Why Use Extraction Ratio as a Comparison Scale?
The well-stirred, parallel-tube, and dispersion liver models describe different assumptions about mixing and concentration gradients within the liver. Their equations may look unlike one another, but each predicts the same kind of observable quantity: the fraction of drug entering the liver that is removed during one passage. Expressing predictions as an extraction ratio puts them on a common, dimensionless scale.
The comparison is useful only when the inputs are matched. In particular, use the same hepatic blood flow \(Q\), the same intrinsic-clearance input \(CL_{\mathrm{int}}\), and consistent units and concentration conventions. The dispersion model additionally requires its dimensionless dispersion number \(D_N\), as defined for that model. The calculation does not establish which model is correct for a particular liver; it makes clear what each model predicts under specified assumptions.
Three Predictions on the Same Scale
For these model comparisons, define the dimensionless input \(x=CL_{\mathrm{int}}/Q\). Both quantities must be expressed in compatible units, so \(x\) has no units. Using the same \(x\) across models is essential: changing \(Q\) or \(CL_{\mathrm{int}}\) between calculations would confound the comparison.
The well-stirred model predicts an extraction ratio of \(x/(1+x)\). The parallel-tube model predicts an outlet-to-inlet concentration ratio of \(\exp(-x)\). The dispersion model predicts its outlet-to-inlet ratio using both \(x\) and \(D_N\). In each case, subtracting the outlet fraction from one gives extraction.
All three extraction ratios are dimensionless fractions. Multiplication by the same \(Q\) converts each prediction to hepatic clearance. An extraction ratio of \(0.30\), for example, means that 30% of the drug entering the liver is predicted to be removed in one passage; it does not mean clearance is \(0.30\) in volume-per-time units.
Why Clearance Equals Flow Times Extraction
The shared scale follows directly from a steady-state mass balance across the liver. Suppose blood flow \(Q\) carries drug into the liver at concentration \(C_{\mathrm{in}}\), and drug leaves at concentration \(C_{\mathrm{out}}\). The amount entering per unit time is \(Q C_{\mathrm{in}}\), and the amount leaving per unit time is \(Q C_{\mathrm{out}}\). With no net production or accumulation, the amount eliminated per unit time is their difference.
Clearance is defined as elimination rate divided by inlet concentration. Therefore,
This identity explains why the extraction ratio is a natural comparison scale: it expresses the concentration change as a fraction of the entering concentration, while \(Q\) supplies the volume-per-time scale. It assumes a steady, single-pass flow representation and consistent definitions of inlet concentration, outlet concentration, and flow. If a model’s clearance and concentration conventions do not match, multiplying by \(Q\) will not repair that mismatch.
Worked Example: Comparing Predictions at High Dimensionless Input
Hypothetical inputs: Let \(Q=90.0\ \mathrm{L/h}\), \(CL_{\mathrm{int}}=180\ \mathrm{L/h}\), and \(D_N=0.500\). The inputs are positive and use compatible units. Compute the three extractions and clearances.
First, \(x=CL_{\mathrm{int}}/Q=180/90.0=2.00\). The well-stirred prediction is \(E_{H,\mathrm{WS}}=2/(1+2)=0.6667\), so \(CL_{H,\mathrm{WS}}=(90.0\ \mathrm{L/h})(0.6667)=60.00\ \mathrm{L/h}\), using the unrounded ratio for the multiplication.
For the parallel-tube model, \(E_{H,\mathrm{PT}}=1-\exp(-2)=1-0.1353353=0.8646647\), or \(0.8647\). Thus \(CL_{H,\mathrm{PT}}=(90.0\ \mathrm{L/h})(0.8646647)=77.82\ \mathrm{L/h}\).
