140g of raw meat was ground thoroughly and divided into 7 equal portions, and on

Question

140g of raw meat was ground thoroughly and divided into 7 equal portions, and one of these portions was run through a tissue homogenizer in order to break open all the cells. This cell lysate was then centrifuged to remove membranes and any insoluble protein, producing 340 mL of a solution containing soluble protein. 15uL of this solution was then subjected to a Bradford assay (linearized regime), yielding A595=0.45, while 15 uL of a standard (10mg/mL) protein solution yielded A595=0.30. How much soluble protein (mg/g) must have been present in the original sample of beef?

Explanation / Answer

Ans. Protein concertation in the finally diluted unknown aliquot (whose OD is taken) is given by-

[Protein] in aliquot = [OD of unknown / OD of standard] x [Protein] in standard

                                    = (0.45 / 0.30) x (10.0 mg/ mL)

                                    = 15 mg/ mL

Therefore, [protein] in finally diluted 340.0 mL aliquot = 15 mg/ mL

Soluble protein content in 340 mL aliquot = [Protein] in aliquot x Vol. of aliquot

                                                = (15 mg/ mL) x 340 mL

                                                = 5100 mg

Since the finally diluted aliquot is prepared from one of the seven portions of beef sample, the soluble protein content of one portion of beef sample = 5100 mg

Now,

Total soluble protein in original sample =

soluble protein per portion x No. of portions

= (5100 mg/ portion) x 7 portion

= 35700 mg

= 35.70 g                                                      ; [1 g = 1000 mg]

Therefore, total soluble protein content in original beef sample = 35.7 g

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