Tutorials › Physiologically-Based Pharmacokinetics (PBPK) › Propagating Uncertainty in IVIVE Scaling Factors

In Vitro to In Vivo Extrapolation of Metabolism · Tutorial 359 of 799

Propagating Uncertainty in IVIVE Scaling Factors

Convert uncertainty in IVIVE inputs into a defensible interval for scaled intrinsic clearance, accounting for positive-valued factors and their correlations.

Advanced 12 min read

What You'll Learn

  • Express scaled intrinsic clearance as a product and ratio of the in vitro estimate and physiological scaling factors.
  • Convert coefficients of variation to log-scale standard deviations for positive, lognormally distributed inputs.
  • Derive and calculate an uncertainty interval for scaled intrinsic clearance.
  • Account for correlation between uncertain inputs, including factors in a denominator.
  • Compare lognormal calculations with first-order approximations and simulation.
  • Diagnose limitations such as bounded unbound fractions, assay uncertainty, and unmodelled bias.

Why Propagate Scaling-Factor Uncertainty?

An IVIVE estimate can depend on several uncertain quantities: in vitro intrinsic clearance, protein or cell content per gram of liver, liver mass, and the unbound fraction in the incubation. Reporting only a single scaled value hides how uncertainty in those inputs combines. It can also make a prediction appear more precise than the measurements and assumptions justify.

This tutorial develops uncertainty propagation for a scaled intrinsic clearance. The introductory material on in vitro scaling establishes the relevant dimensions and scaling relationships; here the focus is how uncertainty in their inputs affects the resulting estimate. The interval developed below represents uncertainty propagated through the stated scaling model. It does not account automatically for systematic IVIVE bias or uncertainty in the model structure.

Definition: For an in vitro apparent intrinsic clearance \(CL_{\mathrm{int,app}}\) measured per unit protein or cell content, a common scaled unbound intrinsic clearance is $$ CL_{\mathrm{int,u,scaled}} = \frac{CL_{\mathrm{int,app}}\;Q\;M_{\mathrm{liver}}}{f_{\mathrm{u,inc}}}. $$ Here \(Q\) is protein or cell content per unit liver mass, \(M_{\mathrm{liver}}\) is liver mass, and \(f_{\mathrm{u,inc}}\) is the fraction unbound in the in vitro incubation. The exact form must match the assay and target quantity: for example, a clearance already corrected for incubation binding must not be divided by \(f_{\mathrm{u,inc}}\) a second time.

With microsomal protein, \(Q\) might be expressed as milligrams of microsomal protein per gram of liver. With hepatocytes, it might be expressed as cells per gram of liver. The units cancel through the product. For example, \((\mu\mathrm{L\,min^{-1}\,mg^{-1}})(\mathrm{mg\,g^{-1}})(\mathrm{g})\) gives \(\mu\mathrm{L\,min^{-1}}\). The unbound fraction is dimensionless.

A standard approach for positive-valued quantities is to represent each uncertain input as lognormally distributed. This is appropriate when an input is positive and uncertainty is naturally expressed proportionally, but it is an assumption to assess, not a universal property of assay data.

Log-Scale Propagation

Let \(C\), \(Q\), \(M\), and \(F\) denote the uncertain values of in vitro apparent intrinsic clearance, protein or cell content, liver mass, and incubation unbound fraction. Write the scaled value as \(Y=CQM/F\). Assume the logarithms of the inputs are jointly normal (so each input is lognormal). Define \(a_C=1\), \(a_Q=1\), \(a_M=1\), and \(a_F=-1\). Then

$$ \ln(Y)=\ln(C)+\ln(Q)+\ln(M)-\ln(F). $$

The signs matter. Increasing any numerator factor increases \(Y\), while increasing \(F\) decreases the binding-corrected value. Let \(\eta_i=E[\ln(X_i)]\) and let \(\Sigma_{ij}=\operatorname{Cov}(\ln(X_i),\ln(X_j))\), where \(X_i\) ranges over the inputs. The log-scale mean and variance are

$$ \mu_{\ln Y}=\sum_i a_i\eta_i,\qquad \sigma_{\ln Y}^{2}=\sum_i\sum_j a_i a_j\Sigma_{ij}. $$

Consequently, \(Y\) is lognormal under these joint-normal assumptions. Its median is \(\exp(\mu_{\ln Y})\), and a central 95% propagated interval is

$$ \left[\exp(\mu_{\ln Y}-1.96\sigma_{\ln Y}),\ \exp(\mu_{\ln Y}+1.96\sigma_{\ln Y})\right]. $$

This is an interval for the propagated scaled value under the specified input distributions and dependence structure. It is not a 95% confidence interval for a fixed unknown parameter, nor does it include bias that was not represented in the inputs.

