K = k_e + k_m + k_n

or

K = k_e + k_m

dA/dt = - KA can be integrated to yield the following equations based on the amount of free drug in the compartment:

ln A = ln A° - Kt                                               

log A = log A° - Kt/2.303 

ln Cp = ln Cp° - Kt        

log Cp = log Cp° - Kt/2.303       

Cp = Cp° e-^Kt             

Cp and Cp° are the plasma concentrations of free drug at time t and initial concentration (t = 0), respectively.

Half-life of Elimination: A useful pharmacokinetic constant of the model is the half-life of elimination. It is the time required for 50% of drug at any initial level to be eliminated from the compartment, i.e.

Therefore

T_1/2 = 0.693/K, or T_1/2 = ln2/K,  or T_1/2 = 2.303log 2/K

Time Constant: Occasionally it is more convenient to use time constant rather than the half-life. The time constant is the reciprocal of the first-order rate constant. The relationship between the time constant and the half-life is as follows:

Time Constant = 1/K                       
1/K = 1/(0.693/T_1/2) = 1.44 T_1/2      

The time constant is also known as turnover time and in non-compartmental analysis of i.v. bolus, 1/K is referred to as Mean Residence Time (MRT). The non-compartmental analysis will be discussed later on in the course.  The time constant or MRT represents the time for 63% of an i.v. bolus dose to be eliminated from the body. The relationship between MRT and T_1/2 is as follows:

MRTi.v. = 1/K = T_1/2/0.693
T_1/2 = 0.693( MRT)_{i.v. bolus}                    

Vd = A_t/Cp_t 

Since the amount of free drug is only known at time zero, a more practical equation would be:

Total Body Clearance: The total body clearance (Cl_t), as it was discussed in chapter five, is the rate of elimination of drug from the body per unit of concentration, i.e.

Cl_t = K x Vd

Therefore, the total body clearance is the product of apparent volume of distribution and the first-order overall elimination rate constant and according to the following equations it is also the sum of metabolic and renal clearances.

Fraction of dose remaining in the body at time t : If equation Cp = Cp° e-^Kt is normalized with respect to Cp° (i.e., dividing both sides of the equation by Cp°) the following equations that represent the fraction of dose in the body (f_b) at time t will be obtained:

Fraction of dose eliminated by all routes of elimination at time t: At time zero the fraction of dose in the body is equal to one. As the fraction of dose in the body declines with time, the fraction of dose eliminated by all routes of elimination increases. Therefore the sum of fraction in the body and fraction eliminated at any time will be equal to 1. Based on this concept of mass balance the following equations define the fraction of dose eliminated (f_el) from the body by all routes of elimination time t:

A_{12 noon} = 2^nd Dose + (Cp_{12 noon of 1st dose} x Vd)

*Notes: 12 noon is just a random time put in to the equation.  If we want to know the concentration when a SECOND dose is given (12 noon), then we have to know what the concentration is in the body BEFORE that second dose is given at 12 noon.   

The regression equation of logCp versus time is:

                LogCp = 0.602 –0.401 t


           What makes up each number? 

                 LogCp = 0.602 –0.401 t

                    0.602 = Cp₀

                 0.401 = K/2.303 

So....Log Cp = Cp₀ -    Kt                                                                      2.303                             

Slope = K/2.303 

(Rate)t = Cl_t x Cp_t