B1 Pharmacokinetics

B1.i Explain the single and multiple compartment pharmacokinetic models

Compartment Models

• Compartment models describe the body as one or more well-mixed spaces the drug moves between, so that plasma concentrations over time can be described mathematically
  • The compartments are mathematical rather than anatomical, although they loosely correspond to groups of tissues with similar blood flow

One-Compartment Model

• The drug distributes instantly and evenly through a single volume ($V_D$), so after an IV bolus the only process is elimination:

$$\frac{dA}{dt}=-CL\cdot C=-k\cdot A\therefore C(t)=C_0{e}^{-kt},k=\frac{CL}{V_D}$$

• A plot of log concentration against time is a straight line with slope $-k$, and extrapolating it back to $t=0$ gives $C_0=\text{Dose}/V_D$

Two-Compartment Model

• The drug is given into a small central compartment ($V_1$: plasma and well-perfused organs such as the brain, heart, liver and kidney) and distributes to a peripheral compartment ($V_2$: poorly perfused tissue such as muscle and fat)
  • Elimination is assumed to happen from the central compartment

    

    Figure 1 Two-compartment model with an effect site. Organs adapted from Servier Medical Art, CC BY 4.0

• Writing the amount of drug in each compartment as $X_1$ and $X_2$:

$$\begin{array}{rl}\frac{d{X}_1}{dt}& =-(k_{10}+k_{12})X_1+k_{21}{X}_2\\ \frac{d{X}_2}{dt}& =k_{12}{X}_1-k_{21}{X}_2\end{array}$$

• The solution after an IV bolus is the sum of two exponentials (biexponential):

$$C(t)=A{e}^{-\alpha t}+B{e}^{-\beta t},\alpha +\beta =k_{10}+k_{12}+k_{21},\alpha \beta =k_{10}{k}_{21}$$

- **Distribution phase ($\alpha$, fast):** plasma concentration falls steeply as drug moves out to the peripheral compartment, even though little has been eliminated yet
- **Elimination phase ($\beta$, slow):** once the compartments have equilibrated, plasma concentration falls at a rate set by clearance; the elimination (terminal) half-life is $0.693/\beta$

What equation describes plasma concentration after an IV bolus in a two-compartment model, and what does each phase represent?

$C(t)=A{e}^{-\alpha t}+B{e}^{-\beta t}$. The fast $\alpha$ phase is distribution from the central to the peripheral compartment; the slow $\beta$ phase is elimination after the compartments have equilibrated (terminal half-life $=0.693/\beta$).

Volumes in a Two-Compartment Model

• With two compartments, “the” volume of distribution depends on when it is measured, since the ratio of amount in the body to plasma concentration changes as the drug distributes:

Volume	Formula	Meaning
Central ($V_1$)	$\frac{\text{Dose}}{A+B}$	Volume at the moment of injection, before any distribution
Steady state ($V_{SS}$)	$V_1+V_2$	Amount in the body divided by plasma concentration at steady state
Area ($V_{\text{area}}$ or $V_{\beta }$)	$\frac{CL}{\beta }=\frac{\text{Dose}}{\beta \cdot \text{AUC}}$	Relates amount to concentration during the elimination phase
Extrapolated ($V_{\text{extrap}}$)	$\frac{\text{Dose}}{B}$	What back-extrapolating the terminal phase gives

• In order of size: $V_1<V_{SS}\le V_{\text{area}}<V_{\text{extrap}}$
• Back-extrapolating the terminal phase to $t=0$ gives its intercept ($B$) rather than the true initial concentration ($A+B$), which overestimates the volume:

  

  Figure 2 Back-extrapolating the elimination phase

• How much it overestimates depends on how fast distribution is compared with elimination. Integrating the biexponential for the AUC:

$$\begin{array}{rl}\text{AUC}& ={\int }_0^{\infty }\left(A{e}^{-\alpha t}+B{e}^{-\beta t}\right)dt=\frac{A}{\alpha }+\frac{B}{\beta }\\ \therefore V_{\text{area}}& =\frac{\text{Dose}}{\beta \cdot \text{AUC}}=\frac{\text{Dose}}{B+A\frac{\beta }{\alpha }}\end{array}$$

- When distribution is much faster than elimination ($\alpha\gg\beta$), the $A\beta/\alpha$ term is negligible and $V_{\text{extrap}}\approx V_{\text{area}}$, which is why the one-compartment back-extrapolation is usually good enough
- When distribution is slow or large (e.g. digoxin), the extrapolated volume overestimates

• The central volume determines the peak concentration straight after a bolus ($\text{Dose}/V_1$), whereas $V_{SS}$ is the one to use for a loading dose

Why does back-extrapolating the terminal phase overestimate the volume of distribution in a two-compartment model?

It gives the intercept of the elimination phase ($B$) rather than the true initial concentration ($A+B$), so $V_{\text{extrap}}=\text{Dose}/B$ is larger than $V_{\text{area}}=\text{Dose}/(B+A\beta /\alpha )$. The error is small when $\alpha \gg \beta$.

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B1.ii Describe the absorption of drugs and factors that influence this

Bioavailability

Main source: Birkett, Pharmacokinetics Made Easy ¹

Definitions

• Absorption is the fraction of the dose absorbed intact from the gut into the portal circulation ($f_g$)
• First-pass clearance is the extent to which the liver removes the drug on its first passage through the liver in the portal blood, before it reaches the systemic circulation; the fraction escaping it is $f_H=1-E_H$
• Bioavailability ($F$) is the fraction of the dose which reaches the systemic circulation as intact drug:

$$\text{Bioavailability }(F)=\text{Fraction Absorbed }(f_g)\times \text{Fraction Escaping First-Pass Clearance }(f_H)$$

Figure 3 Oral bioavailability and first-pass metabolism. Gut and liver adapted from Servier Medical Art, CC BY 4.0

• Two formulations of a drug are bioequivalent if the extent and rate of absorption from them are similar enough that there is no clinically important difference in their effects (either therapeutic or adverse)

What is bioavailability and what determines it?

The fraction of the dose which reaches the systemic circulation as intact drug. It is the fraction absorbed from the gut multiplied by the fraction escaping first-pass clearance by the liver:

$$F=f_g\cdot f_H,f_H=1-E_H$$

When are two formulations of a drug bioequivalent?

