Modeling Microbial Populations in the Chemostathalsmith/MBI2012.pdf · Modeling Microbial Populations in the Chemostat Hal Smith A R I Z O N A S T A T E U N I V E R S I T Y H.L. Smith

Modeling Microbial Populations in the Chemostat

Hal Smith

A R I Z O N A S T A T E U N I V E R S I T Y

H.L. Smith (ASU) Modeling Microbial Populations in the Chemostat MBI, June 13, 2014 1 / 34

Outline

1 Why study microbial ecology?

2 Microbial GrowthGrowth under nutrient limitation, Monod(1942)

3 Continuous Culture: The ChemostatMicrobial Growth in the ChemostatCompetition for NutrientCompetitive Exclusion PrincipleThe Math supporting the CEP35 year old Open Problem


Why study microbial ecology?

Why study microbial ecology?

“The study of the growth of bacterial cultures does not constitute aspecialised subject or a branch of research: it is the basic method ofmicrobiology.” J. Monod, 1949

quantify microbial growth

quantify antibiotic efficacy

microbial ecology

bio-engineering: syntheticbio-fuels

waste treatment

bio-remediation

biofilms, quorum sensing

mammalian gut microflora

food, beverage (beer,wine)


Microbial Growth Growth under nutrient limitation, Monod(1942)

Growth rate and nutrient concentration

J. Monod, The growth of bacterial cultures, Annu. Rev. Microbiol. 1949



Growth under Nutrient Limitation, Monod(1942)

Monod’s experimental data on bacterial growth rate as a function of nutrientconcentration led him to propose:

1 specific growth rate dNNdt varies with S = [glucose] approx. as

dNNdt

=rS

a + S

some r > 0 and a > 0.2 growth rate and nutrient consumption rate are proportional dN = γdS3 This implies that nutrient depletion is:

dSdt

= −1γ

rSNa + S

French biologist and nobelist Jacques Monod led an interesting life. Check it out on:http://en.wikipedia.org/wiki/Main_Page


http://en.wikipedia.org/wiki/Main_Page





dNNdt

=rS

a + S


dSdt

= −1γ

rSNa + S








dNNdt

=rS

a + S


dSdt

= −1γ

rSNa + S








dNNdt

=rS

a + S


dSdt

= −1γ

rSNa + S





The anatomy of Monod’s function G = r Sa+S

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

r = maximum rate

a = half saturation

Alternative functions, having the same general shape, have been proposed. Monod’sfunction can be derived from enzyme kinetics: see wikipedia.r and a can be inferred from least squares applied to 1/G = (a/r)(1/S) + (1/r).




0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

r = maximum rate

a = half saturation





0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

r = maximum rate

a = half saturation




Growth in Batch Culture*

S-Nutrient.N-Bacteria.

dSdt

= −1γ

rSNa + S

dNdt

=rSN

a + S

0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

time in hours

Batch Growth−−No maintenance

SN

*Batch culture is jargon for a closed-system culture, i.e., a covered Petri dish



The principle of conservation of nutrient

dSdt

= −1γ

rSNa + S

dNdt

=rSN

a + S

Total nutrient, nutrient bound up in microbes plus free nutrient, is conserved:

ddt

(Nγ

+ S)

= 0 ⇒N(t)γ

+ S(t) =N(0)γ

+ S(0)


Continuous Culture: The Chemostat

The Chemostat: see Google images



enter “Chemostat” into http://en.wikipedia.org/wiki/Main_Page

Substrate

Biomass V

DS D(S+x) 0




The Old Tank Problem-No Bacteria

V = Volume of chemostat(ml)F = Inflow = Outflow rate (ml/hr)S0 = Concentration of Substrate in Feed (gm/ml).S = Concentration of Substrate in Chemostat (gm/ml).

