Substrate balance of the chemostat

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Reference no: EM1388782

You are the lead engineering in charge of running a single chemostat reactor (75 L) to produce a therapeutic protein from an engineered organism. The protein is a growth associated, extracellular product. The reactor operates at a flow rate of 75 L/h. Your boss asks you to change the reactor conditions to produce 10 g/L-hr of protein product. Assuming all other operating conditions must remain constant. What concentration of input substrate (g substrate/L) is needed to achieve the desired concentration of product leaving the reactor? [Answer should be in format X. That is, some number of digits to the left of the decimal - none to the right.]
Growth parameters for the organism:
td = 0.2 hr
Ks = 1 g/L
kd = 0.05 hr-1
YP/X = 0.1 g protein/g cell
YMX/S = 0.4 g cell/g substrate
Your Answer:
Hint:
If we allow for the extracellular release of a protein product in the chemostat, the protein productivity will be the dilution rate multiplied by the product concentration (i.e., DP). If the protein production is "growth associated", its production is directly proportional to cell mass. Since the protein production is "growth associated", its production is directly proportional to cell growth rate, and we can write:
qp = YP/X ?g = YP/X (D + kd)
This term can be incorporated into the substrate balance of the chemostat (see Eq. 2 on p. 5 of lecture notes on cell growth; see also discussion on bottom of p. 8).
Note also that: YP/S = YP/XYX/S

Reference no: EM1388782

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