When the extent and rate of absorption from them are similar enough that there is no clinically important difference in their therapeutic or adverse effects

Measuring Bioavailability

• Bioavailability is measured by giving the same dose IV and orally and comparing the area under the plasma concentration versus time curves ($\text{AUC}$); clearance is the same both times, so it cancels:

$$\begin{array}{rl}{\text{AUC}}_{\text{oral}}& =\frac{F\times \text{Dose}}{CL}\\ {\text{AUC}}_{\text{IV}}& =\frac{\text{Dose}}{CL}\\ \therefore F& =\frac{{\text{AUC}}_{\text{oral}}}{{\text{AUC}}_{\text{IV}}}\end{array}$$

Figure 4 Measuring bioavailability from the AUC after the same dose given IV and orally

• Similarly, the relative bioavailability of two formulations given by the same route is:

$$\text{Relative Bioavailability}=\frac{{\text{AUC}}_{\text{test}}}{{\text{AUC}}_{\text{reference}}}$$

How is the absolute bioavailability of a drug measured?

Give the same dose IV and orally and compare the areas under the curve. Since $\text{AUC}=F\cdot \text{Dose}/CL$ and clearance is the same both times:

$$F=\frac{{\text{AUC}}_{\text{oral}}}{{\text{AUC}}_{\text{IV}}}$$

Determinants of Drug Absorption

Figure 5 Determinants of oral absorption and first-pass clearance along the path of a dose. Capsule, stomach, gut and liver adapted from Servier Medical Art, CC BY 4.0

• Dissolution: properties of the drug including particle size and form, excipients, special dosage forms (e.g. sustained release), and stomach and small intestine pH
• Gastric emptying rate: affects how long the drug sits in acid (which matters for drugs unstable at acid pH) and how soon it reaches the small intestine; liquids and small particles empty more quickly than solid dosage forms, and emptying is slowed or sped up by antacids, drugs (opioids and anticholinergics slow it, metoclopramide speeds it up) and disease (e.g. autonomic neuropathy)
• Intestinal motility: affects how long slowly soluble drugs (e.g. digoxin, sustained release formulations) have to dissolve, and how long the drug is exposed to chemical degradation or metabolism by gut flora
• Drug interactions in the gut lumen: complexation (e.g. tetracyclines with divalent metal ions), adsorption (e.g. anion exchange resins), and interactions with food (e.g. some antibiotics)
• Passage through the gut wall: physicochemical characteristics of the drug (e.g. quaternary ammonium compounds are permanently charged so cross poorly), and metabolism by enzymes in the intestinal epithelium

What determines how well a drug is absorbed from the gut?

Dissolution, gastric emptying rate, intestinal motility, interactions in the gut lumen (complexation, adsorption, food) and passage through the gut wall (physicochemical properties of the drug and metabolism by the intestinal epithelium)

Determinants of oral absorption and first-pass clearance along the path of a dose. Capsule, stomach, gut and liver adapted from Servier Medical Art, CC BY 4.0

Determinants of First-Pass Clearance

• In low hepatic extraction ratio drugs (e.g. theophylline), nearly all of the dose gets through the liver on its first pass ($f_H\approx 1$), so $F\approx f_g$
  • Doubling or halving the small fraction the liver extracts makes little difference to $F$, e.g. $E_H$ rising from 0.1 to 0.2 only lowers $f_H$ from 0.9 to 0.8
• In high hepatic extraction ratio drugs (e.g. verapamil), only a minor proportion of the dose reaches the systemic circulation intact
  • Inducing or inhibiting the metabolising enzymes has only a small effect on the hepatic extraction ratio, and so on systemic clearance ($CL\approx Q_H$), but has a major effect on the proportion escaping extraction ($f_H$), and so a major effect on bioavailability, e.g. $E_H$ falling from 0.95 to 0.90 doubles $f_H$ from 0.05 to 0.10
  • Hepatic blood flow does change bioavailability in these drugs, as $f_H=\frac{Q_H}{Q_H+f_u\cdot C{L}_{\text{int}}}\approx \frac{Q_H}{f_u\cdot C{L}_{\text{int}}}$, but it changes systemic clearance ($\approx Q_H$) by the same proportion, so the two cancel and the steady state concentration after oral dosing is unaffected (see Figure 1)
• In chronic liver disease, a substantial portion of the portal circulation doesn’t perfuse functional liver because of intrahepatic functional shunts or extrahepatic anatomical shunts, which increases bioavailability markedly; hence high hepatic extraction ratio drugs given to patients with liver disease are particularly liable to cause adverse effects

Cloze

In a low hepatic extraction ratio drug, bioavailability is set mostly by ==absorption from the gut ($f_g$)==, since nearly all of the absorbed dose escapes first-pass clearance

Why does enzyme induction or inhibition change the bioavailability of a high extraction drug much more than its clearance?

Clearance is close to hepatic blood flow, so it barely moves. The fraction escaping first-pass ($f_H=1-E_H$) is small, so a small change in $E_H$ is a large proportional change in $f_H$, e.g. $E_H$ falling from 0.95 to 0.90 doubles $f_H$ (0.05 to 0.10) and so doubles $F$

Why does chronic liver disease increase the bioavailability of high extraction drugs?

Intrahepatic functional shunts and extrahepatic anatomical (portosystemic) shunts let portal blood bypass functioning liver, so more of the dose escapes first-pass extraction. These drugs are therefore more likely to cause adverse effects in liver disease

References

1.

Birkett DJ. Pharmacokinetics made easy. 2.ed., rev. ed. Sydney London: McGraw-Hill; 2005. 131 p.