Rate of change of Substrate (gm/hr)= INFLOW(gm/hr) - OUTFLOW(gm/hr)

ddt

(VS) = FS0 − FS

Let D = F/V be the Dilution Rate. Then

dSdt

= D(S0 − S)

Solution:S(t) = S(0)e−Dt + S0(1 − e−Dt ) → S0






ddt

(VS) = FS0 − FS


dSdt

= D(S0 − S)







ddt

(VS) = FS0 − FS


dSdt

= D(S0 − S)







ddt

(VS) = FS0 − FS


dSdt

= D(S0 − S)



Continuous Culture: The Chemostat Microbial Growth in the Chemostat

Classical Chemostat Model

Novick & Szilard, 1950.

dSdt

= D(S0 − S)︸︷︷︸

dilution

−1γ

rSNa + S

︸︷︷︸

consumption

dNdt

=rSN

a + S︸︷︷︸

growth

− DN︸︷︷︸

dilution

Environmental parameters:1 dilution rate D = F/V .2 nutrient concentration in inflow S0.

Biological parameters:1 maximal growth rate r .2 half-saturation concentration a.3 yield γ.





dSdt

= D(S0 − S)︸︷︷︸

dilution

−1γ

rSNa + S

︸︷︷︸

consumption

dNdt

=rSN

a + S︸︷︷︸

growth

− DN︸︷︷︸

dilution







dSdt

= D(S0 − S)︸︷︷︸

dilution

−1γ

rSNa + S

︸︷︷︸

consumption

dNdt

=rSN

a + S︸︷︷︸

growth

− DN︸︷︷︸

dilution





Break-even nutrient level for survival of microbes

dNNdt

=rS

a + S− D = 0

when

S = λ =aD

r − D

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

D = Dilution rate

Break−even S



Equilibria

0 = D(S0 − S)−1γ

rSNa + S

0 =

(rS

a + S− D

)

N

Washout Equilibrium: N = 0 and S = S0.

Survival Equilibrium: rSa+S = D, i.e., S = λ and N = γ(S0 − λ).

Positive survival equilibrium exists iff rS0

a+S0 > D.



Equilibria

0 = D(S0 − S)−1γ

rSNa + S

0 =

(rS

a + S− D

)

N




a+S0 > D.



Equilibria

0 = D(S0 − S)−1γ

rSNa + S

0 =

(rS

a + S− D

)

N




a+S0 > D.



Equilibria

0 = D(S0 − S)−1γ

rSNa + S

0 =

(rS

a + S− D

)

N




a+S0 > D.



Phase Plane: Survival Equilibrium does not exist

S ’ = D (1 − S) − (2 S N)/(0.3 + S)N ’ = (2 S N)/(0.3 + S) − D N

D = 1.65

0 0.5 1 1.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

S

N



Phase Plane: Survival Equilibrium exists

S ’ = D (1 − S) − (2 S N)/(0.3 + S)N ’ = (2 S N)/(0.3 + S) − D N

D = 1

0 0.5 1 1.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

S

N



Stability of Washout Equilibrium

Small perturbations from the washout equilibrium obey the “linearizedsystem”:

(S,N) = (S0, 0) + (y1, y2), y = (y1, y2) small perturbation

y ′ = Jy

where J is the jacobian matrix at the washout equilibrium:

J =

(−D −f (S0)/γ0 f (S0)− D

)

The eigenvalues are −D and f (S0)− D, where f (S) = rSa+S .

The washout equilibrium is stable if rS0

a+S0 < D and unstable if rS0

a+S0 > D.



Survival or Washout

If the inflow nutrient supply is sufficientfor the microbe to grow:

rS0

a + S0 > D,

then the bacteria survive:

N(t) → γ(S0 − λ), S(t) → λ,

Indeed, they are reproducing at theexponential rate set by dilution rate D.

Otherwise, they are washed out:

N(t) → 0, S(t) → S0

0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Survival

Extinction

D

S0

Operating Diagram

survival boundary: D = rS0

a+S0 .



Phase Plane

S ’ = 1 − S − m S x/(a + S)x ’ = x (m S/(a + S) − 1)

m = 2a = 0.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

S

x



The conservation principle

dSdt

= D(S0 − S)−1γ

rSNa + S

dNdt

=rSN

a + S− DN

Multiply N-eqn. by 1γ

and add to S eqn.:

ddt

[

S(t) +N(t)γ

]

= D(

S0 −

[

S(t) +N(t)γ

])

which implies that[

S(t) +N(t)γ

]

=

[

S(0) +N(0)γ

]

e−Dt + S0(1 − e−Dt)

Solution trajectory approaches line S + Nγ= S0.


