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Mathematica - Cinetica enzimatica, Exercícios de Bioquímica

Resolução de um exercicio no softwere mahematica

Tipologia: Exercícios

2024

Compartilhado em 18/10/2024

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In[]:= params = {μmGlucose 0.662, μmXylose 0.190, νmGlucose 2.005,
νmXylose 0.250, KsGlucose 0.565, KsXylose 3.400, KiGlucose 283.7,
KiXylose 18.1, PmGlucose 129.9, PmXylose 59.04, β 1.29, γ 1.42};
μGlucose[S_] :=μmGlucose *S KsGlucose +S+S^2KiGlucose/. params
μXylose[S_] :=μmXylose *S KsXylose +S+S^2KiXylose/. params
νGlucose[S_] :=νmGlucose *S KsGlucose +S+S^2KiGlucose/. params
νXylose[S_] :=νmXylose *S KsXylose +S+S^2KiXylose/. params
μInhibition[Ethanol_,μ0_, Pm_,β_] :=μ0*1-Ethanol Pm^β
νInhibition[Ethanol_,ν0_, Pm_,γ_] :=ν0*1-Ethanol Pm^γ
fermentationEquations = {glucose'[t]-μGlucose[glucose[t]]*cells[t],
xylose'[t]-μXylose[xylose[t]] * cells[t],
ethanol'[t] νGlucose[glucose[t]]*cells[t] + νXylose[xylose[t]] * cells[t],
cells'[t] μGlucose[glucose[t]]*cells[t] + μXylose[xylose[t]] * cells[t]};
initialConditions = {glucose[0]50, xylose[0]50, ethanol[0]0, cells[0]0.1};
solution =
resolve numéricamente equação diferencial
NDSolve[{fermentationEquations, initialConditions},
{glucose, xylose, ethanol, cells},{t, 0, 48}];
gráf
Plot[
calcula
Evaluate[{glucose[t], xylose[t], ethanol[t], cells[t]} /. solution],{t, 0, 48},
etiquetas de representação
PlotLabels {"Glucose (g/L)", "Xylose (g/L)", "Ethanol (g/L)", "
lista de células
Cells (g/L)"},
legenda do gráfico
PlotLegends "Expressions",
legenda dos eixos
AxesLabel {"Time (hours)", "Concentration (g/L)"}]
Out[]=
Glucose (g/L)
Xylose (g/L)
Ethanol (g/L)
Cells (g/L)
10 20 30 40 Time (hours)
50
100
150
200
Concentration (g/L)
InterpolatingFunction
Domain
Output
InterpolatingFunction
Domain
Output
InterpolatingFunction
Domain
Output
InterpolatingFunction
Domain
Output
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In[ ]:= params = {μ mGlucose 0.662, μ mXylose 0.190, ν mGlucose 2.005, ν mXylose 0.250, KsGlucose 0.565, KsXylose 3.400, KiGlucose 283.7, KiXylose 18.1, PmGlucose 129.9, PmXylose 59.04, β → 1.29, γ → 1.42 } ; μ Glucose [ S _] : = μ mGlucose ***** S   KsGlucose + S + S ^ 2 KiGlucose  /. params μ Xylose [ S _] : = μ mXylose ***** S   KsXylose + S + S ^ 2 KiXylose  /. params ν Glucose [ S _] : = ν mGlucose ***** S   KsGlucose + S + S ^ 2 KiGlucose  /. params ν Xylose [ S _] : = ν mXylose ***** S   KsXylose + S + S ^ 2 KiXylose  /. params μ Inhibition [ Ethanol _ , μ 0 _ , Pm _ , β_] : = μ 0 *** ** 1 -  Ethanol Pm ^ β ν Inhibition [ Ethanol _ , ν 0 _ , Pm _ , γ_] : = ν 0 *** ** 1 -  Ethanol Pm ^ γ fermentationEquations = { glucose ' [ t ] ⩵ - μ Glucose [ glucose [ t **]] *** cells [ t ] , xylose ' [ t ] ⩵ - μ Xylose [ xylose [ t **]] *** cells [ t ] , ethanol ' [ t ] ⩵ ν Glucose [ glucose [ t **]] *** cells [ t ] + ν Xylose [ xylose [ t **]] *** cells [ t ] , cells ' [ t ] ⩵ μ Glucose [ glucose [ t **]] *** cells [ t ] + μ Xylose [ xylose [ t **]] *** cells [ t ]} ; initialConditions = { glucose [ 0 ] ⩵ 50, xylose [ 0 ] ⩵ 50, ethanol [ 0 ] ⩵ 0, cells [ 0 ] ⩵ 0.1 } ; solution = resolve numéricamente equação diferencial

