I began by processing a 500 mg plant sample, grinding it in 5 ml of phosphate buffer. Following this, I centrifuged the mixture for 15 minutes and collected 1 ml of the resulting supernatant. To precipitate the proteins, I added 1 ml of TCA to the 1 ml of supernatant and left it overnight. After centrifuging the sample at 3000 rpm, I discarded the supernatant and retained the pellet. The pellet was then dissolved in 2ml NaOH (0.01 N). From this solution, I took 0.5 ml and added 0.5 ml of water along with 2 ml of Bradford reagent. Finally, I measured the absorbance at 595 nm. The reading for the sample was determined to be 169.82 micrograms/ml.
My question pertains to how to estimate the total protein content in the 500 mg sample, allowing for the determination of the protein percentage within my sample.
The reading for the sample was determined to be 169.82 micrograms/ml.
If this result refers to the Bradford assay measurement, then the value should be multiplied by 2 to represent the dilution into the assay (I took 0.5 ml and added 0.5 ml of water), assuming that the concentrations of the standards used to make the standard curve were not subjected to the same 2-fold dilution. The dilution with Bradford reagent doesn't count because the standards would also have been diluted with it. Therefore, the concentration of the final sample was 2 x 169.82 = 339.6 µg/mL.
The sample on which the Bradford assay was run had a volume of 2 mL (The pellet was then dissolved in 2ml NaOH (0.01 N)), so the sample at this stage contained 2 mL x 339.6 µg/mL = 679 µg of protein.
This sample was obtained from 1 mL of the original 5 mL by TCA precipitation. Dilution with TCA precipitating reagent does not have to be taken into account because the pellet was taken. Assuming 100% yield at this step (no protein left in the supernatant), the original supernatant sample from grinding the plant material contained 5 x 679 = 3396 µg of protein at a concentration of 3396 µg/5 mL = 679 µg/mL.
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