Sensitivity Diagnostics for Predicted Kp

Why Diagnose Sensitivity in a Kp Prediction?

A predicted tissue-to-plasma partition coefficient, \(K_p\), depends on drug properties and tissue inputs as well as on the prediction method. A prediction alone does not show whether its value is strongly influenced by pKa, logP, a tissue-composition fraction, or the plasma unbound fraction. Sensitivity diagnostics address that question by changing one input at a time and measuring the resulting change in the model output.

The aim is not to decide whether a predicted \(K_p\) is correct. It is to determine which inputs have the greatest influence on the prediction near a specified baseline, under specified perturbations. The result is conditional on the chosen method, tissue, baseline inputs, and perturbation sizes. It can help prioritize input verification or further measurement, but it cannot establish that the prediction method itself is appropriate.

Definition: Write a prediction method as \(K_p=f(\boldsymbol{\theta})\), where \(\boldsymbol{\theta}\) is the vector of its inputs and \(f\) is the method's calculation. A local sensitivity describes how \(K_p\) changes for a small change in one input while the other inputs are held fixed. A one-at-a-time (OAT) perturbation recalculates \(K_p\) after changing one input and leaving the others unchanged, except for linked quantities that must be adjusted to keep the input set valid.

For example, a tissue fraction cannot generally be increased while all other fractions are held fixed if the fractions must sum to one. The perturbation must specify which other fraction is reduced, or how the method renormalizes the composition. Similarly, when perturbing an input, retain any model-specific definitions and constraints established by the prediction method. The earlier material on tissue composition and unbound fractions supplies the relevant input meanings; the sensitivity calculation examines their influence rather than redefining those methods.

Local and Scaled Sensitivity

A signed local derivative describes both direction and rate of change. Because \(K_p\) is positive, it is often convenient to take its logarithm: the derivative of \(\ln K_p\) expresses a change on a relative, rather than absolute, output scale. For input \(\theta_i\), define \(g(\boldsymbol{\theta})=\ln K_p\). At the baseline input vector, the local derivative is \(\partial g/\partial\theta_i\). A positive value means that increasing \(\theta_i\) locally increases \(K_p\); a negative value means it locally decreases \(K_p\).

D_i=\frac{\partial \ln K_p}{\partial \theta_i} =\frac{1}{K_p}\frac{\partial K_p}{\partial \theta_i}

The units of \(D_i\) are the reciprocal units of \(\theta_i\). Thus, derivatives with respect to pKa, logP, and a fraction are not directly comparable merely by their numerical magnitudes. To compare input effects on a useful scale, choose a meaningful perturbation size \(a_i\) for each input. The scaled local sensitivity is \(S_i=a_iD_i\). It estimates the change in \(\ln K_p\) associated with a change of size \(a_i\) in input \(i\); for small changes, it is approximately the proportional change in \(K_p\).

S_i \approx \frac{\ln K_p(\theta_i+a_i)-\ln K_p(\theta_i-a_i)}{2}

In this centered finite-difference calculation, all other inputs are held at baseline values, subject to any required composition adjustment. A positive \(S_i\) indicates an increasing local effect over the selected perturbation range; a negative \(S_i\) indicates a decreasing effect. Comparing \(|S_i|\) ranks the influence of the selected perturbations, not an intrinsic importance that is independent of their sizes.

For a positive input, an elasticity is another dimensionless measure: \(E_i=\partial\ln K_p/\partial\ln\theta_i\). It describes the local proportional change in \(K_p\) for a proportional change in that input. It can be useful for a positive \(f_{u,p}\), for example, but it is not generally a good default for pKa or logP: these are logarithmic chemical descriptors, and their zero values are not meaningful reference points for proportional changes. A stated perturbation scale is more interpretable for them.

Why the Centered Calculation Approximates a Local Effect

The finite-difference calculation is justified when \(g=\ln K_p\) is sufficiently smooth near the baseline. For a small perturbation \(a_i\), a Taylor expansion in \(\theta_i\), with the other inputs fixed, gives:

g(\theta_i+a_i)=g(\theta_i)+a_i g'(\theta_i) +\frac{a_i^2}{2}g''(\theta_i)+\frac{a_i^3}{6}g'''(\theta_i)+\cdots

g(\theta_i-a_i)=g(\theta_i)-a_i g'(\theta_i) +\frac{a_i^2}{2}g''(\theta_i)-\frac{a_i^3}{6}g'''(\theta_i)+\cdots

Subtracting the second expansion from the first cancels the baseline term and the even-order terms. Dividing the difference by two gives \(a_i g'(\theta_i)\), plus terms of order \(a_i^3\). Therefore, the centered estimate of \(S_i\) approaches the scaled derivative as the perturbation becomes small, provided the output is smooth. If the estimate changes substantially as \(a_i\) is reduced, the output may be nonlinear over the larger range, or the calculation may be affected by numerical instability.

