Question

Chemostat culture with protozoa Tetrahymena thermophila protozoa have a minimum doubling time of 6.5 hours when grown using bacteria as the limiting substrate. The yield of protozoal biomass is 0.33 g per g of bacteria and the substrate constant is 12 mg l1 . The protozoa are cultured at steady state in a chemostat using a feed stream containing 10 g l1 of nonviable bacteria. (a) What is the maximum dilution rate for operation of the chemostat

Answer 1

The correct answer is - 3.3 g/L.

Step-by-step explanation:

It is known that if endogenous metabolism is not important then,

µnet = µmax = 0.693/doubling time.

Given:

doubling time = 6.5 hr, therefore,

µmax = 0.693/6.5 = 0.1/h

Formula:

First, maximum dilution rate Dmax,

Dmax = µmax*[S0]/{Ks + [S0]}.........(1)

Steady-state cell concentration, X

X = Yx/s{[S0] - KsD/(meumax - D)..........(2)

Solution:

Yx/s = 0.33 g/g-substrate;

Ks = 12 x 10-3 g/L; [S0] = 10g/L,

puting value in (1)

Dmax = 0.1 * 10 / (12 x 10-3 + 10)

= 0.1/h.

Now it is given that operating D is 1/2 of Dmax, therefore,

D in our case will be = 0.05/h

concentration of protozoa cells from equation (2).

Therefore, X = 0.33*{10 - 12 x 10-3 * 0.05/(0.1 - 0.05)} = 3.3 g/L.