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B1.iii Describe the distribution of drugs and factors that influence this

Volume of Distribution

Main source: Birkett, Pharmacokinetics Made Easy ¹

• Volume of distribution is the apparent volume the drug would have to be dissolved in to give the measured plasma concentration, and is defined as the amount of drug in the body divided by the plasma concentration:

$$\text{Volume of Distribution }(V_D)=\frac{\text{Amount of Drug in Body }(A_{\text{body}})}{\text{Plasma Drug Concentration }(C)}$$

- Immediately after an IV bolus, before any drug has been eliminated, the amount in the body is the dose, hence $V_{D}=\text{Dose}/C_{0}$
- It is an apparent volume rather than a real one, so it can be far larger than total body water (e.g. digoxin, amiodarone)

• The volume of distribution depends on where the drug sits, which mostly comes down to binding:
  • If a drug is tightly bound to plasma proteins (e.g. warfarin), most of it stays in the plasma, so the plasma concentration is high and it will appear as though the medication is dissolved in a smaller volume
  • If a drug is bound to tissue (e.g. digoxin to muscle, amiodarone to fat), little of it is left in the plasma, so the plasma concentration is low and the volume appears very large
• For reference, in a 70 kg adult plasma is about 3 L, extracellular fluid about 14 L and total body water about 42 L, so:

$V_D$	Interpretation	Examples (approximate)
$\approx$ Plasma volume	Mostly confined to plasma (highly protein bound or very large molecules)	Warfarin $\approx$ 0.1 L/kg, heparin
$\approx$ Extracellular fluid	Polar drugs that cross capillaries but not cell membranes	Gentamicin $\approx$ 0.25 L/kg
$\approx$ Total body water	Small lipid soluble drugs distributed evenly	Ethanol $\approx$ 0.6 L/kg
$\gg$ Total body water	Concentrated in tissue	Digoxin $\approx$ 7 L/kg, amiodarone $\approx$ 60 L/kg

What is the definition of volume of distribution?

The amount of drug in the body divided by the plasma drug concentration ($V_D=A_{\text{body}}/C$). It is an apparent volume, so it can exceed total body water.

Cloze

A drug highly bound to plasma proteins (e.g. warfarin) has a small volume of distribution, whereas a drug highly bound to tissue (e.g. digoxin) has a large volume of distribution

Plasma and Tissue Binding

• The components of volume of distribution are the plasma volume and the tissue volume, weighted by the ratio of the fraction of the drug unbound in plasma to the fraction of the drug unbound in tissue:

$$V_D=\text{Plasma Volume }(V_P)+\frac{\text{Fraction unbound in plasma }(f_u)}{\text{Fraction unbound in tissue }(f_{u_T})}\times \text{Tissue Volume }(V_T)$$

• This comes from splitting the amount in the body into what is in the plasma and what is in the tissue, then using the fact that only unbound drug crosses between them, so at equilibrium the unbound concentrations are equal:

$$\begin{array}{rl}{f}_u\cdot C& =f_{u_T}\cdot C_T\therefore C_T=\frac{f_u}{f_{u_T}}\cdot C\\ V_D=\frac{A_{\text{body}}}{C}& =\frac{C\cdot V_P+C_T\cdot V_T}{C}=V_P+\frac{f_u}{f_{u_T}}\cdot V_T\end{array}$$

Figure 6 Volume of distribution from plasma and tissue binding. Artery, muscle and fat adapted from Servier Medical Art, CC BY 4.0

• Hence:
  • More plasma protein binding (lower $f_u$) shrinks $V_D$ towards $V_P$
  • More tissue binding (lower $f_{u_T}$) makes $V_D$ larger, without limit

Derive volume of distribution in terms of plasma and tissue binding

Only unbound drug equilibrates, so $f_u\cdot C=f_{u_T}\cdot C_T$. Then $V_D=\frac{C\cdot V_P+C_T\cdot V_T}{C}=V_P+\frac{f_u}{f_{u_T}}\cdot V_T$

Factors Affecting Volume of Distribution

• Drug factors:
  • Lipid solubility: lipid soluble drugs cross membranes and distribute into fat, polar drugs stay largely extracellular
  • Ionisation (pKa relative to the local pH): only the unionised form crosses membranes
  • Plasma protein and tissue binding (see above)
  • Molecular size: very large molecules (e.g. heparin) stay in the plasma
• Patient factors:
  • Body composition: obesity increases $V_D$ for lipophilic drugs, and neonates have a larger proportion of body water
  • Fluid status: fluid resuscitation, oedema and capillary leak in critical illness increase $V_D$ for hydrophilic drugs (e.g. aminoglycosides, β-lactams)
  • Plasma proteins: hypoalbuminaemia increases $f_u$ and hence $V_D$ for drugs bound to albumin
  • Pregnancy increases plasma volume and body water
  • Renal failure: uraemia reduces tissue binding of digoxin, so its $V_D$ falls from about 7 L/kg to about 4–5 L/kg and the loading dose has to be reduced, even though clearance doesn’t enter the loading dose calculation (see Loading in Renal Impairment)

Measuring Volume of Distribution

• Serial plasma concentrations are taken after an IV bolus and plotted as log concentration against time
  • The line of best fit is extrapolated to $t=0$ to find the plasma concentration at time zero ($C_0$)
  • This is used to calculate the volume of distribution ($V_D=\frac{\text{Dose}}{C_0}$)
• This assumes a one-compartment model, i.e. that the drug distributes instantly
  • Most drugs take time to distribute, so the samples are taken after distribution is complete and the extrapolated $C_0$ is lower than the true initial concentration, which overestimates $V_D$ (see Volumes in a Two-Compartment Model)
• Volume of distribution can also be calculated from clearance and the elimination rate constant ($k$), using the AUC (see Elimination Rate Constant):

$$V_D=\frac{CL}{k}=\frac{\text{Dose}}{k\cdot \text{AUC}}$$

Clinical Significance

• Loading dose: the amount of drug needed to fill the volume of distribution to a target concentration straight away, rather than waiting about 4–5 half-lives for steady state:

$$\text{Loading Dose}=V_D\times \text{Target Plasma Concentration}$$

- The loading dose depends only on $V_{D}$, whereas the maintenance dose depends only on clearance ($DR=CL\times C_{SS}$)
- For an oral loading dose, divide by the bioavailability ($F$)

• Half-life: for a given clearance, a larger volume means a longer half-life, since less of the drug is in the plasma where the clearing organs can reach it:

$$t_{1/2}=\frac{0.693\cdot V_D}{CL}$$

• Dialysis and haemofiltration remove little of a drug with a large $V_D$, since only a small fraction of the total amount is in the plasma at any time
  • Plasma concentrations can also rebound after a session as drug redistributes back out of the tissues
• Critical illness: fluid shifts and hypoalbuminaemia increase $V_D$ for hydrophilic drugs, so a larger loading dose is often needed even though the maintenance dose depends on clearance

How is a loading dose calculated, and why does it not depend on clearance?