Worked Example: Ranking Four Inputs at One Baseline

Suppose a hypothetical tissue prediction method gives \(K_p=4.00\) at pKa \(7.4\), logP \(2.0\), tissue fractions \(f_{\mathrm{lipid}}=0.05\), \(f_{\mathrm{water}}=0.70\), and \(f_{\mathrm{protein}}=0.25\), and \(f_{u,p}=0.10\). The three tissue fractions sum to one. The following hypothetical reruns change one input at a time. For the lipid-fraction perturbation, water is adjusted by the opposite amount so the fractions still sum to one.

Input changed	Lower input and predicted \(K_p\)	Upper input and predicted \(K_p\)	Half-range \(a_i\)
pKa	7.2; 3.60	7.6; 4.40	0.20
logP	1.7; 3.20	2.3; 5.00	0.30
\(f_{\mathrm{lipid}}\), with water compensating	0.03; 3.80	0.07; 4.20	0.02
\(f_{u,p}\)	0.09; 3.60	0.11; 4.40	0.01

For pKa, the scaled sensitivity estimate is:

S_{\mathrm{pKa}}\approx\frac{1}{2}\ln\left(\frac{4.40}{3.60}\right) =\frac{1}{2}\ln(1.2222)=0.1003

For logP, it is:

S_{\mathrm{logP}}\approx\frac{1}{2}\ln\left(\frac{5.00}{3.20}\right) =\frac{1}{2}\ln(1.5625)=0.2231

The corresponding calculations for lipid fraction and \(f_{u,p}\) are:

S_{\mathrm{lipid}}\approx\frac{1}{2}\ln\left(\frac{4.20}{3.80}\right)=0.0500, \qquad S_{f_{u,p}}\approx\frac{1}{2}\ln\left(\frac{4.40}{3.60}\right)=0.1003

For these specified perturbations, \(|S_{\mathrm{logP}}|\) is largest, followed by pKa and \(f_{u,p}\), which tie at the reported precision, and then the lipid fraction. The interpretation is limited to these selected input ranges and this baseline. It does not establish that logP always controls this method's prediction, or that the same ranking applies in another tissue.

Respecting Composition Constraints

Tissue fractions are often components of a composition and therefore cannot be varied independently without checking the constraint. If a tissue is represented by water, lipid, and protein fractions, for example, a valid composition may require \(f_{\mathrm{water}}+f_{\mathrm{lipid}}+f_{\mathrm{protein}}=1\). A change to one component then requires a stated change to at least one other component. The sensitivity is consequently the effect of a defined composition change, not an isolated effect of one fraction outside the feasible composition.

Worked Example: A Feasible Tissue-Fraction Perturbation

In a hypothetical prediction, the baseline fractions are \(f_{\mathrm{lipid}}=0.08\), \(f_{\mathrm{water}}=0.67\), and \(f_{\mathrm{protein}}=0.25\), which sum to one. The baseline predicted \(K_p\) is 2.00. The analyst wants to examine a lipid perturbation of \(a=0.02\). Increasing lipid to 0.10 while leaving the other fractions unchanged would produce a total of 1.02, so that is not a valid composition under the stated constraint.

Instead, define the perturbation as a transfer between lipid and water: increase lipid by 0.02 and decrease water by 0.02, or make the reverse transfer. The hypothetical model gives \(K_p=2.30\) for \((0.10,0.65,0.25)\) and \(K_p=1.80\) for \((0.06,0.69,0.25)\). Both compositions sum to one. The scaled centered sensitivity is:

S_{\mathrm{lipid}}\approx\frac{1}{2}\ln\left(\frac{2.30}{1.80}\right) =\frac{1}{2}\ln(1.2778)=0.1226

This value describes the specified lipid-for-water exchange. It should not be reported as the effect of increasing lipid while every other fraction remains fixed. If the prediction method uses a different valid composition constraint or normalization rule, the perturbation should follow that rule and state it explicitly.

Check Locality and Numerical Stability

A single perturbation size can conceal curvature. Repeat the calculation with smaller and larger steps when feasible. Similar estimates across modest step sizes support a locally stable derivative approximation; substantial changes indicate that the response depends on the perturbation range. Also inspect both perturbed predictions, rather than reporting only a sensitivity score: unexpected jumps, nonpositive outputs, or a change in direction can signal a coding error, a discontinuity, an invalid input, or strong nonlinearity.

