A New Model System for Exploring Assembly Mechanisms of the HIV-1 Immature Capsid In Vivo

Abstract

The assembly of the HIV-1 immature capsid (HIC) is an essential step in the virus life cycle. In vivo, the HIC is composed of $\sim 420$ hexameric building blocks, and it takes 5–6 min to complete the assembly process. The involvement of numerous building blocks and the rapid timecourse makes it difficult to understand the HIC assembly process. In this work, we study HIC assembly in vivo by using differential equations. We first obtain a full model with 420 differential equations. Then, we reduce six addition reactions for separate building blocks to a single complex reaction. This strategy reduces the full model to 70 equations. Subsequently, the theoretical analysis of the reduced model shows that it might not be an effective way to decrease the HIC concentration at the equilibrium state by decreasing the microscopic on-rate constants. Based on experimental data, we estimate that the nucleating structure is much smaller than the HIC. We also estimate that the microscopic on-rate constant for nucleation reactions is far less than that for elongation reactions. The parametric collinearity investigation testifies the reliability of these two characteristics, which might explain why free building blocks do not readily polymerize into higher-order polymers until their concentration reaches a threshold value. These results can provide further insight into the assembly mechanisms of the HIC in vivo.

Appendix

Theorem 1

The positive equilibrium point of the model (1) exists and is unique.

Proof

Let the right-hand sides of the model (1) equal zero.

$$\begin{aligned} \left\{ \begin{aligned}\\&-\left( 7k_{7}^{+}{{[H]}^{7}}+6k_{13}^{+}{{[H]}^{6}}[{{H}_{7}}]+\cdots +6k_{\mathrm{nuc}}^{+}{{[H]}^{6}}[{{H}_{nuc-6}}] \right. \\&\left. \quad +\cdots +6k_{N}^{+}{{[H]}^{6}}[{{H}_{N-6}}] \right) +\\&\text { }\left( 7k_{7}^{-}[{{H}_{7}}]+6k_{13}^{-}[{{H}_{13}}]+\cdots +6k_{\mathrm{nuc}}^{-}[{{H}_{\mathrm{nuc}}}]+\cdots +6k_{N}^{-}[{{H}_{N}}] \right) +v=0 \\&k_{7}^{+}{{[H]}^{7}}-k_{13}^{+}{{[H]}^{6}}[{{H}_{7}}]+k_{13}^{-}[{{H}_{13}}]-k_{7}^{-}[{{H}_{7}}]=0 \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \\&k_{\mathrm{nuc}}^{+}{{[H]}^{6}}[{{H}_{\mathrm{nuc}-6}}]-k_{nuc+6}^{+}{{[H]}^{6}}[{{H}_{\mathrm{nuc}}}]+k_{\mathrm{nuc}+6}^{-}[{{H}_{\mathrm{nuc}+6}}]-k_{\mathrm{nuc}}^{-}[{{H}_{\mathrm{nuc}}}]=0 \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \\&k_{N-6}^{+}{{[H]}^{6}}[{{H}_{N-12}}]-k_{N}^{+}{{[H]}^{6}}[{{H}_{N-6}}]+k_{N}^{-}[{{H}_{N}}]-k_{N-6}^{-}[{{H}_{N-6}}]=0 \\&k_{N}^{+}{{[H]}^{6}}[{{H}_{N-6}}]-k_{N}^{-}[{{H}_{N}}]-r[{{H}_{N}}]=0 \\ \end{aligned} \right. \end{aligned}$$

(6)

From the last equation of Eq. (6), we can obtain

$$\begin{aligned} k_{N}^{+}{{[H]}^{6}}[{{H}_{N-6}}]-k_{N}^{-}[{{H}_{N}}]=r[{{H}_{N}}] \end{aligned}$$

(7)

By adding the last two equations of Eq. (6), we can obtain

$$\begin{aligned} k_{N-6}^{+}{{[H]}^{6}}[{{H}_{N-12}}]-k_{N-6}^{-}[{{H}_{N-6}}]=r[{{H}_{N}}] \end{aligned}$$

(8)

In the same way, we can obtain

$$\begin{aligned} \begin{aligned}&k_{N-12}^{+}{{[H]}^{6}}[{{H}_{N-18}}]-k_{N-12}^{-}[{{H}_{N-12}}]=r[{{H}_{N}}] \\&\ \ \ \ \ \ \ \ \ \ \ \vdots \\&k_{13}^{+}{{[H]}^{6}}[{{H}_{7}}]-k_{13}^{-}[{{H}_{13}}]=r[{{H}_{N}}] \\&k_{7}^{+}{{[H]}^{7}}-k_{7}^{-}[{{H}_{7}}]=r[{{H}_{N}}] \\ \end{aligned} \end{aligned}$$

