Lecture 6 - Metabolic Modeling

8/13/2019 Lecture 6 - Metabolic Modeling

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USC-MSChE Course: Bioprocess Technology Metabolic Modeling Engr. Evelyn M. Buque-Taboada

Metabolic Modeling**

Introduction

Basic types of reaction and related enzyme kinetics

Biochemical Networks

Thermodynamic description of chemical networks

** adaptation of the lectures by Prof. J.J. Heijnen, TUDelft, The Netherlands




Micro-organisms contain a network of

enzyme-catalyzed reactions:

Substrate, CiProduct, C j

X 1

X 3

X 2

X 4

E1

E3

E6

E5

E4

E2

Ci,C j extra-cellular substrate or product concentrations

X i intra-cellular metabolite concentrationsEi intra-cellular enzyme concentrations

All reactions are coupled!




Basic types of reactions

Metabolic reactions belong to 2 types:

Uni-uni reaction: A P

Bi-bi reactions: A + B P + Q

Mixed forms can also occur:

Bi-uni reactions: A + B P

Uni-bi reactions: B P + Q




Rates

Rate of any enzyme-catalyzed reaction:

V i = [regulatory term] * [ mass-action term]

Consider the reversible M-M equation for the uni-uni reaction:

A P

eq

P A

P

P A A

iii

K

C C

K C C K

E q

regulatory mass-action

term term




Terms

Mass-action term only contains thedirect reactants and products (A and P).

Regulatory term contains:

direct reactants and products (A, P)

modifier concentrations

enzyme concentration (linear)




Linear expressions

Uni-uni reaction: P Ai C k C k 11

Bi-bi reactions: Q P B Ai C C k C C k 11

Since1

1

k

k K

eq

eq

P Ai

K

C C k 1

eq

Q P

B Ai K

C C C C k 1

uni -uni reaction

bi-bi reactions




Structures in biochemical networks

X1 X2 X3E1 E2

Sequential linkage:

E1

X1 X2

X4 X3

E2

M MGGroup transfer

linkage:

donor couple:

X 1 and X 2

acceptor couple:

X 3 and X 4




Basic structures in metabolic systems

(Hofmeyer)

CHAIN

BRANCH

LOOP

MOEITYCONSERVED

CYCLE




Metabolites in biochemical networks

A X1 X2 X3 Q

E1 E2 E3 E4

Y+

B P




Over-all rate equation

Rates = function of concentrations of: terminal species

modifiers

sum of conserved moeities (enzyme, NAD-/NADH, etc.)

but, not the concentrations of intermediates,which are not moeity conserved.

Concentration of intermediates = function of the concentrations of:

terminal species

modifiers

sum of conserved moeities (enzyme, NAD-, NADH, etc.)




Linear chain of enzymes

A X1 X2 X3 Pk 1 k 2 k 3 k 4

k -1 k -2 k -3 k -4

E1 E2 E3 E4

1)1(1 iiiiii X k X k E Rate of reaction:

K i = k +i/k -i Equilibrium constant:




Conversion rate of substrate A

44321332122111

4321

1111111

/

E k K K K E k K K E k K E k

K K K K C C r

P A A

Here:

44

1

32133

1

2122

1

1

1

11

1111111

E k

k

K K K E k

k

K K E k

k

K E

K

C C E k

r

eq

P A

A

eq K k k k k / k k k k K K K K 432143214321




Effect of enzyme concentration

On the reaction pathway, effect of enzyme concentration varies:

1

2r A

Ei

1. r A proportional to E i2. r A independent of E i




Flux control coefficient

Flux control coefficient, C Ei = relative change in r A upon arelative change in Ei

i

A

i

i

A

A

Ei

E

r

E E

r

r

C

ln

ln

Flux

arginine

Enzyme activity

100%

100%




Control coefficient

By proper mathematical differentiation of the rate equation,

the four (4) control coefficients of the involved enzymesare obtained:

122

2

1

K E Dk C E

2133

3

1

K K E Dk

C E

11

1

1

E Dk C E

32144

4

1

K K K E Dk C E

S CEi = 1

0 < CEi < 1

* Focus on the slowest enzyme !

Why??




Branched pathway

“supply”

branchA

Q

P

Xk 1

k -1k 3

k 2

k -2

k -3E1

E2

E3

two

“output”

branches

reversible branched pathway

Equilibrium constants:

21

21

k k

k k K AP

1

1

k

k K AX

13

31

k k

k k K AQ

23

32

k k

k k K PQ

2

2

k

k K PX

3

3

k

k K QX




Branched pathway: Rates

221133113322

1133

111

11

E k E k K E k E k K E k E k K

K

C C

K E k K

C C

K E k r

QX PX AX

PQ

Q

P QX AP

P

A PX

P

221133113322

1122

111

11


K C C

K E k K C C

K E k r

QX PX AX

PQ

Q

P

QX AQ

Q

A

QX

Q

221133113322

2233

111

11


K

C C

K E k K

C C

K E k r

QX PX AX

AQ

Q

A

QX AP

P A

PX A




Branched pathway

Note: r A = r P + r Q

For the branch flux ratio, R:Q

P

r

r R

PQ

Q

P

QX AQ

Q

A

QX

PQ

Q

P

QX AP

P A

PX

K

C C

K E k K

C C

K E k

K

C C

K E k K

C C

K E k

R

1122

1133

11

11

Finally, for the branched-point metabolite concentration CX,

QX PX AX

Q P A

X

K

E k

K

E k

K

E k

C E k C E k C E k C

332211

332211




Cyclic structures in the pathway

X1 X3

X2

A P

X1 X3

X1 X2 X3

X4

X5

E3

E2E1

cyclic structures

Cyclic structure, the paral lel substrate loop




The parallel substrate loop

X1 X3

X2

A P

E4

E3E2

k -3

k 3k-2

k 2

k -5

k 5

Equilibrium constants:

