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 Dmax
c. Maximum dilution rate
d. Cell productivity at 0.8 Dmax

a)  Cell concentration when the dilution rate is one-half of the maximum is  0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8   is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d)  the cell productivity at 0.8     is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield  on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8    is calculated as:

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore ,  the substrate concentration when the dilution rate is 0.8   is 5.2g/l

c)

Maximum dilution rate is calculated using the expression

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 can be calculated by multiplying the amount  of the cell yield with the amount of the substrate consumed at 0.8  

Cell yield =

Cell productivity at 0.8    = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8     is 2.40g cell/L

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