From Enzyme-Site Inhibition to a Hepatic PBPK Prediction
A competitive inhibitor changes the substrate concentration needed to achieve a given enzyme rate. In a hepatic PBPK model, that change affects the metabolic rate in the liver and, through mass balance, can alter the parent-drug concentration required to maintain a given input rate. The calculation therefore has two parts: evaluate hepatic metabolism at the relevant enzyme-site concentrations, then determine whether the resulting rates balance drug input at steady state.
The calculation here assumes rapid, reversible competitive inhibition and a steady, specified unbound inhibitor concentration at the enzyme site. It does not describe time-dependent enzyme loss or induction. Let \(S\) be unbound substrate concentration at the hepatic enzyme site, \(I_u\) the unbound inhibitor concentration at that site, \(K_i\) the inhibition constant, \(K_m\) the uninhibited Michaelis constant, and \(V_{\max}\) the maximum metabolic rate.
Concentrations in this equation must refer to the enzyme site, and \(I_u\) and \(K_i\) must use the same concentration units. A total plasma inhibitor concentration cannot be substituted for \(I_u\) unless the relationship between plasma and enzyme-site unbound concentration has been established. The earlier material on unbound inhibitor concentrations at the enzyme site addresses that distinction.
Calculating a Rate at Specified Enzyme-Site Concentrations
First calculate \(\alpha\), then use the inhibited rate equation. The uninhibited rate at the same substrate concentration is a useful comparison: it shows the size of the enzyme-rate reduction before considering the rest of the PBPK mass balance.
Worked Example: Rate Reduction at a Fixed Hepatic Substrate Concentration
Suppose a hypothetical hepatic enzyme has \(V_{\max}=80\ \mathrm{mg/h}\) and \(K_m=4\ \mathrm{mg/L}\). The enzyme-site substrate concentration is \(S=4\ \mathrm{mg/L}\), and the unbound inhibitor concentration is \(I_u=3\ \mathrm{mg/L}\), with \(K_i=1.5\ \mathrm{mg/L}\). Assume rapid reversible competitive inhibition and unchanged enzyme abundance.
The inhibition factor is $$ \alpha=1+\frac{I_u}{K_i} =1+\frac{3\ \mathrm{mg/L}}{1.5\ \mathrm{mg/L}}=3. $$
Thus the inhibited hepatic rate is $$ v_{\mathrm{hep}}=\frac{(80\ \mathrm{mg/h})(4\ \mathrm{mg/L})} {(3)(4\ \mathrm{mg/L})+4\ \mathrm{mg/L}} =\frac{320}{16}\ \mathrm{mg/h} =20\ \mathrm{mg/h}. $$ Without inhibitor, the rate at the same \(S\) is $$ v_{\mathrm{hep},0}=\frac{(80)(4)}{4+4}\ \mathrm{mg/h} =40\ \mathrm{mg/h}. $$
In this example the inhibitor halves the hepatic rate at the specified substrate concentration. This is a comparison at fixed \(S\); it does not, by itself, determine the change in steady-state parent concentration, because \(S\) may change when the whole-body mass balance is solved.
Embedding the Rate in a Steady-State Mass Balance
For a transparent steady-state calculation, consider a reduced hepatic PBPK representation. Let \(C_p\) be the parent-drug concentration in plasma, \(R_{\mathrm{in}}\) the constant drug input rate, and \(CL_{\mathrm{other}}\) a linear clearance outside the modeled hepatic metabolic pathway. Assume the enzyme-site substrate concentration is related to plasma concentration by a fixed factor \(a\):
The factor \(a\) summarizes the assumed mapping from plasma concentration to unbound enzyme-site substrate concentration; it is dimensionless when both concentrations use the same units. The amount balance is input equals total elimination at steady state:
This is a reduced steady-state slice of a PBPK model, not a substitute for a dynamic liver compartment when hepatic blood flow, transport, distribution, or changing concentrations must be represented explicitly. Its assumptions are that the stated input and inhibitor concentration are constant, the substrate-enzyme rate is at equilibrium, \(a\) remains fixed, \(V_{\max}\) and \(K_m\) do not otherwise change, and the other clearance is linear in \(C_p\).
Deriving the Concentration Equation
Multiplying the balance by \(\alpha K_m+aC_p\) gives a quadratic in \(C_p\). Showing this step is important: it makes clear how inhibitor concentration changes both the coefficient of \(C_p\) and the constant term.
\(R_{\mathrm{in}}=V_{\max}aC_p/(\alpha K_m+aC_p)+CL_{\mathrm{other}}C_p\).
