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Search for a small egg by spermatozoa in restricted geometries.

Yang J, Kupka I, Schuss Z, Holcman D - J Math Biol (2015)

Bottom Line: In the proposed model the swimmers' trajectories are rectilinear and the speed is constant.Because hitting a small target by a trajectory is a rare event, asymptotic approximations and stochastic simulations are needed to estimate the mean search time in various geometries.We consider searches in a disk, in convex planar domains, and in domains with cusps.

View Article: PubMed Central - PubMed

Affiliation: Applied Mathematics and Computational Biology, Ecole Normale Supérieure, IBENS, 46 rue d'Ulm, 75005, Paris, France.

ABSTRACT
The search by swimmers for a small target in a bounded domain is ubiquitous in cellular biology, where a prominent case is that of the search by spermatozoa for an egg in the uterus. This is one of the severest selection processes in animal reproduction. We present here a mathematical model of the search, its analysis, and numerical simulations. In the proposed model the swimmers' trajectories are rectilinear and the speed is constant. When a trajectory hits an obstacle or the boundary, it is reflected at a random angle and continues the search with the same speed. Because hitting a small target by a trajectory is a rare event, asymptotic approximations and stochastic simulations are needed to estimate the mean search time in various geometries. We consider searches in a disk, in convex planar domains, and in domains with cusps. The exploration of the parameter space for spermatozoa motion in different uterus geometries leads to scaling laws for the search process.

No MeSH data available.


Related in: MedlinePlus

Solid angle subtended at points a over the interior or over the periphery of disk (); b outside disk boundary () (these figures are screenshots from Poxtan’s paper; Paxton 1959)
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Fig7: Solid angle subtended at points a over the interior or over the periphery of disk (); b outside disk boundary () (these figures are screenshots from Poxtan’s paper; Paxton 1959)

Mentions: By definition, the solid angle is given by the integral55\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega = \int \frac{\mathbf {n}_{P} \cdot \mathbf {ds}_P}{R^{2}}, \end{aligned}$$\end{document}Ω=∫nP·dsPR2,where is the area of the projection of the surface element onto the plane perpendicular to the radius , from the north pole to point P (see Fig. 7). Specifically, Integral 55 can be written as:56\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega =\int \frac{ds \cos \theta }{R^{2}}=\int \int \frac{\rho d\beta d\rho \cos \theta }{R^{2}}=\int \int \sin \theta d\theta d\beta \end{aligned}$$\end{document}Ω=∫dscosθR2=∫∫ρdβdρcosθR2=∫∫sinθdθdβin which and . When (see Fig. 7a), Eq. 56 can be written as57\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\Omega }{2}=\int _{0}^{\pi }\int _{0}^{\theta _{s}}\sin \theta d\theta d\beta =\pi -\int _{0}^{\pi }\cos \theta _{s}d\beta \end{aligned}$$\end{document}Ω2=∫0π∫0θssinθdθdβ=π-∫0πcosθsdβAfter a few more steps and re-arrangements, Eq. 57 turns into58\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega = 2 \pi - \frac{2L}{R_{max}} K\left( k \right) - \pi \Lambda _{0} \left( \xi , k \right) \end{aligned}$$\end{document}Ω=2π-2LRmaxKk-πΛ0ξ,kwhere K(k) and are the complete elliptic integrals of the first and third kind, respectively with59\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} k^{2}= & {} 1-(R_{1}/R_{max})^{2} \nonumber \\ R_{1}^{2}= & {} L^{2}+(r_{0}-r_{m})^{2}\\ \xi= & {} \sin ^{-1}(L/R_{1}).\nonumber \end{aligned}$$\end{document}k2=1-(R1/Rmax)2R12=L2+(r0-rm)2ξ=sin-1(L/R1).When (see Fig. 7b), Eq. 56 can be written as60\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\Omega }{2}=\int _{0}^{\beta _{max}}\int _{\theta _{m}}^{\theta _{s}}\sin \theta d\theta d\beta =\int _{0}^{\beta _{max}}\cos \theta _{m} d\beta - \int _{0}^{\beta _{max}}\cos \theta _{s} d\beta \end{aligned}$$\end{document}Ω2=∫0βmax∫θmθssinθdθdβ=∫0βmaxcosθmdβ-∫0βmaxcosθsdβSimilarly, after certain steps and re-arrangements, Eq. 60 can be written as follows:61\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega (\xi , k) = -\frac{2L}{R_{max}}K\left( k \right) + \pi \Lambda _{0}\left( \xi , k \right) \end{aligned}$$\end{document}Ω(ξ,k)=-2LRmaxKk+πΛ0ξ,kIn summary, the expression for the solid angle is62\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega (\xi , k) = \left\{ \begin{array}{ll} 2 \pi - \dfrac{2L}{R_{max}} K\left( k \right) - \pi \Lambda _{0} \left( \xi , k \right) , &{} r_{0}<r_{m} \\ \\ \pi -\dfrac{2L}{R_{max}}K\left( k \right) , &{} r_{0}=r_{m} \\ -\dfrac{2L}{R_{max}}K\left( k \right) + \pi \Lambda _{0}\left( \xi , k \right) , &{} r_{0}>r_{m}. \end{array}\right. \end{aligned}$$\end{document}Ω(ξ,k)=2π-2LRmaxKk-πΛ0ξ,k,r0<rmπ-2LRmaxKk,r0=rm-2LRmaxKk+πΛ0ξ,k,r0>rm. is a function of and as we shall see now. We apply formula 62 to the case of a small hole of radius located at the north pole of the three-dimensional sphere. We obtain for a point of coordinates 63\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} R^2_1= & {} R^2(1-\cos \phi )^2+(R \sin (\phi )-\varepsilon )^2 \end{aligned}$$\end{document}R12=R2(1-cosϕ)2+(Rsin(ϕ)-ε)264\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} R^2_{max}= & {} R^2(1-\cos \phi )^2+(R \sin (\phi )+\varepsilon )^2\end{aligned}$$\end{document}Rmax2=R2(1-cosϕ)2+(Rsin(ϕ)+ε)265\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} k^2(\phi )= & {} \frac{4\varepsilon R \sin (\phi )}{2R^2(1-\cos \phi )+2 \varepsilon R \sin (\phi ) +\varepsilon ^2}\end{aligned}$$\end{document}k2(ϕ)=4εRsin(ϕ)2R2(1-cosϕ)+2εRsin(ϕ)+ε266\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \sin (\xi )= & {} \frac{R(1-\cos \phi )}{R^2(1-\cos \phi )^2+(R \sin (\phi )-\varepsilon )^2}. \end{aligned}$$\end{document}sin(ξ)=R(1-cosϕ)R2(1-cosϕ)2+(Rsin(ϕ)-ε)2.We then used Mathematica to compute the . We obtain for small that67\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \int _{\phi >\varepsilon /R}\Omega (\phi )\sin (\phi )d\phi =\pi \left( 1-\cos \varepsilon \right) . \end{aligned}$$\end{document}∫ϕ>ε/RΩ(ϕ)sin(ϕ)dϕ=π1-cosε.Finally, we show in Fig. 8 that , the ratio for small ,Fig. 8


