Theoretical modeling of mechanical homeostasis of a mammalian cell under gravity-directed vector

Abstract

Translocation of dense nucleus along gravity vector initiates mechanical remodeling of a eukaryotic cell. In our previous experiments, we quantified the impact of gravity vector on cell remodeling by placing an MC3T3-E1 cell onto upward (U)-, downward (D)-, or edge-on (E)- orientated substrate. Our experimental data demonstrate that orientation dependence of nucleus longitudinal translocation is positively correlated with cytoskeletal (CSK) remodeling of their expressions and structures and also is associated with rearrangement of focal adhesion complex (FAC). However, the underlying mechanism how CSK network and FACs are reorganized in a mammalian cell remains unclear. In this paper, we developed a theoretical biomechanical model to integrate the mechanosensing of nucleus translocation with CSK remodeling and FAC reorganization induced by a gravity vector. The cell was simplified as a nucleated tensegrity structure in the model. The cell and CSK filaments were considered to be symmetrical. All elements of CSK filaments and cytomembrane that support the nucleus were simplified as springs. FACs were simplified as an adhesion cluster of parallel bonds with shared force. Our model proposed that gravity vector-directed translocation of the cell nucleus is mechanically balanced by CSK remodeling and FAC reorganization induced by a gravitational force. Under gravity, dense nucleus tends to translocate and exert additional compressive or stretching force on the cytoskeleton. Finally, changes of the tension force acting on talin by microfilament alter the size of FACs. Results from our model are in qualitative agreement with those from experiments.

Appendix 1: Length changes of MT or MF in E

As shown in Fig. 7, we set up a three-dimensional Cartesian coordinate system by taking the y-axis in the opposite direction of gravity and x-y coordinate plane at the contact surfaces between cell and substrate. R is cell radius, r is nucleus radius, and h is the distance from nucleus centroid to the substrate surface. $\theta$ is the angle between MFs or MTs and substrates. $\overline{p^{P}}$ is a radial MT/MF, and $\varphi$ is the angle between the projections of $\overline{p^{P}}$ on x-y coordinate plane and x axis.

Fig. 7

A nucleated cell in E-oriented substrate in the three-dimensional coordinate system before nucleus translocation

Cartesian coordinate for points P and p can be determined from the coordinate transformations

$$\begin{aligned} P = (R\cos \varphi , R\sin \varphi , 0),\quad p = (r\cos \varphi , r\sin \varphi , h). \end{aligned}$$

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Then, the initial length of MFs and MTs can be expressed as

$$\begin{aligned} \ell _0&= \sqrt{(R\cos \varphi -r\cos \varphi )^2 + (R\sin \varphi -r\sin \varphi )^2 + h^2}\nonumber \\&= \sqrt{(R-r)^2 + h^2}. \end{aligned}$$

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1.1 If $\Delta h^{E}=\Delta h_{\parallel }$, the compressed length and additional force of MTs

If nucleus translocation $\Delta h^{E}=\Delta h_{\parallel }$ is parallel to gravity, Cartesian coordinate for point p will be changed into:

$$\begin{aligned} p{\prime } = (r\cos \varphi ,\, r\sin \varphi -\Delta h_{\parallel },\, h). \end{aligned}$$

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The corresponding length of MTs can be expressed as:

$$\begin{aligned} \ell (\Delta h_{\parallel })&=\sqrt{(R\cos \varphi -r\cos \varphi )^{2}+(R\sin \varphi -r\sin \varphi +}\nonumber \\ {}&\qquad \overline{+\Delta h_{\parallel })^{2}+(0-h)^{2}}\nonumber \\&=\sqrt{(R-r)^{2}+h^{2}+\Delta h_{\parallel }^{2}+2(R-r)\Delta h_{\parallel }\sin \varphi }\nonumber \\&=\sqrt{\ell _0^{2}+\Delta h_{\parallel }^{2}+2(R-r)\Delta h_{\parallel }\sin \varphi }. \end{aligned}$$

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Then, the Taylor polynomial for $\ell (\Delta h_{\parallel })$ at zero is defined by

$$\begin{aligned} \ell (\Delta h_{\parallel }) = \ell _0 + \frac{(R-r)\sin \varphi }{\ell _0}\Delta h_{\parallel } + O\big (\Delta h_{\parallel }^2\big ). \end{aligned}$$

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Thus, we have

$$\begin{aligned} \Delta \ell = \ell - \ell _0 \approx \frac{(R-r)\sin \varphi }{\ell _0}\Delta h_{\parallel } = \sin \varphi \cos \theta \Delta h_{\parallel }, \end{aligned}$$

