problem_set_3_answers_v2

problem_set_3_answers_v2 - Name Student ID number...

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Name Student ID number MICROBIOLOGY 140 Due Oct 19 at the beginning of class Problem Set #3 1. A glucose limited chemostat is operated at a dilution rate of 0.15 hr -1 . If Ks for glucose is 1.8 x10 -7 M, the glucose yield constant (Y) is 0.28 and the same medium supports a doubling time of 45 min. in a turbidostat answer the following questions. a) What is the kmax for cells growing in this medium? k=kmax when cells are grown in a turbidostat, therefore, g= 45 min. So k= ln2/g = 0.015 min -1 or 0.92 hr -1 b) what concentration of glucose would you add to the reservoir to maintain a steady-state cell density in the growth vessel of 400μg/ml given: Y = 0.28; x= 400 μg/ml; need c r We know that c= KsD/(Kmax-D) and) x=Y(c r -c), c= KsD/(Kmax-D), and we are given, D = 0.15 hr -1 ; and Ks = 1.8 x10 -7 M c = (1.8 x10 -7 M)(0.15 hr -1 )/[(0.92hr -1 )-(0.15 hr -1 )] = 3.5 x10 -8 M must first convert to μg/ml = [(3.5 x10 -8 mol)/1000ml][( 180 g)/( 1 mol)][(1000000μg)/(1 g) 0.0063 μg/ml have c, now solve for c r c r =x/Y+c c r = (400μg/ml/0.28) + 0.0063 μ g/ml = 1428.5 μg/ml or 0.142 % glucose
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This note was uploaded on 12/02/2009 for the course MIC 140 taught by Professor Meeks,singer during the Spring '09 term at UC Davis.

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problem_set_3_answers_v2 - Name Student ID number...

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