Question

Five ul of a 10-to-1 dilution of a sample were added to 3ml of Bradford reagent. The absorbance at 595 nm was 0.78 and, according to a standard curve, correpsonds to 0.015mg of protein on the x axis. What is the protein concentration of the original solution?

Answer 1

1 out of a total of 10+1=11.

the new concentration = 1/11 of the original concentration.

The concentration of the sample

(0.015mg/3mL+5 x 10^-3mL)=0.0050mg/mL

So what is the concentration of the sample before it was diluted in 3mL of Bradford reagent?

use the following relationship:

C1V1=C2V2

Where

C1=0.0050mg/mL
V1=3mL+5 x 10^-3mL=3.005mL

C2=?
V2=5 x 10^-3mL

Solve for C2:

(0.0050mg/mL)*(3.005mL)=C2*(5 x 10^-3mL)

(0.0050mg/mL)*(3.005mL)/(5 x 10^-3mL)=C2


C2=3.0mg/mL

3.0mg/mL is 1/11 of the original concentration

the original concentration =

x*(1/11)=3.0mg/mL

Solve for x:


x=3.0mg/mL*11

x=32.9mg/mL=33mg/mL