Astrain of escherichia coli has been genetically engineered to produce human protein. a batch culture is started by inoculating 12 g of cells into a 100-litre bubble column fermenter containing 10 g 1^- 1 glucose. the culture does not exhibit a lag phase. the maximum specific growth rate of the cells is 0.9 h^-1; the biomass yield from glucose is 0.575 g g^-1
1. estimate the time required to reach stationary phase.
2. what will be the final cell density if the fermentation is stopped after only 70% of the substrate is consumed?

1. 4.32 h

2. 4.12 g/L

Explanation:

1. For a batch culture, the time (tb) can be calculated by:

tb =  ln (X/X0)/μmax

Where X0 is the initial mass concentration of the cells (12 g/100L = 0.12 g/L), X is the mass concentration of the cells at tb, and μmax is the maximum specific growth rate of the cells.

The biomass yield (Y) is:

Y = (X - X0)/ (S0 - S)

Where S is the mass concentration of the substrate at tb and S0 the initial mass concentration of the substrate (glucose in this case).

Reorganizing:

X = Y*(S0 -S) + X0

Let's assume that at the stationary state all substrate was consumed, so S = 0.

tb = ln[(YS0 + X0)/X0)]/μmax

tb = ln[(0.575*10 + 0.12)/0.12]/0.9

tb = 4.32 h

2.  If 70% of the substrate is consumed, S = 10 - 0.7*10 = 3 g/L

tb = ln[(0.575*(10-3) + 0.12)/0.12]/0.9

tb = 3.93 h

The initial concentration is X0 = 0.12 g/L, the X:

tb =  ln (X/X0)/μmax

3.93 = ln(X/0.12)/0.9

ln(X/0.12) = 3.537

X/0.12 =

X/0.12 = 34.36

X = 4.12 g/L

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