Tissue Profiles During Infusion and Washout

From a Blood Input to a Tissue Profile

A constant-rate infusion does not generally make tissue concentration rise at a constant rate. Blood concentration changes over time, and a perfusion-limited tissue responds to that changing input with its own finite equilibration rate. When the infusion stops, the blood concentration may remain substantial for a while; tissue concentration can therefore continue to rise briefly even though no more drug is being infused.

The calculations below describe one tissue responding to a specified blood concentration profile. In a full PBPK model, blood and tissue concentrations may be coupled and solved together. Here, the blood profile is treated as a known input, such as one obtained from a separately solved blood or whole-body model.

Definition: For a perfusion-limited tissue, let $C_t(t)$ be tissue concentration, $C_b(t)$ be blood concentration, $Q$ be tissue blood flow, $V_t$ be tissue volume, and $K_{t:b}$ be the tissue-to-blood partition coefficient. The tissue equation is $$ \frac{dC_t}{dt}=\lambda_t\left(K_{t:b}C_b(t)-C_t(t)\right), \qquad \lambda_t=\frac{Q}{V_tK_{t:b}}. $$ Here $\lambda_t$ has units of inverse time. The equation assumes perfusion limitation, a constant flow and tissue volume over the period considered, a constant partition coefficient, and instantaneous equilibration between blood leaving the tissue and the tissue concentration implied by $K_{t:b}$.

The term $K_{t:b}C_b(t)$ is the tissue concentration toward which the current blood concentration drives the tissue. If that target exceeds $C_t(t)$, tissue concentration increases; if it is lower, tissue concentration decreases. The rate $\lambda_t$ determines how quickly the tissue responds. Notice that changing $K_{t:b}$ can change both the target concentration and the equilibration rate: with $Q/V_t$ fixed, a larger partition coefficient means a smaller $\lambda_t$.

Deriving the Profile During Infusion

Suppose the tissue starts drug-free, $C_t(0)=0$, and blood concentration during infusion is represented by the specified rising profile $C_b(t)=C_\infty(1-e^{-at})$, where $C_\infty$ is a concentration scale and $a>0$ is a rate constant. This profile is an input assumption for the calculation, not a claim that every constant-rate infusion produces this exact blood trajectory.

Multiplying the tissue equation by the integrating factor $e^{\lambda_t t}$ gives $d(e^{\lambda_t t}C_t)/dt=\lambda_tK_{t:b}e^{\lambda_t t}C_b(t)$. Integrating from zero to $t$, and substituting the specified blood profile, yields

C_t(t)=\lambda_tK_{t:b}C_\infty \int_0^t e^{-\lambda_t(t-s)}(1-e^{-as})\,ds.

For $a\ne\lambda_t$, evaluating the two exponential integrals gives

C_t(t)=K_{t:b}C_\infty \left[ 1-\frac{\lambda_t e^{-at}-a e^{-\lambda_t t}}{\lambda_t-a} \right].

This result depends on the assumed initial condition and input profile. It does not say that tissue concentration immediately equals $K_{t:b}C_b(t)$; finite $\lambda_t$ creates a lag. If $a=\lambda_t$, the displayed formula has a zero denominator, but the solution is finite. Direct integration in that case gives

C_t(t)=K_{t:b}C_\infty \left[1-(1+\lambda_t t)e^{-\lambda_t t}\right].

Worked Example: Tissue Concentration at Infusion Cessation

Consider a hypothetical perfusion-limited tissue with $K_{t:b}=2$, $Q/V_t=0.5\ \text{h}^{-1}$, and therefore $\lambda_t=(Q/V_t)/K_{t:b}=0.25\ \text{h}^{-1}$. Let the supplied blood profile during infusion be $C_b(t)=10(1-e^{-0.5t})\ \text{mg/L}$, and let infusion stop at $T=4\ \text{h}$. Assume $C_t(0)=0$. The rates differ, so use the unequal-rate solution.

At cessation, the blood concentration is $$ C_b(4)=10(1-e^{-0.5(4)})=10(1-e^{-2})\approx 8.647\ \text{mg/L}. $$ The tissue target amplitude in the solution is $K_{t:b}C_\infty=2(10)=20\ \text{mg/L}$. Substitution gives $$ C_t(4)=20\left[1-\frac{0.25e^{-0.5(4)}-0.5e^{-0.25(4)}}{0.25-0.5}\right]. $$ Since $e^{-2}\approx0.1353$ and $e^{-1}\approx0.3679$, the fraction is $(0.25(0.1353)-0.5(0.3679)) / (-0.25)\approx0.6004$. Thus $C_t(4)\approx20(1-0.6004)=7.992\ \text{mg/L}$. The tissue is below its instantaneous target $K_{t:b}C_b(4)\approx17.29\ \text{mg/L}$, so its concentration is still increasing at the moment the infusion stops.

