Early Stages of Bone Fracture Healing: Formation of a Fibrin–Collagen Scaffold in the Fracture Hematoma

Abstract

This work is concerned with the sequence of events taking place during the first stages of bone fracture healing, from bone breakup until the formation of early fibrous callus (EFC). The latter provides a scaffold over which subsequent remodeling processes will eventually result in successful bone repair. Specifically, some mathematical models are proposed to estimate the time required for (1) the formation immediately after fracture of a fibrin clot, described in terms of a phase transition in a polymerization process, and (2) the onset of EFC which is produced when fibroblasts arising from differentiation of chemotactically recruited mesenchymal stem cells remodel a previous fibrin clot by releasing a collagen matrix over it. An attempt has been made to keep models as simple as possible, so that a explicit dependence of the estimates obtained on relevant biochemical parameters involved is obtained.

Appendices

Appendix 1: Experimental Procedure

Sixteen ten-week-old male Sprague-Dawley rats were obtained from the animal facility building of the University of Oviedo. Bone fractures with $2.5$ mm average width were produced by the manual breakage using plate-bending devices, placed at the distal-third of the right tibia. The animal study protocol was approved by the local committee and conformed to Spanish animal protection laws. Rats were anesthetized and sacrificed by cervical dislocation 3, 6, 12 and 24  h after fracture ($n=5$). Tibiae were isolated and cut through the sagittal plane of the metaphyseal fracture region into two equal sized parts and fixed by immersion in 4 % paraformaldehyde at 4 $^{\circ }\hbox {C}$ for 12 h, rinsed in PBS, decalcified in 10 % EDTA (pH 7.0) for 48 h at 4 $^{\circ }\hbox {C}$, dehydrated through a graded ethanol series, cleared in xylene and embedded in paraffin. Sections were serially cut at a thickness of 5 m, mounted on Superfrost Plus slides (Menzel-Glaser), stained with a trichromic method using Weigerts hematoxylin/alcian blue/picrofuchsin and viewed and photographed on a Nikon Eclipse E400 microscope.

Appendix 2: Technical Details

We gather here a number of technical details related to issues that have been mentioned in the main text but were skipped there.

1.1 A General Case in Blood Clot Formation

Case: $k_{p}>0$, $k_{b}>0$, $k_{r}>0$.

When all kinetic rates are different from zero substitution of (19) into (17) leads to

$$\begin{aligned} \frac{dY}{dt}=f(t)Y^{2}+2k_{b}Y; f(t)=4k_{p}Ae^{-k_{r}(t-T_{M})}-k_{b} \end{aligned}$$

(64)

so that $Z=-\frac{1}{Y}$ solves

$$\begin{aligned} \frac{dZ}{dt}+2k_{b}Z=f(t), Z(T_{M})=-\frac{1}{2} \end{aligned}$$

whence

$$\begin{aligned} \begin{aligned} Z(t)=&e^{-2k_{b}(t-T_{M})}\left( -\frac{1}{2}+\int ^{t}_{T_{M}}e^{2k_{b}(s-T_{M})}f(s)ds\right) \\ =&\frac{4Ak_{p}}{2k_{b}-k_{r}}\left( e^{-k_{r}(t-T_{M})}-e^{-2k_{b}(t-T_{M})}\right) -\frac{1}{2}\\ \end{aligned} \end{aligned}$$

(65)

for $t>T_{M}$ when $2k_{b}\ne k_{r}$, and:

$$\begin{aligned} Z(t)=4Ak_{p}(t-T_{M})e^{-2k_{b}(t-T_{M})}-\frac{1}{2} \end{aligned}$$

(66)

for $t>T_{M}$ and $2k_{b}=k_{r}$. Assume now that $2k_{b}>k_{r}$ then, on setting:

$$\begin{aligned} Q=\frac{4Ak_{p}}{2k_{b}-k_{r}}=\frac{4F_{M}(1-e^{-k_{g}C})k_{p}}{2k_{b}-k_{r}} \end{aligned}$$

(67)

it follows that $U(t)=\frac{M_{2}(t)}{M_{1}(t)}$ is given by:

$$\begin{aligned} U(t)=\frac{1+Q\left( e^{-k_{r}(t-T_{M})}-e^{-2k_{b}(t-T_{M})}\right) }{1-2Q\left( e^{-k_{r}(t-T_{M})}-e^{-2k_{b}(t-T_{M})}\right) } \end{aligned}$$

(68)

Let $g(t)=e^{-k_{r}(t-T_{M})}-e^{-2k_{b}(t-T_{M})}$ for $t\ge T_{M}$. Clearly $g(T_{M})=0$, $g(t)>0$ for $t>T_{M}$ and $g(t)$ achieves a maximum at some time $t=t^{*}$. A straightforward computation yields that:

$$\begin{aligned} t^{*}&= T_{M}+\frac{1}{2k_{b}-k_{r}}\log \left( \frac{2k_{b}}{k_{r}}\right) \end{aligned}$$

(69)

$$\begin{aligned} m^{*}&= g(t^{*})=\left( \frac{2k_{b}}{k_{r}}\right) ^{-\frac{k_{r}}{2k_{b}-k_{r}}}-\left( \frac{2k_{b}}{k_{r}}\right) ^{-\frac{2k_{b}}{2k_{b}-k_{r}}} \end{aligned}$$

(70)

Thus, for $U(t)$ in (68) to blow up at time $T_{f}$, we need

$$\begin{aligned} 2Qm^{*}\ge 1 \end{aligned}$$

(71)

where $Q$, $m^{*}$ are respectively given in (67) and (70). If (71) holds, one has that

$$\begin{aligned} T_{f}\le T_{M}+\frac{1}{2k_{b}-k_{r}}\log \left( \frac{2k_{b}}{k_{r}}\right) \end{aligned}$$

