About dynamical system modelling
Illustration in micro-biology Alain Rapaport alain.rapaport@montpellier.inra.fr
Research School AgreenSkills
Toulouse, 29-31 october 2014
About dynamical system modelling Illustration in micro-biology - - PowerPoint PPT Presentation
About dynamical system modelling Illustration in micro-biology Alain Rapaport alain.rapaport@montpellier.inra.fr Research School AgreenSkills Toulouse, 29-31 october 2014 Contents Modeling bacterial growth in batch and continuous cultures
Illustration in micro-biology Alain Rapaport alain.rapaport@montpellier.inra.fr
Research School AgreenSkills
Toulouse, 29-31 october 2014
◮ Modeling bacterial growth in batch and continuous cultures ◮ Simple representations of space ◮ Questions related to biodiversity ◮ Modeling and analysis of growth inhibition ◮ Facing models with experimental data
time number of cells per volume
time number of cells per volume
→ x(t) = xmax 1 + e−rt
x0 − 1
dx dt = rx
x xmax
does not fit the logistic curves...
Hypothesis H0. There exists y s.t. x + ys = m = cste Hypothesis H1a. dx dt = µsx ⇒ dx dt = µm y
x
m
dt = µ(s)x where µ(·) is not necessarily linear:
S µ
feed bootle culture vessel collection vessel pump pump
Monod 1950 – Novick & Szilard 1950
◮ Mechanics: 2nd motion law:
mass × acceleration =
◮ Bio-reaction kinetics: mass balance (for biotic and abiotic):
mass variation = inputs − outputs + growth − death
◮ chemical kinetics:
→ P = AαBβ ⇒ v = dP dt = k[A]α[B]β
◮ microbial kinetics:
→ B + B − → v = dB dt = ✘✘✘
✘ ❳❳❳ ❳
[B][S]k growth is not simply a matter of matching bacteria with molecules of substrate...
dx dt ds dt
= µ(s)x −µ(s) y x + −Q V x Q V (sin − s) growth dilution
Simplification and notations. y = 1 ˙= d dt D = Q V
s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx
˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx ⇒ x⋆ = 0 s⋆ = sin
µ(s⋆) = D x⋆ = sin − s⋆ wash-out positive equilibrium
D Sin S µ S*
˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx
s x
˙ x = 0 ˙ s = 0
˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx
s x
˙ x = 0 ˙ s = 0
˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx
s x
˙ x = 0 ˙ s = 0
For each dilution rate D, one obtains a steady state s⋆ with µ(s⋆) = D:
6 2 1 3
s* s* s* s* D D D
2 3 4 5
D6 µ
1
D D s*s* s s
in 5 4
S µ K µmax µ
max
2
µ(s) = µmaxs K + s :
˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx Ecology of mountain lakes Industrial bioreactors
◮ The mathematical model of the chemostat predicts that the
substrate concentration at equilibrium is independant of the input concentration sin (provided that µ(sin) > D).
◮ Micro-biologists report that this property is not verified when
the tank is not homogeneous or in natural ecosystems such as soil ecosystems. Question: What is the influence of a spatial repartition on output substrate concentration at steady state?
input hot−spots microbial
roots main real soil input input
interconnected chemostats drainage
Q Q Q Q Q Q = Q + Q 1 2 Q2 V 1 V 2 V 1 V 2 V 1 diffusion
OR
serial parallel
with V = V1 + V2
in in in
S =2.5 S =3 S =4 r Sout Sout S = 1.7
in
S = 1.3 S = 1.1
in in
S = 1.5
in
d
in
S =2 S = 2
in
S = 2.2
in in
S =1.5
in
S =1.75
serial parallel
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25
Message 1. For any monotonic µ(·), there exists a threshold ¯ sin such that
◮ for sin > ¯
sin, the serial configuration is more efficient,
◮ for sin < ¯
sin, the parallel configuration is more efficient, Message 2. For the parallel configurations,
◮ for sin > ¯
sin, the map d → s⋆
◮ for sin < ¯
sin, the map d → s⋆
for d⋆ < +∞ Furthermore, there exists another threshold sin < ¯ sin s.t. d⋆ = 0 for sin < sin.
◮ How to model microbial competition? ◮ Can a biodiversity be maintened? ◮ Does spatial structures impact biodiversity? ◮ Is biodiversity is in favor of bioconversion yielding?
˙ s = −µ1(s)x1 − µ2(s)x2 + D(sin − s) ˙ x1 = µ1(s)x1 − Dx1 ˙ x2 = µ2(s)x2 − Dx2 Equilibria: wash-out species 1 only species 2 only species coexistence
sin
s⋆
1
sin − s⋆
1
s⋆
2
sin − s⋆
2
would require µ1(s⋆) = µ2(s⋆) = D non generic condition!
