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Renal and Hepatic Impairment · Tutorial 617 of 799

Incorporating Hemodialysis into Impaired-Renal PBPK

Build a time-aware PBPK representation of hemodialysis and quantify how treatment schedules and dialyzer extraction alter drug concentrations.

Advanced 12 min read

What You'll Learn

  • Define dialyzer extraction and convert it to dialysis clearance using blood or plasma flow.
  • Add a scheduled dialysis sink to residual renal and nonrenal elimination without double counting.
  • Predict concentration changes during and between sessions under a stated one-compartment approximation.
  • Distinguish dialyzer removal from total body clearance and identify when redistribution matters.
  • Check units, extraction measurements, treatment timing, and model assumptions.

Why dialysis must be represented as a timed process

In renal impairment, a drug may still be eliminated by residual renal function and by nonrenal pathways. Hemodialysis adds another removal pathway, but only while treatment is operating. A model that replaces renal clearance with a single “dialysis clearance” can therefore miss both continuing elimination between sessions and the sharp change in elimination during treatment. The key modeling task is to represent dialysis as a scheduled, concentration-dependent sink that operates alongside the patient’s other clearance pathways.

The formulation below uses dialyzer inlet and outlet concentrations and the treatment blood or plasma flow to estimate removal. It then inserts that removal into a PBPK model at the appropriate time. The calculations are simplest when concentrations at the dialyzer inlet correspond to the concentration in a well-mixed, accessible central compartment. If tissue distribution or blood-flow limitations make that approximation poor, dialysis removal should instead be linked to a modeled vascular compartment and the model should represent redistribution.

Definition: For paired dialyzer inlet and outlet concentrations \(C_{\mathrm{in}}\) and \(C_{\mathrm{out}}\), the extraction ratio is \(E=(C_{\mathrm{in}}-C_{\mathrm{out}})/C_{\mathrm{in}}\), defined when \(C_{\mathrm{in}}>0\). Under the usual conditions \(C_{\mathrm{in}}\geq C_{\mathrm{out}}\geq0\), \(E\) lies between 0 and 1. With dialyzer flow \(Q\), the corresponding dialysis clearance is \(CL_{\mathrm{HD}}=QE\), provided flow and concentrations use the same blood or plasma basis.

For example, if whole-blood concentrations are used, \(Q\) must be whole-blood flow through the dialyzer. If plasma concentrations are used, use plasma flow, or convert consistently. Mixing plasma concentration with whole-blood flow without a justified conversion can produce a clearance with the wrong scale. In practice, inlet and outlet samples should be paired closely in time and collected under stable treatment conditions; otherwise the ratio may reflect changing concentrations or sampling variation as well as extraction.

Adding the dialysis sink to a PBPK model

Let \(I_{\mathrm{HD}}(t)\) equal 1 during a dialysis session and 0 otherwise. In a PBPK model, dialysis is represented as removal from the modeled compartment supplying the dialyzer. If that compartment has concentration \(C_{\mathrm{access}}(t)\), the removal rate is \(Q(t)E(t)C_{\mathrm{access}}(t)\). The schedule, flow, and extraction can be time-varying. For a multi-compartment model, this sink belongs in the amount balance for the relevant vascular or central compartment, while intercompartmental exchange and other elimination processes remain in their own equations.

A useful simplified case assumes a single well-mixed accessible compartment with amount \(A(t)\), volume \(V\), and concentration \(C(t)=A(t)/V\). Suppose residual renal clearance \(CL_{\mathrm{renal,res}}\) and nonrenal clearance \(CL_{\mathrm{nonrenal}}\) are linear with respect to this same concentration. With no input during the interval, the amount balance is:

$$ \frac{dA}{dt} = -\left[ CL_{\mathrm{renal,res}} + CL_{\mathrm{nonrenal}} + I_{\mathrm{HD}}(t)Q(t)E(t) \right]C(t). $$

This formulation assumes that the clearances are expressed on a common concentration basis, the parameters are constant within each modeled time segment, and dialysis does not itself change distribution volume or binding during that segment. If renal or nonrenal elimination is nonlinear, or depends on unbound rather than total concentration, use the appropriate rate law instead of treating it as a constant clearance multiplied by total concentration. Likewise, do not include a renal clearance that has already been removed or reduced in the impaired-renal parameterization.

Modeling rule: During dialysis, add \(Q E\) to the clearance acting on the dialyzer-accessible concentration; outside dialysis, set this dialysis term to zero. Retain residual renal and nonrenal elimination throughout unless the model specifically represents a time-dependent change in either pathway.

