Aqueous two-phase extraction is used to purify a recombinant HIV-beta-galactosid

Question

Aqueous two-phase extraction is used to purify a recombinant HIV-beta-galactosidase fusion peptide produced in Escherichia coli. For optimum separation, 450 kg of a mixture of 19.7% w/w polyethylene glycol (PEG) and 17.7% w/w potassium phosphate salt in water is needed. Left over from previous pilot-plant trials is 200 kg of a mixture of 20% w/w PEG in water, and 200 kg of a mixture of 20% w/w PEG and 25% w/w salt in water. Also on hand is 100 kg of an aqueous stock solution of 50% w/w PEG in water, 100 kg of an aqueous stock solution of 40% w/w salt in water, and an unlimited supply of extra water. If all of both stock solution mixtures must be used, how much of each of the leftover mixtures and additional water are required?

Explanation / Answer

PEG required = 450*0.197 = 88.65 Kg, potassium phosphaye, salt = 450*0.177= 79.65, balacne is water = 450-88.65-79.65= 281.7 Kg

Salt from stock solution = 100*0.4= 40 kg , water= 60 kg

PEG in stock solution = 100*0.5= 50 kg , water= 50 kg

Since all the stock solution will have to be used, PEG additionally used =88.65-50= 38.65 Kg

Salt additionally to be used = 79.65-40= 39.65 Kg water = 281.7- 60-50= 171.7 Kg

Let x= PEG to be used from 200 kg of mixture containing 20 wt% PEG and y= 200 kg mixture containing 20 wt% PEG

Writing salt balance 39.65= y*0.25, y= 39.65/0.25= 158.6 kg

PEG avaialble = 158.6*0.2= 31.72, PEG additionally required now 38.65-31.72= 6.93. This has to come from 200 kg of mixtue containing 20 Wt% PEG. Hence PEG of 20 wt% required= 6.93/0.2 =34.65 Kg. Water avilable from the solutions = 34.65*0.8 + 158.6*0.45+100*0.5+100*0.4 =209.09. Additional water required = 281.7-209.09=72.61 Kg

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