Thickness of epithelia on wavy substrates: measurements and continuous models

Abstract

We measured the thickness of MDCK epithelia grown on substrates with a sinusoidal profile. We show that while at long wavelength the profile of the epithelium follows that of the substrate, at short wavelengths cells are thicker in valleys than on ridges. This is reminiscent of the so-called «healing length in the case of a thin liquid film wetting a rough solid substrate. We explore the ability of continuum mechanics models to account for these observations. Modeling the epithelium as a thin liquid film, with surface tension, does not fully account for the measurements. Neither does modeling the epithelium as a thin incompressible elastic film. On the contrary, the addition of an apical active stress gives satisfactory agreement with measurements, with one fitting parameter, the ratio between the active stress and the elastic modulus.

Graphic Abstract

Appendix

1.1 A. Healing length of thin liquid films

Following [10, 11], we describe the properties of a thin film wetting a solid surface by its energy per surface area, which contains three terms, the liquid free surface tension $\gamma _{cm}$, the liquid–solid interfacial tension $\gamma _{cs}$ and a term that depends on the thickness H of the film, P(H). We assume that when the surface of the solid is perfectly flat, defined by the plane $Z=0$, the film has an optimal thickness $H_0$. When the solid surface is rough, with a roughness $h_0(X,Y)$, the position $\zeta (X,Y)$ of the liquid free surface is governed by the minimization of the energy:

$$\begin{aligned} \int \int {dX dY \left[ P(\zeta -h_0)+\frac{1}{2} \gamma _{cm}({\mathbf {\nabla }}\zeta )^2\right] } \end{aligned}$$

(27)

We make the assumption that the curvature of the free surface remains small. For small deformations, P is expanded up to second order around $\zeta -h_0=H_0$. The integral of the zero-order term is a constant, that of the first-order term is equal to zero assuming that the total volume of liquid is constant. Finally the minimization of the energy yields:

$$\begin{aligned} \frac{1}{\xi _e^2}(\zeta -h_0-H_0)-{\mathbf {\nabla }}^2 \zeta = 0 \end{aligned}$$

(28)

where the healing length $\xi _e$ is defined by Eq. 1.

When the substrate shows a sinusoidal profile with wavelength $\lambda$ and amplitude A (as defined by Eq. 9), Eq. 28 yields to $\zeta$ also varying sinusoidally, with amplitude a displayed in Eq. 2. Since $a<A$, the thickness of the liquid film is maximum in the valleys, where $\cos (2\pi X/\lambda )=-1$, and minimum on the ridges, where $\cos (2\pi X/\lambda )=+1$, and the difference in the thickness of the film between the valleys and the ridges is equal to:

$$\begin{aligned} A-a=\frac{A}{1+\left( \frac{\lambda }{2 \pi \xi _e} \right) ^2} \end{aligned}$$

(29)

1.2 B. Resolution of the elastic model

Following [36, 37], we solve the most general linear elastic model described in the main text (Eqs. 10–14 together with 16 and 23) by writing the balance of linear momentum, incompressibility condition and boundary conditions in Fourier space. Introducing the Fourier transform ${\hat{f}}(k,Z)$ of a function f(X, Z) as:

$$\begin{aligned} {\hat{f}}(k,Z)=\frac{1}{\sqrt{2 \pi }} \int ^{\infty }_{\infty } f(X,Z)e^{ikX}\text{ d }k \end{aligned}$$

(30)

Using this definition the unknown fields u and p can be expressed in term of the vertical component of the displacement field. The incompressibility condition first yields:

$$\begin{aligned} {\hat{u}}(k,Z)=\frac{i}{k} \frac{\partial {\hat{v}}}{\partial Z} \end{aligned}$$

(31)

while the X-component of the equilibrium equation implies:

$$\begin{aligned} {\hat{p}}(k,Z)=\mu \left( \frac{1}{k^2} \frac{\partial ^3 {\hat{v}}}{\partial Z^3} - \frac{\partial {\hat{v}}}{\partial Z}\right) \end{aligned}$$

(32)

where ${\hat{u}}$, ${\hat{v}}$ and ${\hat{p}}$ are respectively the Fourier transforms of u, v and p. Using these results, the Y-component of the balance of linear momentum yields a fourth-order ordinary differential equation of ${\hat{v}}$:

$$\begin{aligned} \frac{\partial ^4 {\hat{v}}}{\partial Z^4} - 2k^2 \frac{\partial ^2 {\hat{v}}}{\partial Z^2} + k^4{\hat{v}} = 0 \end{aligned}$$

(33)

This equation is supplemented by four boundary conditions. At $Z=0$ we have:

$$\begin{aligned}&\frac{\partial {\hat{v}}}{\partial Z}=0 \end{aligned}$$

(34)

$$\begin{aligned}&{\hat{v}}(k,0)=\frac{A\sqrt{\pi }}{2\sqrt{2}}\left\{ \delta \left( k-\frac{2\pi }{\lambda }\right) +\delta \left( k+\frac{2\pi }{\lambda }\right) \right\} \end{aligned}$$

(35)

while at the free surface $Z=H$ we have:

$$\begin{aligned}&\mu \frac{\partial ^2 {\hat{v}}}{\partial Z^2}+k^2\left( \mu + \sigma _{act} \right) {\hat{v}}=0 \end{aligned}$$

(36)

$$\begin{aligned}&3 k^2 \mu \frac{\partial {\hat{v}}}{\partial Z}- \mu \frac{\partial ^3 {\hat{v}}}{\partial Z^3}=-\gamma k^4 {\hat{v}} \end{aligned}$$

(37)

Solving the ordinary differential equation for ${\hat{v}}$ together with the above four boundary conditions, we obtain the following solution, here expressed at the free surface for simplicity:

$$\begin{aligned}&{\hat{v}}(k,H)=\frac{A\sqrt{\pi }}{2\sqrt{2}}\left\{ \delta \left( k-\frac{2\pi }{\lambda }\right) +\delta \left( k+\frac{2\pi }{\lambda }\right) \right\} \\&\quad \frac{2(\cosh (Hk)+Hk\sinh (Hk))}{(1+2H^2k^2(1-\frac{\gamma }{\mu H}-\frac{\sigma _{act}}{\mu })+\cosh (2Hk)+\frac{\gamma }{2\mu }\sinh (2Hk)} \end{aligned}$$

where $\delta$ is the Dirac delta distribution. We can now use Fourrier inversion theorem to find the Z-component of the displacement field in real space (again expressed at the free surface for simplicity):

$$\begin{aligned}&v(X,H)=\frac{A}{2}\cos \left( \frac{2\pi X}{\lambda }\right) \\&\quad \frac{2(\cosh (Hk)+Hk\sinh (Hk))}{(1+2H^2k^2\left( 1-\frac{\gamma }{\mu H}-\frac{\sigma _{act}}{\mu }\right) +\cosh (2Hk)+\frac{\gamma }{2\mu }\sinh (2Hk)} \end{aligned}$$

Note that this last result can be rewritten under the form presented in Eqs. 24–25 or, taking $\sigma _{\mathrm{act}}=0$, as 17–18.