Continuous Culture: The Chemostat Competition for Nutrient

Competing Strains of Bacteria

dSdt

= D(S0 − S)−1γ1

r1N1Sa1 + S

−1γ2

r2N2Sa2 + S

dN1

dt=

(r1S

a1 + S− D

)

N1

dN2

dt=

(r2S

a2 + S− D

)

N2

Exploitative Competition: each organism consumes a common resource.



Break-even concentrations

dN1

N1dt=

r1Sa1 + S

− D = 0 ⇔ S = λ1 =a1D

r1 − DdN2

N2dt=

r2Sa2 + S

− D = 0 ⇔ S = λ2 =a2D

r2 − D

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

Break−even values

lambda1

D

lambda2

λ1 = λ2? Coexistence at Equilibrium is Extremely Unlikely!H.L. Smith (ASU) Modeling Microbial Populations in the Chemostat MBI, June 13, 2014 22 / 34


Break-even concentrations

dN1

N1dt=

r1Sa1 + S

− D = 0 ⇔ S = λ1 =a1D

r1 − DdN2

N2dt=

r2Sa2 + S

− D = 0 ⇔ S = λ2 =a2D

r2 − D

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

Break−even values

lambda1

D

lambda2

λ1 = λ2? Coexistence at Equilibrium is Extremely Unlikely!H.L. Smith (ASU) Modeling Microbial Populations in the Chemostat MBI, June 13, 2014 22 / 34


No Coexistence Equilibria if λ1 6= λ2

N1 wins: S = λ1, N1 = γ1(S0 − λ1), N2 = 0

N2 wins: S = λ2, N1 = 0, N2 = γ2(S0 − λ2)


Continuous Culture: The Chemostat Competitive Exclusion Principle

Competitive Exclusion Principle: check it out on wikipedia

Assume each species can survive alone in the chemostat ( ri S0

ai+S0 > D.)

With no loss in generality, assume:

a1Dr1 − D

= λ1 < λ2 =a2D

r2 − D

Then N1 wins:

N1(t) → γ1(S0 − λ1), N2(t) → 0, S(t) → λ1

Winner is the organism that can grow at the lowest nutrient level.The winner of competition for nutrient is determined by quantities which maybe measured by growing each organism separately in the chemostat

Mathematical Proof: Hsu,Hubbell,Waltman (1977);Experimental Test: Hansen & Hubbell (1980)





ai+S0 > D.)


a1Dr1 − D

= λ1 < λ2 =a2D

r2 − D

Then N1 wins:

N1(t) → γ1(S0 − λ1), N2(t) → 0, S(t) → λ1







ai+S0 > D.)


a1Dr1 − D

= λ1 < λ2 =a2D

r2 − D

Then N1 wins:

N1(t) → γ1(S0 − λ1), N2(t) → 0, S(t) → λ1




Continuous Culture: The Chemostat The Math supporting the CEP

Conservation Principle

ddt

[

S(t) +N1(t)γ1

+N2(t)γ2

]

= D(

S0 −

[

S(t) +N(t1)γ1

+N2(t)γ2

])

which implies that

S(t) +N1(t)γ1

+N2(t)γ2

→ S0.

This suggests setting S(t) + N1(t)γ1

+ N2(t)γ2

= S0 and xi =Ni (t)γi

:

dx1

dt=

(f1(S0 − x1(t) − x2(t)) − D

)x1

dx2

dt=

(f2(S0 − x1(t) − x2(t)) − D

)x2

where

fi(S) =riS

ai + S



Conservation Principle

ddt

[

S(t) +N1(t)γ1

+N2(t)γ2

]

= D(

S0 −

[

S(t) +N(t1)γ1

+N2(t)γ2

])

which implies that

S(t) +N1(t)γ1

+N2(t)γ2

→ S0.

