Tutorials › Pharmacokinetic-Pharmacodynamic (PK/PD) Modeling › Mass-Balance Derivation of Biomarker Turnover Equations

Home  /  Tutorials  /  Turnover and Precursor-Pool Models

Turnover and Precursor-Pool Models · Tutorial 150 of 598

Mass-Balance Derivation of Biomarker Turnover Equations

Learn to translate biomarker synthesis and loss assumptions into mass-balance equations for amount and concentration, and verify the equations by checking units and fluxes.

Advanced 12 min read

What You'll Learn

  • Write an amount balance from synthesis and degradation fluxes.
  • Derive the concentration equation using the relationship between amount, concentration, and volume.
  • Distinguish a first-order loss rate from a loss flux per unit volume.
  • Account for dilution or concentration caused by changing volume.
  • Check turnover equations by their units, signs, and conservation assumptions.

Why Start With a Mass Balance?

A biomarker concentration can rise because more biomarker is produced, because less is removed, or because its distribution volume changes. A turnover equation should distinguish these mechanisms rather than treating concentration change as an unexplained response. The starting point is a balance on the amount of biomarker in the modeled pool: the rate of change in amount equals total input flux minus total output flux.

The balance itself does not require a particular drug-effect model or a first-order loss process. Those are additional assumptions about the fluxes. Writing the balance first makes those assumptions visible and provides a direct check on units and conservation.

Definition: Let \(A(t)\) be the amount of biomarker in the modeled pool, \(V(t)\) its effective volume, and \(C(t)=A(t)/V(t)\) its concentration. Let \(S(t)\) denote the total synthesis or input rate, in amount per time, and let \(L(A,t)\) denote the total loss rate, also in amount per time. The mass-balance equation is $$ \frac{dA}{dt}=S(t)-L(A,t). $$

An amount might be measured in milligrams (mg), a concentration in mg/L, and time in hours (h). In that case, \(S\) and \(L\) must have units mg/h, while \(dA/dt\) has units mg/h. The pool boundary matters: synthesis and loss must refer to material crossing the boundary of the same modeled pool. If a measured biomarker has multiple sources or routes of removal, those fluxes must either be included or explicitly treated as negligible for the model's intended use.

From Amount to Concentration

A common turnover assumption is first-order degradation: the loss rate is proportional to the amount present. Write the proportionality constant as \(k_{\mathrm{deg}}\), with units of inverse time. Then

$$ L(A,t)=k_{\mathrm{deg}}A,\qquad \frac{dA}{dt}=S(t)-k_{\mathrm{deg}}A. $$

This form assumes that degradation is linear in amount over the range being modeled and that \(k_{\mathrm{deg}}\) is the applicable loss-rate constant. It does not assert that all real biomarker removal is first order; it specifies a model that can be checked against biological knowledge and data.

To obtain the concentration equation, use \(A=VC\). If volume can vary over time, differentiating this product gives

$$ \frac{dA}{dt} =\frac{d(VC)}{dt} =V\frac{dC}{dt}+C\frac{dV}{dt}. $$

Substitute the amount balance and solve for \(dC/dt\). Under first-order loss, the result is

$$ V\frac{dC}{dt}+C\frac{dV}{dt}=S(t)-k_{\mathrm{deg}}VC, \qquad \frac{dC}{dt}=\frac{S(t)}{V}-k_{\mathrm{deg}}C-\frac{C}{V}\frac{dV}{dt}. $$

If volume is constant, the last term is zero. If synthesis is instead specified as a rate \(s(t)\) per unit volume, with units amount/(volume × time), then the total synthesis rate is \(S(t)=V s(t)\). The constant-volume concentration equation becomes \(dC/dt=s(t)-k_{\mathrm{deg}}C\). These two ways of specifying synthesis are not interchangeable unless the volume factor is handled consistently.

Key result: With a well-mixed pool, \(A=VC\), total synthesis \(S\), and first-order loss \(k_{\mathrm{deg}}A\), the amount equation is \(dA/dt=S-k_{\mathrm{deg}}A\). The concentration equation additionally contains the volume-change term \(-(C/V)(dV/dt)\). When volume is constant, synthesis per unit volume is \(S/V\), and \(dC/dt=S/V-k_{\mathrm{deg}}C\).