\(R_{\mathrm{in}}(\alpha K_m+aC_p)=V_{\max}aC_p+CL_{\mathrm{other}}C_p(\alpha K_m+aC_p)\).
\(CL_{\mathrm{other}}aC_p^2+(V_{\max}a+CL_{\mathrm{other}}\alpha K_m-R_{\mathrm{in}}a)C_p-R_{\mathrm{in}}\alpha K_m=0\).
For positive input, positive parameters, and \(CL_{\mathrm{other}}>0\), the quadratic has one positive root and one negative root: the product of the roots is \(-R_{\mathrm{in}}\alpha K_m/(CL_{\mathrm{other}}a)<0\). The physically meaningful steady-state concentration is therefore the positive root. Also, the elimination function increases with \(C_p\): both the competitive Michaelis–Menten rate and the linear clearance are increasing functions of \(C_p\). Thus there is at most one nonnegative concentration that balances a specified positive input.
Worked Example: Inhibitor Effect on Hepatic Rate and Steady-State Concentration
Consider hypothetical values \(R_{\mathrm{in}}=60\ \mathrm{mg/h}\), \(V_{\max}=100\ \mathrm{mg/h}\), \(K_m=8\ \mathrm{mg/L}\), \(a=0.5\), and \(CL_{\mathrm{other}}=2\ \mathrm{L/h}\). Let \(I_u=4\ \mathrm{mg/L}\) and \(K_i=2\ \mathrm{mg/L}\) for the inhibited condition. Use the same input and other parameters for the uninhibited comparison. Assume the model conditions stated above hold.
First calculate the uninhibited steady state, for which \(\alpha=1\). Substitution into the quadratic gives $$ (2)(0.5)C_p^2+ \big[(100)(0.5)+(2)(8)-(60)(0.5)\big]C_p -(60)(8)=0, $$ so $$ C_p^2+36C_p-480=0. $$ The positive root is $$ C_p=\frac{-36+\sqrt{36^2+4(480)}}{2} =\frac{-36+\sqrt{3216}}{2} \approx 10.3549\ \mathrm{mg/L}. $$ The corresponding enzyme-site substrate concentration is \(S=aC_p=(0.5)(10.3549)\approx5.17745\ \mathrm{mg/L}\). Using the unrounded root in the rate equation, $$ v_{\mathrm{hep},0} =\frac{(100)(5.1774469)}{8+5.1774469} \approx39.2902\ \mathrm{mg/h}. $$ Other clearance is \((2)(10.3548938)\approx20.7098\ \mathrm{mg/h}\). Their sum is \(60.0000\ \mathrm{mg/h}\), matching the input before rounding.
For the inhibited condition, \(\alpha=1+4/2=3\). The quadratic becomes $$ (2)(0.5)C_p^2+ \big[(100)(0.5)+(2)(3)(8)-(60)(0.5)\big]C_p -(60)(3)(8)=0, $$ or $$ C_p^2+68C_p-1440=0. $$ The positive root is $$ C_p=\frac{-68+\sqrt{68^2+4(1440)}}{2} =\frac{-68+\sqrt{10384}}{2} \approx16.95096\ \mathrm{mg/L}. $$ Thus \(S=(0.5)(16.95096)\approx8.47548\ \mathrm{mg/L}\), and $$ v_{\mathrm{hep}}= \frac{(100)(8.4754775)}{(3)(8)+8.4754775} \approx26.0981\ \mathrm{mg/h}. $$ Other clearance is \((2)(16.950955)\approx33.9019\ \mathrm{mg/h}\); their sum is \(60.0000\ \mathrm{mg/h}\) before rounding.
The inhibited-to-uninhibited concentration ratio, calculated from the unrounded roots, is approximately \(16.950955/10.354894=1.637\). In this specified model, competitive inhibition raises the steady-state parent concentration by about 63.7% while reducing the hepatic metabolic rate at steady state. The hepatic rate does not remain fixed at its value at the uninhibited concentration: the higher substrate concentration partly offsets competitive inhibition.
A Second Steady-State Calculation
The same method applies when the input rate differs. This example illustrates why the inhibited concentration must be solved from the balance rather than obtained by multiplying the uninhibited concentration by a fixed factor.
Worked Example: Lower Input With Competitive Inhibition
Use the hepatic parameters \(V_{\max}=100\ \mathrm{mg/h}\), \(K_m=8\ \mathrm{mg/L}\), \(a=0.5\), and \(CL_{\mathrm{other}}=2\ \mathrm{L/h}\). Suppose the hypothetical input is \(R_{\mathrm{in}}=30\ \mathrm{mg/h}\), with \(\alpha=2\), corresponding, for example, to \(I_u/K_i=1\). Assume the same fixed enzyme-site mapping and equilibrium inhibition conditions.
