From a Concentration Time Course to a Cascade Model
A biomarker can respond gradually even when drug concentration changes quickly. A sequence of signal-transduction steps is one explanation: each step transmits part of the incoming signal, so the measured response lags behind the concentration input. Analysis of such data requires more than identifying a delayed curve. The candidate cascade must be fitted, its residuals examined, and its predicted timing interpreted in the context of the exposure and biomarker.
The worked analysis below uses hypothetical data after a drug concentration has reached and remained at a constant level. The measured values are biomarker increases above a stable baseline. The example compares cascade stage counts while holding the transit rate fixed, making the comparison focused on the cascade structure rather than on simultaneous estimation of stage count and rate.
This expression assumes a linear, time-invariant cascade, identical first-order transit rates, a step input beginning at time zero, and a measured output proportional to the final stage. It also assumes that the baseline and initial conditions are known well enough to analyze increases above baseline. The response fraction starts at zero and approaches one; therefore, under a sustained step input, this model predicts a gradual approach to a plateau, not a transient peak.
For observations \(y_i\) at times \(t_i\), and a fixed \(k\), write \(g_i=F_n(t_i;k)\). A least-squares fit chooses \(G\) to minimize the sum of squared residuals, \(SSE=\sum_i(y_i-Gg_i)^2\). This is a descriptive fitting criterion. If residuals are treated as independent, mean-zero observations with common variance, ordinary least squares also has its usual statistical interpretation; those assumptions should not be inferred merely because a curve has been fitted.
Worked Analysis: Fitting Candidate Stage Counts
Worked Example: Estimating the Gain for a Two-Stage Cascade
Hypothetical data. A drug concentration is maintained at 2 mg/L from time zero onward. Biomarker increases above baseline are recorded at 1, 2, 4, 6, and 8 hours as 10, 25, 61, 79, and 92 biomarker units, respectively. For this comparison, take the transit rate as \(k=0.5\ \mathrm{h}^{-1}\), assumed to have been specified independently. Fit a two-stage cascade.
For \(n=2\), \(F_2(t;k)=1-\exp(-kt)(1+kt)\). The dimensionless times \(kt_i\) are 0.5, 1, 2, 3, and 4, so the corresponding response fractions are approximately 0.09020, 0.26424, 0.59399, 0.80085, and 0.90842. Let these be \(g_i\).
For a model with no free intercept because the data are already baseline-subtracted, differentiating \(SSE(G)=\sum_i(y_i-Gg_i)^2\) with respect to \(G\) gives $$ \frac{d\,SSE(G)}{dG}=-2\sum_i g_i y_i+2G\sum_i g_i^2=0, \qquad \widehat{G}=\frac{\sum_i g_i y_i}{\sum_i g_i^2}. $$ The second derivative is \(2\sum_i g_i^2>0\), so this stationary point minimizes the squared error.
Substituting the data and unrounded fractions gives \(\sum_i g_i y_i\approx190.583804\) and \(\sum_i g_i^2\approx1.8973828\). Thus $$ \widehat{G}=\frac{190.583804}{1.8973828}\approx100.45\ \text{biomarker units}. $$
The observed squared values sum to \(10^2+25^2+61^2+79^2+92^2=19{,}151\). Therefore, $$ SSE=19{,}151-\frac{(190.583804)^2}{1.8973828}\approx7.69\ \text{biomarker units}^2. $$ The fitted plateau increase is about 100.45 units for the specified concentration input. The small SSE indicates that the two-stage curve tracks these five observations closely, but by itself it does not establish that the biological pathway literally contains two stages.
The gain formula follows from a one-parameter least-squares problem, not from a special property of cascades. What is specific to the cascade analysis is how the fractions \(g_i\) depend on stage count and transit rate. Holding \(k\) fixed makes each candidate stage count a distinct time-course shape. The fitted gain scales that shape but cannot change its timing.
Worked Example: Comparing One, Two, and Three Stages
Use the same hypothetical observations and \(k=0.5\ \mathrm{h}^{-1}\). For each candidate stage count, calculate its fractions at the five observation times and estimate its gain using the least-squares formula. The table reports rounded gains and the resulting SSE values.
| Candidate stages \(n\) | Fitted gain (biomarker units) | SSE (biomarker units squared) |
|---|---|---|
| 1 | 75.06 | 1,295 |
| 2 | 100.45 | 7.69 |
| 3 | 134.3 | 685.1 |
For three stages, the fractions at \(kt=0.5,1,2,3,4\) are approximately 0.01439, 0.08030, 0.32332, 0.57681, and 0.76190. The fitted gain is approximately 134.3 units. Using the unrounded fractions, the least-squares calculation is $$ SSE=\sum_i y_i^2-\frac{(\sum_i g_i y_i)^2}{\sum_i g_i^2} \approx685.055\ \text{biomarker units}^2, $$ which rounds to 685.1. The one-stage fit has an SSE of about 1,295 units squared. Among these three candidates, the two-stage structure has the smallest SSE by a large margin.
Since all candidates have the same number of fitted parameters in this example (the gain alone), the SSE comparison is not giving the two-stage model a parameter-count advantage. Still, it is a comparison within a specified candidate set and fixed rate, not proof that other structures or rates cannot fit better. If the rate were estimated separately for each structure, that fitting step and its uncertainty would also need to be considered.
Predicting Response Timing
A step response has no finite peak while the input is held constant: it approaches its plateau asymptotically. A useful timing summary is therefore the time to reach a stated fraction of the plateau. This timing is defined by the cascade shape and rate, not by the fitted gain.