D s species 1 species 2
species 1 species 2
D s
species 1 s p e c i e s 2
D s
species 1 s p e c i e s 2
Assume that all µi(·) are increasing Let Λi(D) = {s ≥ 0 | µi(s) > D} and define the break-even concentrations λi(D) = inf Λi(D)
D s
Principle.
◮ Generically, there is at most one species at equilibrium. ◮ The equilibrium (if it exists) with the species that possesses
the smallest break-even concentration is the only attractive equilibrium.
Consider two strains: and the spatial structure:
µ
1
µ2 D* S
Q Q S in Sin S1 S2 + S out
1 2
α (1−α) r (1−r) V = V V = V
Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)
What is the most efficient configuration?
For increasing µ(·), define the conditional break-even conc. ¯ λ(D) =
if D ≤ µ(sin) sin if D > µ(sin) and consider F(α, r) := α¯ λ
α
r D
λ
1 − α
1 − r D
G(α, r) := α¯ λ1
α
r D
λ2
1 − α
1 − r D
Assume there exists D⋆ such that ¯ λ1(D⋆) = ¯ λ2(D⋆).
(α, r) such that α r D < D⋆ < 1 − α 1 − r D < min(µ1(sin), µ2(sin)) that exhibit over-yielding.
DGGE
bioreactors with substrate and biomass concentrations at quasi steady state, while the diversity of the biomass is high and its composition is constantly evolving...
˙ s = −
n
µj(s)xj + D(sin − s) ˙ xi = µi(s)xi − Dxi (i = 1 · · · n) Numerical simulations with n large (over hundred):
◮ random drawing of functions µi(·) among a family of Monod
functions
◮ random drawing of the initial composition of the biomass
Growth curves S
substrate total biomass 50 100 150 200 250 300 350 400 450 500 0.0 0.2 0.4 0.6 0.8 1.0 1.2
time
Let b =
n
xi et pi = xi b ⇒
s = −˜ µ(s, p)b + D(sin − s) ˙ b = ˜ µ(s, p)b − Db with ˜ µ(s, p) =
n
piµi(s) and ˙ pi = (µi(s) − ˜ µ(s, p)) pi
substrate t
a l b i
a s s
m efficient packet of
less efficient species species
˙ s = −¯ µ(s)¯ x −
n
µj(s)xj + D(sin − s) ˙ ¯ x = ¯ µ(s)¯ x − D¯ x ˙ xj = µj(s)xj − Dxj (j = m + 1 · · · n) Competitive Exclusion Principle:
lim
t→+∞ s(t) = s⋆
lim
t→+∞ ¯
x(t) = x⋆ = sin − s⋆ lim
t→+∞ xj(t) = 0
(j = m + 1 · · · n)
for i = 1 · · · m let write µi(s) = ¯ µ(s) + ενi(s) ⇒ ˙ pi = ε
νi(s⋆) −
m
pjνj(s⋆)
pi
Explicit solution: pi(t) = pi(0)eAiεt
m
pj(0)eAjεt with Aj = νj(s⋆)
◮ Each proportion t → pi(t) is
◮ When ε → 0 then Ti → +∞ for all species excepted one. ◮ The “reactional entropy”
E(t) :=
µj(¯ s)pj(t) is increasing toward D.
µ(S) = µmaxS K + S µ(S) = ¯ µS K + S + S2/Ki
e.g. J. F. Andrews, A mathematical model for the continuous culture of microorganisms utilizing inhibitory substrates, Biotech. Bioengrg., 10 (1968), pp. 707-723.
µ(S) = µmaxS K + S (S) D S* S
S S* B
µ(S) = ¯ µS K + S + S2/Ki
D S*
2
S S*
1
S S*
1
S*
2
B
Courbe de croissance S
D > max
s∈[0,Sin] µ(s)
1 equilibrium: wash-out
Courbe de croissance S
µ(sin) < D < max
s∈[0,Sin] µ(s)
3 equilibria : bi-stability
Courbe de croissance S
D < µ(sin) 2 equilibria : stability
INRA LBE (Narbonne) “It happens that some additional species need to be added for the bioprocess to start, but these additional species are no longer present at steady state.”
Hypothesis: Ei(D) = {s > 0 | µi(s) > D} is an interval (λ−
i (D), λ+ i (D))
s D + λ − µ + λ λ − D µ 8 λ =+ s
Proposition. Let Q(D) =
Ei(D) =
(λ−
i (D), λ+ i (D)) ◮ There is generically competitive exclusion. ◮ There are as many species that could win the competition as
the number of connected components of the set Q(D).
