From Signaling to Movement – Mathematical andComputational Problems in Cell Motility

Hans G. OthmerSchool of MathematicsUniversity of Minnesota

IMA – June 2018– 1/50 –

Why understanding cell motility is important ...

• Development – we start as a single cell but ultimately have ∼ 250 cell typescorrectly located in the adult. Morphogenetic movements occur at both thesingle cell and tissue levels.

• The immune system – e.g., neutrophils respond to bacterial invasion.

• Wound healing – some cells move into a wound to fight infection, others toclose the wound and rebuild the tissue.

• Metastasis in cancer – invasion of new sites by active migration and passivetransport in the circulatory system

– 2/50 –

Various modes of cell motility

1. Swimmers

• Bacteria – individually and as swarms, sperm, leukocytes, Dictyosteliumdiscoideum

2. Crawlers• Lot of examples – neutrophils, fibroblasts, macrophages, leukocytes, Dic-

tyostelium — —

– 3/50 –

A model system – Dictyostelium discoideum

– 4/50 –

Dicty’s chemotactic response in a fluid ..

Dictyostelium amoebae and neutrophils can swim N. P. Barry and M. S. Bretscher, PNAS, 2010, 107

11376­11380– 5/50 –

This leads to three major problems ...

• The transduction problem — how areextracellular signals transduced into in-tracellular signals that can be used tocontrol the orientation of the cell. (Mod-ule I)

• The interior problem — how does thecell control the shape changes that giverise to motion. (Module II)

• The exterior problem — how do the shape changes give rise to motion,how fast do they move, and how efficient is the motion. (Module III)

– 6/50 –

Models for the first steps of orientation

Models based on local excitation and global inhibition (LEGI)

H. Meinhardt Orientation of chemotactic cells and growth cones: models and mechanisms J. Cell Science, 1999.

Xiong, et al., Cells navigate with a local­excitation, global­ inhibition­biased excitable network, PNAS, 2010.

– 7/50 –

The signal transduction problem in Dicty

How are extracellular cAMP signals transduced?

A major characteristic is that the system respondsto changes in the extracellular signal, but adaptsto constant signals at the level of Ras.

Y. Cheng and H.G Othmer, A Model for Direction Sensing in

Dictyostelium Discoideum, PLOS Comp. Biol., 2016

du1

dt=

S(t)− (u1 + u2)

te

du2

dt=

S(t)− u2

ta.

– 8/50 –

The major components of the signal transduction step

• This leads to a system of reaction-diffusion equations in the cytosol, reactions on the boundary, and

exchange between the boundary and the cytosol.

• Most parameters can be estimated by matching a subset of the experimental observations.

• The model differs from existing LEGI (local excitation-global inhibition) models in that ALL cytosolic

species diffuse at equal rates. Thus inclusion of details shows where simplified models break down.

– 9/50 –

The reactions involved ..

– 10/50 –

The experimental tests

The new experimental results that have to be captured by the model are ...

• Under a spatially uniform stimulus, Ras is transiently activated, and adaptsimperfectly.

• Under a graded simulus the response is biphasic – uniform followed bysymmetry breaking, independent of the cytoskeleton.

• The system has a well-defined refractory period as shown by the response torepeated stimuli .

• The magnitude of gradient amplification depends on the cAMP amplitudeand gradient

• How cells respond to a passing wave – the ’back-of-the-wave’ problem

– 11/50 –

The response under uniform stimuli

Time(s)

0 10 20 30

RB

Dc

0.6

0.7

0.8

0.9

1

1.1WT

0.1 nM

1 nM

10 nM

100 nM

1 µ M

Time(s)

0 10 20 30

RB

Dc

0.6

0.7

0.8

0.9

1

1.1

gα2

-null

0.1 nM

1 nM

10 nM

100 nM

1 µ M

Takeda K, Shao D, Adler M, Charest PG, Loomis WF, Levine H, et al. Incoherent feedforward control

governs adaptation of activated Ras in a eukaryotic chemotaxis pathway. Cell Biology. 2012.– 12/50 –

Symmetry­breaking under a gradient

Thus the response is biphasic – first uniform around the periphery and then higherat the high end of the gradient

Kortholt A, Keizer­Gunnink I, Kataria R, van Haastert PJM. Ras activation and symmetry breaking

during Dictyostelium chemotaxis. J Cell Sci. 2013– 13/50 –

What cells see in a wave and how they ’remember’

This may be part of a solution for the ’back-of- the- wave’ problem. – 14/50 –

Next the exterior problem ..