$\text{Loading Dose}=V_D\times \text{Target Concentration}$ (divided by $F$ if oral). It only has to fill the volume of distribution to the target once; clearance determines the maintenance dose needed to replace what is eliminated.

Why does the digoxin loading dose need reducing in renal failure, if loading dose doesn't depend on clearance?

Uraemia reduces digoxin’s tissue binding, so $V_D$ falls (about 7 → 4–5 L/kg). Since $\text{LD}=V_D\times \text{target}$, the load drops by about 30–50% in severe CKD or ESRD. The maintenance dose needs a much larger cut, because clearance is about 70% renal.

Why are drugs with a large volume of distribution poorly removed by dialysis?

Only a small fraction of the total amount in the body is in the plasma, which is all that dialysis can reach. Concentrations can also rebound afterwards as drug redistributes from the tissues.

References

1.

Birkett DJ. Pharmacokinetics made easy. 2.ed., rev. ed. Sydney London: McGraw-Hill; 2005. 131 p.

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B1.iv Describe the mechanisms of drug metabolism and clearance

Elimination and Clearance

Main source: Birkett, Pharmacokinetics Made Easy ¹

• Elimination refers to the irreversible excretion of unchanged drug or conversion to a metabolically inactive product
• Clearance refers to the efficiency of elimination and is broadly defined as a volume of plasma that is cleared of the drug per unit time (L/hr or mL/min)

Quantity	Definition	Units
Elimination rate	Amount of drug irreversibly removed per unit time	mg/hour
Clearance ($CL$)	Elimination rate divided by plasma concentration	L/hour
Extraction ratio ($E$)	Fraction of drug entering an organ that is removed in a single pass	none (0 to 1)
Elimination rate constant ($k$)	Elimination rate divided by the amount of drug in the body	hour⁻¹
Volume of distribution ($V_D$)	Amount of drug in the body divided by plasma concentration	L
Half life ($t_{1/2}$)	Time for the plasma concentration to fall by half	hour

Definitions

Elimination Rate

• If $A(t)$ is the amount of drug eliminated by time $t$, the elimination rate is how fast that amount grows:

$$\text{Elimination Rate}=\frac{dA(t)}{dt}\text{(mg/hour)}$$

Clearance

• Clearance is defined as the elimination rate divided by the plasma concentration:

$$\underset{\text{(L/hour)}}{\text{Clearance }(CL)}=\underset{\text{(mg/hour)}}{\text{Elimination Rate}}÷\underset{\text{(mg/L)}}{\text{Drug Concentration }(C)}$$

• The reason clearance is defined as a ratio is that in first-order (linear) kinetics the elimination rate is proportional to concentration, so the ratio stays constant regardless of dose or concentration and describes the drug and the patient rather than the moment
  • Doubling the concentration doubles the elimination rate in mg/hour, but the same volume of plasma is still cleared each hour
  • When elimination saturates (zero-order kinetics, e.g. phenytoin, ethanol) the elimination rate stops rising with concentration, so clearance falls as concentration rises and is no longer a constant
• The “volume cleared per unit time” description is just the interpretation of the units (mg/hour ÷ mg/L = L/hour), i.e. the volume of plasma that would need to be completely emptied of drug each hour to account for the elimination rate

Describe what clearance is?

The volume of plasma completely cleared of drug per unit time (L/hr or mL/min), i.e. elimination rate divided by plasma concentration. In first-order kinetics it is constant, as elimination rate is proportional to concentration.

What is the mathematical definition of clearance?

$$\text{Clearance }(CL)=\frac{\text{Elimination Rate}}{\text{Drug Concentration }(C)}\text{(mg/hour ÷ mg/L = L/hour)}$$

Extraction Ratio

• The extraction ratio refers to ratio of elimination of the drug through the organ of interest and is given by:

$$\text{Extraction Ratio }(E_H)=\frac{C_{\text{in}}-C_{\text{out}}}{C_{\text{in}}}=1-\frac{\text{Concentration Out}}{\text{Concentration In}}$$

• Unlike clearance, the extraction ratio belongs to a single organ and has no units, it is the fraction of drug delivered to the organ that does not come back out in the venous blood
  • $E=0$ means the organ removes nothing, $E=1$ means the organ removes everything delivered to it
  • It is converted into a clearance by multiplying by organ blood flow (see Organ Clearance)

What is the mathematical definition of extraction ratio?

The fraction of drug entering an organ that is removed in a single pass:

$$E=\frac{C_{\text{in}}-C_{\text{out}}}{C_{\text{in}}}=1-\frac{C_{\text{out}}}{C_{\text{in}}}$$

Elimination Rate Constant

• The elimination rate constant ($k$) is the fraction of the drug in the body removed per unit time, i.e. the elimination rate divided by the amount of drug in the body ($A_{\text{body}}$):

$$k=\frac{\text{Elimination Rate}}{A_{\text{body}}}{\text{(hour}}^{-1}\text{)}$$

• Since $A_{\text{body}}=V_D\cdot C$ (the definition of volume of distribution), clearance and $k$ are linked directly:

$$\begin{array}{rl}\text{Elimination Rate}& =CL\cdot C=k\cdot A_{\text{body}}=k\cdot V_D\cdot C\\ \therefore CL& =k\cdot V_D\end{array}$$

Derive the relationship between clearance, elimination rate constant and volume of distribution

Elimination rate $=CL\cdot C=k\cdot A_{\text{body}}$, and $A_{\text{body}}=V_D\cdot C$, so $CL=k\cdot V_D$ (and therefore $t_{1/2}=0.693\cdot V_D/CL$)

Half Life

Half Life and the Elimination Rate Constant

• Fall in plasma concentration after a single IV dose (one-compartment, first-order) is given by:

$$C(t)=C_0\cdot e^{-kt}$$

• Differentiating shows the rate of fall is proportional to the concentration at that time, with $k$ as the proportionality constant:

$$\begin{array}{rl}\frac{dC(t)}{dt}& =-k\cdot C_0\cdot e^{-kt}\\ & =-k\cdot C(t)\end{array}$$

• Solving for when $C(t)=0.5\cdot C_0$ (i.e. when the plasma concentration falls by half):

$$\begin{array}{rl}0.5\cdot C_0& =C_0\cdot e^{-k{t}_{1/2}}\\ \frac{1}{2}& =e^{-k{t}_{1/2}}\\ -k{t}_{1/2}& =-\ln 2\\ \therefore t_{1/2}& =\frac{\ln 2}{k}\approx \frac{0.693}{k}\end{array}$$

