Dynamic Energy Budget Theory - I

Dynamic Energy Budget Theory - I

Tânia Sousa with contributions from : Bas Kooijman

Metabolism in a DEB

individual. Rectangles are state

variables Arrows are flows of food

JXA, reserve JEA, JEC, JES , JEG or structure JVG.

Circles are processes The full circles is the

priority maintenance rule.

A DEB organism

EAJ ME - Reserve

MV - Structure

FeedingXAJ

Assimilation

Mobilization

ECJ

ESJ EGJ

Maintenance

Growth

VGJ

What are the dynamics of the state-variables?

DEB Dynamics

E

V

dMdtdMdt

dEdtdVdt

The dynamics of the state-variables are given

by:

DEB Dynamics

EEA EC

VVG

dM J JdtdM Jdt

A CdE p pdtdVdt

𝑑𝑀𝑉

𝑑𝑡 =[𝑀𝑉 ] 𝑑𝑉𝑑𝑡 = �̇�𝑉𝐺

=𝑦𝑉𝐸 �̇�𝐸𝐺

The dynamics of the state-variables are given

by:

Meaning [EG]?

DEB Dynamics

EEA EC

VVG

dM J JdtdM Jdt

A C

G VE G

V E G

dE p pdt

p y pdVdt M E

[EG]- specific costs of growth

Obtain expressions that depend only on state

variables and parameters for growth for V-1 morph organisms using the following equations

Exercises

The expression that depends only on state

variables and parameters for growth for V1-morph organisms is

What happens at constant food?

Exercises

𝑑𝑉𝑑𝑡 =

𝑀𝐸 �̇�𝐸− [ �̇�𝐸𝑀 ]𝑉𝑀𝐸

𝑉 +[𝑀𝑉 ]𝑦𝑉𝐸


variables and parameters for growth for V1-morph organisms is

At constant food reserve density is constant (weak homeostasis)

Exercises

𝑑𝑉𝑑𝑡 =

𝑀𝐸 �̇�𝐸− [ �̇�𝐸𝑀 ]𝑉𝑀𝐸


- reserve density

Obtain expressions that depend only on state

variables and parameters for growth at constant food (weak homeostasis) for V1-morphs:

Exercises

- reserve density

𝑑𝑉𝑑𝑡 =

𝑀𝐸 �̇�𝐸− [ �̇�𝐸𝑀 ]𝑉𝑀𝐸



variables and parameters for growth at constant food density for V1-morphs (mE is constant) is:

Is this exponential growth?

Exercises

𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉

𝑚𝐸 [𝑀𝑉 ]+[𝑀𝑉 ]𝑦𝑉𝐸

Specific growth rate is constantV ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )


variables and parameters for growth at constant food density for V1-morphs (mE is constant) is:


Exercises

𝑑𝐿𝑑𝑡 =1

3𝑚𝐸 �̇� [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝐿− { �̇�𝐸𝑇 }

𝑚𝐸 [𝑀𝑉 ]+ [𝑀𝑉 ] 𝑦𝐸𝑉

𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


Specific growth rate is constantV ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

𝑑𝑉𝑑𝑡 = ˙𝑟 𝐸𝑉


Yes, with

Exercises

𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


Exponential growth

What is the slope?

Exponential growth in V1-morphs at constant food

𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


V ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

Exponential growth

With a slope:


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


V ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

lnV (t)=lnV (0 )+ ˙𝑟 𝐸 t

Exponential growth

With

What is the relationship between the specific growth rate and the doubling time?

Exercises

𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


V ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

Exponential growth

With

The relationship between the specific growth rate and the doubling time is:

Exercises

𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


V ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

𝑡𝐷=ln 2˙𝑟 𝐸

Exponential growth

With

How does the specific growth rate depends on reserve density?

Exercises

𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


V ( t)=V ( 0 ) exp ( ˙𝑟 𝐸 t )

Exponential growth in DEB theory

DEB theory predicts: increases with the reserve density (food level)


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]



DEB theory predicts: increases with the reserve density (food level)

How does the specific growth rate depends on the specific energy conductance, maintenance needs and on yVE?


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]



DEB theory predicts: increases with the reserve density (food level) decreases with specific maintenance needs and

increases with and


𝑑𝑉𝑑𝑡 =

𝑚𝐸 �̇�𝐸𝑉 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]𝑉


𝑟 𝐸=𝑚𝐸 �̇�𝐸 [𝑀𝑉 ]− [ �̇� 𝐸𝑀 ]


Doubling time:

Doubling time in V1-morphs at constant food

ln 2𝑟 𝐸

Metabolism in a DEB

individual. Rectangles are state

variables Arrows are flows of food

JXA, reserve JEA, JEC, JES , JEG or structure JVG.

