Question

A strain 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?

Answer 1

a) to = 4.3 h

b) xf = 4.1 g/L

Step-by-step explanation:

We have the following data:

So = 10 g*L^-1 = initial concentration

Ys = biomass yield = 0.575 g*g^-1

umax = maximum specific growth rate = 0.9 h^-1

The initial cell concentration is equal to:

xo = 12 g/100 L = 0.12 g/L

a) The batch time it takes to reach the stationary phase equals:

to = (1/umax)*ln(1 + (Ys*(so-sf)/xo))) = (1/0.9)*ln(1+(0.575*(10-0))/0.12)))) = 4.3 h

b) Since the fermentation stops after consuming only 70% of the glucose, we have the following:

sf = (1-07)*so = 0.3*10 = 3 g/L

tb = (1/0.9)*ln(1+(0.575*(10-3))/(0.12)))) = 3.94 h

Finally, the final cell concentration can be found by the following equation:

xf = xo*e^(umax * tb) = 0.12 * e^(0.9 * 3.94) = 4.1 g/L

Answer 2

1. 4.32 h

2. 4.12 g/L

Step-by-step 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 =
`e^(3.537)`

X/0.12 = 34.36

X = 4.12 g/L