Consider the effects of controlled changes in the shape of a swimmer movingthrough a Newtonian, incompressible fluid. Neglect body forces (or absorb them inthe pressure) and scale the Navier-Stokes equations to obtain

Re

[

∂u

∂t+ (u · ∇u)

]

= −∇p+∇2u

where Re = ρLV /µ is the Reynolds number.

– 15/50 –

The Stokes problem for low Reynolds number flows

Thus the problem is to solve the Stokes equations

µ∆u−∇p = 0, ∇ · u = 0.

for a specified sequence of shape changes of the boundary. Notice that timedoes not appear in these equations, and therefore if (u, p) is a solution then so

is (−u,−p).• When can the swimmer propel itself by shape deformations?

It’s easier to say when it can’t – this is Purcell’s scallop theorem —

If the sequence of shapes in a cyclic time-periodic stroke is identical when viewedunder time-reversal, then there is no net motion per cycle.

• Thus if a scallop can only open andclose it’s valves, it cannot swim.

E. Purcell, Life at low Reynolds number, Amer. J.

Physics, 1977.

– 16/50 –

Swimming at low Reynolds number

Let B(t) and V be the boundary and velocity of the swimmer. Set V = v+U where

v defines the intrinsic shape deformation and U is the rigid motion.

A cyclic swimming stroke is a T-periodic sequence of shapes.

The canonical LRN self-propulsion problem is: given a cyclic shape deformationby specifying v, solve the Stokes equations subject to

∑

Fi = 0,∑

Γi = 0, u|B = V, u|x→∞ = 0, (1)

• A general sequence S(t) of shape changes is not an admissable motion,

since it will produce forces and torques, but the swimmer has to be force andtorque free.

• An admissable motion will also include rigid translations and rotations thatgenerate counter flows that exactly cancel the forces and torques.

A. Shapere and F. Wilczek, Geometry of self­propulsion at low Reynolds number, J FLuid Mech, 198,

(1989)– 17/50 –

An abstract setup due to Shapere and Wilczek

Think of the motion and shape changes as living in a fiber bundle in which thebase space is a manifold of shape changes, and the fibers are rigid motions. Thegoverning equations define constraints.

The basic fiber bundle – Is the following possible?

Shapes

Rigid Motions

In other words, if we traverse a cycle in the shape space, do we generate a non-trivial rigid rotation or translation?

– 18/50 –

The simplest problem is 2D ...

In 2D the Stokes equations

µ∆u−∇p = 0, ∇ · u = 0

can be solved by introducing a stream function. Define Λ such that

u(x, y) = ux + iuy =∂Λ

∂y− i

∂Λ

∂x

Then Λ solves the biharmonic equation

∆2Λ = 0.

The general solution to the biharmonic equation in 2D is known –

Λ(z, z) = −Im[zφ(z) + χ(z)]

where φ(z) and χ(z) are called the Goursat functions.

– 19/50 –

The physical variables ...

Once we know Λ we recover the physical variables as follows.

• Velocity: v = φ(z)− zφ′(z)− χ′(z)

• Pressure: p = −4µℜφ′(z)

• Vorticity: ω(

:=∂

∂yℜv − ∂

∂xℑv

)

= −4ℑφ′(z)

• Force: f ≡ σ · n) = 4µℜ(φ′)n− 2µ(zφ′′ + χ′′)n

Along a curve the force becomes

fds = −2iµ[

(φ′ + φ′)dz + (zφ′′ + χ′′)dz]

= −2iµd(

φ+ zφ′ + χ′)

Some basic problems of the mathematical theory of elasticity N.I. Muskhelishvili and JMR Radok,

1953, Cambridge Univ Press – 20/50 –

Pull­back of the fluid domain and boundary data

Suppose the cell occupies Ω(t) ⊂ C, and that the velocity on ∂Ω(t) is V (z, z; t).

z = w(ζ; t) : C/D → C/Ω(t)

• In the z-plane we have to find functions φ(z; t) and ψ(z; t), analytic on C/Ω(t)

and continuous on C/Ω(t), such that

φ(z; t)− zφ′(z; t)− ψ(z; t) = V (z, z; t) (z ∈ ∂Ω(t))

• In the ζ-plane we have to find Φ(ζ; t) = φ(w(ζ; t); t) and Ψ(ζ; t) = ψ(w(ζ; t); t),

analytic on C/D and continuous on C/D at any time t, such that

Φ(σ)− w(σ)

w′(σ)Φ′(σ)−Ψ(σ) = V (σ, t) (σ ∈ S1) – 21/50 –

Reduction to an integral equation

• Extend the boundary data to the entire plane by defining

V (ζ) =1

2πi

∫

S1

V (σ)

