Consider a batch reactor where the substrate lactose is being enzymatically hydrolyzed by B- galactosidase. The data in the table below is for a wild-type enzyme from E.coli and an engineered mutant (REH4). Using the kinetic parameter for both enzymes (on Lactose) in the table below. Assume the following for your calculations: The milk contains 60 g/L of lactose. The molecular weight of the wild type and mutant enzyme are 540,000 Both enzyme preparations are 20% pure (only 20% of the mass is active enzyme) You can source the wild type enzyme for $100/g and the mutant enzyme for $300/g. You process milk in 1000 L batches. . . . . . How much of each type of enzyme would be required per batch to remove 80% of the lactose within one hour? How much to remove 95% of the lactose within three hours? Which enzyme would be most effective under each condition? TABLE I Values of kinetic constants and ratios of rate constants for the B-galactosidases ND indicates that values were not determined in this study. The intercepts are from appkcat versus (kent - appkcat)/(A) plots where A is methanol or 2-mercaptoethanol. The errors given for keat and Km are standard errors which were obtained by the least squares method described by Cornish-Bowden (1976) assuming simple weighting. For the intercepts of the lines with methanol and mercaptoethanol, standard errors of the least squares fits to the plots are also given. The R values (regression coefficients) for each of the plots with acceptor added were all greater than 0.95. Methanol Mercaptoethanol keat K, kg intercept intercept k2 mm $ s-1 ONPG Wild type REH4 750 + 30 285 + 15 0.12 + 0.02 0.16 + 0.03 2,100" 2,320-2,990 1,200 315-325 ND 640 + 2 (R = 0.95) ND 92 +0.7 (R = 0.98) PNPG Wild type REH4 90 + 5 120 + 10 0.039 + 0.003 0.16 + 0.01 90 190-195 1,200 315-325 ND 190 + 3 (R = 0.96) ND 66 + 0.6 (R = 1.00) Lactose Wild type 1.4 + 0.3 2.7 + 0.2 60 >1,500 60 + 10 315 + 40 From Tenu et al., 1971. 1,200 315-325 ND ND ND ND REH4 Consider a batch reactor where the substrate lactose is being enzymatically hydrolyzed by B- galactosidase. The data in the table below is for a wild-type enzyme from E.coli and an engineered mutant (REH4). Using the kinetic parameter for both enzymes (on Lactose) in the table below. Assume the following for your calculations: The milk contains 60 g/L of lactose. The molecular weight of the wild type and mutant enzyme are 540,000 Both enzyme preparations are 20% pure (only 20% of the mass is active enzyme) You can source the wild type enzyme for $100/g and the mutant enzyme for $300/g. You process milk in 1000 L batches. . . . . . How much of each type of enzyme would be required per batch to remove 80% of the lactose within one hour? How much to remove 95% of the lactose within three hours? Which enzyme would be most effective under each condition? TABLE I Values of kinetic constants and ratios of rate constants for the B-galactosidases ND indicates that values were not determined in this study. The intercepts are from appkcat versus (kent - appkcat)/(A) plots where A is methanol or 2-mercaptoethanol. The errors given for keat and Km are standard errors which were obtained by the least squares method described by Cornish-Bowden (1976) assuming simple weighting. For the intercepts of the lines with methanol and mercaptoethanol, standard errors of the least squares fits to the plots are also given. The R values (regression coefficients) for each of the plots with acceptor added were all greater than 0.95. Methanol Mercaptoethanol keat K, kg intercept intercept k2 mm $ s-1 ONPG Wild type REH4 750 + 30 285 + 15 0.12 + 0.02 0.16 + 0.03 2,100" 2,320-2,990 1,200 315-325 ND 640 + 2 (R = 0.95) ND 92 +0.7 (R = 0.98) PNPG Wild type REH4 90 + 5 120 + 10 0.039 + 0.003 0.16 + 0.01 90 190-195 1,200 315-325 ND 190 + 3 (R = 0.96) ND 66 + 0.6 (R = 1.00) Lactose Wild type 1.4 + 0.3 2.7 + 0.2 60 >1,500 60 + 10 315 + 40 From Tenu et al., 1971. 1,200 315-325 ND ND ND ND REH4