Worked Example of Extended Clearance Calculation

From Transporter Parameters to Hepatic Clearance

An extended-clearance calculation connects processes that act at different stages: drug enters hepatocytes from blood, leaves the intracellular unbound pool through efflux or elimination, and may be retained by intracellular binding. The quantities of interest are related but not interchangeable. Intracellular concentration describes the cell pool; net intrinsic clearance describes irreversible removal from blood under the cellular model; hepatic clearance also reflects hepatic blood flow and blood binding.

The calculation below uses a linear, steady-state model. Clearances are hypothetical and are expressed on a consistent blood and unbound-concentration basis. Keeping that basis consistent is essential: inserting a plasma unbound fraction into a model parameterized for blood without accounting for the blood-to-plasma relationship can give an incorrect result.

Definition: Let \(CL_{\mathrm{uptake}}\) be the clearance for uptake from blood into hepatocytes; \(CL_{\mathrm{efflux}}\), \(CL_{\mathrm{met}}\), and \(CL_{\mathrm{bile}}\) be intracellular clearances for return to blood, metabolism, and biliary export, respectively. Each intracellular clearance acts on the unbound intracellular concentration \(C_{\mathrm{cell,u}}\). Let \(C_{\mathrm{b,u}}\) be unbound blood concentration, \(f_{\mathrm{u,cell}}\) the unbound fraction in cells, \(f_{\mathrm{u,b}}\) the unbound fraction in blood, and \(Q_{\mathrm{h}}\) hepatic blood flow. The net intrinsic clearance \(CL_{\mathrm{int,net}}\) is the modelled irreversible hepatic elimination clearance relative to unbound blood concentration.

Steady-State Calculation

For the linear model, uptake input is \(CL_{\mathrm{uptake}}C_{\mathrm{b,u}}\). The intracellular exits are \(CL_{\mathrm{efflux}}C_{\mathrm{cell,u}}\), \(CL_{\mathrm{met}}C_{\mathrm{cell,u}}\), and \(CL_{\mathrm{bile}}C_{\mathrm{cell,u}}\). At steady state, there is no net accumulation in the cell, so uptake equals the sum of these exits. Binding changes the amount of drug represented by a given unbound concentration, but under these assumptions it does not create an additional steady-state elimination route.

CL_{\mathrm{uptake}}C_{\mathrm{b,u}} = \left(CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}}\right)C_{\mathrm{cell,u}}

Solving gives the intracellular unbound concentration. If a total intracellular concentration is requested, the unbound fraction converts between the unbound and total pools.

C_{\mathrm{cell,u}} = \frac{CL_{\mathrm{uptake}}} {CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}}} C_{\mathrm{b,u}}, \qquad C_{\mathrm{cell,tot}}=\frac{C_{\mathrm{cell,u}}}{f_{\mathrm{u,cell}}}

Efflux returns drug to blood, so it is not irreversible elimination. The irreversible intracellular exits are metabolism and biliary export. Their combined rate, divided by unbound blood concentration, gives net intrinsic clearance:

CL_{\mathrm{int,net}} = \frac{(CL_{\mathrm{met}}+CL_{\mathrm{bile}})C_{\mathrm{cell,u}}}{C_{\mathrm{b,u}}} = \frac{CL_{\mathrm{uptake}}(CL_{\mathrm{met}}+CL_{\mathrm{bile}})} {CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}}}

This form makes the competition visible. A larger efflux clearance lowers the proportion of uptake that proceeds to irreversible elimination. The biliary fraction below is defined as the fraction of irreversible hepatic elimination routed to bile, not the fraction of all cellular exits that enter bile.

f_{\mathrm{bile}} = \frac{CL_{\mathrm{bile}}}{CL_{\mathrm{met}}+CL_{\mathrm{bile}}}

Connecting Intrinsic Clearance to the Liver

Under the well-stirred liver model, hepatic clearance depends on hepatic blood flow and on intrinsic clearance adjusted for blood binding. The model assumes a well-mixed liver, constant flow, and that the relevant drug concentration equilibrates with the eliminating compartment as represented by the model. With the clearances defined relative to unbound blood concentration, the relationship is:

