MFA Examples

8/13/2019 MFA Examples

1/17

1

MFASome further examples

Lactic acid bacteriaExample 5.6 Important example!

Lactic acid bacteria are often grown on complex

media rich in amino acids, which provide the carbon

skeleton for biomass synthesis.The catabolic

pathways can therefore be analyzed decoupled from

the anabolic pathways.

There is virtually no drain of precursors in the EMP

pathway, which therefore can be regarded as a linear

pathway.


2/17

2

Lactic acid bacteria

Glucose

NADH

Pyruvate

CO2

ATP

Lactate

Acet yl-

CoA

FormateNADH

Acetyl-P

Acetate

Acetaldehyde

Ethanol

NADH

ATP

NADH

NADH

v1v2

v4

v3

v5 v6

In general we know that

6

5

4

3

2

1

200111

111100

001111

100000

010000

001000

000100

000010

000005.0

0

0

0

v

v

v

v

v

v

r

r

r

r

r

r

NADH

AcCoA

PYR

e

a

f

c

l

g

But we cannot solve the problem using rf, ra and re as measured rates

T r


3/17

3

(Or in C-moles)

6

5

4

3

2

1

1003333.03333.03333.0116667.06667.000

001111

003333.0000

100000

010000

0003333.000

000010

000001

00

0

v

v

v

v

v

v

r

r

r

r

r

r

NADHAcCoA

PYR

f

e

a

c

l

g

T r

6

5

4

3

2

1

200111

111100

001111

100000

010000001000

000100

000010

000005.0

0

0

0

v

v

v

v

v

v

r

rr

r

r

r

NADH

AcCoA

PYR

e

a

f

c

l

g

Assume instead that glucose,lactate and formate are measured

Move rows


4/17

4

The new matrix T becomes

6

5

4

3

2

1

200111111100

001111

001000

000010

000005.0

000100

100000

010000

00

0

v

v

v

v

v

v

r

r

r

r

r

r

NADHAcCoA

PYR

f

l

g

c

e

a

flg

f

f

flg

l

g

f

l

g

rrr

r

r

rrr

r

r

r

r

r

v

v

v

v

v

v

5.02

5.0

2

2

0

0

0

200111

111100

001111

001000

000010

000005.01

6

5

4

3

2

1

Interestingly, the rate of formation of acetate (5) and formate (4)are stoichiometrically directly coupled. Therefore, one cannot

include both of these rates as measured rates, since they are not

independent. (Try it!)

Solving for the fluxes one gets


5/17


6/17

6

A net surplus of NADH results from synthesis of amino acids

E. Albers, Ph.D. thesis, 2000

NADH formation (mmol NADH/mol glucose)

without GOGAT activity with GOGAT activity

Type of cultivation Type of cultivation

Source of NADH batch chemostatD=0.1 h

-1chemostatD=0.4 h

-1batch chemostat

D=0.1 h-1

chemostatD=0.4 h

-1

Amino acids

a

192-228 172-201 221-257 94-228 81-201 105-257Nucleic acids 8-12 9-12 15-21 5-12 5-12 9-21

Organic acids 28-40 26-37 50-83 28-40 26-37 50-83

Total est imated

NADH formation

228-280 207-249 287-362 127-280 112-249 164-362

Experimental

glycerol formation

(various refs)