Search for a small egg by spermatozoa in restricted geometries.

Yang J, Kupka I, Schuss Z, Holcman D - J Math Biol (2015)

Solid angle subtended at points a over the interior or over the periphery of disk (); b outside disk boundary () (these figures are screenshots from Poxtan’s paper; Paxton 1959)
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Related In: Results  -  Collection

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Fig7: Solid angle subtended at points a over the interior or over the periphery of disk (); b outside disk boundary () (these figures are screenshots from Poxtan’s paper; Paxton 1959)
Mentions: By definition, the solid angle is given by the integral55\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega = \int \frac{\mathbf {n}_{P} \cdot \mathbf {ds}_P}{R^{2}}, \end{aligned}$$\end{document}Ω=∫nP·dsPR2,where is the area of the projection of the surface element onto the plane perpendicular to the radius , from the north pole to point P (see Fig. 7). Specifically, Integral 55 can be written as:56\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega =\int \frac{ds \cos \theta }{R^{2}}=\int \int \frac{\rho d\beta d\rho \cos \theta }{R^{2}}=\int \int \sin \theta d\theta d\beta \end{aligned}$$\end{document}Ω=∫dscosθR2=∫∫ρdβdρcosθR2=∫∫sinθdθdβin which and . When (see Fig. 7a), Eq. 56 can be written as57\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\Omega }{2}=\int _{0}^{\pi }\int _{0}^{\theta _{s}}\sin \theta d\theta d\beta =\pi -\int _{0}^{\pi }\cos \theta _{s}d\beta \end{aligned}$$\end{document}Ω2=∫0π∫0θssinθdθdβ=π-∫0πcosθsdβAfter a few more steps and re-arrangements, Eq. 57 turns into58\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega = 2 \pi - \frac{2L}{R_{max}} K\left( k \right) - \pi \Lambda _{0} \left( \xi , k \right) \end{aligned}$$\end{document}Ω=2π-2LRmaxKk-πΛ0ξ,kwhere K(k) and are the complete elliptic integrals of the first and third kind, respectively with59\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} k^{2}= & {} 1-(R_{1}/R_{max})^{2} \nonumber \\ R_{1}^{2}= & {} L^{2}+(r_{0}-r_{m})^{2}\\ \xi= & {} \sin ^{-1}(L/R_{1}).\nonumber \end{aligned}$$\end{document}k2=1-(R1/Rmax)2R12=L2+(r0-rm)2ξ=sin-1(L/R1).When (see Fig. 7b), Eq. 56 can be written as60\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \frac{\Omega }{2}=\int _{0}^{\beta _{max}}\int _{\theta _{m}}^{\theta _{s}}\sin \theta d\theta d\beta =\int _{0}^{\beta _{max}}\cos \theta _{m} d\beta - \int _{0}^{\beta _{max}}\cos \theta _{s} d\beta \end{aligned}$$\end{document}Ω2=∫0βmax∫θmθssinθdθdβ=∫0βmaxcosθmdβ-∫0βmaxcosθsdβSimilarly, after certain steps and re-arrangements, Eq. 60 can be written as follows:61\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega (\xi , k) = -\frac{2L}{R_{max}}K\left( k \right) + \pi \Lambda _{0}\left( \xi , k \right) \end{aligned}$$\end{document}Ω(ξ,k)=-2LRmaxKk+πΛ0ξ,kIn summary, the expression for the solid angle is62\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Omega (\xi , k) = \left\{ \begin{array}{ll} 2 \pi - \dfrac{2L}{R_{max}} K\left( k \right) - \pi \Lambda _{0} \left( \xi , k \right) , &{} r_{0}<r_{m} \\ \\ \pi -\dfrac{2L}{R_{max}}K\left( k \right) , &{} r_{0}=r_{m} \\ -\dfrac{2L}{R_{max}}K\left( k \right) + \pi \Lambda _{0}\left( \xi , k \right) , &{} r_{0}>r_{m}. \end{array}\right. \end{aligned}$$\end{document}Ω(ξ,k)=2π-2LRmaxKk-πΛ0ξ,k,r0<rmπ-2LRmaxKk,r0=rm-2LRmaxKk+πΛ0ξ,k,r0>rm. is a function of and as we shall see now. We apply formula 62 to the case of a small hole of radius located at the north pole of the three-dimensional sphere. We obtain for a point of coordinates 63\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} R^2_1= & {} R^2(1-\cos \phi )^2+(R \sin (\phi )-\varepsilon )^2 \end{aligned}$$\end{document}R12=R2(1-cosϕ)2+(Rsin(ϕ)-ε)264\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} R^2_{max}= & {} R^2(1-\cos \phi )^2+(R \sin (\phi )+\varepsilon )^2\end{aligned}$$\end{document}Rmax2=R2(1-cosϕ)2+(Rsin(ϕ)+ε)265\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} k^2(\phi )= & {} \frac{4\varepsilon R \sin (\phi )}{2R^2(1-\cos \phi )+2 \varepsilon R \sin (\phi ) +\varepsilon ^2}\end{aligned}$$\end{document}k2(ϕ)=4εRsin(ϕ)2R2(1-cosϕ)+2εRsin(ϕ)+ε266\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \sin (\xi )= & {} \frac{R(1-\cos \phi )}{R^2(1-\cos \phi )^2+(R \sin (\phi )-\varepsilon )^2}. \end{aligned}$$\end{document}sin(ξ)=R(1-cosϕ)R2(1-cosϕ)2+(Rsin(ϕ)-ε)2.We then used Mathematica to compute the . We obtain for small that67\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \int _{\phi >\varepsilon /R}\Omega (\phi )\sin (\phi )d\phi =\pi \left( 1-\cos \varepsilon \right) . \end{aligned}$$\end{document}∫ϕ>ε/RΩ(ϕ)sin(ϕ)dϕ=π1-cosε.Finally, we show in Fig. 8 that , the ratio for small ,Fig. 8

Bottom Line: In the proposed model the swimmers' trajectories are rectilinear and the speed is constant.Because hitting a small target by a trajectory is a rare event, asymptotic approximations and stochastic simulations are needed to estimate the mean search time in various geometries.We consider searches in a disk, in convex planar domains, and in domains with cusps.

View Article: PubMed Central - PubMed

Affiliation: Applied Mathematics and Computational Biology, Ecole Normale Supérieure, IBENS, 46 rue d'Ulm, 75005, Paris, France.

ABSTRACT
The search by swimmers for a small target in a bounded domain is ubiquitous in cellular biology, where a prominent case is that of the search by spermatozoa for an egg in the uterus. This is one of the severest selection processes in animal reproduction. We present here a mathematical model of the search, its analysis, and numerical simulations. In the proposed model the swimmers' trajectories are rectilinear and the speed is constant. When a trajectory hits an obstacle or the boundary, it is reflected at a random angle and continues the search with the same speed. Because hitting a small target by a trajectory is a rare event, asymptotic approximations and stochastic simulations are needed to estimate the mean search time in various geometries. We consider searches in a disk, in convex planar domains, and in domains with cusps. The exploration of the parameter space for spermatozoa motion in different uterus geometries leads to scaling laws for the search process.

No MeSH data available.


Related in: MedlinePlus