On the role of calcium diffusion and its rapid buffering in intraflagellar signaling

Abstract

We have considered the realistic mechanism of rapid Ca²⁺ (calcium ion) buffering within the wave of calcium ions progressing along the flagellar axoneme. This buffering is an essential part of the Ca²⁺ signaling pathway aimed at controlling the bending dynamics of flagella. It is primarily achieved by the mobile region of calmodulin molecules and by stationary calaxin, as well as by the part of calmodulin bound to calcium/calmodulin-dependent kinase II and kinase C. We derived and elaborated a model of Ca²⁺ diffusion within a signaling wave in the presence of these molecules which rapidly buffer Ca²⁺. This approach has led to a single nonlinear transport equation for the Ca²⁺ wave that contains the effects brought about by both as necessary buffers for signaling. The presence of mobile buffer calmodulin gives rise to a transport equation that is not strictly diffusive but also exhibits a sink-like effect. We solved straightforwardly the final transport equation in an analytical framework and obtained the implied function of calcium concentration. The effective diffusion coefficient depends on local Ca²⁺ concentration. It is plausible that these buffers' presence can impact Ca²⁺ wave speed and shape, which are essential for decoding Ca²⁺ signaling in flagella. We present the solution of the transport equation for a few specified cases with physiologically reasonable sets of parameters involved.

Appendices

Appendix 1

Starting from Eq. (38) it simply follows.

$$d\Omega =-\nu \left(\frac{{\theta }^{2}+1}{{\theta }^{2}}\right)d\theta ,$$

(74)

so that the first integral reads

$$\Omega =-\nu \left(\theta -\frac{1}{\theta }\right)+{const}_{1}.$$

(75)

If we impose the initial condition $\Omega =0$ for $\theta ={\theta }_{\mathrm{min}}$, it yields

$$\Omega =\frac{d\theta }{d\zeta }=\nu \left(\frac{1}{\theta }-\theta +{\theta }_{\mathrm{min}}-\frac{1}{{\theta }_{\mathrm{min}}}\right).$$

(76)

${\theta }_{min}$ follows from minimal (basal-0.1 $\mathrm{\mu M}$) Ca²⁺ concentration in flagellum

$${\theta }_{\mathrm{min}}=\frac{5+0.1}{\sqrt{5\cdot 10}}=0.72.$$

(77)

By rearranging the last relation we obtain

$$\int \frac{d\theta }{{\left(\theta +\rho \right)}^{2}-{a}^{2}}=-\nu \zeta +{const}_{2};$$

(78)

$$\rho =\frac{1}{2}\left(\frac{1}{{\theta }_{\mathrm{min}}}-{\theta }_{\mathrm{min}}\right); a=\sqrt{1+{\rho }^{2}}.$$

(79)

Introducing a new variable $s$

$$s=\theta +\rho ; ds=d\theta ,$$

(80)

and factorizing the integrand in Eq. (78) we eventually get

$${\left(\theta +\rho +a\right)}^{b+1}{\left(\theta +\rho -a\right)}^{1-b}={const}_{2}\cdot {e}^{-2\nu \zeta }.$$

(81)

The constant of integration could be obtained from condition $\theta ={\theta }_{\mathrm{max}}$ for $\zeta =0$, where ${\theta }_{\mathrm{max}}$ represents the amplitude of Ca²⁺ concentration in the flagellum.

Appendix 2

We seek the solution of linear equation, Eq. (35), in the standard form

$$\Omega \left(\theta \right)=u\left(\theta \right)\cdot v\left(\theta \right),$$

(82)

and we get

$$v\left(\theta \right)=\frac{{\theta }^{2}}{{\theta }^{2}+\gamma };\qquad u\left(\theta \right)=-\nu \left(\theta -\frac{\delta }{\theta }\right)+{const}_{1}.$$

(83)

Inserting these functions in Eq. (82) and imposing the initial condition $\Omega =0$ for $\theta ={\theta }_{\mathrm{min}}$, we get

$$\Omega =\nu \frac{{\theta }^{2}}{{\theta }^{2}+\gamma }\left(\frac{\delta -{\theta }^{2}}{\theta }-\varepsilon \right); \qquad\varepsilon =\frac{\delta -{\theta }_{\mathrm{min}}^{2}}{{\theta }_{\mathrm{min}}}.$$

(84)

Since $\Omega =\frac{d\theta }{d\zeta }$ it safely follows from Eq. (84)

$$\int \frac{d\theta \left({\theta }^{2}+\gamma \right)}{\theta \left({\theta }^{2}+\varepsilon \theta -\delta \right)}=-\nu \zeta +{const}_{2}.$$

(85)

By factorizing the above integrand

$$\frac{{\theta }^{2}+\gamma }{\theta ({\theta }^{2}+\varepsilon \theta -\delta )}=\frac{p}{\theta }+\frac{q}{\theta -{\theta }_{1}}+\frac{r}{\theta -{\theta }_{2}};\qquad {\theta }_{1/2}=\frac{-\varepsilon \pm \sqrt{{\varepsilon }^{2}+4\delta }}{2},$$

(86)

we get the linear system of three algebraic equations for parameters $p$, $q$, $r$ as follows:

$$\begin{array}{c}p+q+r=1 \\ p\left({\theta }_{1}+{\theta }_{2}\right)+q{\theta }_{2}+r{\theta }_{1}=0\\ p{\theta }_{1}{\theta }_{2}=\gamma ,\end{array}$$

(87)

with corresponding solutions:

$$p=-\frac{\gamma }{\delta };\qquad q=1+\frac{\gamma }{\delta }-r;\qquad r=\frac{\frac{\gamma }{\delta }{\theta }_{1}-{\theta }_{2}}{{\left({\varepsilon }^{2}+4\delta \right)}^\frac{1}{2}}.$$

(88)

The final solution of Eq. (85) now reads:

$${\theta }^{p}{\left(\theta -{\theta }_{1}\right)}^{q}{\left(\theta -{\theta }_{2}\right)}^{r}={{const}_{2}\cdot e}^{-\nu \zeta }.$$

(89)

The value of ${const}_{2}$ can be assessed from the condition $\theta ={\theta }_{\mathrm{max}}$ for $\zeta =0$ along with the set of specified parameters involved.