We know, y = ax+b, From BSA curve, a = 0.01017, B = 0.622, In the expt. one absorbance or OD value = 0.765. What will be insoluble protein content?
x = 14.1. You didn't specify the units of the x-axis. It's usually µg.
This only tells you how much protein was in the sample and it assumes that all the protein in the sample was soluble in the dye reagent. If there was protein in the sample that was not soluble in the dye reagent, the protein concentration may not be accurately measured, for 2 reasons. (1) The insoluble protein will not react, but (2) it will scatter light, which will artificially elevate the absorbance. You should be able to tell by visual inspection whether there was any undissolved material in the assay.
Sir, thanks you so much. I am sorry for incomplete question or incomplete data provided.
Ans. This was soluble protein. In the reaction solution used, Sample volume= 5 µl, PBS = 17 µl, G250= 200 µl, mixed and measured at A595nm with ELISA machine. However, 0.2 g plant material (Fresh weight) was grinded or ground,1.6 ml PBS was added during homogenizatin or with homogenate.
Formula: Soluble protein or total protein
= {(x * 20/5)*Vt} / FW, here x is got from standard curve (what you have calculated),
Vt = 1.6 ml, FW= 0.2 g.
If the formula is right, what will be the unit of soluble protein by using this formula.
5 µl out of 1600 µl contained 14.1 µg protein. The total amount in the sample was 14.1 x 1600/5 = 4512 µg. The sample was the soluble protein extracted from 0.2 g of solid plant material: 4.512 mg/0.2 g = 22.5 mg/g.
Respected Sir Adam B Shapiro,
Thanks you so much. Your answer is very easy to understand and very much judicial.
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