hi guys, please help me with my homework Pseudomonas sp. has a mass doubling tim

Question

hi guys, please help me with my homework

Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l (which is unusually high), and cell yield on acetate is 0.46 g cell/g acetate. If we operate a chemostat on a feed stream containing 38 g/l acetate, find the following: a. Cell concentration when the dilution rate is one-half of the maximum b. Substrate concentration when the dilution rate is 0.8 D_max. c. Maximum dilution rate d. Cell productivity at 0.8 D_max [Courtesy of E. Dunlop from "Collected Coursework Problems in Biochemical Engineering, " compiled by H. W. Blanch for 1977 Am. Soc. Eng. Educ. Summer School.]

Explanation / Answer

doubling time t = ln(2)/ µmax

and so knowing doubling time, you can calculate µmax

µ as 0.6931 divided by 2.4000 hrs = 0.2888 hr1 .

The dilution rate for calculating cell concentration is half of the maximum growth rate 0.28881 which is 0.14440.

But notice something - look at the equation for calculating cell concentration

X = YX/S( So – S)

  which can be written

using the expression for

S = KSD /µmaxD

as follows

X = YX/S( So – KSD / µmaxD)

and since D = µmax/ 2

you can write X = YX/S(So KS )

X to be equal to 16.88200 g cell/liter

=====================================================

. When the dilution rate is 0.8Dmax or 0.8µmax (which is 0.23104)

you get S = KS0.8µmax µmax0.8µmax or

simply S = 4KS and you get S as equal to 5.20000 grams/l

Cell productivity is DX which is equal to DYX/S( So – KSD/ µmaxD )

So S = 38.00000 minus 5.20000 = 32.79999

multiplied by DYX/S which is 0.23104 multiplied by 0.46000 which is 0.10628

and hence we get DX to be equal to 3.48606 gcell/(l.hr)

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