A microorganism has a mass doubling time of 2.4 hr when grown on a certain subst

Question

A microorganism has a mass doubling time of 2.4 hr when grown on a certain substrate. The saturation constant using this substrate K_s is 1.3 g/L and maximum growth rate mu_m of 0.4 hr^-1. The cell yield on for this substrate is 0.46 g cell/g substrate. The feed stream to the chemostat contains 38 g/L substrate calculate the following. a. Cell concentration when the dilution rate is one half of the maximum b. Substrate concentration when the dilution rate is one half of the maximum c. Cell productivity when the dilution rate is one half of the maximum

Explanation / Answer

Cell concentration when the dilution rate is one-half the maximum

d= 2.4 hr

Ks = 1.3 g/L

Yx/s = 0.46 g cells/g Subs

Sin = 38 g/La)

Because td is min, µ is µminµm= (ln 2) / td = (ln 2) / 2.4 = 0.29 hr-1

Dc = µm= 0.29 hr

µ = D,

X=Yx/s(Sin -S)

=DKs/(µm- D)

µ = D =µm/2 = 0.15 (hr^-1)

X = Yx/s(Sin - DKs/(µm- D)) =16.8 g/L---------------answer

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S Substrate concentration = DKs/(µm- D) =1.39 g/L--------answer

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