σ − ζdσ

This leads to two analytic functions V + and V − such that

V (σ) = V −(σ)− V +(σ) on S1 and

V −(ζ) = λ0 + λ1ζ + λ2ζ2 + · · · for |ζ| ≤ 1;

−V +(ζ) =λ−1

ζ+

λ−2

ζ2+

λ−3

ζ3+ · · · for |ζ| ≥ 1

• Apply the operator 12πi

∫

S1

·σ−ζ

dσ for |ζ| > 1 to the velocity

boundary condition

Φ(σ)− w(σ)

w′(σ)Φ′(σ)−Ψ(σ) = V (σ) (σ ∈ S1)

– 22/50 –

Fredholm integral equation

Then Φ(ζ) satisfies

Φ(ζ) +1

2πi

∫

S1

w(σ)

w′(σ)

Φ′(σ)

σ − ζdσ = −V +(ζ) (|ζ| ≥ 1)

If the conformal mapping z = w(ζ) has only finitely many terms then the integral

equation can be reduced to a finite system of complex linear equations for the

coefficients An, λn of Φ and −V +, resp

Φ(ζ) =A−1

ζ+

A−2

ζ2+ · · · + A−n

ζn+ · · ·

−V +(ζ) =λ−1

ζ+

λ−2

ζ2+ · · · + λ−n

ζn+ · · ·

Q. Wang and H.G. Othmer, Modeling of Amoeboid Swimming at Low Reynolds Number, J

Math Biol, (2016).– 23/50 –

Symmetric cyclic shape deformations

Consider a map of the form

w(ζ; t) = C(t)ζ +1

M

[

a1sin(t)

ζ+ a2

sin(t+ ϕ)

2ζ2+ · · ·+ aN

sin(t+ (N − 1)ϕ)

NζN

]

For example, consider w(ζ; t) = 3ζ + cos(2πt)ζ−1 − sin(2πt)ζ−2.

Characteristics of rigid motions

• Mean velocity within a period: U := Tr(T )/T .

• Mean power within a period: P := T−1∫ T

0P(t)dt.

• Variance of power within a period: V ar(P) := T−1∫ T

0[P − P ]2dt.

• Efficiency: E := U2/P. – 24/50 –

A computational experiment

Fix ϕ, test randomly generated a(∼ N (0, 5)N ) where M =∑N

k=1 |ak|, C(t) is deter-

mined by area conservation and ϕ.

– 25/50 –

Shape approximations for Dictyostelium

Suppose that we have an n-polygon Γ in the z-plane with vertices z1, · · · , zn, andthe corresponding exterior angles are θ1, · · · , θn. Let Ω be the interior regionbounded by Γ. A Schwarz-Christoffel transformation mapping the exterior of theunit disk D in the ζ-plane to Ωc is given by

z = w(ζ) = A+ C

∫ ζ 1

ξ2

n∏

k=1

(

ξ − ζk)θk−1

dξ (2)

where zk = w(ζk) and ζk is the prevertex to the vertex zk under w.

– 26/50 –

What can we say about Dictyostelium’s swimming ?

van Haastert Barry

Maximum cell body length ∼ 25µ ∼ 22µ

Average cell body width ∼ 6µ ∼ 4µ

Maximum protrusion height ∼ 2µ ∼ 4µ

Average protrusion width ∼ 2µ ∼ 2µ

Period of a stroke ∼ 1 min ∼ 1.5 min– 27/50 –

How various factors affect the results ..

1 2 3 4 51

2

3

4

5

6

7

Max Bump Height

Mean

Velo

cit

y

(a) Mean Velocity

Barry

Haastert

1 2 3 4 52000

4000

6000

8000

10000

12000

Max Bump Height

Mean

Po

wer

(b) Mean Power

Barry

Haastert

1 2 3 4 52

4

6

8

10x 10

ï

Max Bump Height

Perf

orm

an

ce

(d) Performance

Barry

Haastert

2 4 6 8 10 121

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

Length/Width

Me

an

ve

loci

ty (

µm

/min

)

(a)

2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5x 10

4

µ

Me

an

po

we

r (µ

m /

min

)-1

22

Length/Width

(b)

2 4 6 8 10 121

1.2

1.4

1.6

1.8

2

2.2x 10

ï

µ E

ci

en

cy

(c)

Length/Width

But this is just one mode of swimming! – 28/50 –

Other modes of cell motility

Charras 1 Liu 1 Liu2 Paluch 1

How do we begin to understand these different modes of movement?

– 29/50 –

Does cortical flow drive movement?

Observations: There is a strong cortical reward cortical flow in a number of celltypes, which in some cells leads to a thicker, more contractile cortex at the rear.

