Question 3 continued

(e) The two-stage chemostat system is being used for the production of antibiotics. The volumes of the first and second reactors are V1 = 500 l and V2 = 300 l, respectively. The first reactor is used for recombinant E. coli biomass production, and the second is for the formation of the secondary metabolite LL-37 (an antimicrobial peptide). The feed flow rate to the first reactor is F = 100 l h-1, and the glucose concentration in the feed is [S] = 5.0 g l-1. For the first reactor, the maximum specific growth rate (m) is 0.3 h-1, the saturation constant (Ks) is 0.1 g l-1, and the cell yield with respect to glucose is 0.4 g of cell per g of glucose.

Assuming that the feed medium to the first chemostat is sterile (X0 = 0), the cell death rate is negligible compared to the growth rate, and that the system is at steady state (dX/dt = 0):

(e.1) Determine the cell and glucose concentration in the effluent of the first stage.

[6 Marks]

(e.2) Assume that the growth is negligible in the second stage. In this second reactor, the cells metabolise the substrate to obtain the product, with a specific rate of product formation qp = 0.02 g of product per gram of cell per hour. The yield of product with respect to glucose is 0.6 g of antibiotic per g of glucose. Determine the glucose and LL-37 antimicrobial peptide concentrations in the effluent of the second reactor.