For the dispersion model, \(z=\sqrt{1+4(0.500)(2.00)}=\sqrt{5}=2.2360679\). Substitution into the outlet-fraction equation gives
Therefore \(E_{H,\mathrm{disp}}=1-0.2485516=0.7514484\), or \(0.7514\). Its clearance is \((90.0\ \mathrm{L/h})(0.7514484)=67.6304\ \mathrm{L/h}\), or \(67.63\ \mathrm{L/h}\). For these inputs, extraction ranks as well-stirred \(0.6667\), dispersion \(0.7514\), then parallel-tube \(0.8647\). These are predictions under the stated models, not measurements.
Worked Example: Comparing Lower Extraction Predictions
Hypothetical inputs: Let \(Q=75.0\ \mathrm{L/h}\), \(CL_{\mathrm{int}}=30.0\ \mathrm{L/h}\), and \(D_N=0.200\). Compute each extraction ratio on the same input scale.
Here \(x=30.0/75.0=0.400\). The well-stirred prediction is \(E_{H,\mathrm{WS}}=0.400/1.400=0.2857143\), or \(0.2857\).
The parallel-tube prediction is \(E_{H,\mathrm{PT}}=1-\exp(-0.400)=1-0.6703200=0.3296800\), or \(0.3297\).
For dispersion, \(z=\sqrt{1+4(0.200)(0.400)}=\sqrt{1.32}=1.1489125\). Substituting,
Thus \(E_{H,\mathrm{disp}}=1-0.685861=0.314139\), or \(0.3141\). The three predictions are \(0.2857\) (well-stirred), \(0.3141\) (dispersion), and \(0.3297\) (parallel-tube). If the flow is \(75.0\ \mathrm{L/h}\), their corresponding clearances are, respectively, \(21.43\), \(23.56\), and \(24.73\ \mathrm{L/h}\), calculated as \(Q E_H\). The dispersion result lies between the other two for these specified inputs; that observed ranking should not be substituted for calculation at different inputs.
Worked Example: From Extraction to Outlet Concentration and Clearance
Hypothetical inputs: Let \(Q=60.0\ \mathrm{L/h}\), \(CL_{\mathrm{int}}=60.0\ \mathrm{L/h}\), and \(D_N=0.250\). The inlet concentration is \(C_{\mathrm{in}}=4.00\ \mathrm{mg/L}\). Find the extraction ratios, and illustrate how the dispersion prediction determines outlet concentration and clearance.
The common input is \(x=60.0/60.0=1.00\). The well-stirred extraction is \(1/(1+1)=0.5000\); the parallel-tube extraction is \(1-\exp(-1)=0.6321206\), or \(0.6321\).
For dispersion, \(z=\sqrt{1+4(0.250)(1.00)}=\sqrt{2}=1.4142136\). The outlet fraction from the dispersion equation is approximately \(0.423923\), giving \(E_{H,\mathrm{disp}}=1-0.423923=0.576077\), or \(0.5761\).
The dispersion model therefore predicts \(C_{\mathrm{out}}=(4.00\ \mathrm{mg/L})(0.423923)=1.69569\ \mathrm{mg/L}\), or \(1.696\ \mathrm{mg/L}\). Its clearance is \(CL_{H,\mathrm{disp}}=(60.0\ \mathrm{L/h})(0.576077)=34.5646\ \mathrm{L/h}\), or \(34.56\ \mathrm{L/h}\). As a check, the mass removed per unit time is \(Q(C_{\mathrm{in}}-C_{\mathrm{out}})\approx(60.0)(4.00-1.69569)=138.26\ \mathrm{mg/h}\). Dividing by \(C_{\mathrm{in}}=4.00\ \mathrm{mg/L}\) gives approximately \(34.56\ \mathrm{L/h}\), matching the clearance calculation.
Assumptions, Checks, and Common Pitfalls
The comparison is meaningful when each model receives the same \(Q\) and \(CL_{\mathrm{int}}\), both expressed on a consistent basis. Each model’s assumptions must also be appropriate to the quantity being compared: the well-stirred representation assumes uniform hepatic mixing, the parallel-tube representation treats blood as passing through tubes with longitudinal concentration change, and the dispersion representation includes axial mixing. The dispersion prediction additionally depends on \(D_N\). A numerical agreement among models does not show that their internal assumptions are equivalent.