210 172 217 210 172 217

There are also other sources of net NADH formation, but protein

synthesis is the most important

E. Albers, Ph.D. thesis, 2000


7/17

7

Glycerol formation in yeast

DHAP G-3-P Glycerol

NAD+NADH

Regeneration of NAD+ is accomplished by glycerol formation

Stoichiometry

03/13/2:5

03/13/1O:4

03/13/13/2:3

03/13/1:2

042.215.012.012.1:1

25.023/4

382

25.033/4

3/42

212.06.074.12

COOCHOCH

ATPNADH-O-CHCH

CONADHOCHOCH

ATPNADHOCHOCH

ATPNADHCONOCHOCH

/


8/17

8

The stoichiometric matrix

-1.12 1 0 0 0.12 0 0.15 -2.42 0

-1 0 0 0 0 0 0.333 0.333 1

0 0 0 0.667 0.333 0 -0.333 0 -1

-1 0 0 0 0 1 -0.333 -0.333 0

0 0 0.667 0 0.333 0 0 0 -1

=

s x acet eth CO2 glyc NADH ATP pyr

-1.12 -1 0 -1 0

1 0 0 0 0

0 0 0 0 0.667

0 0 0.667 0 0

0.12 0 0.333 0 0.333

0 0 0 1 0

0.15 0.333 -0.333 -0.333 0

-2.42 0.333 0 -0.333 0

0 1 -1 0 -1

T =

T2

T1


9/17

9

0.3937 -0.1311 0 -0.3937 0.1311

2.8611 0.047244 0 0.14187 0.95276

3.0385 -1.0118 -3.003 -0.035469 1.0118

0 1 0 0 0

-0.177341.0591 3.003 0.17734 -1.0591

T2-1 =

glycc

glyc

glycc

glycc

glycc

rr

r

rr

rr

rr

0591.11773.0

0118.10385.3

0472.08611.2

13.03937.0

02

02

02

02

5

4

3

2

1

mc rTTTr1

211

What about calculated net formation rates?

Fluxes


10/17

10

-1.1200 -1.0000 0 -1.0000 0

1.0000 0 0 0 0

0 0 0 0 0.6670

0 0 0.6670 0 0

T1 =

-3.3021 -0.9004 0 0.2991 -1.0996

0.3937 -0.1311 0 -0.3937 0.1311

-0.1183 0.7064 2.0030 0.1183 -0.7064

2.0267 -0.6749 -2.0030 -0.0237 0.6749

T1* T2-1 =

glycc

glycc

glycc

glycc

e

a

x

s

rr

rr

rrrr

r

r

rr

6749.00267.2

7064.01183.0

13.03937.09004.03.3021-

02

02

02

02

Perhaps difficult to see but re is in fact proportional to rx

The stoichiometric matrix - rearranged

=

s xacet eth CO2 glyc NADH ATPpyr

0 0 0.12 0 -1.12 1 0.15 -2.42 0

0 0 0 0 -1 0 0.333 0.333 10 0.667 0.333 0 0 0 -0.333 0 -1

0 0 0 1 -1 0 -0.333 -0.333 0

0.667 0 0.333 0 0 0 0 0 -1

rcalc rmeas 0


11/17

11

T=

T2

0 0 0 0 0.667

0 0 0.667 0 0

0.12 0 0.333 0 0.333

0 0 0 1 0

-1.12 -1 0 -1 0

1 0 0 0 0

0.15 0.333 -0.333 -0.333 0

-2.42 0.333 0 -0.333 00 1 -1 0 -1

T2-1 =

0 1 0 0 0

-0.5 3.0736 0 1.5015 0

0 7.7177 -3.003 3.003 0

-0.5 -4.1936 0 -1.5015 0

-0.5 -4.6441 3.003 -1.5015 -1

xs

xs

x

xs

x

rr

rr

r

rr

r

6441.45.0

1936.45.0

7177.7

0736.35.0

5

4

3

2

1


12/17

12

-0.3335 -3.0976 2.0030 -1.0015 -0.6670

0.0000 5.1477 -2.0030 2.0030 0.0000

-0.1665 1.1435 0 0.5000 -0.3330

-0.5000 -4.1936 0.0000 -1.5015 -0.0000

T1 * T2-1 =

xs

xs

x

xs

glyc

c

e

a

rr

rr

r

rr

r

r

r

r

1936.40.5-

1435.10.1665-

147.5

0976.30.333-

Now it is easy to see that re is proportional to rx

Assume no acetaldehyde excretion

xs rr5.0

6441.405

Given this assumption all yield coefficients can be calculated!