(9)

By substituting Eqs.  (9) , (8) and (7) into the first equation of Equations (6), we obtain

$$\begin{aligned} r[{{H}_{N}}]=\frac{v}{N} \end{aligned}$$

(10)

By substituting Eq.  (10) into the last equation of Eq.  (9), we obtain

$$\begin{aligned}{}[{{H}_{7}}]=\frac{k_{7}^{+}{{[H]}^{7}}-r[{{H}_{N}}]}{k_{7}^{-}}=\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{7}}-\frac{1}{k_{7}^{-}}\frac{v}{N} \end{aligned}$$

(11)

By substituting Eqs.  (10) and (11) into the penultimate equation of Eq.  (9), we obtain

$$\begin{aligned} =\frac{k_{13}^{+}{{[H]}^{6}}[{{H}_{7}}]-r[{{H}_{N}}]}{k_{13}^{-}}=\frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{13}}-\frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{6}}\frac{v}{N}-\frac{1}{k_{13}^{-}}\frac{v}{N}\qquad \end{aligned}$$

(12)

In the same way, we can obtain

$$\begin{aligned} \begin{aligned}{}[H_{19}]=&\frac{k_{19}^{+}}{k_{19}^{-}}\frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{19}}-\frac{k_{19}^{+}}{k_{19}^{-}}\frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{12}}\frac{v}{N}-\frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}{{[H]}^{6}}\frac{v}{N}-\frac{1}{k_{19}^{-}}\frac{v}{N}\\&\ \ \ \ \ \ \ \ \ \ \ \vdots \\ [{{H}_{\mathrm{nuc}}}]=&\frac{k_{\mathrm{nuc}}^{+}}{k_{\mathrm{nuc}}^{-}}\frac{k_{\mathrm{nuc}-6}^{+}}{k_{\mathrm{nuc}-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{\mathrm{nuc}}}-\frac{k_{\mathrm{nuc}}^{+}}{k_{\mathrm{nuc}}^{-}}\frac{k_{\mathrm{nuc}-6}^{+}}{k_{\mathrm{nuc}-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{\mathrm{nuc}-7}}\frac{v}{N}- \\&\text { }\frac{k_{\mathrm{nuc}}^{+}}{k_{\mathrm{nuc}}^{-}}\frac{k_{\mathrm{nuc}-6}^{+}}{k_{\mathrm{nuc}-6}^{-}}\cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}{{[H]}^{\mathrm{nuc}-13}}\frac{v}{N}-\cdots -\frac{k_{\mathrm{nuc}}^{+}}{k_{\mathrm{nuc}}^{-}}\frac{1}{k_{\mathrm{nuc}-6}^{-}}{{[H]}^{6}}\frac{v}{N}-\frac{1}{k_{\mathrm{nuc}}^{-}}\frac{v}{N} \\&\ \ \ \ \ \ \ \ \ \ \ \vdots \\ \text { }[{{H}_{N}}]=&\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{N}}-\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{N-7}}\frac{v}{N}- \\&\text { }\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}{{[H]}^{N-13}}\frac{v}{N}-\cdots -\frac{k_{N}^{+}}{k_{N}^{-}}\frac{1}{k_{N-6}^{-}}{{[H]}^{6}}\frac{v}{N} \\&\quad -\frac{1}{k_{N}^{-}}\frac{v}{N} \\ \end{aligned}\nonumber \\ \end{aligned}$$

(13)

By substituting the last equation of Eq.  (13) into Eq.  (10), we obtain

$$\begin{aligned} \begin{aligned}&\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{N}}-\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{N-7}}\frac{v}{N}-\\&\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}{{[H]}^{N-13}}\frac{v}{N}-\cdots -\frac{k_{N}^{+}}{k_{N}^{-}}\frac{1}{k_{N-6}^{-}}{{[H]}^{6}}\frac{v}{N}-\left( \frac{1}{k_{N}^{-}}+\frac{1}{r} \right) \frac{v}{N}=0 \\ \end{aligned} \end{aligned}$$

(14)

Let the right-hand side of Eq.  (14) be f([H])

$$\begin{aligned} \begin{aligned}&f([H])=\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}{{[H]}^{N}}-\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}{{[H]}^{N-7}}\frac{v}{N}- \\&\text { }\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}{{[H]}^{N-13}}\frac{v}{N}-\cdots -\frac{k_{N}^{+}}{k_{N}^{-}}\frac{1}{k_{N-6}^{-}}{{[H]}^{6}}\frac{v}{N}-\left( \frac{1}{k_{N}^{-}}+\frac{1}{r} \right) \frac{v}{N}. \qquad \\ \end{aligned} \end{aligned}$$

(15)

Because $f'([H])$ has both positive and negative terms, it is possible that $f'([H])$ is not a monotone function. It is difficult to proof its monotony by $f'([H])>0$ or $f'([H])<0, H\in (0,+\infty )$, directly.