From Coefficients of Variation to Log-Scale Uncertainty

A coefficient of variation (CV) is the standard deviation divided by the arithmetic mean. For a lognormal input \(X\), if \(s_{\ln X}\) is the standard deviation of \(\ln(X)\), then

$$ \operatorname{CV}(X)^2=\exp(s_{\ln X}^{2})-1, \qquad s_{\ln X}=\sqrt{\ln\!\left(1+\operatorname{CV}(X)^2\right)}. $$

To justify the relationship, write \(\ln(X)\sim N(\eta,s_{\ln X}^{2})\). The lognormal moments are \(E[X]=\exp(\eta+s_{\ln X}^{2}/2)\) and \(\operatorname{Var}(X)=(\exp(s_{\ln X}^{2})-1)\exp(2\eta+s_{\ln X}^{2})\). Dividing the variance by the squared mean gives \(\operatorname{CV}(X)^2=\exp(s_{\ln X}^{2})-1\), and rearranging gives the conversion above. In particular, the conversion uses \(\ln(1+\mathrm{CV}^2)\), not \(\ln((1+\mathrm{CV})^2)\).

If a supplied central value is an arithmetic mean \(m\), its lognormal median is \(m/\sqrt{1+\mathrm{CV}^2}\). If it is a geometric mean or median, it can be used directly as the lognormal median. Make this distinction explicit: multiplying reported arithmetic means does not, in general, give the median of the scaled result.

Worked Example: Independent Uncertainty in All Four Inputs

Suppose hypothetical assay and physiological inputs are reported as lognormal medians: \(C=0.80\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\), \(Q=40\ \mathrm{mg\,g^{-1}}\), \(M=1500\ \mathrm{g}\), and \(F=0.80\). Assume their log values are independent, and their respective CVs are 20%, 10%, 8%, and 15%. Calculate the scaled median and central 95% propagated interval.

First calculate the scaled median, using the stated input medians: $$ \operatorname{median}(Y)= \frac{(0.80)(40)(1500)}{0.80} =60000\ \mu\mathrm{L\,min^{-1}} =60.0\ \mathrm{mL\,min^{-1}}. $$

Convert the four CVs to log standard deviations using \(s_{\ln X}=\sqrt{\ln(1+\mathrm{CV}^2)}\): $$ s_{\ln C}=\sqrt{\ln(1+0.20^2)}=0.1980,\quad s_{\ln Q}=\sqrt{\ln(1+0.10^2)}=0.09975, $$ $$ s_{\ln M}=\sqrt{\ln(1+0.08^2)}=0.07987,\quad s_{\ln F}=\sqrt{\ln(1+0.15^2)}=0.1492. $$

Independence makes the log variances add. The minus sign on \(\ln(F)\) does not change its variance: $$ \sigma_{\ln Y}^{2}=0.1980^2+0.09975^2+0.07987^2+0.1492^2 \approx0.07780,\qquad \sigma_{\ln Y}\approx0.2789. $$ Thus the interval is \(60.0\exp(\pm1.96\times0.2789)\), or approximately \([34.7,\ 103.7]\ \mathrm{mL\,min^{-1}}\). Under the assumed distributions, the propagated scaled value has a median of \(60.0\ \mathrm{mL\,min^{-1}}\) and a central 95% interval from about \(34.7\) to \(103.7\ \mathrm{mL\,min^{-1}}\).

Correlation Can Increase or Reduce the Interval

Uncertain inputs need not be independent. When two input estimates tend to move together, their covariance contributes to the variance of the scaled result. The coefficients \(a_i\) in the general variance formula determine whether that covariance increases or reduces uncertainty. For a numerator \(C\) and denominator \(F\), the pair’s contribution is

$$ \operatorname{Var}(\ln C-\ln F) =\operatorname{Var}(\ln C)+\operatorname{Var}(\ln F) -2\operatorname{Cov}(\ln C,\ln F). $$

Positive correlation between \(\ln(C)\) and \(\ln(F)\) reduces uncertainty in their ratio: high values of one tend to coincide with high values of the other, partly offsetting each other. Negative correlation increases uncertainty in the ratio. Correlation should reflect how the quantities were measured or estimated; it should not be chosen merely to narrow an interval.

Worked Example: A Correlated Clearance and Unbound Fraction

Consider hypothetical medians \(C=0.50\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\) and \(F=0.70\), with CVs of 20% and 30%. Assume \(\ln(C)\) and \(\ln(F)\) are jointly normal, with log-scale correlation \(+0.60\). Hold the other scaling factors fixed. The median corrected clearance is \(0.50/0.70=0.7143\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\).

The log variances are \(\ln(1+0.20^2)=0.03922\) and \(\ln(1+0.30^2)=0.08618\). The log covariance is $$ 0.60\sqrt{(0.03922)(0.08618)}\approx0.03488. $$

Because \(F\) is in the denominator, the variance is $$ \sigma_{\ln(C/F)}^2=0.03922+0.08618-2(0.03488)=0.05564, \qquad \sigma_{\ln(C/F)}\approx0.2359. $$ The 95% propagated interval is \(0.7143\exp(\pm1.96\times0.2359)\), approximately \([0.450,\ 1.134]\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\). If the correlation were incorrectly ignored, the variance would be \(0.1254\), the log standard deviation \(0.3541\), and the interval approximately \([0.357,\ 1.430]\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\). In this hypothetical case, the positive correlation narrows the ratio’s uncertainty.