2

2

2

k

k K

3

3

3

k

k K

5

5

5

k

k K

Equilibrium equation for X 1 and X 3:

K 2K 3 = K 5




The parallel loop

5

31

3322255

111

1

1

1

K

C C

E k K E k E k

r X X

Proper derivation of the rate equation at steady-state

and 1st-order kinetics:

X1 X3

k+

k-

The rate equation shows that a parallel cyclic structure can be

lumped into an equivalent linear structure:




Moeity conserved cycle

A P

Q B

X1 X2

v1

v2

moeity

conserved

cycle

Coupling of two

different processes:

A to P and

B to Q

Rates ??

1

2111

K

C C C C k v X P

X A

2

1

222 K

C C C C k v X Q

X B

Equilibrium constants?

K 1 = ?K 2 = ?






Moeity conserved cycle

A P

Q B

X1 X2

v1

v2

Rate equations for the intermediates:

0211 vv

dt

dC X

0212 vv

dt

dC X

In steady-state: v1 = v2

Under all conditions: 021

dt

dC

dt

dC X X

CX1 + CX2 = constant = T

“conserved moeity sum”




Moeity conserved cycles

Using the steady-state condition (v1 = v2 ) and the

conserved moeity sum (C X1 + C X2 = T), the followingrates are obtained:

2

22

1

11

2

1

1

1

K

C k C k K

C k C k

C k K

C k

T

C

Q

B P

A

B P

X

2

22

1

11

2

21

2

K

C k C k K

C k C k

K

C k C k

T

C

Q

B P

A

Q

A

X

2

2

1

1

2121

K

C C k

K

C C k

K K

C C

C C T k k r

Q

B P

A

Q P

B A

A

B A

B A A

C k C k

C C T k k r

21

21

Simplified to:

When??




Metabolite control coefficients

Metabolite control coefficient, CX,ij

= relative change in

metabolite concentration X i with respect to

the relative change in each enzyme

concentration E j.

S CX,ij = 0 Why??

For example: for 3 enzymes

CX1,1 + CX1,2 + CX1,3 = 0

j

i

i j

ji

j

j

i

i

ij X E

X

X E

E X

E

E

X

X

C ln

ln,




Thermodynamic Driving Force

Consider a simple reaction:

A B

The kinetic driving force is:

K

B A where K = equilibrium constant

Thermodynamic theory states that for DGR :

Gibbs energy of reaction

At reference (standard) state:

DD

B

A RT GG o

R R ln

K RT Go

R lnD




Relation of Kinetics & Thermodynamics

D

RT

G

K

A

B

Rexp11

Combing kinetic and thermodynamic equations:

Close to equilibrium: 1D RT G R using for x << 1, the approximation

)()exp(1 x x

Thus,

'1 G RT

G

K

A

B

R D

D

D

D RT

GG R'where dimensionless




Consider an example: moeity cycle

A P

Q B

X1 X2

v1

v2

Using mass-action kinetics:

1

2111

K

PX AX k V X AP K K

X

X

A

P K

1

21

Then, the rate expression is:

X AP AP K

X X

K

P k

K

P A X k V 2

11

111

kinetic kinetic driving force




Transformation of driving forces

Transforming kinetic and thermodynamic driving forces:

'1 AP

AP

G K

A P

D

'1

2

1 X

X

G K

X X

D

Then, for V1 and V2:

'

11

'

111 X

AP

AP G K

P X k G X k V D

D

'

12

'

222 X

BQ

X BQ G

K

K Q X k G B X k V D

D




Steady-state condition

At steady-state: V1 = V2

Then,

BQ

X

AP

BQ AP

X

K

Q K X k

K

P X k

G B X k G A X k G

1211

'

22

'

11'

DDD

Over-all rate equation of the coupled process:

P K K k

Q K k

G K Q

B X G K

P

A K X k k

r AP

X

BQ

BQ BQ AP AP X

21

'

2

'

121

DD




Equilibrium condition

Close to equilibrium:

1

1

21

BQ

AP

X

K Q

B

K P A

X K X

At this condition, the rate equation is reduced to an expressiondescribing the sum driving force of the moiety cycle reactions:

A + B P + Q

P

K K k

Q

K k

GG X k k r AP

X

BQ

BQ AP

21

''

221

DD

(simplified form of the over-all rate equation close to equilibrium and at steady-state!!)




FINAL EXERCISE

Metabolic Modeling

Consider the pathway for penicillin production:

AA ACV IPN Penv1

O2

IPN

O ACV

ACV

X ek v

C X ek v

K

X ek v

333

2222

1

111 1

Rate kinetics:

2.0

30

150

1

1.0

2

33

22

11

1

O

o

o

o

C

ek

ek

ek

K

At reference steady-state ( = 0.03 h -1 ):

mmol ACV per C-mol X

mmol ACV per C-mol X per h

L.mmol -1.h-1

h-1

mM















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Lecture 6 - Metabolic Modeling

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