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.

· 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/ t_{1/2}

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 = x _{0}exp^{(}**

x_{0} = concentration at time t = 0

To convert the exponential equation to linear

Ln X/X_{0} = -kt

Ln X = Ln X_{0} -
kt

Or

log(X/x_{0}) = -kt/2.303

log X = log x_{0} – kt/2.303

for X = x0/2

k = 0.693/t_{1/2}

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