HW2_2011_sol - K S = 1.008 (mM) y = 1.2565x + 1.2463 5 10...

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Problem 2 S f =10 mM D (hr -1 ) X (g/L) S (mM) 1/D 1/S 0.05 0.248 0.067 20 14.92537 0.5 0.208 1.667 2 0.59988 5 0 10 0.2 0.1 Notice: wash out; exclude this row of data for analysis a) ܻ ௑/ௌ ܺ ܵ ௜௡ െܵ Use data from the first row Yx/s = 0.248 /(0.01 - 0.000067) Yx/s = 24.97 (g biomass/mole glucose)
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b) Using Lineweaver-Burk equation in the following form: 1 ܦ ܭݏ μ ݉ܽݔ 1 ܵ 1 ܵ Y-intercept = 1 / μmax μ max = 0.8024 (h -1 ) Slope = Ks/ μmax
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Unformatted text preview: K S = 1.008 (mM) y = 1.2565x + 1.2463 5 10 15 20 25 5 10 15 20 1/D 1/S Lineweaver Burk plot Alternate solution, Problem #4 Assuming no product generation in tank #1. P1=0 0 = F(P 2 )+r P V 0 = P 2 +r P /D r P = (0.02 gproduct/(gcells*hr))*X 1 r P = (0.02 gproduct/(gcells*hr))*1.92gcells/L r P = 0.0383 gproduct/(L*hour) P 2 = r P /D = [0.0383 gproduct/(L*hr)]/[(100L/hr)/300L] P 2 = 0.1152 gproduct/L...
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This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.

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HW2_2011_sol - K S = 1.008 (mM) y = 1.2565x + 1.2463 5 10...

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