• The half life is the same whatever the starting concentration, which is only true in first-order kinetics (in zero-order kinetics a fixed amount is removed per unit time, so the time to halve gets longer at higher concentrations)

Derive the relationship between half life and the elimination rate constant

$C(t)=C_0{e}^{-kt}$. Setting $C(t)=0.5\cdot C_0$ gives $\frac{1}{2}=e^{-k{t}_{1/2}}$, so $k{t}_{1/2}=\ln 2$ and $t_{1/2}=\frac{\ln 2}{k}\approx \frac{0.693}{k}$

Determinants of Half Life

• Substituting $k=CL/V_D$ (see Elimination Rate Constant):

$$t_{1/2}=\frac{\ln 2}{k}=\frac{0.693\cdot V_D}{CL}$$

• So half life is a dependent variable, set by clearance and volume of distribution rather than the other way round
  • A drug can have a long half life either because it is cleared slowly or because most of it sits outside the plasma where the clearing organs can’t get to it
  • Volume of distribution depends on the drug’s lipid solubility and its binding to plasma proteins and tissues, not on the dose (see Volume of Distribution)
  • e.g. in critical illness, a rise in $V_D$ (fluid overload, hypoalbuminaemia) can prolong half life even when clearance is unchanged

What Half Life Determines

• Duration of action after a single dose
• Time to reach steady state with repeated dosing or a constant infusion, which takes about 4–5 half lives (94–97% of the final concentration), independent of the dose rate
• Time for a drug to be eliminated after it is stopped, which also takes about 4–5 half lives
• Dosing interval needed to avoid large fluctuations in plasma concentration between doses

What does half life determine clinically?

Duration of action after a single dose, time to reach steady state (about 4–5 half lives), time to eliminate the drug once stopped, and the dosing interval needed to limit peak–trough fluctuation

Clearance at Steady State

• The amount of drug in the body changes by whatever goes in minus whatever is eliminated:

$$\frac{d{A}_{\text{body}}}{dt}=\text{Dose Rate}-CL\cdot C$$

• At steady state the amount in the body is no longer changing, so $d{A}_{\text{body}}/dt=0$ and the dose rate in equals the elimination rate out:

$$\begin{array}{rl}\text{Maintenance Dose Rate }(DR)& =\text{Elimination Rate}\\ & =\text{Clearance }(CL)\times \text{Steady State Drug Concentration }(C_{SS})\\ \therefore C_{SS}& \propto \frac{1}{CL}\text{(where }DR\text{ is constant)}\end{array}$$

Measuring Clearance

• Every method below is the definition $CL=\text{Elimination Rate}/C$ applied to a situation where the elimination rate can be worked out

Steady State Infusion

• Whole body clearance be determined using the steady state formula:

$$CL=\frac{DR}{C_{SS}}$$

How is clearance mathematically defined at steady state?

$$\text{Clearance }(CL)=\frac{\text{Maintenance Dose Rate }(DR)}{\text{Steady State Drug Concentration }(C_{SS})}$$

Renal Clearance

• Renal clearance can be derived from the definition of clearance based on elimination rate, where the elimination rate is the amount of drug appearing in urine per unit time:

$$\begin{array}{rl}\text{Clearance }(CL)& =\frac{\text{Elimination Rate}}{\text{Drug Concentration}}\\ & =\frac{\text{Urine Concentration }(U)\times \text{Urine Flow }(V)}{\text{Plasma Concentration}}\end{array}$$

Figure 7 Renal handling of a drug along the nephron. Kidney and nephron adapted from Servier Medical Art, CC BY 4.0

Single IV Bolus and AUC

Figure 8 Clearance from a single IV bolus

• An IV bolus can be used to determine clearance where after given enough time, all of it is eliminated therefore:

$$\begin{array}{rl}\text{Dose}& =CL\cdot \text{AUC}\\ CL& =\frac{\text{Dose}}{\text{AUC}}\end{array}$$

How is clearance calculated from a single IV dose?

$CL=\frac{\text{Dose}}{{\text{AUC}}_{0\to \infty }}$, because total amount eliminated $={\int }_0^{\infty }CL\cdot C(t)dt=CL\cdot \text{AUC}$, and after an IV dose that equals the dose.

Mathematical Derivation

• Let $A(t)$ be the amount of drug eliminated by time $t$, its rate of change is the elimination rate:

$$\frac{dA(t)}{dt}=CL\cdot C(t)$$

• To get the total amount ever eliminated, integrate from 0 to infinity:

$$\begin{array}{rl}{A}_{\text{total}}& ={\int }_0^{\infty }CL\cdot C(t)\cdot dt\\ & =CL\cdot {\int }_0^{\infty }C(t)\cdot dt\\ & =CL\cdot \text{AUC}\end{array}$$

• After an IV dose all of the drug is eventually eliminated, so $A_{\text{total}}=\text{Dose}$
  • This doesn’t assume any compartment model, only that $CL$ is constant so it can come out of the integral, hence it is the “model-independent” way of measuring clearance
  • For an oral dose only the bioavailable fraction ever reaches the plasma, so $F\cdot \text{Dose}=CL\cdot \text{AUC}$
• For a one-compartment IV bolus, $C(t)=C_0{e}^{-kt}$ where $C_0=\text{Dose}/V_D$ and $k$ is the elimination rate constant, which gives the same $CL=k\cdot V_D$ result as Elimination Rate Constant:

$$\begin{array}{rl}\text{AUC}& ={\int }_0^{\infty }{C}_0{e}^{-kt}\cdot dt=C_0{\left[-\frac{e^{-kt}}{k}\right]}_0^{\infty }=\frac{C_0}{k}\\ \therefore CL& =\frac{\text{Dose}}{C_0/k}=k\cdot \left(\frac{\text{Dose}}{C_0}\right)\\ & =k\cdot V_D\end{array}$$

Organ Clearance

• Whole body clearance is the sum of the clearances of each eliminating organ, since each organ’s elimination rate adds to the total and they all divide by the same plasma concentration:

$$C{L}_{\text{total}}=C{L}_{\text{hepatic}}+C{L}_{\text{renal}}+C{L}_{\text{other}}$$

• Each organ’s share of total clearance is therefore its share of the elimination rate, so the fraction of the dose excreted unchanged in the urine and the fraction metabolised are:

$$\begin{array}{rl}\text{Fraction Excreted Unchanged}& =\frac{C{L}_R}{C{L}_{\text{total}}}\\ \text{Fraction Metabolised}& =\frac{C{L}_H}{C{L}_{\text{total}}}=\frac{C{L}_{\text{total}}-C{L}_R}{C{L}_{\text{total}}}\end{array}$$

- The second form assumes all clearance which isn't renal is hepatic metabolism

How are the fraction of a dose excreted unchanged and the fraction metabolised related to clearance?