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where $\varphi \in (\pi ,2\pi )$ since MTs mainly resist compression. We can find the y component resultant force $F_y$ by the algebraic summing of the components of the forces in the y direction,

$$\begin{aligned} F_{y}&=-\int _{\pi }^{2\pi }\frac{n_{{t}}k_{{t}}}{2\pi }\sin \varphi \cos \theta \Delta h_{\parallel }\cos \theta \mathrm {d}\varphi \nonumber \\&=-\frac{n_{{t}}k_{{t}}}{2\pi }\Delta h_{\parallel }\cos ^{2}\theta \int _{\pi }^{2\pi }\sin \varphi \mathrm {d}\varphi \nonumber \\&=\frac{n_{{t}}k_{{t}}}{\pi }\Delta h_{\parallel }\cos ^{2}\theta \nonumber \\&\approx \frac{n_{{t}}k_{{t}}}{\pi }\Delta h_{\parallel }. \end{aligned}$$

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1.2 If $\Delta h^{E}=\Delta h_{\perp }$, the stretched length and additional force of MFs

If nucleus migration $\Delta h^{E}=\Delta h_{\perp }$ is perpendicular to MT, Cartesian coordinate for point p will be changed into

$$\begin{aligned} p' = (r\cos \varphi ,\, r\sin \varphi -\Delta h_{\perp }\sin \theta ,\, h+\Delta h_{\perp }\cos \theta ). \end{aligned}$$

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The corresponding length of MFs can be expressed as

$$\begin{aligned} \ell (\Delta h_{\perp })&=\sqrt{(R\cos \varphi -r\cos \varphi )^{2}+(R\sin \varphi -r\sin \varphi + }\nonumber \\&\qquad \overline{+\Delta h_{\perp }\sin \theta )^{2}+(h+\Delta h_{\perp }\cos \theta )^{2}}\nonumber \\&=\sqrt{(R-r)^{2}+2(R-r)\Delta h_{\perp }\sin \varphi \sin \theta +} \nonumber \\&\qquad \overline{+h^{2}+\Delta h_{\perp }^{2} +2h\Delta h_{\perp }\cos \theta }\nonumber \\&=\sqrt{2\big [(R-r)\sin \varphi \sin \theta +h\cos \theta \big ]\Delta h_{\perp }+}\nonumber \\&\qquad \overline{+\ell _{0}+\Delta h_{\perp }^{2}}. \end{aligned}$$

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Then, the Taylor polynomial for $\ell (\Delta h_{\perp })$ at zero is defined by

$$\begin{aligned} \ell (\Delta h_{\perp }) = \ell _0^2&+ \frac{(R-r)\sin \varphi \sin \theta +h\cos \theta }{\ell _0}\Delta h_{\perp }\nonumber \\&+ O\big (\Delta h_{\perp }^2\big ). \end{aligned}$$

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Thus, we have

$$\begin{aligned} \Delta \ell&= \ell - \ell _0 \approx \frac{(R-r)\sin \varphi \sin \theta +h\cos \theta }{\ell _0}\Delta h_{\perp }\nonumber \\&= \sin \theta \cos \theta (1+\sin \varphi )\Delta h_{\perp }. \end{aligned}$$

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We can find y and z components resultant force $F_y$ and $F_z$ by the algebraic summing of the components of the forces in y and z directions,

$$\begin{aligned} F_{y}&=\int _{0}^{2\pi }\frac{n_{{a}}k_{{a}}}{2\pi }\sin \theta \cos \theta (1+\sin \varphi )\Delta h_{\perp }\cos \theta \sin \varphi \mathrm {d}\varphi \nonumber \\&=\frac{n_{{a}}k_{{a}}}{2\pi }\Delta h_{\perp }\sin \theta \cos ^{2}\theta \int _{0}^{2\pi }(1+\sin \varphi )\sin \varphi \mathrm {d}\varphi \nonumber \\&=\frac{n_{{a}}k_{{a}}}{2}\Delta h_{\perp }\sin \theta \cos ^{2}\theta \nonumber \\&\approx \frac{1}{2}n_{{a}}k_{{a}}\theta \Delta h_{\perp }, \end{aligned}$$

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$$\begin{aligned} F_{z}&=\int _{0}^{2\pi }\frac{n_{{a}}k_{{a}}}{2\pi }\sin \theta \cos \theta (1+\sin \varphi )\Delta h_{\perp }\sin \theta \mathrm {d}\varphi \nonumber \\&=\frac{n_{{a}}k_{{a}}}{2\pi }\sin ^{2}\theta \cos \theta \Delta h_{\perp }\int _{0}^{2\pi }(1+\sin \varphi )\mathrm {d}\varphi \nonumber \\&=n_{{a}}k_{{a}}\sin ^{2}\theta \cos \theta \Delta h_{\perp }\nonumber \\&\approx n_{{a}}k_{{a}}\theta ^{2}\Delta h_{\perp }. \end{aligned}$$

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