Worked Example: Time to Half of the Plateau Response
For the selected two-stage model, the fraction of plateau at time \(t\) is \(F_2(t;k)=1-\exp(-kt)(1+kt)\). Set this fraction to 0.5. With \(x=kt\), the equation to solve is $$ 1-\exp(-x)(1+x)=0.5, \qquad \exp(-x)(1+x)=0.5. $$
Numerical solution gives \(x\approx1.67835\). Since \(k=0.5\ \mathrm{h}^{-1}\), the predicted time is $$ t_{50}=\frac{x}{k}=\frac{1.67835}{0.5}\approx3.357\ \mathrm{h}. $$ At the fitted gain of 100.45 biomarker units, half the plateau increase is approximately 50.22 units. Under the stated step-input model, the prediction is that the biomarker increase reaches about 50.22 units roughly 3.36 hours after the input begins. This is a model-based timing prediction, not an observed time unless it is confirmed by measurements.
The same calculation applies to another target fraction by solving \(F_n(t;k)=q\), where \(q\) is between zero and one. For a fixed stage count, increasing \(k\) shortens the time scale because the equation depends on time through the product \(kt\). Conversely, changing the gain changes the plateau magnitude but leaves the fraction-of-plateau timing unchanged.
Diagnostics and Model Choice
A small total SSE can conceal a systematic misspecification, especially with sparse data. Inspect the fitted values and residuals \(e_i=y_i-\widehat{G}g_i\) in time order. A suitable candidate should not show a persistent run of positive residuals followed by a run of negative residuals, curvature, or a pattern linked to concentration history. Such structure suggests the candidate response shape is missing something, even if its overall SSE appears acceptable.
For the selected two-stage fit, the predicted increases are approximately 9.06, 26.54, 59.66, 80.44, and 91.25 units at 1, 2, 4, 6, and 8 hours. The residuals are therefore about 0.94, -1.54, 1.34, -1.44, and 0.75 units. They alternate in sign rather than showing a sustained directional pattern. This is reassuring for this small example, but five observations provide limited ability to detect subtler structure.
Worked Example: Reading Residuals Against the Candidate Models
Suppose the one-stage and three-stage fits are compared with the two-stage fit in the table above. The one-stage model has a much larger SSE, and its fitted gain is only about 75.06 units. Its rapid initial rise and early approach to its plateau cannot reproduce the gradual middle portion of these observations as well as the two-stage shape. The three-stage model also has a larger SSE, about 685.1 units squared; its response rises too slowly over much of the observed interval and requires a larger fitted plateau to compensate.
The residual pattern should be checked directly rather than inferred solely from SSE. For the two-stage candidate, the approximate residual sequence is \(0.94,-1.54,1.34,-1.44,0.75\) units; it has no long run in one direction. If instead residuals were systematically positive at early times and negative later, the modeled response would likely be too delayed or too rapid in the wrong part of the curve. That pattern would motivate checking the assumed transit rate, input timing, baseline, or cascade shape rather than simply reporting the fitted gain.
The assumptions warrant substantive checks. A concentration that rises gradually rather than stepping immediately may make the modeled biomarker appear delayed if the input history is ignored. A drifting baseline, a concentration-dependent nonlinear transfer, unequal transit rates, or an additional direct response can also produce systematic residuals. Replicate measurements, if available, help distinguish observation variability from a reproducible time pattern. Where observation variances differ across times, an unweighted SSE gives equal importance to each point and may not be appropriate.
The main alternative is not always another stage count. A different transit rate can alter response timing; a model with unequal rates can change the shape; a direct-effect component can add an earlier response; and a nonlinear transfer can change how exposure propagates. Those alternatives should be considered when scientifically plausible and when the data can identify their additional features. The present fixed-rate, equal-stage comparison is useful for illustrating structural fit, but it cannot adjudicate among all such possibilities.
Why the Analysis Matters
Cascade structure influences predictions between observed times and beyond the sampling window. In this example, the two-stage candidate is clearly preferred among the three fixed-rate options by SSE and has no obvious residual run. Its estimated half-plateau time translates the fitted shape into an interpretable timing quantity. That conclusion is conditional: it depends on the sustained input, baseline subtraction, linearity, fixed rate, and candidate set stated above.
Check Your Understanding
Answer each question using the definitions and assumptions above. Brief solutions follow the questions.
- For the two-stage model, what is the dimensionless time \(kt\) at \(t=6\) hours when \(k=0.5\ \mathrm{h}^{-1}\), and what response fraction does the model predict there?
- If the fitted gain doubled while \(n\) and \(k\) remained unchanged, would the predicted time to half the plateau change? Explain.
- Why does a positive residual early and a negative residual later raise concern about the response shape?
- What limitation follows from comparing stage counts while fixing \(k\) at one specified value?
Solutions. (1) \(kt=(0.5\ \mathrm{h}^{-1})(6\ \mathrm{h})=3\). Thus \(F_2=1-\exp(-3)(1+3)\approx0.8009\), or about 80.09% of the plateau. (2) No. The response fraction is \(F_n(t;k)\), which does not depend on \(G\); doubling \(G\) doubles the absolute response but leaves the fraction-based timing unchanged. (3) It suggests the model is underpredicting early and overpredicting later, or otherwise has the wrong temporal shape; gain scaling alone may not correct that pattern. (4) The comparison cannot show whether another transit rate would fit better, and the preferred stage count is conditional on the fixed rate rather than a global comparison of stage-rate combinations.