Worked Examples: Following the Fluxes

Worked Example: Calculating the Instantaneous Amount Change

Suppose a hypothetical biomarker pool contains \(A=12.0\) mg. At this time, synthesis is \(S=3.00\) mg/h and degradation is first order with \(k_{\mathrm{deg}}=0.100\) h\(^{-1}\). Find the amount rate of change.

First calculate the degradation flux:

$$ L=k_{\mathrm{deg}}A =(0.100\ \mathrm{h}^{-1})(12.0\ \mathrm{mg}) =1.20\ \mathrm{mg/h}. $$

The mass balance then gives

$$ \frac{dA}{dt}=S-L =3.00\ \mathrm{mg/h}-1.20\ \mathrm{mg/h} =1.80\ \mathrm{mg/h}. $$

At the stated time, the amount is increasing at 1.80 mg/h. The units agree because both fluxes, and therefore their difference, are amounts per time. This is an instantaneous rate, not a claim that the rate remains constant over a later interval; as \(A\) changes, the first-order loss flux changes too.

Worked Example: Converting an Amount Balance to Concentration

Consider a hypothetical biomarker pool with constant volume \(V=4.00\) L, concentration \(C=5.00\) mg/L, synthesis \(S=2.00\) mg/h, and \(k_{\mathrm{deg}}=0.0800\) h\(^{-1}\). Calculate \(dC/dt\) using both the amount and concentration equations.

The amount in the pool is

$$ A=VC=(4.00\ \mathrm{L})(5.00\ \mathrm{mg/L})=20.0\ \mathrm{mg}. $$

Thus the loss flux is \(k_{\mathrm{deg}}A=(0.0800\ \mathrm{h}^{-1})(20.0\ \mathrm{mg})=1.60\) mg/h, and the amount rate is

$$ \frac{dA}{dt}=2.00\ \mathrm{mg/h}-1.60\ \mathrm{mg/h} =0.400\ \mathrm{mg/h}. $$

Because the volume is constant, \(dC/dt=(1/V)(dA/dt)\). Therefore

$$ \frac{dC}{dt} =\frac{0.400\ \mathrm{mg/h}}{4.00\ \mathrm{L}} =0.100\ \mathrm{mg/(L\,h)}. $$

The concentration equation gives the same result:

$$ \frac{dC}{dt} =\frac{S}{V}-k_{\mathrm{deg}}C =\frac{2.00\ \mathrm{mg/h}}{4.00\ \mathrm{L}} -(0.0800\ \mathrm{h}^{-1})(5.00\ \mathrm{mg/L}) =0.500-0.400 =0.100\ \mathrm{mg/(L\,h)}. $$

Agreement between the two calculations checks the volume conversion and the loss term. In particular, the loss flux in the amount equation is 1.60 mg/h, whereas the concentration loss rate is 0.400 mg/(L·h); they describe different quantities.

Worked Example: A Changing Volume Can Reverse the Concentration Trend

Suppose a hypothetical pool has \(V=3.00\) L, \(C=8.00\) mg/L, and an expanding volume with \(dV/dt=0.150\) L/h. Synthesis is \(S=1.80\) mg/h, and first-order degradation has \(k_{\mathrm{deg}}=0.0500\) h\(^{-1}\). Find the instantaneous rates of change in amount and concentration.

The amount is \(A=VC=(3.00\ \mathrm{L})(8.00\ \mathrm{mg/L})=24.0\) mg. The degradation flux is

$$ L=k_{\mathrm{deg}}A =(0.0500\ \mathrm{h}^{-1})(24.0\ \mathrm{mg}) =1.20\ \mathrm{mg/h}. $$

Therefore the amount is increasing:

$$ \frac{dA}{dt}=1.80-1.20=0.600\ \mathrm{mg/h}. $$

For concentration, include all three terms:

$$ \frac{dC}{dt} =\frac{1.80\ \mathrm{mg/h}}{3.00\ \mathrm{L}} -(0.0500\ \mathrm{h}^{-1})(8.00\ \mathrm{mg/L}) -\frac{8.00\ \mathrm{mg/L}}{3.00\ \mathrm{L}}(0.150\ \mathrm{L/h}) $$
$$ \frac{dC}{dt}=0.600-0.400-0.400 =-0.200\ \mathrm{mg/(L\,h)}. $$

The amount is increasing while the concentration is decreasing. The volume expansion dilutes the pool enough to outweigh the positive amount change. This is not a contradiction: concentration is amount divided by volume, and both quantities are changing.

