A Mathematical Model for Amyloid-𝜷 Aggregation in the Presence of Metal Ions: A Timescale Analysis for the Progress of Alzheimer Disease

Abstract

The aggregation of amyloid-𝛽 (A𝛽) proteins through their self-assembly into oligomers, fibrils, or senile plaques is advocated as a key process of Alzheimer’s disease. Recent studies have revealed that metal ions play an essential role in modulating the aggregation rate of amyloid-𝛽 (A𝛽) into senile plaques because of high binding affinity between A𝛽 proteins and metal ions. In this paper, we proposed a mathematical model as a set of coupled kinetic equations that models the self-assembly of amyloid-𝛽 (A𝛽) proteins in the presence of metal ions. The numerical simulations capture four timescales in the A𝛽 dynamics associated with three important events which include the formation of the amyloid–metal complex, the homogeneous aggregation of the amyloid–metal complexes, and the non-homogeneous aggregation of the amyloid–metal complexes. The method of singular perturbation is used to identify these timescales in the framework of slow–fast systems.

Appendix

Proof of Theorem 2.1.

• Evaluation of ${P_h(t)}$

Setting $i=j$ in Eq. (2), applying the summation over i and using the definition of $P_h$ yields,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d} P_h(t)}{\mathrm{d}t}&=k_nf(t,1,1)^2+ k_hf(t,1,1)\left[ \sum \limits _{i=3}^\infty f(t,i-1,i-1)-\sum \limits _{i=2}^\infty f(t,i,i)\right] \\&\quad -k_{nh}m(t)\sum \limits _{i=2}^\infty f(t, i,i). \end{aligned} \end{aligned}$$

Since, $\sum \limits _{i=3}^\infty f(t,i-1,i-1)=\sum \limits _{i=2}^\infty f(t,i,i)$, it is simplified to:

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d} P_h(t)}{\mathrm{d}t}=k_nf(t,1,1)^2-k_{nh}m(t)P_h(t) \end{aligned} \end{aligned}$$

• Evaluation of P(t)

Observe that,

$$\begin{aligned} \begin{aligned} \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1}\frac{\partial f(t,i,j)}{\partial t}&=k_{nh}m(t)\left[ \sum \limits _{i=3}^\infty \sum \limits _{j=1}^{i-1}f(t,i-1,j) -\sum \limits _{i=3}^\infty \sum \limits _{j=1}^{i-1}f(t,i,j)\right] \\&=k_{nh}m(t)\left[ f(t,2,2)+\sum \limits _{i=3}^\infty \left( \sum \limits _{j=2}^i f(t,i,j) -\sum \limits _{j=2}^{i-1}f(t,i,j)\right) \right] \\&=k_{nh}m(t)P_h(t), \end{aligned} \end{aligned}$$

and since,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d} P(t)}{\mathrm{d} t}&=\frac{\mathrm{d} P_h(t)}{\mathrm{d}t}+ \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1}\frac{\partial f(t,i,j)}{\partial t}, \end{aligned} \end{aligned}$$

it implies that,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d} P(t)}{\mathrm{d} t}=k_nf(t,1,1)^2. \end{aligned} \end{aligned}$$

• Evaluation of $M_h(t)$

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d} M_h(t)}{\mathrm{d}t}&=2k_nf(t,1,1)^2 +k_hf(t,1,1)\left[ \sum \limits _{i=3}^\infty i\Big [f(t,i-1,i-1)-\sum \limits _{i=2}^\infty i f(t,i,i)\right] \\&\quad -k_{nh}m(t)\sum \limits _{i=2}^\infty i f(t, i,i) \end{aligned} \end{aligned}$$

(31)

Since,

$$\begin{aligned} \begin{aligned} \sum \limits _{i=3}^\infty if(t,i-1,i-1)&=\sum \limits _{i=2}^\infty if(t,i,i)+\sum \limits _{i=2}^\infty f(t,i,i),\\ \end{aligned} \end{aligned}$$

we get:

$$\begin{aligned} \frac{\mathrm{d} M_h(t)}{\mathrm{d} t}=2k_nf(t,1,1)^2+k_hf(t,1,1)P_h(t)-k_{nh}m(t)M_h(t) \end{aligned}$$

• Evaluation of $M_1(t)$

Observe that,

$$\begin{aligned} \begin{aligned} \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} i\frac{\partial f(t,i,j)}{\partial t}&= k_nm(t)\left[ \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1}if(t,i-1,j) -\sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} if(t,i,j)\right] \\&= k_nm(t)\left[ \sum \limits _{i=2}^\infty \sum \limits _{j=2}^{i-1}if(t,i,j) +\sum \limits _{i=2}^\infty \sum \limits _{j=2}^{i-1} f(t,i,j) -\sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} if(t,i,j)\right] \\&= k_nm(t)\left[ P(t)+M_h(t)\right] , \end{aligned} \end{aligned}$$

and since,

$$\begin{aligned} \frac{\mathrm{d} M_1(t)}{\mathrm{d} t}=\frac{\mathrm{d} M_h(t)}{\mathrm{d}t}+ \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} i\frac{\partial f(t,i,j)}{\partial t} \end{aligned}$$

we get:

$$\begin{aligned} \frac{\mathrm{d} M_1(t)}{\mathrm{d} t}=2k_nf(t,1,1)^2+k_hf(t,1,1)P_h(t)+k_{nh}m(t)P(t) \end{aligned}$$

• Evaluation of $M_2(t)$

Observe that,

$$\begin{aligned} \begin{aligned} \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} j\frac{\partial f(t,i,j)}{\partial t}&= k_nm(t)\left[ \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1}jf(t,i-1,j) -\sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} jf(t,i,j)\right] \\&= k_nm(t)\left[ \sum \limits _{i=2}^\infty \sum \limits _{j=2}^{i}jf(t,i,j) -\sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} jf(t,i,j)\right] \\&= k_nm(t)M_h(t), \end{aligned} \end{aligned}$$

and since,

$$\begin{aligned} \frac{\mathrm{d} M_2(t)}{\mathrm{d} t}=\frac{\mathrm{d} M_h(t)}{\mathrm{d}t}+ \sum \limits _{i=3}^\infty \sum \limits _{j=2}^{i-1} j\frac{\partial f(t,i,j)}{\partial t}, \end{aligned}$$

we get,

$$\begin{aligned} \frac{\mathrm{d} M_2(t)}{\mathrm{d} t}=2k_nf(t,1,1)^2+k_hf(t,1,1)P_h(t) \end{aligned}$$