From Turnover Equations to Predictions
Expression-turnover models turn a proposed mechanism into a time course. The central task is to keep the states and fluxes distinct: transcription changes the supply of mRNA, translation changes protein production, and protein degradation removes protein. A change to one flux need not cause an immediate change in every state. This distinction is especially useful when predicting the time course after a drug perturbation.
The worked problems below use first-order turnover and step changes in rates. Assume each compartment is well mixed; the stated rates are constant within each treatment interval; the drug effect acts through the specified rate only; and there is no feedback, saturation, or cell-count change. Initial steady states are identified explicitly. These assumptions make the equations linear and permit closed-form solutions. Units must be consistent: a first-order rate has units of inverse time, while a production rate has units of state units per time.
A transcriptional inhibitor that reduces transcription by a fraction \(I\), where \(0\leq I\leq1\), changes \(k_s\) to \((1-I)k_s\). A translation inhibitor instead changes \(k_{tr}\). In either case, calculate the new steady state from the altered rate, but do not assume the state jumps instantly to that steady state. The differential equations determine how quickly the states approach it.
Worked Example: Transcriptional Suppression
Worked Example: Transcriptional Suppression
Suppose an expression system is initially at steady state with \(M(0)=20\) mRNA units and \(P(0)=100\) protein units. Let \(k_m=0.500\ \text{day}^{-1}\), \(k_p=0.100\ \text{day}^{-1}\), and \(k_{tr}=0.500\) protein units per mRNA unit per day. Transcription is reduced by 60% at time zero. Find the post-treatment steady states and predict both states at day 4.
Before treatment, the transcription rate required to maintain the mRNA baseline is \(k_s=k_mM(0)=(0.500)(20)=10.0\) mRNA units/day. Protein production at baseline is \(k_{tr}M(0)=(0.500)(20)=10.0\) protein units/day, equal to protein loss \(k_pP(0)=(0.100)(100)=10.0\) protein units/day. Thus both stated initial values are consistent with steady state.
After suppression, transcription is \(0.40(10.0)=4.00\) mRNA units/day. The new mRNA steady state is \(4.00/0.500=8.00\) mRNA units. Protein production at that state is \((0.500)(8.00)=4.00\) protein units/day, so the new protein steady state is \(4.00/0.100=40.0\) protein units.
For a constant rate after the step, mRNA approaches its new steady state exponentially:
At day 4, \(e^{-2.00}=0.135335\), so \(M(4)=8.00+12.0(0.135335)=9.624\) mRNA units.
Protein is driven by the changing mRNA state, so its trajectory is not simply a single exponential toward 40.0. The coupled solution is derived below; applying it gives \(P(4)=40.0+75.0e^{-0.400}-15.0e^{-2.00}\). Using \(e^{-0.400}=0.670320\) and \(e^{-2.00}=0.135335\), \(P(4)=40.0+50.2740-2.03003=88.24\) protein units. The protein remains well above its new steady state at day 4 because protein turnover is slower than mRNA turnover in this example.
Why the Coupled Time Course Has Two Exponentials
The protein solution can be obtained by substituting the mRNA trajectory into the protein equation. This derivation also shows why a protein response can have a delayed shape even when the transcriptional perturbation begins immediately.
For the example, \(\Delta M=12.0\) mRNA units and \(k_{tr}\Delta M=6.00\) protein units/day. The coefficient is \(6.00/(0.100-0.500)=-15.0\), giving \(P(t)=40.0+60.0e^{-0.100t}-15.0(e^{-0.500t}-e^{-0.100t})\), or \(P(t)=40.0+75.0e^{-0.100t}-15.0e^{-0.500t}\). At \(t=0\), this returns \(P(0)=100\); its initial derivative is zero, as it must be: transcription inhibition does not instantly change mRNA, so protein production and loss are initially still balanced. These checks help catch sign errors in the coupled solution.
Worked Example: A Temporary Translation Blockade
Worked Example: A Temporary Translation Blockade
Suppose baseline protein abundance is 100 units and \(k_p=0.200\ \text{day}^{-1}\). Translation is completely blocked for 3 days, then restored to baseline. Assume mRNA remains at its baseline steady state throughout. Predict protein abundance at the end of the blockade and at day 8.
At baseline, protein production must equal loss: \((0.200)(100)=20.0\) protein units/day. During the blockade, translation is zero, so \(dP/dt=-0.200P\). The initial condition is \(P(0)=100\), giving \(P(t)=100e^{-0.200t}\) for \(0\leq t\leq3\).
At day 3, \(P(3)=100e^{-0.600}=100(0.548812)=54.88\) units. Once translation resumes, production returns to 20.0 units/day, and the steady state returns to 100 units. For the following 5 days, \(P(8)=100+(54.8812-100)e^{-0.200(5)}\). Since \(e^{-1.00}=0.367879\), \(P(8)=100+(-45.1188)(0.367879)=83.40\) units.
