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One-, Two- and Three-Compartment Structural Models · Tutorial 28 of 797

Mapping Two-Compartment Microconstants to Macroparameters

Derive the mapping from two-compartment IV-bolus microconstants to biexponential exponents and amplitudes, and check the mapping with numerical examples.

Beginner 10 min read

What You'll Learn

  • Derive the characteristic equation connecting the rate constants to alpha and beta.
  • Calculate the biexponential exponents from k10, k12, and k21.
  • Derive the IV-bolus amplitudes A and B using the initial concentration and slope.
  • Check amplitude, ordering, and unit relationships implied by the model.
  • Recognize boundary cases and common errors in mapping rates to macroparameters.

From Compartment Rates to a Concentration Curve

A two-compartment IV-bolus model describes drug moving between a central compartment, where concentration is measured, and a peripheral compartment. Its rate constants describe elimination and exchange pathways. A measured concentration-time curve, by contrast, is often expressed using two exponential terms. The mapping between these descriptions matters because the exponential rates and amplitudes are not individual compartment rate constants: each is determined by the coupled system.

Here the dose \(D\) enters the central compartment instantaneously at time zero. The central and peripheral amounts are \(X_c(t)\) and \(X_p(t)\), and the measured concentration is \(C(t)=X_c(t)/V_c\), where \(V_c\) is the central volume. The rate constants \(k_{10}\), \(k_{12}\), and \(k_{21}\) have units of inverse time: elimination from the central compartment, transfer from central to peripheral, and transfer from peripheral to central, respectively.

Model and assumptions: The model is linear and time-invariant, with constant, nonnegative rate constants; drug is eliminated only from the central compartment; concentration is proportional to central amount through a fixed \(V_c\); and an IV bolus gives \(X_c(0)=D\) and \(X_p(0)=0\). For the usual two-phase curve, assume \(k_{10}>0\), \(k_{12}>0\), and \(k_{21}>0\). The amount-balance equations are

\(\frac{dX_c}{dt}=-(k_{10}+k_{12})X_c+k_{21}X_p,\quad \frac{dX_p}{dt}=k_{12}X_c-k_{21}X_p.\)

For positive rates, the central concentration can be written as \(C(t)=A e^{-\alpha t}+B e^{-\beta t}\), with \(\alpha>\beta>0\). The parameters \(\alpha\) and \(\beta\) are exponential rates, in inverse-time units; \(A\) and \(B\) are concentration amplitudes, in concentration units. This tutorial derives how all four are determined by the microconstants and the bolus initial conditions.

Deriving the Exponential Rates

The coupled amount equations have a two-by-two system matrix. An exponential mode with time dependence \(e^{rt}\) is possible when \(r\) satisfies the characteristic equation of that matrix. Equivalently, substituting the matrix into its characteristic determinant gives

$$ \begin{aligned} \det\begin{pmatrix} r+k_{10}+k_{12} & -k_{21}\\ -k_{12} & r+k_{21} \end{pmatrix} &=(r+k_{10}+k_{12})(r+k_{21})-k_{12}k_{21}\\ &=r^2+(k_{10}+k_{12}+k_{21})r+k_{10}k_{21}=0. \end{aligned} $$

The stable modes have \(r=-\alpha\) and \(r=-\beta\). Therefore the corresponding positive exponential rates are the roots of the polynomial obtained by changing the sign of the mode variable:

$$ x^2-(k_{10}+k_{12}+k_{21})x+k_{10}k_{21}=0. $$
Rate mapping: The sum and product of the two macro-exponential rates are \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\). Explicitly,

\(\alpha,\beta=\frac{(k_{10}+k_{12}+k_{21})\pm\sqrt{(k_{10}+k_{12}+k_{21})^2-4k_{10}k_{21}}}{2},\)

where the plus sign gives \(\alpha\), the larger rate, and the minus sign gives \(\beta\), the smaller rate.

The discriminant is positive when all three rates are positive. To see why, write \(S=k_{10}+k_{12}+k_{21}\) and \(P=k_{10}k_{21}\). The discriminant also equals \((k_{10}+k_{12}-k_{21})^2+4k_{12}k_{21}\), which is strictly positive. The two roots are real and distinct. Their sum and product are positive, so both roots are positive.

A further useful ordering follows by evaluating \(f(x)=x^2-Sx+P\) at \(k_{21}\). This gives \(f(k_{21})=-k_{12}k_{21}<0\). A quadratic with positive leading coefficient is negative between its two real roots, so \(\beta<k_{21}<\alpha\). This ordering will also show that both amplitudes are positive for a positive bolus.