Our Hypothesis: This creates a tension gradient with higher tension at the rear.Assuming that the membrane does not flow, this creates a reverse tension gradientin the membrane, and mechanical balance implies a tension gradient in the exteriorfluid that drives cell movement.

– 30/50 –

The mathematical problem

Let Ω ⊂ R3 denote the volume occupied by the cell and let S denote its boundary.

Assume that S is a smooth, compact, two-dimensional manifold without boundary,parameterized by the map

Φ : D ⊂ R2 → Sdefined so that the position vector x to any point on the membrane is given by

x = x(u1, u2) for a coordinate pair (u1, u2) ∈ D.

Let n denote the outward normal on S, and define basis vectors on the surface via

ei =∂x

∂uii = 1, 2.

– 31/50 –

The Canham­Helfrich free energy

This description only takes into account the bending energy, not torsion or shear.

FB =

∫

S

2kc(H − C0)2dS +

∫

S

kGKdS. (3)

κ1, κ2 are the principal curvatures H = −(κ1 + κ2)/2 is the mean curvature

K = κ1κ2 is the Gaussian curvature C0 is the spontaneous curvature

If we add area and volume constraints then

F = FB +

∫

D

Λ (√g −√

g0) du1du2 + P

(∫

Ω

dV − V0

)

Absent other forces a critical point is a stationary shape, and a minimizer is prob-ably a stable stationary state.

– 32/50 –

Shape variation and the force equations

Any infinitesimal deformation can be writtenin terms of tangential and normal compo-nents as

x = x0 + φiei + ψn

This leads to normal and tangential compo-nents of the force

F n = −δFδψ

= −∆s [kB (2H − C0)]

−kB (2H − C0)(

2H2 + C0H − 2K)

− ∆s kG + 2ΛH − P

F ti = − δF

δφi=

1

2(2H − C0)

2 ∇ikB +K∇ikG +∇iΛ

i = 1, 2.

– 33/50 –

Shapes of cells under normal and tangential forces

µd

dψ(u1, u2)

dτ= F n(H,K, P,Λ, kB, kG, u

1, u2) + fn

µd

dφi(u1, u2)

dτ= F t

i (H,K, P,Λ, kB, kG, u1, u2) + f ti i = 1, 2.

Here Γ =A

π(

L2π

)2 =4πA

L2,

is the reduced area, which expresses the ratio of the area of a given 2D shape tothat of a circle of circumference L and a larger area. – 34/50 –

The fluid­cell interactions at low Reynolds number

v(x) = − 1

4πµ(1 + λ)

∫

∂Ω(t)

G(x, ξ) · Fm(ξ)dS(ξ)

+1− λ

4π(1 + λ)

∫

∂Ω(t)

v(ξ) ·T(x, ξ) · dS(ξ)

where G is the Green’s function

G(x, ξ) =1

r

[

δ +rr

r2

]

r = x − ξ and r = |r|. Fm is the force exerted by the membrane on the fluid,

T is the third-rank stress tensor or stresslet, and λ is the ratio of the interior toexterior viscosities. We assume continuity of the interior, exterior and membranevelocities, and mechanical equilibrium at the membrane, and therefore the forcebalance reads

Fm ≡[

(σin − σext) · n]

m= fm.

– 35/50 –

The swimming velocity

Cortical contraction

Fluid drag

Fluid drag

reaction

Membrane

Reality

Model

Fluid drag

Fluid dragreaction

Membrane

-1

-0.5

0

0.5

1

6 7 8 9

y/R

0

x/R0

The swimming velocity as a function of the applied forces. The cells move to the right. In both

diagrams the cell has a reduced area Γ = 0.6.– 36/50 –

Module II – Actin waves –the intracellular dynamics

A big question is

‘How does a cell generate the forces necessary to produce movement by controlledre-modeling of the cytoskeleton?’

In the absence of directional signals neutrophils and Dictyostelium discoideum ex-plore their environment randomly, and thus the intracellular biochemical networksthat control the mechanics must be tuned to produce signals that generate thisrandom movement.

The rebuilding of the actin cortex during migration or following treatment with la-trunculin provides some insights.

Latrunculin sequesters monomer with high affinity and leads to de-polymerizationof the network. Following washout one can track the re-building of the network withTIRF and confocal miscroscopy.

