Question: Matlab Modeling Of Product Inhibition
gravatar for jzabilansky
6.8 years ago by
jzabilansky60 wrote:

I am trying to model the Product Inhibition of Beta-Galactosidase by the product glucose which is model by the diagram E + S <==> ES ==> E + P <==> EP where E + S goes to ES with the constants K1 and K_1 while ES goes to E + P with constant K_cat and E + P goes to EP with constants K2 and K_2 The ordered differential equations I obtained from this model are:

dE/dt = -K1[E]o[S] + K_1[ES] - K_cat[ES]

dS/dt = -K1[E]o[S] + K_1[ES] - K_cat[ES]

dES/dt = K1[E]o[S] - K_1[ES] - K_cat[ES]

dP/dt = K_cat[ES] + K_2[E][P] + K_2[EP]

dEP/dt = K2[E]o[P] - K_2[EP]

when I try to solve and plot these equations using Matlab I obtain the error that my initial vector is of different length than the vector my function returns can anyone help me out? Here is my code:

k_1 = 50;
k_cat = 10;
k2 = 100;
k_2 = 6;

e_o = 0.05;
s_o = 100;
es_o = 0; 
p_o = 0;
ep_o = 0;

x0 = [e_o s_o 300 es_o p_o 0];
p = [k1 k_1 k_cat k2 k_2];
[T1 Y1] = ode15s(@(t,y)competitive(t, y, p), [0 50], x0);

plot(T1, Y1(:,5), 'b');
xlabel('Time [s]');
ylabel('Product Concentration [mM]');
title('Competitive Product Inhibition');

function xdot = competitive(t,x,k)
% x = [E S ES P EP]
% k = [k1 k_1 k_cat k2 k_2]

xdot = [-k(1)*x(1)*x(2) + k(2)*x(3) + k(3)*x(3);
    -k(1)*x(1)*x(2) + k(2)*x(3) - k(3)*x(3);
    k(1)*x(1)*x(2) - k(2)*x(3) - k(3)*x(3);
    k(3)*x(3) + k(4)*x(1)*x(4) + k(5)*x(4);
    k(4)*x(1)*x(4) - k(5)*x(4);
matlab • 1.9k views
ADD COMMENTlink written 6.8 years ago by jzabilansky60

This sounds suspiciously like a homework question.

ADD REPLYlink written 6.8 years ago by Devon Ryan95k

Even if it is, the OP has clearly defined the question, shown that he tried to solve the problem himself, and is now asking how to solve a particular issue with his approach. Fine.

ADD REPLYlink written 6.8 years ago by Michael Schubert7.0k

it might be but can you help me

ADD REPLYlink written 6.8 years ago by jzabilansky60
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