Each is that organ’s share of total clearance, since every organ’s elimination rate is its clearance times the same plasma concentration:

$$\text{Fraction excreted unchanged}=\frac{C{L}_R}{C{L}_{\text{total}}},\text{Fraction metabolised}=\frac{C{L}_H}{C{L}_{\text{total}}}$$

Clearance = Blood Flow × Extraction Ratio

• For blood ($Q$) flowing through an organ, the organ’s elimination rate is the drug delivered minus the drug leaving (Fick principle), and blood flow in equals blood flow out as blood isn’t stored in the organ:

$$\text{Elimination Rate}=Q\cdot C_{\text{in}}-Q\cdot C_{\text{out}}=Q(C_{\text{in}}-C_{\text{out}})$$

• Dividing by the concentration delivered to the organ (the definition of clearance) gives:

$$\begin{array}{rl}C{L}_{\text{organ}}& =\frac{Q(C_{\text{in}}-C_{\text{out}})}{C_{\text{in}}}\\ & =Q\cdot E\end{array}$$

Figure 9 Organ clearance from the Fick principle

• Clearance of an organ is dependent on the blood flow to that organ
• So the maximum possible clearance of an organ is its blood flow ($E=1$), which for the liver is roughly 1.5 L/min (about 90 L/hour)
• Strictly $Q\cdot E$ is a blood clearance ($C{L}_b$), since it uses blood flow and blood concentrations. The elimination rate is the same whichever concentration it is divided by, so using the blood:plasma ratio ($\lambda$, see Pharmacology Basics):

$$CL\cdot C=C{L}_b\cdot C_b=C{L}_b\cdot \lambda \cdot C\therefore CL=\lambda \cdot C{L}_b$$

• Hence plasma clearance can appear to exceed hepatic blood flow for drugs that concentrate in red cells ($\lambda >1$)

Derive organ clearance in terms of blood flow and extraction ratio

Elimination rate $=Q(C_{\text{in}}-C_{\text{out}})$ (Fick principle), so $CL=\frac{Q(C_{\text{in}}-C_{\text{out}})}{C_{\text{in}}}=Q\cdot E$

Hepatic Extraction Ratio (Well-Stirred Model)

• The well-stirred model treats the liver as a single well-mixed compartment, so the concentration the hepatocytes see is the same as the concentration leaving in the hepatic vein ($C_{\text{out}}$)
• At steady state, drug entering the liver must leave by one of two routes:
  • Washed out in hepatic venous blood at a rate of $Q_H\cdot C_{\text{out}}$
  • Metabolised by hepatocytes, which only see unbound drug ($f_u\cdot C_{\text{out}}$) and clear it at their intrinsic clearance ($C{L}_{\text{int}}$, the clearance of the liver if blood flow were not limiting), at a rate of $f_u\cdot C{L}_{\text{int}}\cdot C_{\text{out}}$

$$\begin{array}{rl}{Q}_H\cdot C_{\text{in}}& =Q_H\cdot C_{\text{out}}+f_u\cdot C{L}_{\text{int}}\cdot C_{\text{out}}\\ \therefore C_{\text{out}}& =\frac{Q_H\cdot C_{\text{in}}}{Q_H+f_u\cdot C{L}_{\text{int}}}\end{array}$$

• Substituting into the definition of extraction ratio:

$$\begin{array}{rl}{E}_H& =1-\frac{C_{\text{out}}}{C_{\text{in}}}=1-\frac{Q_H}{Q_H+f_u\cdot C{L}_{\text{int}}}\\ & =\frac{f_u\cdot C{L}_{\text{int}}}{Q_H+f_u\cdot C{L}_{\text{int}}}\\ \therefore C{L}_H& =Q_H\cdot E_H=\frac{Q_H\cdot f_u\cdot C{L}_{\text{int}}}{Q_H+f_u\cdot C{L}_{\text{int}}}\end{array}$$

Figure 10 Well-stirred model of the liver. Liver adapted from Servier Medical Art, CC BY 4.0

• The denominator is not hepatic blood flow, it is the sum of the two exits from the liver
  • Each exit is a rate divided by $C_{\text{out}}$, so both have units of L/hour and can be added: $Q_H$ is the volume of liver blood emptied of drug per hour by being carried away, $f_u\cdot C{L}_{\text{int}}$ is the volume emptied per hour by metabolism
  • $E_H$ is therefore the share of drug that takes the metabolism exit rather than the washout exit
  • e.g. with $Q_H=90$ L/hour, $f_u\cdot C{L}_{\text{int}}=10$ L/hour and $C_{\text{out}}=1$ mg/L, 90 mg/hour leaves in hepatic venous blood and 10 mg/hour is metabolised, so 100 mg/hour must be entering ($C_{\text{in}}\approx 1.11$ mg/L) and $E_H=10/100=0.1$
• Assumptions of the model:
  • The liver is well-stirred (other models such as the parallel-tube model give a different formula)
  • Only unbound drug is metabolised
  • Kinetics are linear, so $C{L}_{\text{int}}$ is constant (roughly $V_{\max }/K_m$ at concentrations well below $K_m$, where $V_{\max }$ is the maximal velocity of the reaction at saturating substrate concentration and $K_m$ is Michaelis constant which expresses how tightly the enzyme binds the drug substrate)

What is the hepatic extraction ratio in the well-stirred model, and what does its denominator represent?

$$E_H=\frac{f_u\cdot C{L}_{\text{int}}}{Q_H+f_u\cdot C{L}_{\text{int}}}$$

The denominator is the two competing routes for drug to leave the liver: washout in hepatic venous blood ($Q_H$) and metabolism of unbound drug ($f_u\cdot C{L}_{\text{int}}$)

