Implications of immune-mediated metastatic growth on metastatic dormancy, blow-up, early detection, and treatment

Abstract

Metastatic seeding of distant organs can occur in the very early stages of primary tumor development. Once seeded, these micrometastases may enter a dormant phase that can last decades. Curiously, the surgical removal of the primary tumor can stimulate the accelerated growth of distant metastases, a phenomenon known as metastatic blow-up. Recent clinical evidence has shown that the immune response can have strong tumor promoting effects. In this work, we investigate if the pro-tumor effects of the immune response can have a significant contribution to metastatic dormancy and metastatic blow-up. We develop an ordinary differential equation model of the immune-mediated theory of metastasis. We include both anti- and pro-tumor immune effects, in addition to the experimentally observed phenomenon of tumor-induced immune cell phenotypic plasticity. Using geometric singular perturbation analysis, we derive a rather simple model that captures the main processes and, at the same time, can be fully analyzed. Literature-derived parameter estimates are obtained, and model robustness is demonstrated through a time dependent sensitivity analysis. We determine conditions under which the parameterized model can successfully explain both metastatic dormancy and blow-up. The results confirm the significant active role of the immune system in the metastatic process. Numerical simulations suggest a novel measure to predict the occurrence of future metastatic blow-up in addition to new potential avenues for treatment of clinically undetectable micrometastases.

Appendix

Here we prove Proposition 3 and provide additional details concerning the parameter sensitivity analysis reported in Fig. 4. We begin by proving Proposition 3.

Table 3 Details of the parameter sensitivity analysis depicted in Fig. 4

Proof

Consider $(u,x(u), y(u))\in {\mathcal {M}}$. We compute the derivatives $x_{u}(u)$ and $y_{u}(u)$. We begin with $x_{u}$:

$$\begin{aligned} \begin{aligned} x_{u}(u)&= \partial _{u}\left( \frac{\alpha }{-\left( \lambda (u) - \rho u - \omega - ed(u)\right) }\right) \\&= \frac{\alpha \left( \lambda '(u) - \rho - ed'(u)\right) }{\left( \lambda (u) - \rho u - \omega - ed(u)\right) ^{2}}. \end{aligned} \end{aligned}$$

(14)

The sign of $x_{u}$ depends entirely on the sign of

$$\begin{aligned} \lambda '(u) - \rho - ed'(u). \end{aligned}$$

With the choices we made for $\lambda (u)$ and ed(u), we arrive at the following chain of equivalent conditions:

$$\begin{aligned} \begin{aligned} x_{u}(u)&> 0\\ \lambda '(u) - \rho - ed'(u)&> 0\\ \frac{a_{1}b_{1}}{(b_{1}+u)^{2}} - \rho - \chi&> 0\\ \frac{a_{1}b_{1}}{\rho + \chi }&> (b_{1}+u)^{2}\\ 0&> u^{2} + 2b_{1}u + b_{1}^{2} - \frac{a_{1}b_{1}}{\rho + \chi }. \end{aligned} \end{aligned}$$

(15)

The roots to the above quadratic are given by

$$\begin{aligned} u_{+,-} = -b_{1} \pm \sqrt{\frac{a_{1}b_{1}}{\rho + \chi }}. \end{aligned}$$

(16)

This gives distinct, real roots (assuming that $a_{1}b_{1}\ne 0$) with at least one of them negative. If $u_{+} > 0$, then we see that $x_{u} > 0$ for $u < u_{+}$ and negative otherwise. If $u_{+} < 0$, then we simply have $x_{u} < 0$ for all non-negative u.

Next, we consider $y_{u}$. We show simply that $y_{u}$ is non-negative, as

$$\begin{aligned} \begin{aligned} y_{u}(u)&= \partial _{u}\left( \frac{\psi + ed(u)x(u)}{-\left( f(u) - \tau \right) }\right) \\&= \frac{-(ed'(u)x(u) + ed(u)x_{u}(u))(f(u) - \tau ) + (q\psi + ed(u)x(u))f'(u)}{(f(u) - \tau )^{2}}\\&\ge \frac{-(ed'(u)x(u) + ed(u)x_{u}(u))(f(u) - \tau )}{(f(u) - \tau )^{2}}. \end{aligned}\nonumber \\ \end{aligned}$$

(17)

The final inequality results from the fact that $(\psi + ed(u)x(u))f'(u) \ge 0$ since f is increasing (A1). Now, we can use the expressions for x(u) and $x_{u}$, as well as our choice of $ed(u) = \chi u$ to arrive at

$$\begin{aligned} \begin{aligned} (17) =&\left[ \frac{\chi \alpha }{-(\lambda (u) - \rho u - \omega - ed(u))} + \frac{\chi u \alpha (\lambda '(u) - \rho - ed'(u))}{(\lambda (u) - \rho u - \omega - ed(u))^{2}}\right] (\tau - f(u)) \\ =&\frac{\chi \alpha }{-(\lambda (u) - \rho u - \omega - ed(u))}(\tau - f(u))\left[ 1 - \frac{u(\lambda '(u) - \rho - ed'(u))}{\lambda (u) - \rho u - \omega - ed(u)}\right] . \end{aligned}\nonumber \\ \end{aligned}$$

(18)

The sign of this expression depends only on the sign of the term

$$\begin{aligned} \left[ 1 - \frac{u(\lambda '(u) - \rho - ed'(u))}{\lambda (u) - \rho u - \omega - ed(u)}\right] = \frac{\lambda (u) - \omega - ed(u) - \lambda '(u)u + ed'(u)u}{\lambda (u) - \rho u - \omega - ed(u)}. \end{aligned}$$

By assumption (A2), the denominator is always negative, and therefore the sign of the above expression is determined by the sign of its numerator. With the choices for ed(u) and $\lambda (u)$ from Sect. 2.3, the numerator simplifies to

$$\begin{aligned} \frac{u^{2}(a_{1} - \omega ) - 2b_{1}\omega u - \omega b_{1}^{2}}{(b_{1}+u)^{2}}. \end{aligned}$$

Using the fact that $a_{1}-\omega < 0$ (A2) guarantees that the quadratic is negative for all $u \ge 0$, and therefore $y_{u}(u) \ge 0$ for all $u \ge 0$. $\square$