How Partitioning Changes a Finite-Infusion Profile

A comparison of tissues must keep track of which quantities are held fixed. In the next example, $Q/V_t=0.5\ \text{h}^{-1}$, the blood input, and the four-hour infusion duration are held fixed. Thus, $\lambda_t=(Q/V_t)/K_{t:b}$ changes with $K_{t:b}$. These are illustrative predictions under the same perfusion-limited model, not a claim that partitioning alone always determines observed tissue concentrations.

Worked Example: Three Partition Coefficients Under the Same Blood Input

Use $C_b(t)=10(1-e^{-0.5t})\ \text{mg/L}$, $T=4\ \text{h}$, and $C_t(0)=0$. Compare hypothetical tissues with $K_{t:b}=0.5,\ 1,$ and $2$, all with $Q/V_t=0.5\ \text{h}^{-1}$.

For $K_{t:b}=0.5$, $\lambda_t=1\ \text{h}^{-1}$. Using the unequal-rate expression, $$ C_t(4)=5\left[1-\frac{1e^{-2}-0.5e^{-4}}{1-0.5}\right] \approx5(1-0.2524)=3.738\ \text{mg/L}. $$

For $K_{t:b}=1$, $\lambda_t=0.5\ \text{h}^{-1}$, equal to $a$. Use the equal-rate form: $$ C_t(4)=10[1-(1+0.5(4))e^{-0.5(4)}] =10(1-3e^{-2})\approx5.940\ \text{mg/L}. $$

For $K_{t:b}=2$, $\lambda_t=0.25\ \text{h}^{-1}$; the preceding example gives $C_t(4)\approx7.992\ \text{mg/L}$. Under these specified conditions, the predicted tissue concentrations increase across the three partition coefficients. The result reflects both the larger tissue target associated with larger $K_{t:b}$ and the slower response rate that accompanies it; it should not be generalized without specifying the input, flow, tissue volume, and duration.

After the Infusion Stops

To calculate washout, let $u$ denote time since cessation. Assume blood concentration after cessation follows $C_b(T+u)=C_b(T)e^{-bu}$, with $b>0$, and tissue concentration at cessation is $C_t(T)$. The tissue concentration remains continuous at the cessation time: stopping the infusion changes its subsequent input trajectory, not the amount already in the tissue.

Applying the same integrating-factor argument with the new initial value gives, for $\lambda_t\ne b$,

C_t(T+u)=C_t(T)e^{-\lambda_t u} +\lambda_tK_{t:b}C_b(T) \frac{e^{-bu}-e^{-\lambda_t u}}{\lambda_t-b}.

The first term is the contribution from drug already in the tissue at cessation. The second is the continued contribution from drug remaining in blood. The initial post-cessation slope follows directly from the differential equation: $$ \left.\frac{dC_t}{du}\right|_{u=0} =\lambda_t\left(K_{t:b}C_b(T)-C_t(T)\right). $$ Consequently, stopping the infusion does not imply that tissue concentration immediately decreases. It initially rises if $C_t(T)<K_{t:b}C_b(T)$, falls if $C_t(T)>K_{t:b}C_b(T)$, and has zero instantaneous slope if they are equal. If $b=\lambda_t$, the washout expression is replaced by its finite limit, $e^{-\lambda_tu}[C_t(T)+\lambda_tK_{t:b}C_b(T)u]$.

Worked Example: A Temporary Rise After Cessation

Continue the hypothetical example with $K_{t:b}=2$, $\lambda_t=0.25\ \text{h}^{-1}$, $C_t(T)=7.992\ \text{mg/L}$, and $C_b(T)=8.647\ \text{mg/L}$. Suppose blood concentration after cessation declines with $b=0.5\ \text{h}^{-1}$. First, the initial tissue slope is $$ 0.25[2(8.647)-7.992] =0.25(9.302)\approx2.325\ \text{mg/L/h}. $$ The positive slope is consistent with tissue concentration being below its target at cessation.

