Deterministic Models

 

Compartmental modeling

 

Principle of compartmental modeling

A biological system can be divided into a number of entities called compartments and represented by squares. The interactions and relationships between these entities are represented by arrows from one to the other entity or from both. These compartments (entities) can be cells, organs, organisms or even ecosystems.

 

 

Classification of the Compartment Systems

 

·       Open: A system of compartments with exit to the outside

·       Closed: A system of compartments with no exit to the outside

·       Initial compartment: The compartment in which the material has been introduced at time t = 0

·       Connected: If it is possible to reach all other compartments from initial compartment

·       Strongly connected ( strong ): If it is possible to reach every other compartment from any compartment

·       Symmetric: If two adjacent compartments are connected in a reversible way

·       Asymmetric: if no two adjacent compartments are connected in a reversible way

·       Mammillary System: Symmetric system formed by a central compartment reversibly connected with a number of other compartments each adjacent to the central one only

·       Catenary System: Each compartment is connected only with the preceding and the successive compartment

 

 

 

 

 

One Compartment Models

 

Two cases will be examined

 

1)   Absorption

2)   Elimination

 

1)   Absorption

 

Absorption can be modeled based on the fact that the amount of a substance that enters through a biological membrane in one unit of time is proportional to the difference in concentration between the external environment and internal environment

 

First order reaction

 

dx/dt = k ( Q – X)

 

X = concentration of substance in internal environment

 

Q = concentration of substance in internal environment

 

k = absorption constant

 

 

 

 

 

 

At t = 0 , X = 0

 

 

and integrating both sides of the equation

 

 

X = Q ( 1- exp(-kt))

 

and

 

k = dx/dt  /  Q – X

 

k corresponds to the ratio between the slope at any given point on the curve and the distance of that point from the value of Q

 

Converting the exponential form of the equation to the linear form by taking the natural logarithm of both sides

 

ln (Q – X) = ln Q – kt

 

 or

 

log (Q – X) = logQ – kt/2.303

 

 

for X = Q/2

 

log(Q-X) = log Q/2 = log Q – log2 = log Q – 0.301

 

So,

 

k = 0.693/ t1/2

 

 

 

 

In some cases of absorption it is possible that the asymptotic value of X is different from the value of Q (two different factors affect absorption)

 

In this case the original equation becomes

 

 

dx/dt = k1Q – k2X

 

at equilibrium as dx/dt = 0 ,

 

X = k1/k2 Q          ( Partition coefficient )

 

In this case two coefficients control absorption

 

To obtain the values of k1 and k2

 

W = M – NZ   where,

 

W = dx/dt and Z = X

 

k1 = M/Q

 

k2 = N

 

 

Note that

 

X = k1/k2 Q(1- exp(-k2t))

 

 

 

 

 

 

 

 

2)   Elimination

 

 

Elimination can be modeled based on the fact that the decrease in the concentration of X is proportional to the concentration at that moment

So,

 

dx/dt = -kX

 

 

if k = 0.1 then 10% of the substance present at any given time is eliminated in one unit time

 

By integrating both sides

 

 

X = x0exp(-kt)

 

x0 = concentration at time t = 0

 

To convert the exponential equation to linear

 

 

Ln X/X0 = -kt

 

Ln X = Ln X0 - kt

 

Or

 

log(X/x0) = -kt/2.303

 

log X = log x0 – kt/2.303

 

for X = x0/2

 

k = 0.693/t1/2

 

 

 

If elimination is controlled by two processes not one (if there are two processes make X decrease)

 

 

Then,

 

 

-dX/dt = k1X + k2X

 

 

-dX/dt = (k1+k2) X

 

 

Therefore, by integrating both sides

 

 

 

 X = x0exp-(k1+k2) t

 

 

 

 

 

 

 

 

 

 

 

 

 

Suppose that  elimination is proportional to the difference between X and a fixed value Xa

 

Then,

 

 

dX/dt =  -k ( X – Xa )

 

 

with X = X0 for t = 0  and integrating both sides

 

 

X = (X0 – Xa) exp(-kt)  + Xa

 

 

The value of X tends toward Xa when t ® ¥

 

It is possible to calculate the value of k from the amount of a given substance that has been eliminated in a series of uniform intervals

 

Let Y indicate the total quantity of a substance present in the organism

 

Then

 

dY/dt = the quantity disappearing in one unit of time

 

Let U indicate the total quantity eliminated from the organism

 

Then

 

dU/dt = the quantity eliminated in the same interval of time

 

and,

 

dY/dt  / -k  =  dU/dt  / k  =  Y

 

In linear form

 

 

log (dU/dt) = - kt / 2.303  + K

 

 

If more than one process intervene to decrease the value for Y

 

 

Then,

 

dY/dt =  - ( k1 + k2 )Y

 

 

dU/dt = k1Y

 

 

Thus, by integrating

 

 

U = k1Y0 / k1+k2  ( 1 – exp(-(k1+k2)t))

 

 

Y0 = the quantity of substance to be eliminated (amount supplied)

 

At t = ¥

 

U =  k1Y0  / k1+k2

 

The linear equation becomes

 

Log(dU/dt)  =  - (k1+k2) t / 2.303   + log k1Y0