The quadratic coefficients are \(CL_{\mathrm{other}}a=(2)(0.5)=1\), \(V_{\max}a+CL_{\mathrm{other}}\alpha K_m-R_{\mathrm{in}}a=50+32-15=67\), and \(-R_{\mathrm{in}}\alpha K_m=-(30)(2)(8)=-480\). Therefore, $$ C_p^2+67C_p-480=0, \qquad C_p=\frac{-67+\sqrt{67^2+4(480)}}{2} =\frac{-67+\sqrt{6409}}{2} \approx6.5281\ \mathrm{mg/L}. $$
Then \(S=(0.5)(6.5281)\approx3.2641\ \mathrm{mg/L}\). The hepatic rate is $$ v_{\mathrm{hep}}= \frac{(100)(3.2641)}{(2)(8)+3.2641} \approx16.9438\ \mathrm{mg/h}, $$ and other clearance is \((2)(6.5281)\approx13.0562\ \mathrm{mg/h}\). Together these are approximately \(30.0000\ \mathrm{mg/h}\), as required by steady state. Rounding the concentration or rate earlier can create small apparent discrepancies, so retain unrounded values through the balance check.
Diagnostics, Alternatives, and Limitations
A reliable calculation should satisfy both the enzyme-rate equation and the whole-body balance. Check that \(\alpha\geq1\), the positive concentration root is selected, \(S=aC_p\) uses the stated mapping, and hepatic plus other elimination equals input. Concentration units must match between \(S\), \(K_m\), \(I_u\), and \(K_i\); rates must use compatible amount-per-time units. If the computed inhibited rate exceeds the uninhibited rate at the same \(S\), or if the calculated elimination does not balance input, recheck the inhibition factor, denominator, and arithmetic.
| Approach | What it represents | When it is useful |
|---|---|---|
| Competitive rate equation at fixed \(S\) | Enzyme-rate change while substrate concentration is held constant | Comparing enzyme rates or interpreting an assay condition |
| Steady-state PBPK mass balance | Rate change plus the resulting parent concentration needed to balance input | Predicting a steady-state concentration under specified model assumptions |
| Dynamic liver-compartment model | Changes over time, potentially including hepatic blood flow, distribution, and transport | When rapid equilibrium, fixed substrate mapping, or steady state is not adequate |
A common mistake is to apply the fixed-\(S\) rate reduction directly as a concentration multiplier. For example, a 50% lower enzyme rate does not imply a 100% increase in steady-state concentration. Concentration is determined by the total elimination function, including other clearance and the way the hepatic rate rises as substrate concentration increases. Another error is to use total inhibitor concentration in \(\alpha\) without establishing its relationship to unbound enzyme-site concentration.
The reduced calculation is limited by its assumptions. If inhibitor concentration changes over time, if hepatic substrate concentration is not proportional to plasma concentration, or if flow, transport, enzyme abundance, or other clearance changes materially, the stated algebraic balance may not represent the system. In such cases, use an appropriately specified PBPK model and solve its time-dependent balances. Competitive equilibrium inhibition also should not be used to represent enzyme inactivation, which involves a different mechanism.
Check Your Understanding
Use the stated assumptions in each question. Numerical answers may differ slightly from rounding.
- If \(I_u=2\ \mathrm{mg/L}\) and \(K_i=2\ \mathrm{mg/L}\), what is \(\alpha\), and what happens to apparent \(K_m\)?
- At fixed \(S=6\ \mathrm{mg/L}\), let \(V_{\max}=90\ \mathrm{mg/h}\), \(K_m=3\ \mathrm{mg/L}\), and \(\alpha=2\). Calculate the inhibited rate.
- Why is the negative root of the quadratic in the steady-state example not a physical parent concentration?
- In the worked example with \(R_{\mathrm{in}}=60\ \mathrm{mg/h}\), why does the hepatic rate under inhibition not remain equal to its uninhibited rate evaluated at \(C_p=10.3549\ \mathrm{mg/L}\)?
Solutions. (1) \(\alpha=1+2/2=2\); the apparent \(K_m\) is doubled. (2) \(v_{\mathrm{hep}}=(90)(6)/[(2)(3)+6]=540/12=45\ \mathrm{mg/h}\). (3) A negative concentration is not physically meaningful, and the mass-balance model has a unique positive root for positive input and parameters. (4) At steady state, inhibition raises \(C_p\) and therefore raises \(S=aC_p\); the rate must be evaluated at this new substrate concentration. The higher substrate concentration partly offsets the competitive inhibition.