Time-dependent Changes in Smooth Muscle Cell Stiffness and Focal Adhesion Area in Response to Cyclic Equibiaxial Stretch

Abstract

Observations from diverse studies on cell biomechanics and mechanobiology reveal that altered mechanical stimuli can induce significant changes in cytoskeletal organization, focal adhesion complexes, and overall mechanical properties. To investigate effects of short-term equibiaxial stretching on the transverse stiffness of and remodeling of focal adhesions in vascular smooth muscle cells, we developed a cell-stretching device that can be combined with both atomic force and confocal microscopy. Results demonstrate that cyclic 10%, but not 5%, equibiaxial stretching at 0.25 Hz significantly and rapidly alters both cell stiffness and focal adhesion associated paxillin and vinculin. Moreover, measured changes in stiffness and focal adhesion area from baseline values tend to correlate well over the durations of stretching studied. It is suggested that remodeling of focal adhesions plays a critical role in regulating cell stiffness by recruiting and anchoring actin filaments, and that cells rapidly remodel in an attempt to maintain a homeostatic biomechanical state when perturbed above a threshold value.

Appendix

Assume that, relative cylindrical coordinates, when a membrane is stretched equibiaxially (μ _r  = μ _θ  = μ), it has an isochoric, axisymmetric, and homogeneous deformation of the form, r = μR, θ = Θ, z = λZ, where (R, Θ, Z) and (r, θ, z) are particle coordinates in an unstretched reference and a stretched current configuration, respectively, with μ and λ constants at each level of stretch. Hence, the physical components of deformation gradient F are

$$ [{\mathbf{F}}] = {\left[ {\begin{array}{*{20}c} {{\partial r/\partial R}} & {0} & {0} \\ {0} & {{r/R}} & {0} \\ {0} & {0} & {\lambda } \\ \end{array} } \right]} = {\left[ {\begin{array}{*{20}c} {\mu } & {0} & {0} \\ {0} & {\mu } & {0} \\ {0} & {0} & {\lambda } \\ \end{array} } \right]}, $$

(A1)

with $\lambda = 1/\mu ^{2}$ from incompressibility (i.e., detF = 1). Since the membrane is thin, we can reduce F from 3D to 2D. Hence, consider the equibiaxial field

$$ [{\mathbf{F}}_{{2{\text{D}}}} ] = {\left[ {\begin{array}{*{20}c} {\mu } & {0} \\ {0} & {\mu } \\ \end{array} } \right]}. $$

(A2)

Cartesian coordinates are obtained by rotating cylindrical coordinates through an angle $\hat{\theta }$ about the axis perpendicular to the membrane. Thus,

$$ {\mathbf{e}}_{x} = {\text{cos}}\,\hat{\theta }{\mathbf{e}}_{r} + {\text{sin}}\,\hat{\theta }{\mathbf{e}}_{\theta } ,\quad {\mathbf{e}}_{y} = - {\text{sin}}\,\hat{\theta }{\mathbf{e}}_{r} + {\text{cos}}\,\hat{\theta }{\mathbf{e}}_{\theta } . $$

(A3)

Components of the 2D left Cauchy-Green tensor, $\overline{{\mathbf{B}}} = \overline{{\mathbf{F}}} _{{2{\text{D}}}} \overline{{\mathbf{F}}} ^{{\text{T}}}_{{2{\text{D}}}}$, defined relative to a Cartesian coordinate system are thus determined as

$$ \overline{{\mathbf{B}}} = {\mathbf{ABA}}^{{\text{T}}} \quad {\text{or}}\quad \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{ij}} = A_{{ik}} A_{{jl}} B_{{kl}} , $$

(A4)

where $[{\mathbf{A}}] = {\left[ {\begin{array}{*{20}c} {{{\text{cos}}\,\hat{\theta }}} & {{{\text{sin}}\,\hat{\theta }}} \\ {{ - {\text{sin}}\,\hat{\theta }}} & {{{\text{cos}}\,\hat{\theta }}} \\ \end{array} } \right]}$ and B is the left Cauchy-Green tensor “for” the cylindrical coordinate system. Substituting A into Eq. (A4) gives

$$ {\left[ {\begin{array}{*{20}c} {{\ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{11}} }} & {{\ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{12}} }} \\ {{\ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{21}} }} & {{\ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{22}} }} \\ \end{array} } \right]} = {\left[ {\begin{array}{*{20}c} {{{\text{cos}}\,\hat{\theta }}} & {{{\text{sin}}\,\hat{\theta }}} \\ {{ - {\text{sin}}\,\hat{\theta }}} & {{{\text{cos}}\,\hat{\theta }}} \\ \end{array} } \right]}{\left[ {\begin{array}{*{20}c} {{B_{{11}} }} & {{B_{{12}} }} \\ {{B_{{21}} }} & {{B_{{22}} }} \\ \end{array} } \right]}{\left[ {\begin{array}{*{20}c} {{{\text{cos}}\,\hat{\theta }}} & {{ - {\text{sin}}\,\hat{\theta }}} \\ {{{\text{sin}}\,\hat{\theta }}} & {{{\text{cos}}\,\hat{\theta }}} \\ \end{array} } \right]}. $$

(A5)

Thus,

$$ \begin{aligned}{} \overline{B} _{{11}} = & B_{{11}} \cos ^{2} \hat{\theta } + B_{{22}} \sin ^{2} \hat{\theta }, \\ \overline{B} _{{12}} = & -B_{{11}} \sin \hat{\theta }\cos \hat{\theta } + B_{{22}} \sin \hat{\theta }\cos \hat{\theta }, \\ \overline{B} _{{22}} = & B_{{11}} \sin ^{2} \hat{\theta } + B_{{22}} \cos ^{2} \hat{\theta }, \\ \end{aligned} $$

(A6)

with $\ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{21}} = \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{12}}$.

Since $B_{{11}} = B_{{22}} = \mu ^{2}$ in this problem,

$$ \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{11}} = B_{{11}} = B_{{22}} = \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{22}} \quad {\text{with}}\quad \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{12}} = \ifmmode\expandafter\bar\else\expandafter\=\fi{B}_{{21}} = 0. $$

(A7)

Therefore, components of deformation gradient relative to cylindrical coordinates are the same as those relative any Cartesian coordinate system, thus the deformation is equibiaxial independent of coordinate system.