Assumptions, Checks, and Failure Modes

The derivation is a conservation statement only for the chosen pool and its boundary. Its useful application requires several checks:

  • Consistent pool definition. The amount \(A\), volume \(V\), and measured concentration \(C\) must refer to the same effective pool. A plasma concentration cannot automatically be multiplied by an unrelated tissue volume to infer total biomarker amount.
  • Complete flux accounting. Each relevant source and sink must be represented in \(S\) or \(L\). An omitted route of degradation, transport, or production can appear as unexplained model error.
  • Consistent units. Total fluxes in the amount equation have units amount/time. If a rate is defined per unit volume, multiply it by volume before using it as a total flux. In the concentration equation, every term must have units concentration/time.
  • Appropriate loss law. The form \(k_{\mathrm{deg}}A\) assumes loss proportional to amount. If loss is saturable or otherwise nonlinear, retain a general \(L(A,t)\) instead of forcing a first-order term.
  • Appropriate volume treatment. Omitting the volume-change term is justified only when volume is constant (or when its change is negligible for the question being studied). A changing volume affects concentration even if it does not itself create or destroy biomarker amount.

A practical diagnostic is to reconstruct the balance at a given time: calculate the input and output fluxes in the same units, take their difference, and check whether the implied direction and scale of change are plausible. If the model predicts negative amounts, that generally signals an unsuitable rate law, parameterization, or numerical implementation for the stated domain. If concentration residuals track volume changes or fluid shifts, a constant-volume equation may be missing a dilution effect.

An amount formulation is often the clearest way to enforce conservation and combine fluxes with different physical origins. A concentration formulation can be convenient when observations are concentrations and volume is fixed or known. They are equivalent when \(A=VC\) is respected and the volume term is included as needed. Treating \(k_{\mathrm{deg}}C\) as an amount loss without multiplying by volume is a common dimensional error: \(k_{\mathrm{deg}}C\) has units concentration/time, not amount/time.

Common mistake: “The amount is rising, so the concentration must also be rising.” This is false if volume is changing. Since \(C=A/V\), the concentration derivative contains both the amount-change contribution and the volume-change contribution. Use \(dC/dt=(1/V)(dA/dt)-(C/V)(dV/dt)\), not just \((1/V)(dA/dt)\), unless volume is constant.

Check Your Understanding

For each question, use amount and time units consistently. Solutions follow each problem.

  1. A pool contains 10 mg of biomarker, with synthesis 1.5 mg/h and first-order degradation \(k_{\mathrm{deg}}=0.10\) h\(^{-1}\). What is \(dA/dt\)?
    Solution: The loss flux is \((0.10\ \mathrm{h}^{-1})(10\ \mathrm{mg})=1.0\) mg/h. Thus \(dA/dt=1.5-1.0=0.50\) mg/h.
  2. For a constant 2.0 L volume, total synthesis is 0.80 mg/h. What synthesis rate per unit volume does this represent?
    Solution: Divide the total rate by volume: \((0.80\ \mathrm{mg/h})/(2.0\ \mathrm{L})=0.40\) mg/(L·h).
  3. A concentration is 6 mg/L in a 5 L pool. If volume is increasing at 0.20 L/h while amount is momentarily constant, what is \(dC/dt\)?
    Solution: With \(dA/dt=0\), \(dC/dt=-(C/V)(dV/dt)=-(6/5)(0.20)=-0.24\) mg/(L·h).
  4. Why is \(k_{\mathrm{deg}}C\) not, by itself, the total degradation flux in an amount balance?
    Solution: Its units are concentration/time. The total first-order loss flux is \(k_{\mathrm{deg}}A=k_{\mathrm{deg}}VC\), which has units amount/time.

Previous

Next

Back to Tutorials
Clinical Biostats