Biomedical Engineering Reference
In-Depth Information
the volume of the reactor. The consumption of a substrate is expressed similarly to
(7.26) with a negative sign to the production rate. For growth-limiting substrates,
the Monod growth kinetic model is also used.
dC
−
kSX
=
max
−
pS
(7.30)
dt
K
+
S
S
where
k
max
is the maximum substrate consumption rate, S is the substrate con-
centration, and
K
s
is the Monod growth constant. Perfusion bioreactors provide
a number of advantages, including: (1) less nonproductive time expended in emp-
tying, filling, and sterilizing the reactor, (2) reduced labor expense, (3) consistent
product quality due to invariable operating parameters, (4) decreased toxicity risks
to staff, due to automation, and (5) reduced stress on instruments due to steril-
ization. The disadvantages of perfusion bioreactors include: (1) higher investment
costs in control and automation equipment and increased expenses for continuous
sterilization of the medium, (2) minimal flexibility, since only slight variations in
the process are possible (throughput, medium composition, oxygen concentration,
and temperature), and (3) greater processing costs with continuous replenishment
of nonsoluble components.
EXAMPLE 7.5
Thrombopoietin (TPO) is a hematopoietic growth factor that induces the proliferation of
megakaryocytes and the differentiation of immature megakaryocytes. TPO is a potential
therapeutic glycoprotein for the amelioration of thrombocytopenia associated with
chemotherapy, irradiation, and bone marrow transplantation. To produce TPO, Sung et
al. [2] transfected a Chinese hamster ovary (CHO) with the hTPO gene and cultured it in a
50-mL medium at 2
×
10
5
cells/mL. These cells can produce 4
μ
g/10
6
cells per day of TPO.
If an environmental condition supporting a specific growth rate constant is 0.39 day
−
1
and
a death rate constant is 0.04 day
−
1
, how much TPO is generated in 3 days?
Solution:
To determine the production of TPO, we need to solve (7.28) and (7.29) simultaneously.
dX
(
)
=
0.39
−
0.04
X
dt
dC
TPO
=
4
X
dt
Solving the two equations simultaneously in the Mathsolver with
X
(
t
= 0) = 0.2 × 10
6
cells/mL.
X
(
t
=
3 days)
=
0.571
×
10
6
cells/mL
C
=
4.246
μ
g/mL
For 50 mL,
C
= 0.213 mg