In Eq.  (15), we find that only the coefficient of the highest-order term of f([H]) is positive, and the other terms are all negative. If we convert the only positive term into a constant, its derivative will become zero. As a result, all terms in its derivative will become negative and the monotonicity of this new function will be easy to proof.

Based on the above analysis, let

$$\begin{aligned} \begin{aligned}&F([H])=\frac{f([H])}{{{[H]}^{N}}}=\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{k_{7}^{+}}{k_{7}^{-}}-\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{7}}}- \\&\text { }\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{13}}}-\cdots -\frac{k_{N}^{+}}{k_{N}^{-}}\frac{1}{k_{N-6}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{N-6}}} \\&\quad -\left( \frac{1}{k_{N}^{-}}+\frac{1}{r} \right) \frac{v}{N}\frac{1}{{{[H]}^{N}}} \\ \end{aligned} \end{aligned}$$

(16)

Obviously, $F([H])=0$ and $f([H])=0$ for $H>0$ have the same solutions.

The derivative of F([H]) is

$$\begin{aligned} \begin{aligned}&{F}'([H])=0+7\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}}\cdots \frac{k_{13}^{+}}{k_{13}^{-}}\frac{1}{k_{7}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{8}}}+13\frac{k_{N}^{+}}{k_{N}^{-}}\frac{k_{N-6}^{+}}{k_{N-6}^{-}} \\&\quad \cdots \frac{k_{19}^{+}}{k_{19}^{-}}\frac{1}{k_{13}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{14}}}+\cdots \\&\text { }+(N-6)\frac{k_{N}^{+}}{k_{N}^{-}}\frac{1}{k_{N-6}^{-}}\frac{v}{N}\frac{1}{{{[H]}^{N-5}}}+N\left( \frac{1}{k_{N}^{-}}+\frac{1}{r} \right) \frac{v}{N}\frac{1}{{{[H]}^{N+1}}} \\ \end{aligned} \end{aligned}$$

(17)

Because $k_{i}^{\pm }>0,v>0,r>0$, ${F}'([H])>0$, for $[H]\in (0,+\infty )$. As a result, $y=F([H])$ is a strictly monotone increasing function in $(0,+\infty )$. In addition,

$$\begin{aligned} \underset{[H]\rightarrow 0+}{\mathop {\lim }}\,F([H])<0\ \text {and}\ \underset{[H]\rightarrow +\infty }{\mathop {\lim }}\,F([H])>0. \end{aligned}$$

Therefore, there is a unique positive solution denoted by $[{{H}^{*}}]$ for $F([H])=0$. Because $F([H])=0$ and $f([H])=0$ in $(0,+\infty )$ have the same solution, $[{{H}^{*}}]$ is also the unique positive solution of $f([H])=0$. Therefore, the positive equilibrium point of the model (1) exists and is unique.

Theorem 2

At equilibrium, the concentration of $H_N$ is independent of two microscopic on-rate constants $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$, and the concentration of free hexamers increases as $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$ decrease.

Proof

From Eq. (10), we can obtain

$$\begin{aligned}{}[{{H}_{N}}]=\frac{v}{rN} \end{aligned}$$

We can obviously see that the concentration of $H_N$ at equilibrium does not depend on two microscopic on-rate constants $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$.

Based on Eq. (2), we can obtain

$$\begin{aligned} \frac{k_{i}^{+}}{k_{i}^{-}}=\frac{{{S}_{i}}}{{{\sigma }_{i}}K_{\mathrm{Acon}}^{-{{c}_{i}}}}, \end{aligned}$$

which is independent of $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$. So, in Eq. (16), the constant term of F([H]) is independent of $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$. But the coefficients of the other terms all have the factor $-\frac{1}{k_{i}^{-}}$. Because $k_{i}^{-}={{f}_{\mathrm{nuc}}}{{\sigma }_{i}}K_{\mathrm{Acon}}^{-{{c}_{i}}}$ or $k_{i}^{-}={{f}_{\mathrm{elong}}}{{\sigma }_{i}}K_{\mathrm{Acon}}^{-{{c}_{i}}}$ in Eq. (2), the coefficients of these terms in F([H]) will decrease as $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$ decrease. As a result, F(H) will decrease as $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$ decrease. In addition, $y=F([H])$ is a strictly monotone increasing function in $(0,+\infty )$. Therefore, the unique positive solution of $F([H])=0$ will increase as $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$ decrease. So, at equilibrium, the concentration of free building blocks increases as $f_{\mathrm{nuc}}$ and $f_{\mathrm{elong}}$ decrease.