NDSolve [{ fermentationEquations, initialConditions } ,

{ glucose, xylose, ethanol, cells } , { t, 0, 48 }] ;

gráf⋯

Plot [ calcula

Evaluate [{ glucose [ t ] , xylose [ t ] , ethanol [ t ] , cells [ t ]} /. solution ] , { t, 0, 48 } ,

etiquetas de representação

PlotLabels → { "Glucose ( g / L ) ", "Xylose ( g / L ) ", "Ethanol ( g / L ) ", " lista de células

Cells ( g / L ) " } ,

legenda do gráfico

PlotLegends "Expressions", legenda dos eixos

AxesLabel → { "Time ( hours ) ", "Concentration ( g / L ) " }]

Out[ ]=

Glucose (g/L)

Xylose (g/L)

Ethanol (g/L)

Cells (g/L)

10 20 30 40 Time^ (hours)

50

100

150

200

Concentration (g/L)

InterpolatingFunction Domain Output

InterpolatingFunction Domain Output

InterpolatingFunction Domain Output

InterpolatingFunction Domain Output

In[ ]:= (* Parameters for the kinetic model ) params = ( Maximum specific growth rate and Monod constants max 0.313, ( 1 h ) Ks 47.51, ( g / L )( Substrate and product inhibition constants ) KIS 308.13, ( g / L ) KIP 299.67, ( g / L ) KSP 28.39, ( g / L ) ( Maximum ethanol concentrations for growth and production ) PXmax 83.35, ( g / L ) PPmax 107.79, ( g / L )( Maximum specific ethanol production rate ) qmax 3.69, ( g / g h )( Yields and maintenance coefficients ) YXS 0.48, ( g / g ) YPS 0.50, ( g / g ) m 0.001 ( 1 h *) ;

(* Define the specific growth rate function modified Monod ) μ[* S _ , PE _] : = μ max S    Ks + S + S ^ 2 KIS   1 - ( PE / PXmax ) ^ 1.53  /. params;

(* Define the ethanol production rate function Andrew - Levenspiel )* qP [ S _ , PE _] : =  qmax S    KSP + S + S ^ 2 KIP   1 - ( PE / PPmax ) ^ 1.53  /. params;

(* Define the system of differential equations ) eqs = ( Substrate S consumption ) S ' [ t ] ⩵ -  1 YXS X [ t ] μ[ S [ t ] , PE [ t ]] +  1 YPS X [ t ] qP [ S [ t ] , PE [ t ]] + m X [ t ] /. params, ( célula

Cell growth ( X ))* X ' [ t ] ⩵ X [ t ] μ[ S [ t ] , PE [ t ]] ,

(* Ethanol production ( PE ))* PE ' [ t ] ⩵ X [ t ] qP [ S [ t ] , PE [ t ]] ;

(* Initial conditions for substrate S ,cell mass ( X ) ,and ethanol ( PE ))* initConditions = { S [ 0 ] ⩵ 225, (* g / L initial sugar concentration ) X [ 0 ] ⩵ 0.01, ( g / L initial cell concentration ) PE [ 0 ] ⩵ 0 ( g / L initial ethanol concentration *)} ;

resolve

Solve the system of differential equations over time 0 to 100 hours )*

solution = resolve numéricamente equação diferencial

NDSolve [{ eqs, initConditions } , { S, X, PE } , { t, 0, 100 }] ;

gráfico

Plot the results:Substrate, lista de células

Cells,and Ethanol over time *)

gráf⋯

Plot [ calcula

Evaluate [{ S [ t ] , X [ t ] , PE [ t ]} /. solution ] , { t, 0, 100 } ,

etiquetas de representação

PlotLabels → { "Substrate ( g / L ) ", " lista de células

Cells ( g / L ) ", "Ethanol ( g / L ) " } ,

legenda do gráfico

PlotLegends "Expressions", legenda dos eixos

AxesLabel → { "Time ( hours ) ", "Concentration ( g / L ) " } ,

estilo do gráfico

PlotStyle → { azul

Blue, ve⋯

Red, verde

Green }]

2 Cinetica Bioquimica ART 1-2.nb