Two hours after cessation, $$ C_t(T+2)=7.992e^{-0.25(2)} +(0.25)(2)(8.647) \frac{e^{-0.5(2)}-e^{-0.25(2)}}{0.25-0.5}. $$ The first term is $7.992e^{-0.5}\approx4.847\ \text{mg/L}$. The ratio in the second term is $(e^{-1}-e^{-0.5})/(-0.25)\approx0.9546$, so the second term is approximately $(0.5)(8.647)(0.9546)=4.127\ \text{mg/L}$. Therefore $C_t(T+2)\approx4.847+4.127=8.974\ \text{mg/L}$, which is higher than the cessation value of $7.992\ \text{mg/L}$. This does not mean the tissue will rise indefinitely: by two hours, blood concentration has fallen to $8.647e^{-1}\approx3.181\ \text{mg/L}$, making the current target $2C_b\approx6.362\ \text{mg/L}$, below the tissue concentration. The tissue slope at that time is consequently negative.

Checks, Alternatives, and Common Pitfalls

A reliable profile should pass several checks. Concentrations must have concentration units, while $a$, $b$, and $\lambda_t$ must have inverse-time units. The infusion solution must give $C_t(0)=0$, and the washout solution at $u=0$ must return $C_t(T)$. If a finite infusion is followed by a continuous blood decline, tissue concentration should also be continuous at cessation. The derivative should agree with $\lambda_t(K_{t:b}C_b-C_t)$ on each side, using the blood concentration applicable at that time.

For computation, the analytical expressions are useful when the input is exponential over each interval. A numerical ODE solver is a practical alternative for more complicated inputs or multiple intervals, provided it uses the same initial condition and parameter units. Another exact approach is to integrate the tissue response to the blood input directly, as in the convolution integral above. Whichever approach is used, compare results at interval boundaries and check whether the concentration initially rises or falls against the sign of the tissue equation.

Missing the partition coefficient in the target. The target is $K_{t:b}C_b$, not $C_b$. Omitting $K_{t:b}$ changes the predicted scale of tissue concentration.
Assuming tissue concentration falls as soon as infusion stops. The infusion rate is not the tissue slope. The slope depends on the difference $K_{t:b}C_b-C_t$, so residual blood drug can continue to drive uptake.
Using the unequal-rate formula when the rates are equal. A zero denominator signals that the limiting equal-rate expression is needed, not that the concentration is undefined.
Changing partition coefficient while silently holding equilibration rate fixed. If $Q/V_t$ is fixed, $\lambda_t=(Q/V_t)/K_{t:b}$; changing $K_{t:b}$ changes both the target and the response rate.

These expressions rely on the perfusion-limited assumptions stated above and on the specified blood input. If flow, tissue volume, or partitioning changes over time, or if the tissue cannot be represented by one rapidly mixed compartment, the equations may not describe the tissue profile adequately. A poor match between the assumed and observed blood profile also propagates directly into the tissue prediction.

Key takeaway: During infusion and washout, perfusion-limited tissue concentration is a lagged response to the blood-driven target $K_{t:b}C_b(t)$. Calculate the slope from the difference between that target and the current tissue concentration; cessation alone does not determine whether tissue concentration rises or falls.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

For a tissue with $Q/V_t=0.6\ \text{h}^{-1}$ and $K_{t:b}=3$, what is $\lambda_t$?
At infusion cessation, $K_{t:b}C_b(T)=12\ \text{mg/L}$ and $C_t(T)=9\ \text{mg/L}$. Is the initial post-cessation tissue slope positive, negative, or zero?
Why is the equal-rate infusion formula required when $a=\lambda_t$?
For the three tissues in the partition-coefficient example, which one has the smallest $\lambda_t$, and why?

Solutions. (1) $\lambda_t=(Q/V_t)/K_{t:b}=0.6/3=0.2\ \text{h}^{-1}$. (2) The slope is $\lambda_t(12-9)=3\lambda_t\ \text{mg/L/h}$, which is positive for $\lambda_t>0$; tissue concentration initially rises. (3) The unequal-rate expression has denominator $\lambda_t-a$, which is zero when the rates match; integration, or taking the limit, gives the finite expression $K_{t:b}C_\infty[1-(1+\lambda_t t)e^{-\lambda_t t}]$. (4) The $K_{t:b}=2$ tissue has the smallest rate, $0.25\ \text{h}^{-1}$, because $\lambda_t=(Q/V_t)/K_{t:b}$ with $Q/V_t$ held fixed.