This suggests setting S(t) + N1(t)γ1

+ N2(t)γ2

= S0 and xi =Ni (t)γi

:

dx1

dt=

(f1(S0 − x1(t) − x2(t)) − D

)x1

dx2

dt=

(f2(S0 − x1(t) − x2(t)) − D

)x2

where

fi(S) =riS

ai + S



Stability of Equilibria: 0 < λ1 < λ2 < S0

The Jacobian Matrix evaluated at the non-zero equilibria.

At x̂1 = S0 − λ1, x̂2 = 0

J =

(−x̂1f ′1 −x̂1f ′1

0 f2(λ1)− D

)

f2(λ1)− D < f2(λ2)− D = 0 sodiagonal entries are negative.Asymptotically Stable.

At x̃1 = 0, x̃2 = S0 − λ2

Jacobian =

(f1(λ2)− D 0

−x̃2f ′2 −x̃2f ′2

)

f1(λ2)− D > f1(λ1)− D = 0 so redentry is positive. Unstable Saddle.



No Periodic Orbits

Dulac’s Criterion satisfied:

∂

∂x1

(f1 − D)x1

x1x2+

∂

∂x2

(f2 − D)x1

x1x2= −

(f ′1x2

+f ′2x1

)

< 0

Learn about Dulac’s Criterion at:

http://www.scholarpedia.org/article/Encyclopedia_of_dynamical_systems


http://www.scholarpedia.org/article/Encyclopedia_of_dynamical_systems


The Phase Plane of Competition

x1 ’ = 1.2 x1 (1 − x1 − x2)/(1.01 − x1 − x2) − x1x2 ’ = 1.5 x2 (1 − x1 − x2)/(1.1 − x1 − x2) − x2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x1

x2



Winner may depend on dilution rate

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Nutrient S

Gro

wth

Rat

e

Break−even values

lambda1

D

lambda2

λ2 < λ1 so N2 wins.By increasing D a bit, green line goes up, so λ1 < λ2 and N1 wins!



Species 1 Wins when D = 1.16

x ’ = 1.2 x (1 − x − y)/(1.01 − x − y) − D xy ’ = 1.5 y (1 − x − y)/(1.1 − x − y) − D y

D = 1.16

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y



Species 2 Wins when D = 1.17

x ’ = 1.2 x (1 − x − y)/(1.01 − x − y) − D xy ’ = 1.5 y (1 − x − y)/(1.1 − x − y) − D y

D = 1.17

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y


Continuous Culture: The Chemostat 35 year old Open Problem

Famous Open Problem

General competition model with species-specific removal rates Dj

S′ = D(S0 − S)−n∑

j=1

fj(S)xj

x ′

j = (fj (S)− Dj)xj , 1 ≤ j ≤ n.

Hypotheses:1 S0,D,Dj > 0.2 fj(0) = 0, f ′j (S) > 0.3 Break-even nutrient values λj : fj (λj) = Dj .

Conjecture: If λ1 < λ2 ≤ λj < S0, j ≥ 2, then:

x1(t) → S0 − λ1, xj(t) → 0, S(t) → λ1.



Proofs of the Competitive Exclusion Principle.

Author(s) and Date HypothesesHsu, Hubbel, Waltman 1977 D = Di and fi MonodHsu 1978 fi MonodArmstrong & McGehee 1980 Di = D, fi monotoneButler & Wolkowicz 1985 Di = D, fi mixed-monotoneWolkowicz & Lu 1992 Di 6= D, fi mixed-monotone, add. assumptionsWolkowicz & Xia 1997 Di − D small, fi monotoneLi 1998,1999 Di − D small, fi mixed-monotoneLiu et al 2013 allows time delays for nutrient assimilationYour Name, Date Di 6= D, fi monotone

mixed-monotone means one-humped.



References

The Theory of the Chemostat, Smith and Waltman, Cambridge Studies inMathematical Biology, 1995Microbial Growth Kinetics, N. Panikov, Chapman& Hall, 1995Resource Competition, J. Grover, Chapman& Hall, 1997


Documents

Modeling Microbial Populations in the Chemostathalsmith/MBI2012.pdf · Modeling Microbial Populations in the Chemostat Hal Smith A R I Z O N A S T A T E U N I V E R S I T Y H.L. Smith