CL_{\mathrm{h}} = \frac{Q_{\mathrm{h}}f_{\mathrm{u,b}}CL_{\mathrm{int,net}}} {Q_{\mathrm{h}}+f_{\mathrm{u,b}}CL_{\mathrm{int,net}}}

The term \(f_{\mathrm{u,b}}CL_{\mathrm{int,net}}\) is the effective intrinsic clearance relative to total blood concentration. The result cannot exceed \(Q_{\mathrm{h}}\), because blood flow limits delivery to the liver in this model. The calculation presumes that the supplied uptake, efflux, metabolism, and biliary clearances are linear and mutually compatible in units and concentration basis.

Worked Example: Calculating All Four Quantities

Consider hypothetical clearances \(CL_{\mathrm{uptake}}=90\) L/h, \(CL_{\mathrm{efflux}}=30\) L/h, \(CL_{\mathrm{met}}=45\) L/h, and \(CL_{\mathrm{bile}}=15\) L/h. Let \(C_{\mathrm{b,u}}=0.20\) mg/L, \(f_{\mathrm{u,cell}}=0.25\), \(f_{\mathrm{u,b}}=0.10\), and \(Q_{\mathrm{h}}=90\) L/h. Assume all clearances are linear and use the definitions above.

1. Intracellular concentration. The sum of intracellular clearances is \(30+45+15=90\) L/h. Therefore:

C_{\mathrm{cell,u}} = \frac{90\ \mathrm{L/h}}{90\ \mathrm{L/h}} (0.20\ \mathrm{mg/L}) = 0.20\ \mathrm{mg/L}

The total intracellular concentration is \(C_{\mathrm{cell,tot}}=0.20/0.25=0.800\) mg/L. This is higher than the unbound concentration because only one quarter of the total cell concentration is unbound under the assumed binding model.

2. Net intrinsic clearance. Metabolism plus biliary export is \(45+15=60\) L/h:

CL_{\mathrm{int,net}} = \frac{90(60)}{30+45+15} = 60\ \mathrm{L/h}

3. Hepatic clearance. First, \(f_{\mathrm{u,b}}CL_{\mathrm{int,net}}=0.10(60)=6.00\) L/h. Thus:

CL_{\mathrm{h}} = \frac{90(6.00)}{90+6.00} = 5.625\ \mathrm{L/h}

This is below the assumed flow of 90 L/h, as required by the well-stirred expression.

4. Biliary fraction. Among irreversible elimination routes, \(15\) L/h is biliary and \(45\) L/h is metabolic:

f_{\mathrm{bile}} = \frac{15}{45+15} = 0.250

Thus 25.0% of modelled irreversible hepatic elimination is biliary. This is not the fraction of all intracellular exits that enter bile; that alternative fraction would be \(15/(30+45+15)=0.167\), because its denominator includes return to blood.

Why the Net-Clearance Formula Has This Form

The derivation follows directly from the steady-state balance. Dividing the balance equation by the positive sum of intracellular clearances gives the ratio of unbound intracellular to unbound blood concentration. Multiplying that ratio by the combined irreversible clearance gives net intrinsic clearance:

\frac{C_{\mathrm{cell,u}}}{C_{\mathrm{b,u}}} = \frac{CL_{\mathrm{uptake}}} {CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}}}

\frac{(CL_{\mathrm{met}}+CL_{\mathrm{bile}})C_{\mathrm{cell,u}}} {C_{\mathrm{b,u}}} = \frac{CL_{\mathrm{uptake}}(CL_{\mathrm{met}}+CL_{\mathrm{bile}})} {CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}}}

The units are also a useful check: the concentration ratio is dimensionless, and multiplying it by an intracellular clearance in L/h leaves L/h. Intracellular binding does not appear in the net-clearance equation because all modeled exits act on the same unbound concentration and the calculation is at steady state. Binding does matter when converting that unbound concentration to total intracellular concentration.

