Suppose you had 85µl of a protein sample. Of this 12 µl were diluted into 108 µl

Question

Suppose you had 85µl of a protein sample. Of this 12 µl were diluted into 108 µl of water. Of this dilution 25 µl were submitted to Bradford protein assay and produced an OD = 0.63. This is a thought question: for pure molecules, one would use the Lambert-Beer equation and need the extinction coefficient. But, some biochemical assays do not contain pure molecules under defined conditions, thereby require a standard curve. What does the dilution from the original sample that 25 µl measured in Bradford assay represent? How much protein is in the original protein sample?

Explanation / Answer

The orginal protein sample was diluted 1/10 times. This is done by mixing 12 ul of original protein sample with 108 ul of water. The dilited sample gave an OD of 0.63.

The microplate Bradford assay achieves an absorbance of 0.1 at about 6 micrograms/ml

So the protein concentration is 0.63* 10 * 6 micrograms/ml

= 37.8 micrograms/ml is the concentration of the protein or

= 0.037 micrograms/ul

So for 85 ul it would be 0.037 * 85 = 3.145 micrograms

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