108.0288.9

x

x

s

xsx

r

r

r

rY

554.0288.9

147.5

x

x

s

ese

r

r

r

rY

290.0288.9

)1435.1288.91665.0(

x

x

s

csc

r

r

r

rY

C-mol/C-mol

048.0...sgY

C-mol/C-mol

C-mol/C-mol

C-mol/C-mol


13/17

13

Lactic acid bacteria (revisited)

Glucose

NADH

Pyruvate

CO2

ATP

Lactate

Acet yl-

CoA

FormateNADH

Acet yl-P

Acetate

Acet aldehy de

Ethanol

NADH

ATP

NADH

NADH

v1v2

v4

v3

v5 v6

Biomass

ATP

ATP

CO2 + NADHSink for ATP (or ratherthe reason for ATP

synthesis)

7

0 0 0 0 0 0 1

0 1 0 0 0 0 0

0 0 0 0.3333 0 0 01 -1 -1 -1 0 0 0

0 0 0.6667 0.6667 -1 -1 0

0.3333 -0.3333 0.3333 0 0 -1 0.1

0.3333 0 0 0 0.5 0 -2

T2 =


14/17

14

6.1500 0.0000 -1.5000 0.5000 1.5000 -1.5000 3.0000

0 1.0000 0 0 0 0 0

6.1500 -1.0000 -4.5000 -0.5000 1.5000 -1.5000 3.0000

-0.0000 0.0000 3.0000 -0.0000 -0.0000 0.0000 -0.0000

-0.1000 -0.0000 1.0000 -0.3333 -1.0000 1.0000 0.0000

4.2000 -0.6667 -2.0000 0.0000 1.0000 -2.0000 2.0000

1.0000 0 0 0 0 0 0

T2-1 =

1,3-propanediolProblem 5.5

Terephtalic acid and 1,3-propanediol (3G) can form the

polyester Sorona

(If you exchange 3G for ethylene glycol you get PET!)

1,3-propanediol can be produced by recombinant E. coli

(a process developed by Du Pont/Genencor)


15/17

15

Glycerol utilization by Klebsiella

DHAP G-3-P Glycerol

NAD+NADH

Pyruvate

Acet yl-CoA

Acetyl-P Acetaldehyde

Ethanol

Formic acid

CO2

+ H2

Acet ic acid

ADPATP

Lactic acid

NADHNAD+

NADHNAD+

(1)(3)

(4) (6)

(5) NADHNAD+

(7)

Biomass

3G

NADNADHATP

2 ATP+NADH

Undesired consumption ofNADH

Stoichiometry1. Anaerobic fermentation

2. Product formation

3. Biomass formation

- CH8/3O + 2/3 CH2O (=HAc) + 1/3(H2+CO2(or HCOOH)) + 2/3 NADH +2/3 ATP = 0- CH8/3O + CH8/3O2/3 1/3 NADH =0

- 1.1 CH8/3O + X + 0.1 CO2+ 0.467 NADH 2.42 ATP =0


16/17

16

mc rTTTr1

211

0 0 1.0000

0.6667 -0.3333 0.4670

0.6667 0 -2.4200

T2=

-1.0000 -1.0000 -1.1000

0.6667 0 0

0.3333 0 0

0 0 0.1000

0 1.0000 0

T1=

-13.3910 3.0000 -4.5000

2.4200 0 1.0000

1.2100 0 0.50000.1000 0 0

8.6610 -3.0000 3.0000

121TT

x

x

x

x

x

G

CO

HCOOH

HAc

s

r

r

r

r

r

r

r

r

r

r

66.8

1.0

21.1

42.2

39.13

3

2

647.039.13

66.833

x

x

s

G

Gsr

r

r

rY

The yield of 3G (under the most favorable fermentation route) is thus obtained as


17/17

StoichiometryE. coli case

ATPO

P

ONADHv

ATPNADHCOOCHv

ATPNADHOCHOCHv

ATPNADHCOXOCHv

24

223

3

2

3

822

221

2

1

:

3

12:

3

1

3

2:

42.21.01.01.1:

MFA Examples

Documents

Transcript of MFA Examples