Flow-Limited vs Capacity-Limited Clearance

• The two extremes of the well-stirred model (Figure 10) are where it becomes clinically useful:

	High extraction ($E>0.7$)	Low extraction ($E<0.3$)
When	$f_u\cdot C{L}_{\text{int}}\gg Q_H$	$f_u\cdot C{L}_{\text{int}}\ll Q_H$
$C{L}_H$ simplifies to	$\approx Q_H$ (flow-limited)	$\approx f_u\cdot C{L}_{\text{int}}$ (capacity-limited)
Clearance changed by	Hepatic blood flow (shock, low cardiac output, β-blockers, ageing)	Enzyme induction or inhibition, protein binding
First-pass metabolism	Large, so low oral bioavailability ($F_H=1-E_H$)	Small, so high oral bioavailability
Examples	Propofol, lignocaine, morphine, fentanyl, GTN, propranolol, verapamil	Phenytoin, warfarin, diazepam, theophylline

Figure 11 Hepatic clearance against unbound intrinsic clearance

• e.g. lignocaine can accumulate in low cardiac output states as its clearance follows hepatic blood flow, whereas enzyme induction (e.g. rifampicin) speeds up the clearance of warfarin but makes little difference to high extraction drugs

Cloze

High extraction drugs have flow-limited hepatic clearance, whereas low extraction drugs have capacity-limited hepatic clearance

References

1.

Birkett DJ. Pharmacokinetics made easy. 2.ed., rev. ed. Sydney London: McGraw-Hill; 2005. 131 p.

Link to original

Steady State

Main source: Birkett, Pharmacokinetics Made Easy ¹

• At steady state the dose rate going in equals the elimination rate coming out (see Clearance at Steady State), so for intravenous dosing:

$$\text{Maintenance Dose Rate }(DR)=CL\times \text{Steady State Concentration }(C_{SS})$$

• The unbound steady state concentration is $C{u}_{SS}=f_u\cdot C_{SS}$, and it’s the unbound concentration which drives the effect
• The rest of this note is for a drug cleared by the liver, where only unbound drug is metabolised, so clearance follows the well-stirred model (see Hepatic Extraction Ratio (Well-Stirred Model)):

$$CL=\frac{Q_H\cdot f_u\cdot C{L}_{\text{int}}}{Q_H+f_u\cdot C{L}_{\text{int}}}$$

Determinants of Steady State Concentration

Figure 12 Determinants of total and unbound steady state concentration for a hepatically cleared drug

• For drugs given orally, the unbound concentration depends only on intrinsic clearance (and the fraction absorbed), and the total concentration depends on intrinsic clearance and fraction unbound, whatever the hepatic extraction ratio
• For drugs given intravenously, it depends on the hepatic extraction ratio:
  • Low hepatic extraction ratio: the unbound concentration depends on intrinsic clearance, and the total concentration depends on intrinsic clearance and fraction unbound
  • High hepatic extraction ratio: the unbound concentration depends on hepatic blood flow and fraction unbound, and the total concentration depends on hepatic blood flow
• Hence hepatic blood flow only matters for a high extraction ratio drug given intravenously

Intravenous Dosing

• For low hepatic extraction ratio drugs $CL\approx f_u\cdot C{L}_{\text{int}}$, so:

$$\begin{array}{rl}{C}_{SS}& \approx \frac{DR}{f_u\cdot C{L}_{\text{int}}}\\ C{u}_{SS}& =f_u\times C_{SS}\approx \frac{DR}{C{L}_{\text{int}}}\end{array}$$

• For high hepatic extraction ratio drugs $CL\approx Q_H$, so:

$$\begin{array}{rl}{C}_{SS}& \approx \frac{DR}{Q_H}\\ C{u}_{SS}& \approx \frac{f_u\cdot DR}{Q_H}\end{array}$$

What determines the total and unbound steady state concentration of a low hepatic extraction ratio drug given IV?

$CL\approx f_u\cdot C{L}_{\text{int}}$, so $C_{SS}\approx \frac{DR}{f_u\cdot C{L}_{\text{int}}}$ (intrinsic clearance and fraction unbound) and $C{u}_{SS}\approx \frac{DR}{C{L}_{\text{int}}}$ (intrinsic clearance only)

What determines the total and unbound steady state concentration of a high hepatic extraction ratio drug given IV?

$CL\approx Q_H$, so $C_{SS}\approx \frac{DR}{Q_H}$ (hepatic blood flow only) and $C{u}_{SS}\approx \frac{f_u\cdot DR}{Q_H}$ (hepatic blood flow and fraction unbound)

Oral Dosing

• Only the bioavailable fraction of each dose reaches the systemic circulation (see Bioavailability), so:

$$\begin{array}{rl}F\times DR& =CL\times C_{SS}\\ \therefore C_{SS}& =\frac{F\times DR}{CL}=\frac{f_g\cdot f_H\cdot DR}{CL}\end{array}$$

• Substituting the well-stirred model for both $CL$ and $f_H=1-E_H=\frac{Q_H}{Q_H+f_u\cdot C{L}_{\text{int}}}$:

$$\begin{array}{rl}{C}_{SS}& =(f_g\cdot f_H\cdot DR)\times \frac{Q_H+f_u\cdot C{L}_{\text{int}}}{Q_H\cdot f_u\cdot C{L}_{\text{int}}}\\ & =(f_g\cdot DR)\times \frac{Q_H}{Q_H+f_u\cdot C{L}_{\text{int}}}\times \frac{Q_H+f_u\cdot C{L}_{\text{int}}}{Q_H\cdot f_u\cdot C{L}_{\text{int}}}\\ & =\frac{f_g\cdot DR}{f_u\cdot C{L}_{\text{int}}}\\ \therefore C{u}_{SS}& =\frac{f_g\cdot DR}{C{L}_{\text{int}}}\end{array}$$

• Hepatic blood flow drops out because it changes bioavailability and clearance by the same proportion, so a fall in blood flow lowers clearance but lets proportionally less drug through the liver on its first pass, and the two cancel
• This holds whatever the extraction ratio, as the approximations for low and high extraction drugs were never needed

Derive the unbound steady state concentration of a hepatically cleared drug given orally

$F\cdot DR=CL\cdot C_{SS}$ with $F=f_g\cdot f_H$. Substituting $f_H=\frac{Q_H}{Q_H+f_u\cdot C{L}_{\text{int}}}$ and $CL=\frac{Q_H\cdot f_u\cdot C{L}_{\text{int}}}{Q_H+f_u\cdot C{L}_{\text{int}}}$, the $Q_H+f_u\cdot C{L}_{\text{int}}$ and $Q_H$ terms cancel:

$$C_{SS}=\frac{f_g\cdot DR}{f_u\cdot C{L}_{\text{int}}},C{u}_{SS}=\frac{f_g\cdot DR}{C{L}_{\text{int}}}$$

Cloze

After oral dosing of a hepatically cleared drug, the unbound steady state concentration depends only on intrinsic clearance, whatever the hepatic extraction ratio

Why doesn't hepatic blood flow affect the steady state concentration of a high extraction drug given orally?

