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Unformatted text preview: BME 339/ CHE 339/BIO 335 Fall 2011 Homework #2 Assigned: September 8, 2011 Due: Thursday, September 15, 2011 1. From Kargi and Shuler. 2. Consider a continuous, aerobic bacterial culture in a chemostat with sterile feed. Three different dilution rates D are tested for a glucose feed concentration Sf = 10 mM, and the biomass concentration x and glucose concentration S in the exit stream are measured. The results are as follows: D (h‐1)
0.05 0.5 5 X (g/L) 0.248 0.208 0 S (mM)
0.067 1.667 10 a. Estimate the glucose yield coefficient YX/S (g biomass/mole glucose). b. Assuming Monod growth kinetics, estimate the maximum specific growth rate μmax (h‐1) and the Monod constant KS (mM). 3. In the production of bacterial toxins the growth rate depends on the concentration of the toxin product P and the concentration of the limiting substrate S. In some cases (for low dilution rates), the following equations hold: K
K ,where Kp, Ks are constants (units g product/ liter and g substrate/liter, respectively). The rate of product formation (g product/ liter/ hr) is: RP = A .X ,where A is a constant (g product/ g cells/hr). Derive an expression for the steady state cell concentration X, product concentration P and substrate S in a continuous culture as a function of dilution rate and the known parameters µmax, A, KP, KS, and Sin. Assume Pin=Xin=0 4. In a two stage chemostat system the volumes of the first and second reactors are V1=500 L and V2=300 L, respectively. The first tank is used for biomass production and the second reactor is for secondary metabolite production. The flow rate, F=100 L/hr and the glucose concentration in the feed is Sin=5.0 g/L. The second tank is kept at a temperature above the maximum temperature for cell growth so that µ=0 in the second tank. Using the following constants determine the concentration of product in the effluent of the second tank. Sin = 5.0 g/L YX/S= 0.4 g cell/g substrate KS = 0.1 g/L µmax= 0.3 hr‐1 RP = 0.02 (g product/g cells hr) × X ...
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- Spring '11