IL-8 Response of Cyclically Stretching Alveolar Epithelial Cells Exposed to Non-fibrous Particles

Abstract

Using a cell stretcher device, we have previously shown that A549 cells exposed to asbestos fibers gave significantly increased cytokine responses (IL-8) when they were cyclically stretched [Tsuda, A., B. K. Stringer, S. M. Mijailovich, R. A. Rogers, K. Hamada, and M. L. Gray. Am. J. Respir. Cell Mol. Biol. 21(4):455–462, 1999]. In the present study, cell stretching experiments were performed using non-fibrous riebeckite particles, instead of fibrous particles. Riebeckite particles are ground asbestos fibers with the size of a few microns and non-fibrous shape, and are often used as “non-toxic” control particles in the studies of fibrous particle-induced pathogenesis. Although it is generally assumed that riebeckite particles do not elicit strong biological responses, in our studies in cyclically stretched cell cultures, the riebeckite particles coated with adhesion proteins induced significant IL-8 responses, but in static cell cultures the treatment with adhesion protein-coated riebeckite did not induce comparable cytokine responses. To interpret these data, we have developed a simple mathematical model of adhesive interactions between a cell layer and rigid fibrous/non-fibrous particles that were subjected to external tensile forces. The analysis showed that because of considerable dissimilarity in deformations (i.e., strain mismatch) between the cells and particles during breathing, the attachment of particles as small as 1 μ in size could induce significant mechanical forces on the cell surface receptors, which may trigger subsequent adverse cell response under dynamic stretching conditions.

Appendix A

The mathematical model described here was developed to demonstrate how mechanical force is generated at the interface between the cell and an adhering object, such as a particle.

Interfacial Micromechanics

Similar to the analysis of Dembo et al.,15 adhesion receptor molecules are treated as spring-like bridges (Fig. 3), namely, the receptor force, f _recp (force generated by receptor) is taken to be linearly dependent on the stretch of receptor ${(\Delta \ell =\ell - \ell_{\rm o}}),$

$$f_{\rm recp} =\kappa \Delta \ell$$

(1)

where κ (≈1 pN/nm74) represents the stiffness of the receptor, ℓ_o (≈15 nm74) and ℓ denotes unstretched and stretched receptor length, respectively.

We assume that, (1) the gap between the cell and particle remains constant, h (≈10 nm6), and (2) the unstrained receptor with length, ℓ_o, is generally tilted with the equilibrium angle, α_o (Fig. 3). When the receptor angle α is larger than α_o, the receptor is stretched ${(\Delta \ell > 0}),$ thus the receptor force f _recp is positive. This indicates that cell and particle are pulling each other. If on the other hand, α < α_o the receptor is compressed ${(\Delta \ell < 0}),$ thus the receptor force f _recp is negative. This indicates that cell and particle are pushing apart each other.6

Cell–particle Macromechanics

Let the displacement of the cell (apical surface) at position x (the origin of the coordinate, x = 0, fixed at the center of the particle) be u(x,t) at time t (Fig. 3). The strain, ɛ(x,t), along the cell is given by the gradients of the displacement:

$$\varepsilon (x,t)=\partial u(x,t)/\partial x$$

(2)

In contrast, the strain and displacements in the particle are assumed to be equal to zero because the particle is almost rigid comparing to the soft cell.

Denoting the stiffness (force per unit strain) of the cell ⁶ as k _cell, the tensile force in the cell, F _cell, is obtained from the constitutive equation:

$$F_{\rm cell} (x,t)=k_{\rm cell} \varepsilon (x,t)$$

(3)

It is important to note that due to the presence of receptors that mediate the cell–particle adhesion, the tension and hence the strains along the cell (i.e., along x) are not uniform. Note that the tension may generally be a function of time t, but in the simplified analysis considered here, time is only considered as a parameter. The change in F _cell from a position x to x + dx is equal to the force, which corresponds to the traction generated by the sum ⁷ of nanoscale receptor forces around position x per unit length, and is equal to τ (x,t)dx. Here τ denotes traction which has a dimensionality of force per unit length in one-dimensional case. The equilibrium equation for the cell is therefore given by

$${\partial F_{\rm cell}}/{\partial x}=\tau (x,t)$$

(4)

In addition, the macroscale force equilibrium implies that at any location (x), the sum of the tension in the cell, F _cell(x,t), and the force transferred to a rigid particle F _part (x,t), equals to the total tensile force applied to the system, i.e., F(t) = F _cell (x,t) +  F _part (x,t).

Link Between Nano- and Macrointerfacial Mechanics

Since our analysis is one-dimensional, only the longitudinal component of f _recp contributes to the traction. The traction, τ (x,t), can be expressed as,

$$\tau (x,t)=f_{\rm recp} \sin (\alpha)n(x,t)=c(\ell -\ell_{\rm o})\sin (\alpha)$$

(5)

where n(x,t) is the density of the attached receptors (number per unit area) at the location, x, and c = κ n(x,t). For simplicity, in this analysis we assumed that the distribution of n(x,t) is uniform and constant over the stretching cycle ⁸ (n _r =  400/μm²) and independent of time. Substitution of Eqs. (2), (3), (5) and the relation ${\sin \alpha =(u+u_{\rm o})/\ell}$ into Eq. (4) yields to the field equation

$$\partial ^2u/\partial ^2x=\Omega^2\;\left({1-\frac{\ell_{\rm o}}{\ell}} \right)\left({u+u_{\rm o}}\right)$$

(6)

where Ω² = κ n _r/k, ${\ell =\sqrt {h^2+(u_{\rm o} +u)^2}},$ and u _o is the longitudinal component of ℓ_o, i.e., ${u_{\rm o} =\ell_{\rm o} \sin \alpha_{\rm o} =\sqrt {\ell_{\rm o}^2 -h^2}}.$ Boundary conditions are:

$$ u=0\;\;\hbox{at}\;\;x=0\;\;\hbox{(symmetry)}$$

(7a)

$${\partial u}/{\partial x}=\varepsilon_{\rm o}\;\;\hbox{at}\;\;x=L/2$$

(7b)

where ɛ_o and L are input parameters in this analysis, denoting the strain in the cell and the length of the particle, respectively.

Numerical Solution

Since we consider the receptor–ligand bond density approximately constant over stretching cycle, the time, t, only appears as a parameter. Thus, Eq. (6) can be reduced to the second order ordinary differential equation (O.D.E) with respect to x. Substitution of ${p=\frac{du}{dx}}$ and also ${\frac{d^{2}u}{dx^2}=\frac{dp}{du}},$ further simplifies Eq. (6) to the first order O.D.E. with respect to u

$$p\frac{dp}{du}=\Omega^2\left({1-\frac{\ell_{\rm o}}{\sqrt {h^2+(u_{\rm o} +u)^2} }}\right)\left({u+u_{\rm o}}\right)$$

(8)

Equation 8 can be integrated by separation of variables. After analytical integration of Eq. 8 followed by back substitution of ${\frac{du}{dx}=p},$ the solution was obtained from the integral:

$$x=\int\limits_0^{u(x)} {\frac{du}{\Omega \sqrt {\ell ^2-2 \ell \ell _{\rm o}+B}}}$$

(9)

where B is the integration constant that must satisfy boundary condition at the edge of the particle (Eq. 7b). Note that the boundary condition at the center of the particle (Eq. 7a) is included in the lower integration limit. The solution of the integral (Eq. 9) was obtained numerically by Newton–Raphson iterative procedure4 with respect to B in order to satisfy the boundary condition at x = L/2 (Eq. 7b).