Deriving the concentration change over a session

For the one-compartment case, suppose the input is zero and all clearances, flow, and extraction are constant during a treatment interval of length \(T\). Since \(A=VC\), the amount equation becomes a first-order differential equation in concentration. Dividing by \(V\) gives the concentration decline rate:

$$ \frac{dC}{dt} = -\frac{CL_{\mathrm{renal,res}}+CL_{\mathrm{nonrenal}}+QE}{V}C. $$

Separating variables and integrating from the start concentration \(C_0\) to the concentration \(C(T)\) yields \(\ln[C(T)/C_0]=-(CL_{\mathrm{renal,res}}+CL_{\mathrm{nonrenal}}+QE)T/V\). Exponentiating gives:

$$ C(T)=C_0\exp\left[ -\frac{CL_{\mathrm{renal,res}}+CL_{\mathrm{nonrenal}}+QE}{V}T \right]. $$

The same expression applies between sessions with \(QE=0\). For multiple sessions or changing inputs, apply the balance piecewise at the actual session start and stop times, or integrate the full PBPK differential equations numerically. A concentration measured after treatment may not be the end-of-session concentration: tissue-to-plasma redistribution can cause rebound, which a single well-mixed compartment cannot represent.

Worked Example: Calculating Extraction and Dialysis Clearance

Hypothetical data. During a stable session, whole-blood flow through a dialyzer is \(300\ \mathrm{mL/min}\). Paired, simultaneous whole-blood concentrations are \(C_{\mathrm{in}}=8.0\ \mathrm{mg/L}\) and \(C_{\mathrm{out}}=2.0\ \mathrm{mg/L}\). Assume the samples represent the same blood matrix and that the inlet concentration is positive.

Convert flow to liters per hour: \(300\ \mathrm{mL/min}\times60\ \mathrm{min/h}\div1000\ \mathrm{mL/L}=18.0\ \mathrm{L/h}\). The extraction ratio is:

$$ E=\frac{8.0-2.0}{8.0}=0.750. $$

Thus, under these sampling conditions, the dialyzer removes 75.0% of the drug concentration in the blood passing through it. The estimated dialysis clearance is:

$$ CL_{\mathrm{HD}}=(18.0\ \mathrm{L/h})(0.750)=13.5\ \mathrm{L/h}. $$

The instantaneous removal rate at this inlet concentration is \(Q(C_{\mathrm{in}}-C_{\mathrm{out}})=18.0\ \mathrm{L/h}\times6.0\ \mathrm{mg/L}=108\ \mathrm{mg/h}\). This rate is not necessarily constant over the session, because inlet concentration and extraction may change. The calculation estimates dialyzer clearance; it does not by itself give the fraction of the body’s total drug amount removed.

Worked concentration predictions with treatment timing

Worked Example: A Four-Hour Session and the Subsequent Off-Dialysis Interval

Hypothetical model. A one-compartment accessible pool has \(V=20.0\ \mathrm{L}\) and concentration \(10.0\ \mathrm{mg/L}\) at session start. Residual renal clearance is \(0.40\ \mathrm{L/h}\), nonrenal clearance is \(0.60\ \mathrm{L/h}\), and dialysis runs for 4.00 h at \(250\ \mathrm{mL/min}\) with constant extraction \(E=0.600\). Assume no input, linear clearances, and no redistribution. The concentration used in the model is on the same basis as the flow-derived dialysis clearance.

Convert the flow: \(250\ \mathrm{mL/min}\times60/1000=15.0\ \mathrm{L/h}\). Then \(CL_{\mathrm{HD}}=15.0\times0.600=9.00\ \mathrm{L/h}\), while residual renal plus nonrenal clearance is \(0.40+0.60=1.00\ \mathrm{L/h}\). The during-session total is therefore \(10.0\ \mathrm{L/h}\). The session-end concentration is:

$$ C(4.00)=10.0\exp\left[-\frac{(10.0\ \mathrm{L/h})(4.00\ \mathrm{h})}{20.0\ \mathrm{L}}\right] =10.0e^{-2}=1.353\ \mathrm{mg/L}. $$

For the next 8.00 h, dialysis is off, so only \(1.00\ \mathrm{L/h}\) clearance remains. The concentration after that interval is:

$$ C(12.0)=1.353\exp\left[-\frac{(1.00\ \mathrm{L/h})(8.00\ \mathrm{h})}{20.0\ \mathrm{L}}\right] =0.907\ \mathrm{mg/L}. $$

The timing matters: using the during-session total clearance for all 12 hours would incorrectly treat dialysis as continuous. This example also shows why a predialysis concentration and an immediate postdialysis concentration need not describe the same phase of the concentration-time profile.

Worked Example: Comparing a Dose Given Before or After a Session

Hypothetical model. Under the same linear one-compartment assumptions, a dose produces an immediate \(5.00\ \mathrm{mg/L}\) increase in the accessible concentration. During a scheduled 4.00 h session, \(V=20.0\ \mathrm{L}\), residual plus nonrenal clearance is \(1.00\ \mathrm{L/h}\), and \(CL_{\mathrm{HD}}=9.00\ \mathrm{L/h}\). Compare the dose-related concentration contribution at session end if the dose is given immediately before treatment versus immediately after treatment. This comparison isolates the dose contribution; any pre-existing concentration follows the same elimination model.