Worked Example: Correlated Physiological Scaling Factors

Suppose hypothetical lognormal medians are \(C=1.20\ \mu\mathrm{L\,min^{-1}\,mg^{-1}}\), \(Q=35\ \mathrm{mg\,g^{-1}}\), \(M=1600\ \mathrm{g}\), and \(F=0.75\). Their CVs are 10%, 20%, 5%, and 10%, respectively. Assume the log inputs are jointly normal, with all log-scale correlations zero except for a correlation of \(-0.30\) between \(\ln(Q)\) and \(\ln(M)\). This example illustrates how dependent physiological inputs can be handled; the assumed correlation would need empirical or design-based justification in an application.

The scaled median is $$ \frac{(1.20)(35)(1600)}{0.75} =89600\ \mu\mathrm{L\,min^{-1}} =89.6\ \mathrm{mL\,min^{-1}}. $$

The log variances for \(C,Q,M,F\), in that order, are \(\ln(1.01)=0.009950\), \(\ln(1.04)=0.03922\), \(\ln(1.0025)=0.002497\), and \(\ln(1.01)=0.009950\). The covariance between \(\ln(Q)\) and \(\ln(M)\) is \(-0.30\sqrt{(0.03922)(0.002497)}\approx-0.002969\). Both factors are in the numerator, so twice this covariance is added: $$ \sigma_{\ln Y}^{2} =0.009950+0.03922+0.002497+0.009950+2(-0.002969) \approx0.05568. $$

Therefore, \(\sigma_{\ln Y}\approx0.2360\). The interval is \(89.6\exp(\pm1.96\times0.2360)\), approximately \([56.4,\ 142.3]\ \mathrm{mL\,min^{-1}}\). The negative correlation reduces the variance relative to the same marginal uncertainties with independence. The calculation is valid only if the stated joint lognormal model and covariance are defensible.

Practical Checks, Alternatives, and Limitations

Before reporting an interval, verify the model inputs and the calculation. Confirm that a binding correction has not been applied twice, that each scaling factor uses compatible units, and that the unbound fraction corresponds to the relevant incubation. Check whether central values are arithmetic means or medians, and whether estimated correlations are on the original or log scale. For the log-scale formula, the covariance matrix must be valid; in particular, it must be positive semidefinite so that the computed variance cannot be negative.

A useful diagnostic is to examine each input’s contribution to log variance, including covariance terms. A large contribution from a single factor identifies where improved measurement might most reduce uncertainty. Also inspect the resulting interval for plausibility. If it implies values far outside reasonable physiological or assay ranges, that may indicate an inappropriate distribution, a coding or unit error, or substantial uncertainty that should be acknowledged rather than hidden.

The lognormal approach is convenient because products and ratios of jointly lognormal inputs remain lognormal. A first-order delta-method approximation is an alternative when relative uncertainty is small; it approximates the transformed distribution locally and can be less reliable for wide or skewed uncertainty. Monte Carlo propagation is more flexible: draw inputs from specified distributions, preserve their dependence, calculate the scaled result for each draw, and take the desired quantiles. Simulation does not fix misspecified input distributions or omitted correlations, but it can accommodate bounded distributions and nonlinear transformations more readily.

Unbound fractions are bounded between zero and one, whereas a lognormal distribution is not. A lognormal approximation for \(F\) is most defensible when its uncertainty is small and its distribution is not materially affected by the upper bound. If the bound matters, a beta or logit-normal model, or a suitable truncated distribution, may better represent uncertainty; the resulting interval may require simulation. Likewise, uncertainty from assay quantification limits, unstable measurements, or systematic differences between in vitro and in vivo conditions is not resolved by a propagation formula. Those limitations must be represented separately or stated when interpreting the prediction.

Common pitfall: Adding the CVs, or adding the original-scale standard deviations, does not generally give the uncertainty of a product or ratio. Under independent lognormal inputs, add log variances; with dependence, add the signed covariance terms as well. Also use \(s_{\ln X}=\sqrt{\ln(1+\mathrm{CV}^2)}\): replacing \(1+\mathrm{CV}^2\) with \((1+\mathrm{CV})^2\) gives an incorrect conversion.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. If an in vitro apparent clearance is already corrected for incubation binding, should it be divided by the incubation unbound fraction again?
  2. For an independent lognormal input with CV 25%, what is its log-scale standard deviation to three significant figures?
  3. What does positive log-scale correlation between a numerator and denominator generally do to the variance of their ratio?
  4. In the first worked example, what variance would result from the rough sum of squared CVs, and why is it not the exact log variance?
  5. Name one limitation of using a lognormal distribution for an unbound fraction.

Solutions. (1) No; that would apply the same binding correction twice. (2) \(s_{\ln X}=\sqrt{\ln(1+0.25^2)}=\sqrt{\ln(1.0625)}\approx0.246\). (3) It generally reduces the ratio’s variance because the covariance is subtracted. (4) The rough sum is \(0.20^2+0.10^2+0.08^2+0.15^2=0.07890\). It is an approximation based on squared CVs; the exact independent log variance is the sum of \(\ln(1+\mathrm{CV}_i^2)\), approximately \(0.07780\). (5) A lognormal draw can exceed one, even though an unbound fraction cannot.