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### #1 Pangea

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Posted 31 October 2013 - 11:44 AM

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?

### #2 Missle

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Posted 01 November 2013 - 04:40 AM

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.

### #3 Pangea

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Posted 11 December 2013 - 12:21 AM

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?

Edited by Pangea, 11 December 2013 - 12:38 AM.