find out the concentration of the unknown. We know that this is a linear equation and the general expression: y = a + bx where y is absorbance and x the concentration:

0.08 = 2.86x10-3 + 0.44857(c)

0.07714 = 0.44857c

c = 0.1719687 mg/mL

But this value of concentration is the result of diluting the innitial concentration so:

C innitial = 0.1719687 * (100 / 20) = 0.85984 mg/mL

Hope this helps