– 37/50 –

An example of spontaneous actin waves

A continuum model of actin waves in Dictyostelium discoideum, PLoS One, 8,24pp, (2013), V. Khamviwath and J. Hu and H. G. Othmer

– 38/50 –

Actin dynamics ­ the dendritic network

– 39/50 –

The simplified feedback loop

– 40/50 –

A schematic of the stochastic model for actin waves

– 41/50 –

The evolution of the mobile cytosolic species

In the cytosol

G− actin :∂[g]

∂t= Dg∇

2[g] +Rg

Arp2/3 :∂[arp]

∂t= Darp∇

2[arp] +Rp1

Coronin :∂[cp]

∂t= Dcp∇

2[cp] +Rcp

Cappingprotein :∂[cor]

∂t= Dcor∇

2[cor]−Rp2 +Rp1

and on the membrane Ω2d

−Dg∂

∂z[g]|z=0 = −k+

bk[g]|z=0 · Fbkfree

−k+gan[g]|z=0 · [npf∗_arp] + k−gan[npf∗_arp_g]

−Darp∂

∂z[arp]|z=0 = −k+an[arp]|z=0 · [npf∗] + k−an[npf

∗_arp]

−Dcp∂

∂z[cp]|z=0 = −k+cap[cp]|z=0 · Fbkfree

−Dcor∂

∂z[cor]|z=0 = 0

and homogeneous Neumann data elsewhere on the boundary.

– 42/50 –

The evolution of mobile species on the surface ...

∂[npf ]

∂t= Dnpf∇

2[npf ]− kactvFbrfree · [npf ] + kdeg [npf∗] + krecov [npf

∗∗]

∂[npf∗]

∂t= Dnpf∗∇2[npf∗] + kactvFbrfree · [npf ]

−kdeg [npf∗]− k+an[arp]|z=0 · [npf∗] + k−an[npf

∗_arp]

∂[npf∗_arp]

∂t= k+an[arp]|z=0 · [npf∗]− k−an[npf

∗_arp]

−k+gan[g]|z=0 · [npf∗_arp] + k−gan[npf∗_arp_g]

∂[npf∗_arp_g]

∂t= k+gan[g]|z=0 · [npf∗_arp]− k−gan[npf

∗_arp_g]− knucl[npf∗_arp_g] · Fbtot

∂[npf∗∗]

∂t= Dnpf∗∗∇2[npf∗∗] + knucl[npf

∗_arp_g] · Fbtot − krecov [npf∗∗]

– 43/50 –

The evolution of the nucleation sites and filamentsattached at the membrane

∂Sf

∂t= − k+nuk[g]|z=0Sf + k+cap[cp]|z=0

∑

n≥2

fk(n) + k−nukfk(1)

∂fk(1)

∂t= k+nuk[g]|z=0Sf − k−nukfk(1) + k−pkfk(2)− k+bk[g]|z=0fk(1)

∂fk(n)

∂t= k+bk[g]|z=0fk(n− 1) + k−pkfk(n+ 1)

− (k+bk[g]|z=0 + k−pk)fk(n)− k+cap[cp]|z=0fk(n), (n ≥ 2)

... and there are equations for branched fillaments, etc ... I said this is a stochasticprocess but these look like deterministic equations!

We convert these equations to ODEs by discretizing the spatial derivatives, andtreat diffusion between ’boxes’ as a jump process.

– 44/50 –

A computational TIRF sequence

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (m

icro

ns)

t = 5 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (m

icro

ns)

t = 10 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (m

icro

ns)

t = 20 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (m

icro

ns)

t = 40 sec

0

10

20

30

40

50

60

70

– 45/50 –

A 3D depiction of the temporal evolution of a wave

– 46/50 –

TIRF images of colliding actin waves

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 10 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 20 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 25 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 30 sec

0

10

20

30

40

50

60

70

– 47/50 –

Long term dynamics after the collision of waves

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 50 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 80 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 100 sec

0

10

20

30

40

50

60

70

1 2 3 4 5

1

2

3

4

5

X (microns)

Y (

mic

ron

s)

t = 120 sec

0

10

20

30

40

50

60

70

– 48/50 –

Conclusion

We have arrived at the stage where models are useful to sug-gest experiments, and the facts of the experiments in turn leadto new and improved models that suggest new experiments.By this rocking back and forth between the reality of experi-mental facts and the dream world of hypotheses, we can moveslowly toward a satisfactory solution of the major problems ofdevelopmental biology.

John Bonner

On Development

– 49/50 –

Acknowledgments

Collaborators

• Yougan Cheng – Directionsensing

• Qixuan Wang – swimming cells

• Varunyu Khamviwath, Jifeng Hu– Actin waves

• Hao Wu, Marco Ponce de Leon– Cell shape analysis

Funding provided by the National Institutes of Health, the National Science Foun-dation, the Newton Institute and the Simons Foundation

– 50/50 –