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HW1_2011_sol

# HW1_2011_sol - Problem 2 → 2 2 Conservation of mass C 5a...

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Unformatted text preview: Problem 2 → 2 2 Conservation of mass: C: 5a = 1 + d H: 10a + 5c = 1.66 + 2e O: 5a + 2b + c = 0.4 + 2d + e N: c = 0.13 (2 X Eq 3) - Eq 2: 4b - 3c = -0.86 +4d replace c = 0.13 & d = 1.4b into the above equation 4b - 0.39 = -0.86 + 5.6b 1.6b = 0.47 b = 0.29 Then, d = 1.4 X 0.29 = 0.41 Eq 1, 5a = 1 + 0.41 ; a = 0.28 Eq 2, 10 X 0.28 + 5 X 0.13 = 1.66 + 2e; e = 0.90 a = 0.28 b = 0.29 c = 0.13 d = 0.41 e = 0.90 . . 1.4 ; Eq 1 Eq 2 Eq 3 . 1.4 a) / . . . 12 . 14 ∗ 0.13 0.28 ∗ 12 ∗ 5 . / 1.66 16 ∗ 0.4 10 16 ∗ 5 b) ∆ ∆ / 0.5 / 10 / 0 0.52 X = 5.7 g/L Problem 3 → . . . Stoichiometric ratio: 0.3 25.2 / 1000 / 15 2 Conservation of mass: C: 6a = 1 + d H: 12a + 3c = 1.6 + 2e O: 6a + 2b = 0.55 + 2d + e N: c = 0.20 1 0.79 ; Eq 1 Eq 2 Eq 3 Eq2 – 2 X Eq3: 3c - 4b = 0.5 -2d Replace b = 1.27, c = 0.20 3(0.20) – 4(1.27) = 0.5 -4d ; d =1.245 Replace d = 1.88 in Eq1: 6a = 1 + 1.245 ; a = 0.374 Replace a = 0.374, c = 0.2 in Eq2: 12(0.374) + 3(0.2) = 1.6 + 2e ; e = 1.744 a = 0.374, b = 1.27, c = 0.20, d = 1.245, e = 1.744 a) To generate 0.3g/15.2g/mol of yeast 0.0119 mol we need from stoichiometry, 1 1 0.0119 ; 0.374 Then YX/S is 0.3 / 0.00445 Amount glucose left: b) 0.00445 . / 10mmol – 4.45mmol = 5.55 mmol From stoichiometry, 1 1 1.245 0.0119 ; 0.00148 Amount of CO2 produced: 14.8 mmol 2 1.27 Problem 4 Substrate (Glucose) limited growth, therefore, use Monod Model Use shaking flasks to grow bacteria at various initial substrate levels Measure biomass concentration (X) at early time points and use dX/dt=uX to find u After obtaining the various data points [substrate levels (S), growth rates (u)], plot their inverse values to fit the Lineweaver Burk equation 1/u = Ks/umax * 1/S + 1/umax Y-intercept gives 1/umax and Slope gives Ks/umax ---> Ks ...
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