Worked Example: Calculating the Two Exponential Rates

Suppose a hypothetical model has \(k_{10}=0.30\ \text{min}^{-1}\), \(k_{12}=0.80\ \text{min}^{-1}\), and \(k_{21}=0.40\ \text{min}^{-1}\). Find \(\alpha\) and \(\beta\).

First calculate the sum and product: \(S=0.30+0.80+0.40=1.50\ \text{min}^{-1}\), and \(P=(0.30)(0.40)=0.120\ \text{min}^{-2}\). Thus the discriminant is \(S^2-4P=(1.50)^2-4(0.120)=1.77\ \text{min}^{-2}\), with square root \(1.3304\ \text{min}^{-1}\).

The larger rate is \(\alpha=(1.50+1.3304)/2=1.4152\ \text{min}^{-1}\); the smaller is \(\beta=(1.50-1.3304)/2=0.08479\ \text{min}^{-1}\). As a check, their sum is \(1.5000\ \text{min}^{-1}\) and their product is approximately \(0.1200\ \text{min}^{-2}\), matching the rate-constant sum and product.

Deriving the IV-Bolus Amplitudes

The exponential rates describe the system's modes, but the initial conditions determine how strongly each mode contributes to central concentration. Let \(C_0=D/V_c\), the concentration immediately after the bolus. At time zero, the macroparameter form gives \(A+B=C_0\). From the central amount equation and \(X_p(0)=0\), the initial concentration slope is \(C'(0)=-(k_{10}+k_{12})C_0\). Differentiating the biexponential form gives \(C'(0)=-\alpha A-\beta B\). Consequently,

$$ A+B=C_0,\qquad \alpha A+\beta B=(k_{10}+k_{12})C_0. $$

Solving these two equations gives \(A=C_0(k_{10}+k_{12}-\beta)/(\alpha-\beta)\) and \(B=C_0(\alpha-k_{10}-k_{12})/(\alpha-\beta)\). Using \(\alpha+\beta=k_{10}+k_{12}+k_{21}\), these expressions can be written directly in terms of \(k_{21}\):

$$ A=\frac{D}{V_c}\frac{\alpha-k_{21}}{\alpha-\beta}, \qquad B=\frac{D}{V_c}\frac{k_{21}-\beta}{\alpha-\beta}. $$

This is the amplitude mapping for the specified central IV bolus. The amplitude attached to the faster exponential is \(A\), and the amplitude attached to the slower exponential is \(B\). Since \(\beta<k_{21}<\alpha\), both fractions are positive. They sum to one, so \(A+B=D/V_c\), as required by the initial concentration.

Worked Example: Calculating Rates and Amplitudes

Use the hypothetical rate constants from the first example, and suppose \(D=100\ \text{mg}\) and \(V_c=20\ \text{L}\). Calculate the full biexponential concentration curve.

The initial concentration is \(C_0=D/V_c=100/20=5.00\ \text{mg/L}\). The rates are \(\alpha=1.4152\ \text{min}^{-1}\) and \(\beta=0.08479\ \text{min}^{-1}\), so \(\alpha-\beta=1.3304\ \text{min}^{-1}\).

The fast amplitude is \(A=5.00(1.4152-0.4000)/1.3304=3.815\ \text{mg/L}\). The slow amplitude is \(B=5.00(0.4000-0.08479)/1.3304=1.185\ \text{mg/L}\). Rounding explains the last digit; the amplitudes sum to \(5.000\ \text{mg/L}\).

Thus \(C(t)=3.815e^{-1.4152t}+1.185e^{-0.08479t}\ \text{mg/L}\), where \(t\) is in minutes. The initial slope provides another check: the rate equations give \(-(k_{10}+k_{12})C_0=-(1.10)(5.00)=-5.50\ \text{mg/L/min}\). The macroparameter expression gives \(-\alpha A-\beta B\), approximately \(-5.50\ \text{mg/L/min}\) as well.

Worked Example: A Symmetric Amplitude Case

For a separate hypothetical model, let \(k_{10}=0.20\ \text{h}^{-1}\), \(k_{12}=0.30\ \text{h}^{-1}\), \(k_{21}=0.50\ \text{h}^{-1}\), and let \(D/V_c=20.0\ \text{mg/L}\). Find the rates and amplitudes.

Here \(S=1.00\ \text{h}^{-1}\) and \(P=0.100\ \text{h}^{-2}\). The square root of the discriminant is \(\sqrt{1.00^2-4(0.100)}=\sqrt{0.600}=0.77460\ \text{h}^{-1}\). Therefore \(\alpha=(1.00+0.77460)/2=0.88730\ \text{h}^{-1}\) and \(\beta=(1.00-0.77460)/2=0.11270\ \text{h}^{-1}\).