Several checks catch common errors. The input \(x\) must be dimensionless; a unit mismatch between \(CL_{\mathrm{int}}\) and \(Q\) corrupts every model prediction. For elimination without net production, the outlet fraction should be between zero and one, and so should the extraction ratio. The two must add to one: \(E_H+C_{\mathrm{out}}/C_{\mathrm{in}}=1\). Finally, \(CL_H=QE_H\) must have units of volume per time. Retain sufficient precision through the calculations and round only the reported results; premature rounding can alter a comparison when predictions are close.
A frequent mistake is to compare a well-stirred clearance directly with a parallel-tube extraction ratio and call one larger. These quantities have different units. First compare \(E_H\) with \(E_H\); if clearance is the desired scale, convert every extraction using the same \(Q\). Another error is to assume that \(C_{\mathrm{out}}/C_{\mathrm{in}}\) is the extraction ratio. It is the fraction that remains; extraction is one minus that fraction.
| Model | Prediction to calculate | Additional input | Conversion to clearance |
|---|---|---|---|
| Well-stirred | \(E_H=x/(1+x)\) | None beyond \(x\) | \(CL_H=QE_H\) |
| Parallel-tube | \(E_H=1-\exp(-x)\) | None beyond \(x\) | \(CL_H=QE_H\) |
| Dispersion | Compute the outlet fraction, then \(E_H=1-C_{\mathrm{out}}/C_{\mathrm{in}}\) | \(D_N\) | \(CL_H=QE_H\) |
The main alternatives are to compare clearances or outlet concentrations instead. Clearance is often useful for pharmacokinetic interpretation, but it combines the model’s extraction prediction with flow. Outlet fraction is useful when concentration remaining after a passage is the quantity of interest. Neither removes the need to align inputs and units. Extraction ratio makes the fractional removal explicit; clearance and outlet fraction can then be recovered directly.
A final limitation is that matched parameter values do not guarantee matched real-world predictions or identify the best model. The extraction ratio summarizes a single-pass fractional outcome; by itself it does not reveal a spatial concentration profile or establish a unique mechanism. Use the common scale to make predictions comparable, then interpret differences in light of the model assumptions and available evidence.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- If \(C_{\mathrm{in}}=5.0\ \mathrm{mg/L}\) and \(C_{\mathrm{out}}=3.5\ \mathrm{mg/L}\), what is \(E_H\)?
- For \(Q=50\ \mathrm{L/h}\) and \(CL_{\mathrm{int}}=50\ \mathrm{L/h}\), calculate the well-stirred and parallel-tube extraction ratios.
- A dispersion calculation gives \(C_{\mathrm{out}}/C_{\mathrm{in}}=0.40\). What is extraction? If \(Q=80\ \mathrm{L/h}\), what clearance corresponds to that extraction?
- Why is it incorrect to compare a clearance in \(\mathrm{L/h}\) directly with an extraction ratio?
- If a calculated outlet fraction is \(1.08\) for an elimination-only liver model, name a check to perform before interpreting the result.
Solutions. (1) \(E_H=(5.0-3.5)/5.0=0.30\), meaning 30% extraction. (2) \(x=50/50=1\); well-stirred \(E_H=1/(1+1)=0.500\), while parallel-tube \(E_H=1-\exp(-1)=0.6321\). (3) \(E_H=1-0.40=0.60\), and \(CL_H=(80\ \mathrm{L/h})(0.60)=48\ \mathrm{L/h}\). (4) They have different units: extraction is dimensionless, while clearance is volume per time; convert with \(CL_H=QE_H\). (5) Check the inputs, formula, units, and arithmetic. An outlet fraction above one implies negative extraction, inconsistent with ordinary elimination without net production under the stated model.