# Bradford assay Standard Curve Calculation - (Dec/06/2012 )

Hi all

1) BSA Stock 1mg/ml dissolved in deionised water 2) Bradford reagent from Sigma

I took 5µl of stock BSA and add 995µl of Bradford reagent and checked 0.D at 595 nm and also took 10µl, 25µl, 30µl, 40µl, 50µl, 60µl, 70µl of stock BSA and add 990, 975, 970, 960, 950, 940, 930 µl of Bradford reagent and did O.D at 595 nm.

BSA (1mg/ml) Bradford reagent 0.D 595
5 µl 995 µl 0.106
10 µl 990 µl 0.236
25 µl 975 µl 0.544
30 µl 970 µl 0.690
40 µl 960 µl 0.791
50 µl 950 µl 0.861
60 µl 940 µl 0.882
70 µl 930 µl 0.911

The value I got from the graph was

y = 16.04x

R² = 0.7773

Now I have my protein sample (unknown concentration), I did in the following way.

Protein Sample 1 Bradford 0.D 595

10µl 990 µl 0.117
40 µl 960 µl 0.726
50 µl 940 µl 0.813

Protein Sample 2 Bradford O.D 595
10 µl 990 µl 0.135
40 µl 960 µl 0.545
50 µl 940 µl 0.821

Protein Sample 3 Bradford 0.D 595
10 µl 990 µl 0.05
40 µl 960 µl 0.321
50 µl 940 µl 0.671

Then I did calculation in this way: Y = 16.04 x
x= y/16.04 and I got 0.958 mg/ml for sample 1, for sample 2 I got 0.904 mg/ml and for sample 3 I got 0.516 mg/ml

Now I need up to 350 µl for gels, but I need same concentration, so I took the lowest one which is sample 3 (516 µg in 1000 µl) so in 350 µl I got 180.6 µg.

Then I calculated for sample 1 (958µg in 1000 µl) so I got 180 µl and for sample 2 (904 µg in 1000 µl), so I got 199.7 µl

Is this the right way I did to check the known concentration and required concentration? Please help

-chandch-

Your logic seems correct. Just recheck sample 1, there may be a small typo somewhere (958 ug/mL x 0.18mL is not 180ug).

The R square is also high; probably you need to restrict the standard curve to the lowest four/five readings of the BSA standard as it looks to be curving off at the top end.

-DRT-

i'm not crazy about changing the volume of sample and reagent in the standard curve and samples, that adds a couple of more variables to the assay. linearity will suffer.

normally, sample volume is adjusted to some constant with water or buffer and the reagent volume remains constant.

also, your r2 value is low. linearity is suspect. as drt said, you may only be able to use the first few values of the standard due to loss of linearity (loss would be later if volumes were constant, bradford is not really linear).

-mdfenko-