# Bradford Assay Calculation! Help! - (Oct/31/2013 )

Dear All,

I have simple problem about Bradford Assay Calculation. I not designed this and would never do it in this way.

Stock Solution: 1.48 mg/ml

1: 5.92 ug/ml (from Stock 4 ul into 796 ul H2O + 200 ul Dye Reagenz )   OD595 : 0.434

2. 8.88 ug/ml (from Stock 6 ul into 794 ul H2O + 200 ul Dye Reagenz)    OD595 : 0.657

3: 11.84 ug/ml (from Stock 8 ul into 792 ul H2O + 200 ul Dye Reagenz)  OD595 : 0865

4: 14.8 ug/ml (from Stock 10 ulinto 790 ul H2O + 200 ul Dye Reagenz)   OD595 : 0.989

y = 0.23x + 0.177

R2 = 0.99

Lets take 6 ul an unkown sample without dilution and a absorbence of OD595 0.295:

0.295 = 0.23 x +0.177 ----> (0.295-0.177)/0.23 = x ---> 0.513 mg/ml ATTENTION: We dived by 6 ul to obtain the true conc. : 0.513/6 = 0.085 mg/ml.

Can you confirm the calculation neglecting to way it was generated.?

And we do not have a sample 0 the blank is not in the graph.!

Must we dived by 6?

Can you send me a protocol?

-Pangea-

Hi.  The algebra is correct but the answer does not make sense to me.  The output of a bradford assay is mass so you do need to divide by volume to get your concentration but your answer, 0.085mg/ml or 85ug/ml, is above the highest point of your graph while the OD is below the lowest.  Also, idealy your samples should fall within the standard curve data points.  I quickly graphed the data and got a different equation entirely with an answer of 3.5 ug/ml.  I'd double check your standard curve.

-Missle-

Missle thank alot can you explain me again why ı should not divide by the volume? I also remember that is not necessary. Can you give me an example? How much of the unkown sample in volume must be pipeted into the 1ml Cuvet?

-Pangea-