Author Topic: Bradford assay. Please check if I am doing it right way  (Read 6539 times)

chandhch

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Bradford assay. Please check if I am doing it right way
« on: July 28, 2013, 11:43:48 AM »
Hi all
     I extracted plant protein using phenol extraction method by methanol precipitation. Could you please check if I am doing bradford assay standard curve and also checking protein concentration from unknown sample.


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 I 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


2) 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


Thank you
chandch

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