3
$\begingroup$

The formula in textbooks for $t_{\frac{1}{2}}$ of a drug following first order elimination is generally given as $$t_{\frac{1}{2}}= \frac {\ln(2).V_d}{Cl}$$where $V_d$ is the volume of distribution and $Cl$ is the clearance.

Shown below is my attempt to derive the formula but I don't know where I'm going wrong.

Assuming first order kinetics of elimination, the rate of elimination ($R$) would be proportional to the plasma concentration ($Cp$)

$$ R = Cp . k$$

Where $k$ is the rate constant.

For calculation of $t_{\frac{1}{2}}$, since this is a first order elimination, $$ t_{\frac{1}{2}} = \ln(2)/k$$ Substituting, $$ t_{\frac{1}{2}} = \frac{\ln(2).Cp}{R}$$ Since $ Cl = R/Cp$, substituting would give, $$ t_{\frac{1}{2}} = \ln(2)/Cl$$

Since both the formulae don't match (this one doesn't have a $V_d$ term at all!), where have I gone wrong? Also then how is $Cl$ different from $k$?

According to the correct formula, $k=Cl/V_d$. How is that so?

EDIT

Volume of distribution is the apparent volume of blood the drug takes up and is given by $$V_d = \frac {Dose}{Pc}$$

Clearance is the volume of blood that has been cleared of the drug in unit time and is given by $ Cl = R/Cp$

Correct me if the basic definitions themselves are wrong.

$\endgroup$
  • $\begingroup$ Should this be asked in chemistry? $\endgroup$ – Polisetty Nov 5 '16 at 7:56
  • 1
    $\begingroup$ Can you define the terms, "distribution volume" and "clearance"? The problem certainly has to do with the way you have defined these terms. $\endgroup$ – WYSIWYG Nov 5 '16 at 11:18
  • 1
    $\begingroup$ $Cl = R/Cp$ seems incorrect. Cl unit is $volume \cdot time^{-1}$, R is $weight \cdot volume^{-1} \cdot time^{-1}$ , Cp is $weight \cdot volume^{-1}$. R/Cp is $time^{-1}$ which is imcompatible with Cl. $\endgroup$ – Eliane B. Nov 5 '16 at 16:55
5
$\begingroup$

Well, $Cl = R/Cp$ is correct, but $R$ is not $Cp \cdot k$.

Given $X(t)$, the actual quantity (ie weight or moles) of drug in the system, $R$ should have the dimension of $\frac{dX(t)}{dt}$, that is $weight \cdot time^{-1}$ or $mole \cdot time^{-1}$ because the base definition of clearance is given by $$- \frac{dX(t)}{dt} = Cl \cdot C(t)$$.

The volume of distribution $Vd$ comes into play when converting from $X(t)$ to $C(t)$. Following the same steps as you did :

$$ R = \frac{dX(t)}{dt} = \frac{dC(t)}{dt} * Vd = C_p \cdot k \cdot V_d$$ because (due to 1st order kinetics) $$\frac{dC(t)}{dt} = Cp \cdot k$$

Substituting, $$ t_{\frac{1}{2}} = \frac{\ln(2)}{k}$$ $$ t_{\frac{1}{2}} = \frac{\ln(2) \cdot C_p \cdot V_d}{R}$$ Since $ Cl = \frac{R}{Cp}$, substituting would give, $$ t_{\frac{1}{2}} = \ln(2) \cdot \frac{1}{Cl} \cdot V_d$$ $$ Cl = \ln(2) \cdot \frac{1}{t_{\frac{1}{2}}} \cdot V_d$$ $$ Cl = k \cdot V_d$$ $$ k = \frac{Cl}{V_d}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.