The chemical basis of morphogenesis
Severin Bang
Paper by A. M. Turing about pattern formation
29 November 2016
Motivation
Periodic pattern in nature, despite homogeneous stage in embryo genesis.
Too many possibilities to be encoded in genes.
Question: How? Why?
Idea
This Paper: Pattern due to diffusion and reaction.
Differentmorphogenes Inhibitory and excitatory Unstable equilibrium
Diffusion, production and destruction
Illustration
Forest in California.
Random fire outbreaks.
Fast “diffusing”
firefighters.
Burned area as pattern.
Inhibitor must be faster.
Setting
Two morphogenes c1,2(x,y,t), change in concentration due to:
Reaction:f(c1,c2),g(c1,c2).
Diffusion:D1,2∇2c1,2.
⇒ Reaction-diffusion systems:
˙
c1=f(c1,c2) +D1∇2c1
˙
c2=g(c1,c2) +D2∇2c2
Notation
˙
u=γf(u,v) +∇2u
˙
v=γg(u,v) +d∇2v γ=L2/D1T scaling factor
Lspatial scale T time scale d =D2/D1
Properties
To be a Turing-Mechanismit must hold, that:
(I) There is a (homogeneous and stationary) positive solution (u0,v0) forf(u0,v0) =g(u0,v0) = 0
(II) Stable ifnodiffusion.
(III) Instable under diffusion.
The clue: diffusion creates patterns!
Condition (I)
Determine solution for f(u0,v0) =g(u0,v0) = 0 Does a positive solution exists? Fine! next!
Condition (II)
No diffusion: ˙u=γf(u,v), v˙=γg(u,v).
Consider small perturbation around stable state (u0,v0):
~ w=
u−u0
v−v0
A=
fu fv
gu gv
u0,v0
⇒ w˙ =γAw Linear system: w∝eλt
Condition (II)
Solution:λ1,2 for which det(γA−λ1) = 0 λ1,2=1
2γ
(fu+gv)± q
(fu+gv)2−4(fugv−fvgu)
For Re(λ1,2)<0:
fu+gv =Tr(A)<0, fugv−fvgu= det(A)>0
Condition fulfilled? Fine! next!
Condition (III)
Bringing back the diffusion:
˙
w=γAw+D∇2w, D= 1 0
0 d
Separation ansatz:
w(r,t) =
∑
k
ckeλktWk(r)
⇒ λ(k)Wk =γAWk+D∇2Wk ck from start statew(r,0).
With∇2W(r) +k2W(r) = 0 time independent solution of spatial eigenvalue problem:
⇒ λ(k)WK=γAWk−Dk2Wk
Condition (III)
Again:λ1,2 from det(λ1−γA+Dk2) = 0:
0 =λ2+λ
k2(1 +d)−γ(fu+gv)
+h(k2) h(k2) =dk4−γ(dfu+gv)k2+γ|A|
Instable when Re(λ(k))>0, that means that either:
[k2(1 +d)−γ(fu+gv)]<0 h(k2)<0
Condition (III)
[k2(1 +d)−γ(fu+gv)]<0:
k2(1 +d)>0∀k6= 0, (fu+gv)<0
⇒not possible
h(k2) =dk4−γ(dfu+gv)k2+γ|A|<0 it’s necessary, that:
dfu+gv>0
Condition (III)
dfu+gv >0, but from first condition:fu+gv<0.
Implications:
d 6= 1
fu>0 because activator activates itself⇒gv<0 fu,gv different signs.
d >1 ⇒ D2>D1: Inhibitor diffuses faster.
Condition (III)
dfu+gv <0 by itself not sufficient. With:
hmin=γ2
|A| −(dfu+gv)2 4d
, kmin2 =γ(dfu+gv) 2d it holds, that h(k2)<0 for:
(dfu+gv)2 4d >|A|
Pattern Formation
Critical for:
|A|=(dcfu+gv)2 4dc
kc2=γdcfu+gv 2dc =γ
s
|A|
dc
Ford >dc:
roots atk1,k2, instable for k∈[k1,k2]
Ford =dc:
root atkc, instable forkc k∈/[k1,k2]:
no pattern!
source: J.D. Murray, Mathematical Biology II: Spatial Models and Biomedical Applications
Recap
Two component reaction–diffusion system.
Three conditions restrict the solution set:
(I) Positive solution forg(u0,v0) =f(u0,v0) = 0 (II) fu+gv <0, fugv−fvgu>0
(III) dfu+gv>0, (dfu+gv)2>4d|A|
Linear approximation.
No instabilities for small and large k.
Waves withk∈[k1,k2] create patterns.
Diffusion creates patterns!
Example
Explicit example for f,g:
f(u,v) =a−u+u2v g(u,v) =b−u2v
˙
u=γ(a−u+u2v) +uxx
˙
v=γ(b−u2v) +dvxx
With a,b parameters describing the reaction.
Condition I:
0 =a−u+u2v 0 =b−u2v
⇒ u0=a+b, v0= b
(a+b)2, b>0, a+b>0
At the stationary state:
fu=b−a
a+b, fv = (a+b)2>0 gu= −2b
a+b, gv=−(a+b)2<0
Condition II:
fu+gv<0 ⇒ (a+b)3>b−a fugv−fvgu>0 ⇒ (a+b)2>0
Condition III:
dfu+gv>0⇒d(b−a)>(a+b)3 (dfu+gv)2−4d(fugv−fvgu)>0⇒
d(b−a)−(a+b)32
>4d(a+b)4
The conditions on f(u,v),g(u,v) set the Turing-Space.
source: J.D. Murray, Mathematical Biology II: Spatial Models and Biomedical Applications
source: J.D. Murray, Mathematical Biology II: Spatial Models and Biomedical Applications
Take home message
One mechanism for pattern formation.
Reaction Diffusion system with conditions:
(I) Homogeneous state, which is:
(II) Stable ifD1=D2= 0 (III) Instable ifD1,26= 0
d≥dc>1: short range activation, long range inhibition.
Spotted animals can have striped tails,notvice versa!