A fall in hepatic blood flow lowers clearance ($CL\approx Q_H$) but also lowers the fraction escaping first-pass ($f_H\approx \frac{Q_H}{f_u\cdot C{L}_{\text{int}}}$) by the same proportion, so $F/CL$, and hence $C_{SS}=F\cdot DR/CL$, is unchanged

Clinical Consequences

• These follow from the figure above (Figure 12):
  • Enzyme induction or inhibition (changing $C{L}_{\text{int}}$) changes the unbound steady state concentration in every case except a high extraction ratio drug given intravenously
  • Displacement from plasma proteins (raising $f_u$) lowers the total concentration but leaves the unbound concentration unchanged for low extraction ratio drugs and for oral dosing, so it’s mostly of no clinical consequence; the exception is a high extraction ratio drug given intravenously, where the unbound concentration rises
  • A fall in hepatic blood flow (e.g. low cardiac output) only raises the steady state concentration of a high extraction ratio drug given intravenously, e.g. a lignocaine infusion

When does displacement of a drug from plasma proteins raise its unbound steady state concentration?

Only for a high hepatic extraction ratio drug given IV ($C{u}_{SS}\approx f_u\cdot DR/Q_H$). For low extraction drugs and oral dosing, clearance rises with $f_u$, so the total concentration falls and the unbound concentration is unchanged

Which steady state concentrations change with a fall in hepatic blood flow?

Only those of a high hepatic extraction ratio drug given IV, as $C_{SS}\approx DR/Q_H$ (e.g. lignocaine in low cardiac output states). Low extraction drugs and oral dosing are unaffected

References

1.

Birkett DJ. Pharmacokinetics made easy. 2.ed., rev. ed. Sydney London: McGraw-Hill; 2005. 131 p.

Link to original

B1.v Describe the concept of effect-site concentration

Effect-Site Concentration

Main source: Birkett, Pharmacokinetics Made Easy ¹

• Drug effect follows the concentration at the site of action (the effect site), not the plasma concentration
  • Whether the two match depends on how quickly the effect site equilibrates with plasma
• The effect site is modelled as an extra compartment of negligible volume linked to the central compartment (see Two-Compartment Model), which equilibrates at a rate set by $k_{e0}$:

$$\frac{d{C}_e}{dt}=k_{e0}(C_p-C_e)$$

- When $k_{e0}$ is large, $C_{e}$ follows $C_{p}$ almost immediately, so effect follows plasma concentration
- When $k_{e0}$ is small, $C_{e}$ lags behind $C_{p}$, rising while $C_{p}>C_{e}$ and peaking when $C_{e}=C_{p}$ (where $dC_{e}/dt=0$)
- The equilibration half-time ($t_{1/2}k_{e0}=0.693/k_{e0}$) describes how quickly the effect catches up with a change in plasma concentration

Figure 13 Plasma and effect-site concentrations after an IV bolus

What equation links plasma and effect-site concentration, and when does effect-site concentration peak?

$\frac{d{C}_e}{dt}=k_{e0}(C_p-C_e)$. After a bolus, $C_e$ rises while $C_p>C_e$ and peaks when $C_e=C_p$.

Rapidly Equilibrating Effect Site

• e.g. diazepam used intravenously in status epilepticus:
  • The brain is highly perfused, so brain concentrations and the anticonvulsant effect follow plasma concentrations
  • Plasma concentrations fall rapidly during the distribution phase as diazepam redistributes to other tissues, so the effect falls with them and fitting can recur within 2–4 hours if more drug is not given, even though the elimination half-life is very long
  • The duration of action of a single dose is therefore set by redistribution rather than elimination
• If the site of a toxic effect is rapidly accessible, the peak concentration at the end of a bolus (before distribution, $\approx \text{Dose}/V_1$) can cause toxicity
  • The injection is given slowly so distribution happens while the drug is being given, e.g. intravenous theophylline and lignocaine

Cloze

After a single IV dose of diazepam, the anticonvulsant effect wears off due to redistribution, not elimination, so fitting can recur within 2–4 hours

Slowly Equilibrating Effect Site

• e.g. digoxin given intravenously:
  • Digoxin distributes slowly to its site of action (the inotropic effect) in cardiac muscle, so the effect increases while plasma concentrations are decreasing, both being driven by the same movement of drug from plasma into tissue
  • It takes about 6 hours to reach full effect even when given intravenously, so IV dosing gives no advantage in time to full effect, and loading is best done with divided oral doses at least 6 hours apart so the full effect of each dose can be assessed before more is given
  • In the first 6–8 hours after a dose, plasma concentrations bear no relationship to effect, so levels must be taken at least 6 hours after a dose, preferably as a trough just before the next dose

Why must digoxin levels be taken at least 6 hours after a dose?

Digoxin equilibrates slowly with its site of action in cardiac muscle, so for the first 6–8 hours plasma concentrations are high while the effect is still rising and bear no relationship to effect. Ideally take a trough level just before the next dose.

References

1.

Birkett DJ. Pharmacokinetics made easy. 2.ed., rev. ed. Sydney London: McGraw-Hill; 2005. 131 p.

Link to original

B1.vi Describe the concept of context sensitive half-time

B1.vii Compare the kinetics of an intravenous bolus dose (and dosing regimen) and a continuous infusion

B1.viii Explain clinical drug monitoring including the relevance of

B1.viii.a Peak and trough concentrations

B1.viii.b Minimum therapeutic concentrations