Then \(A=20.0(0.88730-0.5000)/(0.88730-0.11270)=10.0\ \text{mg/L}\), and \(B=20.0(0.5000-0.11270)/(0.88730-0.11270)=10.0\ \text{mg/L}\). Equal amplitudes occur here because \(k_{10}+k_{12}=k_{21}\), which makes the initial slope correspond to the average of the two exponential rates. This equality is a feature of these chosen rates, not a general two-compartment property.

Checks, Alternatives, and Limitations

There are two complementary ways to organize the mapping. The characteristic-polynomial approach gives the exponential rates through their sum and product, then the initial concentration and slope give the amplitudes. Alternatively, one can solve the coupled differential equations directly by representing the solution as a sum of exponential modes. Both approaches give the same mapping; the polynomial-and-slope route makes the roles of the rates and bolus initial conditions especially transparent.

Several checks help identify arithmetic or modeling errors:

  • Ordering: with positive exchange and elimination rates, \(\alpha>k_{21}>\beta>0\). A result that violates this ordering suggests a calculation or labeling error.
  • Sum and product: verify \(\alpha+\beta=k_{10}+k_{12}+k_{21}\) and \(\alpha\beta=k_{10}k_{21}\), allowing for rounding.
  • Initial concentration: verify \(A+B=D/V_c\); both sides must have concentration units.
  • Initial slope: verify \(\alpha A+\beta B=(k_{10}+k_{12})D/V_c\); both sides must have concentration-per-time units.
  • Units: \(k_{10}, k_{12}, k_{21}, \alpha,\beta\) must use the same inverse-time unit. The amplitudes use the same units as \(D/V_c\).

The mapping also depends on the stated structural assumptions. A peripheral elimination pathway, time-varying rates, nonlinear processes, or a non-bolus input changes the equations or initial conditions, and the formulas for this bolus model need not apply unchanged. With no exchange (\(k_{12}=0\)), one mode may have zero amplitude: the mathematical system has two rates, but a central bolus does not necessarily excite both modes. For example, if \(k_{10}=0.25\ \text{h}^{-1}\), \(k_{21}=0.60\ \text{h}^{-1}\), and \(k_{12}=0\), the roots are \(0.60\) and \(0.25\ \text{h}^{-1}\), but the amplitude for the \(0.60\ \text{h}^{-1}\) mode is zero. The central concentration is just \(C_0e^{-0.25t}\).

A common mistake is to identify \(\alpha\) with \(k_{10}+k_{12}\), or \(\beta\) with \(k_{21}\). Neither equality generally holds: the observed rates are the roots of a coupled-system polynomial. Another mistake is to calculate the roots correctly but assign the larger rate to \(B\). By convention here, \(\alpha\) is the larger, faster rate and has amplitude \(A\); \(\beta\) is the smaller, slower rate and has amplitude \(B\). The formulas map a specified microparameter set to a curve; they do not imply that the macroparameters themselves are individual physiologic rate constants.

Check Your Understanding

Try each question, then compare your work with the solutions below.

  1. For the positive-rate model, state the sum and product of \(\alpha\) and \(\beta\) in terms of the microconstants.
  2. Suppose \(k_{10}=0.10\ \text{min}^{-1}\), \(k_{12}=0.20\ \text{min}^{-1}\), and \(k_{21}=0.30\ \text{min}^{-1}\). Calculate \(\alpha\) and \(\beta\).
  3. For the rates in question 2 and \(D/V_c=8.00\ \text{mg/L}\), calculate \(A\) and \(B\), and check their sum.
  4. Why is a negative amplitude inconsistent with the usual positive-rate two-compartment bolus model?

Solutions. (1) \(\alpha+\beta=k_{10}+k_{12}+k_{21}\), and \(\alpha\beta=k_{10}k_{21}\). (2) The sum is \(0.60\ \text{min}^{-1}\), the product is \(0.0300\ \text{min}^{-2}\), and the discriminant is \(0.60^2-4(0.0300)=0.240\ \text{min}^{-2}\). Thus \(\alpha=(0.60+\sqrt{0.240})/2=0.5449\ \text{min}^{-1}\) and \(\beta=(0.60-\sqrt{0.240})/2=0.05505\ \text{min}^{-1}\). (3) \(A=8.00(0.5449-0.3000)/(0.5449-0.05505)=4.000\ \text{mg/L}\), and \(B=8.00(0.3000-0.05505)/(0.5449-0.05505)=4.000\ \text{mg/L}\); their sum is \(8.00\ \text{mg/L}=D/V_c\). (4) For positive rates, \(\beta<k_{21}<\alpha\), so both amplitude fractions are positive. A negative amplitude therefore signals an error in calculation or labeling, or that the fitted curve does not satisfy the assumptions of this model.