Worked Example: Detecting Curvature in a pKa Response

A hypothetical method gives baseline \(K_p=6.00\) at pKa \(7.0\). With all other inputs fixed, its predictions are 5.70 at pKa \(6.9\) and 6.30 at pKa \(7.1\). For the half-range \(a=0.1\):

S_{\mathrm{pKa},\,0.1} \approx\frac{1}{2}\ln\left(\frac{6.30}{5.70}\right) =\frac{1}{2}\ln(1.1053)=0.0500

Now consider a wider range: the model predicts 4.20 at pKa \(6.5\) and 8.40 at pKa \(7.5\). For \(a=0.5\):

S_{\mathrm{pKa},\,0.5} \approx\frac{1}{2}\ln\left(\frac{8.40}{4.20}\right) =\frac{1}{2}\ln(2)=0.3466

The wider-range value is much larger. This does not mean the local sensitivity at pKa \(7.0\) is 0.3466; it indicates a larger average log-scale response across the broader interval. The local result should be described using the smaller perturbation, while the difference between the two results is a diagnostic of range dependence. Further step-size checks or examination of the response curve would be appropriate before summarizing the effect.

Choosing Perturbations and Comparing Approaches

Perturbation sizes should be justified by the question being asked. They may reflect plausible uncertainty in an input, a practical range of values, or a deliberately small step to estimate a local derivative. These choices answer different questions. A small step estimates local behavior; a plausible uncertainty range indicates how much that uncertainty might move the prediction; a broad range probes behavior over that interval. Report the baseline, the changed input, the perturbation size, and any linked inputs that were adjusted.

OAT analysis is transparent and useful for identifying influential inputs near a baseline. Its main limitation is that it holds other inputs fixed, so it does not reveal interactions. For example, the effect of changing pKa may depend on logP or on tissue pH. A local derivative or OAT ranking can therefore miss a combined effect that appears when inputs change together. A broader global sensitivity analysis varies multiple inputs jointly and can characterize interactions, but requires more computation and a defined distribution or range for each input. It is an alternative when the question concerns the full input space rather than a local, controlled perturbation.

Analytical derivatives, when available, can efficiently estimate local slopes. Finite differences are more broadly applicable because they require only repeated model evaluations, but their accuracy depends on step size and numerical precision. A forward difference uses a baseline and one perturbed prediction; a centered difference uses predictions on both sides and generally cancels more low-order approximation error for a smooth response. Neither approach removes the need to check that perturbed inputs are valid.

Sensitivity is not the same as uncertainty. A large \(|S_i|\) means that the output responds strongly to the selected change in input \(i\); it does not show that the input is uncertain. A small sensitivity may still matter if the input uncertainty is large. Conversely, a highly sensitive input may be well characterized. Sensitivity rankings also do not validate a prediction method or demonstrate which biological mechanism is responsible for an output.

Common Mistakes and Practical Diagnostics

Comparing raw derivatives across unlike units. A derivative per pKa unit and one per unit of a fraction are not directly comparable. Use stated perturbation scales or a suitable elasticity, and report the scale.
Changing a tissue fraction without preserving composition. The result may represent an impossible input set. Specify the compensating fraction or the model's normalization rule.
Calling an OAT ranking universal. A ranking is conditional on the baseline, method, tissue, and perturbation sizes. Recheck it when those conditions change.
Treating sensitivity as evidence of input uncertainty or model accuracy. Sensitivity measures output response; it does not measure input uncertainty or predictive validity.
Ignoring nonlinear response or numerical artifacts. Compare more than one step size, inspect both perturbed predictions, and investigate abrupt or direction-changing responses.

Key takeaway: A useful Kp sensitivity diagnosis defines a valid baseline, changes one input under an explicit perturbation rule, calculates the output response on a comparable scale, and checks that the conclusion is stable over a relevant local range.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

If the centered predictions are \(K_p=3.0\) at the lower input and \(K_p=5.0\) at the upper input, what is the scaled sensitivity estimate?
Why is it not generally appropriate to compare a derivative per unit of pKa directly with a derivative per unit of tissue fraction?
A tissue has fractions that sum to one. What must be specified when increasing one fraction in an OAT perturbation?
What does a large scaled sensitivity establish, and what does it not establish?
What might it mean if a centered sensitivity estimate changes substantially when the perturbation size is reduced?

Solutions. (1) \(S\approx\frac{1}{2}\ln(5.0/3.0)=0.2554\). (2) The derivatives have different input units and scales, so their numerical magnitudes are not directly comparable; use justified perturbation scales or a suitable dimensionless measure. (3) Specify which other fraction decreases, or state the valid normalization rule, so the composition constraint remains satisfied. (4) It establishes a strong output response to that particular input change near the stated baseline; it does not establish that the input is uncertain, that the method is accurate, or that the ranking is universal. (5) The response may be nonlinear over the original range, or numerical instability may be affecting the calculation; inspect predictions and perform further step-size checks.