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# Basic Concepts Related to Metabolic Pathway Analysis by 4G8D0j

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```									    Basic Concepts Related to
Metabolic Pathway Analysis
• Closed System vs. Open System
– A closed system is a system that no material
(and / or energy) flow-in to / -out from the
system.
• E.g. test tube (has no material flow, but has energy
flow), batch reactor, etc.
– An open system is a system with both
material and energy flow-in to / -out from the
system.
• E.g. CSTR, plug-flow reactor, cell, etc.
Basic Concepts Related to
Metabolic Pathway Analysis
A state where none of any variable changes vs.
time.
• Equilibrium:
A condition where competing influences are
balanced, e.g. chemical equilibrium – a condition
that the concentrations of reactants and products
have no net change over time and in they are in
equilibrium.
•    You should pay attention to the fact that, in reaction
engineering, equilibrium is one type of the steady
state but steady state is not necessarily equilibrium.
This will be demonstrated in following examples.
Irreversible Reaction
in Closed System
• Consider an irreversible reaction that carried out
in test tube environment.
AB
• We put 100 M A and 0 M B into the test tube
initially.
• Simulate this reaction in Polymath:
Irreversible Reaction
in Closed System
• If we have long enough in time, all A will be converted into B, and
the rate of reaction (rF) will be zero when A is consumed.

Irreversible Reaction                                           Irreversible Reaction
in Test Tube                                                    in Test Tube

100                                                            1000
90                                                             900
80                                                             800
70                                                             700

Conc. (M)
Conc. (M)

60                                                             600
A
50                                                             500                                     rF
B
40                                                             400
30                                                             300
20                                                             200
10                                                             100
0                                                               0
0     0.2    0.4         0.6   0.8       1                      0      0.2   0.4     0.6    0.8        1
t (min)                                                         t (min)
Reversible Reaction
in Closed System
• Consider a reversible reaction that carried out in
test tube environment.
A  B
• We put 100 M A and 0 M B into the test tube
initially.
• Simulate this reaction in Polymath:
Reversible Reaction in a Closed
System
Beq
• Equilibrium constant, Keq, is equal to:              K eq 
Aeq

• At equilibrium: forward reaction rate = backward reaction
rate (they are not zero).                rFeq  rBeq

• At equilibrium:     rFeq  kF   Aeq   rB eq  kB  B eq

Beq       kF
• Therefore:     K eq          
Aeq       kB

• For this example, Keq = 5, i.e. at equilibrium, [B]eq will be
5 times of [A]eq.
Reversible Reaction
in a Closed System
•         At equilibrium, [B]eq = 83.3 M and [A]eq = 16.7 M ( [B]eq / [A]eq = Keq = 5.0 ).
•         And rF = rB = 166.67 M / min.
•         This system is in equilibrium and steady state, where [A], [B], rF and rB have no
further change.
Reversible Reaction                                                Reversible Reaction
in Test Tube                                                       in Test Tube

100                                                               1000
90                                                                900
80                                                                800
70                                                                700
Conc. (M)

Conc. (M)
60                                                                600
A                                                              rF
50                                                                500
B                                                              rB
40                                                                400
30                                                                300
20                                                                200
10                                                                100
0                                                                  0
0    0.2    0.4             0.6   0.8       1                      0     0.2   0.4         0.6   0.8        1
t (min)                                                        t (min)
Reversible Reaction
in Open System with Feeding
• Consider the same reversible reaction, but it is carried out in cellular
environment.
Ax  (( A  B ))
• Ax is A outside the cell (external). A is transported into the cell by a
unidirectional transport enzyme.
• Initially, we have 100 M A and 0 M B inside the cell, and we have 10
M Ax (that Ax always keeps constant).
• Simulate this reaction in Polymath:
Reversible Reaction
in Open System with Feeding
• Note that as the cell keeps uptaking A into the cell , A
will “accumulate” inside the cell and this system can
never achieve steady state.
Reversible Reaction                                           Reversible Reaction
in Cell, with Feeding                                         in Cell, with Feeding

500
120                                                           450
rF
100                                                           400
rB
350
rIN

Conc. (M)
Conc. (M)

80                                                           300
A
250
60                                       B
200
40                                                           150
100
20
50
0                                                             0
0      1      2             3   4       5                     0     1       2             3   4         5
t (min)                                                       t (min)
Reversible Reaction
in Open System with Withdrawal
•   Consider the same reversible reaction, but it is carried out in cellular
environment.
(( A  B ))  Bx
•   Bx is B outside the cell (external). B is transported out from the cell by a
unidirectional transport enzyme.
•   Initially, we have 100 M A and 0 M B inside the cell, and we have 10 M Bx
(that Bx always keeps constant).
•   Simulate this reaction in Polymath (note that Bx will not affect this system):
Reversible Reaction
in Open System with Withdrawal
• Note that as the cell keeps losing B, this system can be at steady
state when all A converted into B and all B transported out from the
cell. (I.e., [A] = [B] = 0 M.)
Reversible Reaction                                           Reversible Reaction
in Cell, with Withdraw                                        in Cell, with Withdraw

100                                                           300
90
250
80
70
200
Conc. (M)

Conc. (M)
60                                                                                                     rF
A
50                                                           150                                       rB
B
40                                                                                                     rOUT
100
30
20
50
10
0                                                             0
0      1      2             3   4       5                     0      1      2             3   4        5
t (min)                                                       t (min)
Reversible Reaction
in Open System
with Feeding and Withdrawal
•   Consider the same reversible reaction, but it is carried out in cellular
environment.
Ax  (( A  B ))  Bx
•   A is transported into the cell and B is transported out from the cell by
unidirectional transport enzymes.
•   Initially, we have 100 M A and 0 M B inside the cell, and we have 10 M Ax, 10
M Bx (that Ax and Bx always keeps constant).
•   Simulate this reaction in
Polymath (note that Bx
will not affect this
system):
Reversible Reaction
in Open System
with Feeding and Withdraw
•   Note that the system finally achieves steady state, where rIN = rOUT = (rF - rB).
•   [A]final = 3.0 M, [B]final = 10.0 M, [B]final / [A]final = 3.33 ≠ Keq (where is 5 in this example).
•   Therefore, steady state is not always equivalent to equilibrium, but equilibrium is
always equivalent to steady state.
Reversible Reaction                                            Reversible Reaction
in Cell, with Feeding                                          in Cell, with Feeding
and Withdraw                                                   and Withdraw

100                                                            200
90                                                            180                                       rF
80                                                            160                                       rB
70                                                            140                                       rIN
Conc. (M)

Conc. (M)
60                                       A                    120
rOUT
50                                       B                    100
40                                                             80
30                                                             60
20                                                             40
10                                                             20
0                                                              0
0     2       4             6   8       10                     0     2       4             6   8         10
t (min)                                                        t (min)
Steady State vs. Equilibrium
in Reaction Pathway
• In a reaction pathway that consists many
reactions in series, and we have a reversible
reaction (A  B) in this pathway.
• At steady state, rF = rB and we have net
reaction rate = 0.
• In order to have net forward reaction rate (at
steady state), [A]ss > [A]eq, and/or [B]ss < [B]eq.
• Therefore, [B]ss / [A]ss must be less than Keq.

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