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339 HW 2 Solutions

# 339 HW 2 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#2...

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Fall 2010 BME339/CHE339/BIO335 HW#2 Solutions Problem 1 Run 1 S = 1 g/L time (hr) X (mg/L) X (g/L) Tot S added (g) 0 100 0.1 1 24 333 0.333 1.466 Run 2 S = 0.1 g/L time (hr) X (mg/L) X (g/L) Tot S added (g) 0 100 0.1 No Data 24 124 0.124 No Data Using µ = [ln(X) – ln (X o )] /t to determine the specific growth rates for each run µ1 = 0.0501 (h -1 ) µ2 = 0.00896 (h -1 ) Plot using Lineweaver-Burk equation u (hr^-1) S (g) 1/u 1/S 0.050124 1 19.95052 1 0.008963 0.1 111.5698 10 Y-intercept = 1 / µmax barb2right µmax = 0.102 (h -1 ) Slope = Ks/ µmax barb2right Ks = 1.04 (g/L) Using Run 1 information, g1851 g3051/g3046 = ∆g1850 −∆g1845 = (0.333−0.1) (1.466−1) =0.5 Y X/S =0.5 (g biomass/ mole glucose) y = 10.18x + 9.7706 0 20 40 60 80 100 120 0 2 4 6 8 10 12 1 / u 1/S Lineweaver-Burk Plot

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Fall 2010 BME339/CHE339/BIO335 HW#2 Solutions 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) Using equation from Handout #5 g1851 g3025/g3020 = g1850 g1845 g3036g3041 −g1845 Use data from the first row Yx/s = 0.248 /(0.01 - 0.000067) Yx/s = 24.97 (g biomass/mole glucose) b) Using Lineweaver-Burk equation in the following form: 1 g1830 = g1837g1871 μg1865g1853g1876 1 g1845 + 1 g1845 Y-intercept = 1 / µmax barb2right µmax = 0.8024 (h
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339 HW 2 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#2...

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