(72)

the case where $2k_{b}<k_{r}$ is similarly dealt with. Finally, when $2k_{b}=k_{r}$ blow up for $U(t)$ occurs if there exists $T_{f}>T_{M}$ such that

$$\begin{aligned} 8Ak_{p}(T_{f}-T_{M})e^{-2k_{p}(T_{f}-T_{M})}=1 \end{aligned}$$

(73)

It is readily seen that (73) holds if $m^{*}=\max _{t>T_{M}}\left( (t-T_{M})e^{-2k_{b}(T-T_{M})}\right)$ is such that $8Ak_{p}m^{*}\ge 1$. Since $m^{*}=(2ek_{b})^{-1}$ in this case, the resulting condition reads:

$$\begin{aligned} 4Ak_{p}>ek_{b} \end{aligned}$$

(74)

1.2 On Neglecting MSC Diffusion

There seems to be no evidence supporting random motility (the underlying microscopical process responsible for diffusion) in MSCs or any other cell lineage involved in bone repair. Experimental studies indicate instead that most cellular events during bone healing are mainly regulated by local factors that may originate from a variety of cells. These factors produce a biochemical environment that varies at different stages of the process and determines precise cellular responses including directional cell progression, proliferation and differentiation (Cho et al. 2002).

In addition to this biological argument, it is easy to see on mathematical terms that adding a diffusive term in Eq. (36) has virtually no impact on the inward motion of MSC fronts for any reasonable choice of the corresponding diffusion coefficient D. The corresponding argument can be sketched as follows:

To avoid technicalities, we just show here that adding diffusion in (36) results in a small correction in the inward motion of MSC front. More precisely, let us replace (36) by

$$\begin{aligned} \frac{\partial m}{\partial t}+\frac{\partial }{\partial x}\left( \mu m\frac{\partial g}{\partial x}-D_{m}\frac{\partial m}{\partial x}\right) =0;\quad -\infty <x<\infty \,,\ t>T_{M} \end{aligned}$$

(75)

where $D_{m}>0$ with initial conditions (37), (38) as before. Using again the approximation $\frac{\partial g}{\partial x}\approx -\frac{2a}{L}x$ in (75), the resulting equation can be written in the form

$$\begin{aligned} \frac{\partial m}{\partial t}-\frac{2\mu a x}{L}\frac{\partial m}{\partial x}-\frac{2\mu a}{L}m-D_{m}\frac{\partial ^{2} m}{\partial x^{2}}=0 \end{aligned}$$

(76)

We first estimate the characteristic time $\tau _{\epsilon }^\mathrm{tr}$ needed to travel a distance $\frac{L}{2}$ under transport effects. The time to fill an infinitesimal interval of the form $\left[ x-dx,x\right]$ is then given by:

$$\begin{aligned} \frac{\mathrm {d}x}{-2\mu a x/L} = \mathrm {d}t\,. \end{aligned}$$

Integrating this expression between $L/2$ and a characteristic size epsilon ($\epsilon$) (which can be interpreted as an intercellular space), we find that $\tau _{\epsilon }^\mathrm{tr}$ satisfies the following relationship:

$$\begin{aligned} \int ^{\tau _{\epsilon }^\mathrm{tr}}_{0}\mathrm {d}t=\int ^{\frac{L}{2}}_{\epsilon }\frac{L}{2\mu a x} \mathrm {d}x \end{aligned}$$

and we then have that:

$$\begin{aligned} \tau _{\epsilon }^\mathrm{tr}=\frac{L}{2\mu a}\text {log}\left( \frac{L}{2\epsilon }\right) \end{aligned}$$

(77)

Analogously, we estimate the characteristic time $\tau _{\epsilon }^\mathrm{diff}$ for gap filling due to diffusion as follows:

$$\begin{aligned} \tau _{\epsilon }^\mathrm{diff}\sim \frac{\left( \frac{L}{2}-\epsilon \right) ^{2}}{D_{m}} \end{aligned}$$

(78)

Comparing (77) and (78) we obtain:

$$\begin{aligned} \frac{\tau ^\mathrm{tr}}{\tau ^\mathrm{diff}}=\frac{LD_{m}}{2\mu a}\frac{\text {log}\left( \frac{L}{2\epsilon }\right) }{\left( \frac{L}{2}-\epsilon \right) ^{2}}\approx \frac{LD_{m}}{2\mu a}\frac{\text {log}\left( \frac{L}{2\epsilon }\right) }{\left( \frac{L}{2}\right) ^{2}}=\frac{2D_{m}}{\mu a L}\text {log}\left( \frac{L}{2\epsilon }\right) \end{aligned}$$

(79)

On selecting $\epsilon =\frac{L}{100}$, $L=2$ mm, it follows that for a typical diffusion coefficient $D_{m}=10^{-6}$ $\hbox {cm}^{2}/\hbox {s}$, (79) yields:

$$\begin{aligned} \frac{\tau ^\mathrm{tr}}{\tau ^\mathrm{diff}}=\frac{2D_{m}}{\mu a L}\text {log}\left( \frac{L}{2\epsilon }\right) =\frac{2\cdot 10^{-6}}{0.5\cdot 10^{-2}\cdot 1.25\cdot 0.2}\text {log}(50)\approx 6\cdot 10^{-3} \end{aligned}$$

which shows the low impact of diffusion in the process under consideration.