Worked Example: Increased Efflux Reduces Net Elimination

Use the first example's parameters, but increase \(CL_{\mathrm{efflux}}\) from 30 to 60 L/h. Keep \(CL_{\mathrm{uptake}}=90\) L/h, \(CL_{\mathrm{met}}=45\) L/h, \(CL_{\mathrm{bile}}=15\) L/h, and \(C_{\mathrm{b,u}}=0.20\) mg/L. The steady-state assumption still applies, and the clearances remain linear.

The total intracellular clearance is now \(60+45+15=120\) L/h. Therefore:

C_{\mathrm{cell,u}} = \frac{90}{120}(0.20) = 0.150\ \mathrm{mg/L}

The combined irreversible clearance remains \(45+15=60\) L/h, but the lower intracellular concentration reduces its rate relative to blood concentration:

CL_{\mathrm{int,net}} = \frac{90(60)}{120} = 45\ \mathrm{L/h}

With \(f_{\mathrm{u,b}}=0.10\) and \(Q_{\mathrm{h}}=90\) L/h, the hepatic clearance is:

CL_{\mathrm{h}} = \frac{90(0.10)(45)}{90+0.10(45)} = \frac{405}{94.5} = 4.286\ \mathrm{L/h}

The biliary fraction remains \(15/(45+15)=0.250\). Increasing efflux lowers total irreversible clearance here, but does not change the relative split between metabolism and bile because their clearances were unchanged.

Worked Example: Binding Conversion and Flow Limitation

For a separate hypothetical case, let \(CL_{\mathrm{uptake}}=120\) L/h, \(CL_{\mathrm{efflux}}=20\) L/h, \(CL_{\mathrm{met}}=60\) L/h, and \(CL_{\mathrm{bile}}=20\) L/h. Take \(C_{\mathrm{b,u}}=0.050\) mg/L, \(f_{\mathrm{u,cell}}=0.20\), \(f_{\mathrm{u,b}}=0.080\), and \(Q_{\mathrm{h}}=100\) L/h. Assume the linear steady-state model and consistent clearance units.

The intracellular clearance sum is \(20+60+20=100\) L/h, so:

C_{\mathrm{cell,u}} = \frac{120}{100}(0.050) = 0.0600\ \mathrm{mg/L}, \qquad C_{\mathrm{cell,tot}} = \frac{0.0600}{0.20} = 0.300\ \mathrm{mg/L}

Net intrinsic clearance is \(120(60+20)/100=96.0\) L/h. The blood-binding-adjusted intrinsic term is \(0.080(96.0)=7.68\) L/h, giving:

CL_{\mathrm{h}} = \frac{100(7.68)}{100+7.68} = 7.137\ \mathrm{L/h}

The biliary fraction is \(20/(60+20)=0.250\). Although the total intracellular concentration is five times its unbound value, it is the unbound intracellular concentration that enters the specified clearance-rate equations.

Checking the Calculation and Comparing Approaches

A compact workflow helps catch errors before interpreting the result. The assumptions should be checked against the intended use: a steady-state calculation is not a time course, and a linear clearance model is not automatically appropriate when transport or metabolism is saturated.

Confirm the basis.
Check that concentrations and clearances refer to compatible blood or plasma conventions, that unbound fractions use the same matrix, and that all clearances have consistent units.

Compute intracellular unbound concentration.
Divide uptake clearance by the sum of efflux, metabolism, and biliary clearances, then multiply by unbound blood concentration.

Convert only if a total cell concentration is needed.
Divide by \(f_{\mathrm{u,cell}}\); do not substitute total cell concentration into clearances defined to act on unbound concentration.

Calculate elimination and liver clearance.
Use metabolism plus bile for net intrinsic clearance, then apply the well-stirred equation with \(f_{\mathrm{u,b}}\) and \(Q_{\mathrm{h}}\).

State the biliary-fraction denominator.
For the fraction of irreversible elimination, divide biliary clearance by metabolism plus biliary clearance; exclude efflux because it returns drug to blood.