For a dose given just before dialysis, the added concentration is reduced by the during-session total clearance:

$$ \Delta C_{\mathrm{before,end}} =5.00\exp\left[-\frac{(1.00+9.00)(4.00)}{20.0}\right] =5.00e^{-2} =0.677\ \mathrm{mg/L}. $$

For a dose given just after treatment, there is no dialysis removal of that dose during the session, so its immediate post-dose contribution is \(5.00\ \mathrm{mg/L}\). The difference at that comparison time is \(5.00-0.677=4.323\ \mathrm{mg/L}\). This does not establish a universally preferred dosing time: the appropriate timing depends on the drug, target exposure, treatment schedule, and other clinical considerations. The calculation simply quantifies the consequence of the stated model and schedule.

Assumptions, diagnostics, and alternatives

A useful dialysis representation requires a credible session schedule, compatible flow and concentration measurements, and an extraction estimate that is physically and analytically plausible. For nonnegative concentrations, require \(C_{\mathrm{in}}>0\) and \(C_{\mathrm{out}}\leq C_{\mathrm{in}}\) before interpreting the usual extraction ratio as lying between 0 and 1. If \(C_{\mathrm{in}}=C_{\mathrm{out}}=0\), the ratio is undefined, not zero. A negative estimate or one above 1 calls for review of assay error, sampling, units, sample timing, and the extraction definition; it should not be silently clipped to the interval from 0 to 1.

Compare predicted and observed concentrations at times that identify the phases of interest: before dialysis, near session end, and after a defined post-treatment interval. A systematic mismatch that grows during treatment may indicate a poor extraction estimate, changing flow, or an incorrectly specified dialyzer-accessible concentration. A good end-session fit followed by a poor post-session fit may point to distributional rebound or an incorrect off-dialysis elimination rate. These patterns are diagnostic clues, not proof of a single mechanism.

A time-averaged dialysis clearance can sometimes approximate average exposure over a repeated schedule, but it loses the concentration changes associated with treatment starts and stops. It is most defensible when the question concerns a sufficiently long-term average and the schedule is regular; it is unsuitable for predicting immediate postdialysis concentrations or the effect of a dose close to treatment. A time-varying clearance in a one-compartment model preserves the schedule but still assumes instant mixing. A multi-compartment PBPK approach is more appropriate when tissue distribution, blood-to-plasma differences, or rebound materially affect concentrations. Directly representing dialyzer inlet and outlet compartments can be useful when flow and extraction vary substantially or when dialyzer behavior is the modeling focus.

Common pitfall: “Extraction is between 0 and 1 whenever inlet and outlet concentrations are nonnegative and outlet is no higher than inlet.” This reasoning overlooks the case \(C_{\mathrm{in}}=0\), where the extraction formula divides by zero. Require \(C_{\mathrm{in}}>0\); if both concentrations are zero, extraction is undefined. A second frequent error is treating \(QE\) as a clearance that operates all day. It is a dialysis clearance only during the modeled treatment schedule.

Dialysis removal should be interpreted as one pathway in the amount balance, not as a replacement for residual renal or nonrenal elimination. The principal limitations of the simple calculation are its dependence on representative extraction measurements and its assumptions about mixing, linearity, and concentration basis. The schedule-aware formulation makes those assumptions explicit and provides a starting point for adding dialysis to a more detailed PBPK model.

Check Your Understanding

Answer each question using the stated assumptions. Brief solutions follow so the calculations can be checked.

  1. With \(C_{\mathrm{in}}=5.0\ \mathrm{mg/L}\), \(C_{\mathrm{out}}=1.5\ \mathrm{mg/L}\), and flow \(12.0\ \mathrm{L/h}\), calculate \(E\) and \(CL_{\mathrm{HD}}\).
  2. What is the extraction ratio when both inlet and outlet concentrations are zero? Explain why.
  3. A model has residual renal clearance \(0.30\ \mathrm{L/h}\), nonrenal clearance \(0.70\ \mathrm{L/h}\), and \(CL_{\mathrm{HD}}=8.0\ \mathrm{L/h}\) during treatment. What clearance applies outside dialysis under the stated model?
  4. For \(V=10.0\ \mathrm{L}\), \(C_0=4.0\ \mathrm{mg/L}\), and a 2.00 h session with total during-session clearance \(5.0\ \mathrm{L/h}\), calculate the session-end concentration.

Solutions. 1. \(E=(5.0-1.5)/5.0=0.700\), so \(CL_{\mathrm{HD}}=(12.0)(0.700)=8.40\ \mathrm{L/h}\). 2. It is undefined because \(C_{\mathrm{in}}=0\) makes the denominator zero; it is not valid to report zero extraction from this ratio. 3. Outside dialysis, \(CL_{\mathrm{HD}}=0\), leaving \(0.30+0.70=1.00\ \mathrm{L/h}\). 4. \(C(2.00)=4.0\exp[-(5.0)(2.00)/10.0]=4.0e^{-1}=1.47\ \mathrm{mg/L}\), to three significant figures.