**5.7 Case Studies**

Example 5.2

You are part of a tech service team asked to evaluate if the available 10,000 liter fermentor is adequate to produce 10 kg/day of a recombinant protein using a strain of E. coli that expresses the protein as 20 % of cellular protein. In order to enhance plasmid stability, the nutrients are manipulated to give a low specific growth rate is 0.2 h

^{-1}. The oxygen demand is 0.15 g O^{2}/g cell - h. Assume that the r-protein formation is cell growth associated.

Data: The lag phase is 4 hours. Typical clean-up time following a fermentation batch and preparation for the next batch is 8 hours. The plant runs three shifts. Cell yield on substrate is 0.55 g cell/g substrate. Available support services can supply inoculum of a maximum of 6 kg of cells every 24 hour period. Maximum K_{La}for the available fermentor is 500 h^{-1}. Fermentor accessories are capable of handling cell concentrations of 60 g/L. Assume any other parameters you need to complete the calculation.Assumption: Critical oxygen conc. is 0.2 mg/L and DO at air saturation is 6.4 mg/L

Solution A:Lag phase and clean-up/ prep time is given as 12 h. If a batch is to completed within each 24 h period, production is limited to 12 h per day. If this is not a limitation, one can optimize production by varying batch time. Let us first evaluate assuming 12 h batch times.If max. cell concentration of 20.6 g/L is obtained, amount of r-protein produced is = (0.2) (0.5) (20.6) = 2.06 g/L. 50% of cell dry matter was assumed to be protein. Hence in 10,000 liters, we will produce 20.6 kg.

Next to determine the inoculum level. The maximum batch growth phase is 12 h. Substitute in growth eqn, and assuming nutrients are present to support exponential growth during the 12 h period,

For 10,000 liters, we will need 18.7 kg every 12 h. Since only 6 kg is available, max. protein that can be produced is

{(0.2)(0.5)[0.6 Exp((0.2)(12)] 10,000 = 6.61 kg

Solution B:Now let us allow batch times to be longer than 12 h, meaning that there might not be a harvest every day. Since it is advantageous to use the max. inoculum concentration, select X0 = 0.6 g/L. This value is obtained by diving 6 kg of cells in 10,000 L. Max. cell concentration is fixed due to aeration requirements. Use the batch growth eqn to find the batch growth time of 17.7 h. Hence 20.6 kg or r-protein will be produced every 29.7 h which gives a 24 h production rate of 16.6 kg.What alternative way of running reactor would you recommend to achieve the production target?