The steady-state algebraic calculation is useful for tracing how parameter values combine, while a hepatic ODE model represents time-varying concentrations and accumulation explicitly. The well-stirred model adds a flow-limited organ relationship to the intrinsic-clearance calculation. These approaches answer different questions: an ODE can describe transients, whereas the algebraic result gives a steady-state relation under its assumptions. Neither makes nonlinear transporter kinetics linear; if parameters vary with concentration, the constant-clearance formulas above need not hold.

Quantity	What it describes	Key denominator or adjustment
\(C_{\mathrm{cell,u}}\)	Steady-state unbound intracellular concentration	Efflux + metabolism + bile
\(CL_{\mathrm{int,net}}\)	Irreversible elimination relative to unbound blood concentration	Intracellular exits, with efflux reducing the eliminated share
\(CL_{\mathrm{h}}\)	Organ clearance in the well-stirred model	Blood binding and hepatic flow
\(f_{\mathrm{bile}}\)	Share of irreversible elimination routed to bile	Metabolism + bile

Common Pitfalls and Limits

Counting efflux as elimination. Efflux is a cellular exit but returns drug to blood. Include it in the steady-state intracellular denominator, not in the irreversible-elimination numerator.
Using total intracellular concentration in an unbound-clearance equation. The rate terms specified here use \(C_{\mathrm{cell,u}}\). Convert to total concentration only for reporting or another model component that requires it.
Applying binding twice. \(f_{\mathrm{u,cell}}\) converts intracellular total to unbound concentration; \(f_{\mathrm{u,b}}\) adjusts net intrinsic clearance in the well-stirred equation. They refer to different compartments and serve different steps.
Calling the bile share \(CL_{\mathrm{bile}}/(CL_{\mathrm{efflux}}+CL_{\mathrm{met}}+CL_{\mathrm{bile}})\) the biliary fraction of elimination. That ratio is bile's share of all intracellular exits. The fraction of irreversible elimination is \(CL_{\mathrm{bile}}/(CL_{\mathrm{met}}+CL_{\mathrm{bile}})\).
Ignoring model limits. The calculation assumes constant clearances, steady state, and the specified compartment structure. Saturation, concentration-dependent binding, multiple cell zones, or time-varying concentrations may require a more detailed model.

Key takeaway: First balance uptake against every intracellular exit to obtain \(C_{\mathrm{cell,u}}\). Then count metabolism and biliary export—but not return efflux—as irreversible elimination, and only afterward apply blood binding and hepatic flow to calculate \(CL_{\mathrm{h}}\).

Check Your Understanding

Use the linear steady-state definitions in this tutorial. Give units where appropriate.

Let uptake be 50 L/h, efflux 10 L/h, metabolism 30 L/h, and biliary export 10 L/h. If \(C_{\mathrm{b,u}}=0.40\) mg/L, calculate \(C_{\mathrm{cell,u}}\).
For the parameters in question 1, calculate net intrinsic clearance and the biliary fraction of irreversible elimination.
If \(f_{\mathrm{u,cell}}=0.50\) in question 1, what is total intracellular concentration?
For question 2, take \(f_{\mathrm{u,b}}=0.20\) and \(Q_{\mathrm{h}}=80\) L/h. Calculate hepatic clearance.

Solutions. (1) The intracellular clearance sum is \(10+30+10=50\) L/h, so \(C_{\mathrm{cell,u}}=(50/50)(0.40)=0.400\) mg/L. (2) \(CL_{\mathrm{int,net}}=50(30+10)/50=40.0\) L/h, and \(f_{\mathrm{bile}}=10/(30+10)=0.250\), or 25.0%. (3) \(C_{\mathrm{cell,tot}}=0.400/0.50=0.800\) mg/L. (4) The binding-adjusted intrinsic term is \(0.20(40.0)=8.00\) L/h, so \(CL_{\mathrm{h}}=80(8.00)/(80+8.00)=7.273\) L/h.