˙ s = −µ1(s)x1 − µ2(s)x2 + D(sin − s) ˙ x1 = µ1(s)x1 − Dx1
˙ x2 = µ2(s)x2 − Dx2
S
case 1
S
case 2
S
case 3
D Sin
max
[0,sin] µ < µ(sin) < D
⇒ the wash-out is attractive... What is the effect of adding strain “blue” or “green” (in small quantity)?
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.00 0.01 0.02 0.04 0.05 0.06D Sin
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.00 0.01 0.02 0.04 0.05 0.06D Sin
100 200 300 400 500 600 700 800 900 1000 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
substrate time
100 200 300 400 500 600 700 800 900 1000 0.0 0.5 1.0 1.5
substrate time
100 200 300 400 500 600 700 800 900 1000 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
substrate time
Let E = (λ−, λ+) for the red species. Let λ be the break-even conc. for the additional species.
◮ λ ∈ E: stabilization of the red species ◮ λ /
∈ E: possibility of invasion
Consider D = Q V s.t. µ(sin) < D < max
s∈[0,Sin] µ(s).
Q Q Q Q Q Q = Q + Q 1 2 Q2 V 1 V 2 V 1 V 2 V 1
OR
serial parallel
with V = V1 + V2
Q Q V 1 V 2
V1 < V ⇒ Q V1 > D , V2 < V ⇒ Q V2 > D The wash-out equilibrium is attractive in both tanks.
Q Q Q = Q + Q 1 2 Q2 V 1 V 2 1
Q1 V1 < µ(sin) ⇒ Q2 V2 > µ(sin) (and vice-versa) The wash-out equilibrium is attractive in at least one tank.
V 1 V 2 Q1 Q2 Q2 Q Q = Q + Q 1 2 Q Q V
with V = V1 + V2
V 1 V 2 Q1 Q2 Q2 Q Q = Q + Q 1 2
Two parameters: (α, r) with Q2 V2 = αD et V1 = rV
◮ buffer tank: classical chemostat
⇒ unique positive equ. if αD < µ(sin)
◮ main tank: chemostat with double inputs:
µ(s⋆
1) = D
r − αD
r
sin − s⋆
2
sin − s⋆
1
1
S D S µ
1
S D S µ
˙ S = −1 y R(S, X) + MS + DSin ˙ X = R(S, X) + MX + DX in where D is a diagonal matrix, and M is a compartmental matrix:
i = j,
Mij ≤ 0
Q d Q d
ik ji ij ik
Qi Qi V
i
Vj
k
V
in
that fulfills the Kirchoff law of mass conservation.
Principle: one fixes a dilution rate Di and “wait” for the corresponding equilibrium s⋆
i .
6 2 1 3
s* s* s* s* D D D
2 3 4 5
D6 µ
1
D D s*s* s s
in 5 4
Principle: one fixes a dilution rate Di and “wait” for the corresponding equilibrium s⋆
i .
6 5
D
2
s* s
in 6 5 4 3
D D D
2 3 4 1
D D µ µ
1
s s* s* s* s* s*
One fixes a reference value s⋆ and look for stabilizing the chemostat at s = s⋆ adjusting the dilution rate D.
dilution chemostat model input
concentration substrate rate
closed loop:
controller chemostat model
input
Fix ¯ D as a value of the dilution rate.
◮ Step 1: With the feedback law
D(¯ D, s) = ¯ D + G1(s⋆ − s) the closed-loop chemostat admits an equilibrium with s = ¯ s such that µ(¯ s) = D(¯ D,¯ s).
◮ Step 2: With the adaptation law for ¯
D: d ¯ D dt = G2(s⋆ − s) the closed-loop chemostat admits the equilibrium with s = s⋆ and ¯ D⋆ = µ(s⋆).
controller input
chemostat model controller input
see Gostomski, Muhlemann, Lin, Mormino & Bungay. Auxostats for continuous culture research. Journal of Biotechnology, 1994
G1 = 1, G2 = 1 :
5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Concentration en substrat mesurée temps
5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8Taux de dilution appliqué temps
G1 = 10, G2 = 10:
5 10 15 20 25 30 35 40 45 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9Concentration en substrat mesurée temps
5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0Taux de dilution appliqué temps
◮ A model is ONE representation of the reality. There exist
modelS.
◮ A model is generally built for a question of interest
⇒ different representations for the same object.
◮ A model is a representation ALONG WITH HYPOTHESES
⇒ a validity domain is associated to a model.
◮ Simple models can represent complex systems, depending on
the scale of representation.
model inputs
inputs model
known known unknown prediction known unknown known identification unknown known known control
◮ J. Haefner. Modeling Biological Systems, Springer 2005. ◮ C. Taubes. Mathematical Differential Equations in Biology,
Cambridge 2008.
◮ L. Edelstein-Keshet. Mathematical Models in Biology,
SIAM 2005.
◮ N. Britton. Essential Mathematical Modelling, Springer
2005.