Circles are processes The full circles is the

priority maintenance rule.

A DEB organismAssimilation, dissipation and growth

EAJ ME - Reserve

MV - Structure

FeedingXAJ

Assimilation

Mobilization

ECJ

ESJ EGJ

Maintenance

Growth

VGJ

Assimilation: X(substrate)+M E(reserve) +

M + P linked to surface area

Dissipation: E(reserve) +M M somatic maintenance: linked to surface area &

structural volume Growth: E(reserve)+M V(structure) + M Compounds:

Organic compounds: V, E, X and P Mineral compounds: CO2, H2O, O2 and Nwaste

3 types of aggregated chemical transformations

E - Reserve

V - Structure

=1Catabolism: Cp

Maintenance: Mp Growth: Gp

Assimilation: Ap

Klebsiella Aerogenes in DEB Theory

• Characteristics: Gram-negative bacteria and a facultatively anaerobic rod (V1-morph). T=35ºC

pH: 6.8

O2, NH3

XpX – GlycerolC3H8O3

Dp

CO2, H2O, and sensible heat

Dissipation:

Biomass: E+ V

CH1.64O0.379N0.198

Reserve Turnover Rate: E=2.11h-1

CH1.66O0.422N0.312

yXE=1.345

yVE=0.904M=0.021h-1

Maintenance Rate Coefficient:

Energy Investment Ratio: g=1

Obtain the aggregated chemical reactions for

assimilation, dissipation and growth for klebsiella aerogenes in a chemostat (see next slide)

Identify in these equations yXE, yPE and yVE. Constraints on the yield coeficients Degrees of freedom

Exercises

What is the relationship between these

equations and , , , ,, , and ?

Exercises


equations and , , ,, , and ? How would you obtain the aggregate chemical

transformation?

Exercises



transformation? Compute the total consumption of O2.

Write it as a function of , and .

Exercises



transformation? Compute the total consumption of O2.

Write it as a function of , and .

Exercises

The stoichiometry of the aggregate chemical transformation that describes the organism has 3 degrees of freedom: any flow produced or consumed in the organism is a weighted average of any three other flows

Write the energy balance for each chemical

reactor (assimilation, dissipation and growth)

Exercises


reactor (assimilation, dissipation and growth) Compute the total metabolic heat production

as a function of , and .

Exercises


reactor (assimilation, dissipation and growth) Compute the total metabolic heat production

as a function of , and .

Exercises

Indirect calorimetry (estimating heat production without measuring it): Dissipating heat is weighted sum of three mass flows: CO2, O2 and nitrogeneous waste (Lavoisier in the XVIII century).

T EA T A EG T G ED T Dp J p J p J p

Dissipating heat

Steam from a heap of moist Prunus serotina litter illustrates metabolic heat production by aerobic bacteria, Actinomycetes, fungi and other organisms

E - Reserve

V - Structure

=1Catabolism: Cp

Maintenance: Mp Growth: Gp

Assimilation: Ap

Klebsiella Aerogenes in DEB Theory

• Characteristics: Gram-negative bacteria and a facultatively anaerobic rod (V1-morph). T=35ºC

pH: 6.8

O2, NH3

XpX – GlycerolC3H8O3

Dp

CO2, H2O, and sensible heat

Dissipation:

Biomass: E+ V

CH1.64O0.379N0.198

Reserve Turnover Rate: E=2.11h-1

CH1.66O0.422N0.312

yXE=1.345

yVE=0.904M=0.021h-1

Maintenance Rate Coefficient:

Energy Investment Ratio: g=1

D(h-1)

Measurements (points) and DEB model results (lines).

Comparison with experimental data I

yield (C-molWoutput.C-molX-1)

O2 (molO2.C-molWoutput-1.h-1)

CO2 (molCO2.C-molWoutput-1.h-1)

Esener et al. (1982, 1983)

Measurements (points) and DEB model results (lines).

Comparison with experimental data II

nHW (molH.C-molW-1)

nOW (molO.C-molW-1)

nNW (molN.C-molW-1)

Esener et al. (1982, 1983)

D(h-1)

Heat Production vs. Dilution rates

kJ per mol O2 consumed

kJ per C-mol biomass inside the chemostat per hour

kJ per C-mol biomass formed

Thornton’s coefficient

D(h-1)

• Irreversibilities are equal to the amount of heat released• Production of biomass becomes more efficient

Dynamic Energy Budget Theory - I

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