for determination of total protein concen S Explain principal of Bradford colori
ID: 207469 • Letter: F
Question
for determination of total protein concen S Explain principal of Bradford colorimetric assay Problem The absorbance of sample of crude protein extraction of a cancer tissue is 1.8 at 280 nm wavelength and 0.6 at 260 nm You purified and cleaned up the sample from any non-protein contamination. At this point your sample showed an absorbance of 0.9 at 280 nm and 0.01 at 260. Calculate the concentration of your sample in both crude and purified states. (15) Draw your conclusion (10) Hint. If the correction formula is reliable to eliminate the interference of other non-protein molecules in measuring protein concentration by direct spectrometric assay?Explanation / Answer
Hi,
The unknown protein concentration can be calculated using the correction formulae. The question does not mention about the path length and molar extinction coefficient. So most suitable formula is
= Protein concentration(mg/ml) = (1.55* OD 280) - (0.76*OD 260).
Now in the crude extract, OD280 = 1.8; OD 260 = 0.6; so applying in the formula we get. protein crude (mg/ml) = 2.334
In the purified sample, the OD280 = 0.9; OD260 = 0.01; so applying in formula we get,
protein pure (mg/ml) = 1.38
The protein after purifiation has considerably reduced the absorbance at 260 nm. The absorbane at 280nm is reduced by half. The purification process has removed the contaminants which otherwise were giving absorbance at 260nm. The higher the absorption at 280nm, the higher the protein concentration. The major amino acids which absorb at 280 are tyrosine, tryptophan and phenylalanine residue.
The presence of certain contaminants can interfere in the absorption at 280nm, such as nuclei acids, nucleotides, detergents...etc. In the formula, the correction for possible nucleic acid contamination is included, still, the contaminants do exert substantial absorption at 280 and 260.
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