The solution is continuous at the time of release: protein does not jump from 54.88 to 100. Instead, its production rate changes and its abundance then recovers gradually. The assumption of unchanged mRNA is appropriate only if the perturbation affects translation and not transcription or mRNA degradation.
Worked Example: Drug-Modulated Indirect Response Production
Worked Example: Drug-Modulated Indirect Response Production
Let \(R(t)\) be a measured response with baseline \(R(0)=40.0\) response units and first-order loss rate \(k_{out}=0.300\ \text{day}^{-1}\). Drug inhibits response production according to \(I(C)=I_{max}C/(IC_{50}+C)\), where \(I_{max}=0.750\), \(C=4.00\) concentration units, and \(IC_{50}=1.00\) concentration unit. Assume concentration is constant from time zero. Calculate the inhibited steady state and \(R(4)\).
The baseline production rate is \(k_{in}=k_{out}R(0)=(0.300)(40.0)=12.0\) response units/day. The fractional inhibition at the stated concentration is \(I(C)=0.750(4.00)/(1.00+4.00)=0.600\). Thus residual production is \(12.0(1-0.600)=4.80\) response units/day, and the inhibited steady state is \(R_{ss}=4.80/0.300=16.0\) response units.
Under constant production and first-order loss, \(dR/dt=4.80-0.300R\). Its solution is \(R(t)=16.0+(40.0-16.0)e^{-0.300t}\). At day 4,
The prediction lies between the initial value of 40.0 and the inhibited steady state of 16.0, as expected for a stable first-order system. It is a time-course prediction under the assumed constant concentration, not a prediction that the response immediately reaches 16.0.
Checks, Alternatives, and Limitations
Before interpreting a model prediction, check the balances and the shape of the solution. At a claimed steady state, production must equal loss in every state equation. At the start of a perturbation, a state can change instantaneously only if the model includes an instantaneous jump mechanism; a change in a rate alone does not create such a jump. For a stable first-order turnover equation with positive loss rate, a constant new input drives the state toward its new steady state without crossing it. In a coupled model, the driven protein state may show a more complex transient, but its initial value and derivative must still agree with the equations.
The analytic step solutions are efficient when rates are constant over intervals and the initial conditions are known. For time-varying drug concentrations, multiple dosing, nonlinear degradation, feedback, or several linked states, numerical integration is usually more practical. It should use the same equations and initial conditions; it does not repair a mistaken mechanism or inconsistent units. A quasi-steady-state approximation for mRNA may simplify the model when mRNA adjusts much faster than protein, but it can miss the transient after a rapid perturbation. Likewise, a model that directly changes protein degradation can produce a different time course from one that changes transcription, even if both are tuned to the same final protein level.
Common diagnostic warning signs include a predicted steady state that fails the production-loss balance, discontinuous state values at a rate switch, a response moving away from a stable new steady state under constant rates, or negative abundance produced by a model intended to represent nonnegative states. Recheck signs, initial conditions, rate units, and whether the drug effect was applied to production or loss. A mismatch between predicted and observed paired mRNA and protein trajectories may indicate that the assumed mechanism is incomplete; it does not by itself establish which alternative mechanism is correct.
These models are deliberately compact. First-order turnover assumes a constant fractional loss rate, and the Emax-style inhibition example assumes an instantaneous concentration effect with no delay or tolerance. Real systems may violate these assumptions. Treat a good numerical fit as evidence that a model can describe the observations, not as proof that its biological mechanism is unique.
Check Your Understanding
Try each problem, then compare with the solutions.
- An mRNA state has \(k_s=6.00\) units/day and \(k_m=0.300\ \text{day}^{-1}\). What is its steady state?
- A protein starts at 80 units, with \(k_p=0.100\ \text{day}^{-1}\). Its production is completely blocked for 2 days. What is its abundance at day 2?
- A response has baseline 30 units and \(k_{out}=0.200\ \text{day}^{-1}\). Production is reduced by 50% at time zero. What is its new steady state, and what is its value after 3 days?
- Why does a translation blockade not necessarily change mRNA abundance in the model used here?
Solutions. (1) \(M_{ss}=k_s/k_m=6.00/0.300=20.0\) units. (2) With zero production, \(P(2)=80e^{-(0.100)(2)}=80e^{-0.200}=65.50\) units. (3) Baseline production is \((0.200)(30)=6.00\) units/day; the new production is 3.00 units/day, so \(R_{ss}=3.00/0.200=15.0\) units. The time course is \(R(t)=15.0+(30.0-15.0)e^{-0.200t}\), hence \(R(3)=15.0+15.0e^{-0.600}=23.23\) units. (4) Translation blockade changes \(k_{tr}\), not the transcription or degradation terms in the mRNA equation. With those rates unchanged, the mRNA equation has no direct drug effect; a real system could still show an mRNA change through additional mechanisms not included in this model.