Problem #2: checkpoint mechanism
Li et al Trends Cell Biol 13 553 2003
Shah and Cleveland Cell 103 997 2000
Facts:
F
t a single
i l unattached
tt h d KT prevents
t anaphase;
h
rapid
id reactions
ti
off involved
i
l d proteins
t i diffusing
diff i iin th
the
cytoplasm; in fused cells with two spindles in the same cytoplasm, anaphase can initiate in one spindle
even if the other has unattached chromosomes: Presumably, this reflects competition between the short
half-life of Mad2* and its finite diffusion rate from the last unattached KT
Questions: 1) What is the composition of the reactions in the “wait anaphase” signal?
2) How is the tight inhibition of APC is maintained throughout the cell/nucleus?
3) How is this inhibition removed fast upon attachment of the last chromosome?
What is really happening at the kinetochore?
Shah and Cleveland Cell 103 997 2000
Ibrahim et al 2008 PLoS ONE 3: e1555
d [ APC : Cdc ]
dt
= −k−8 [ APC : Cdc ] + k8 [ APC ] × [Cdc ]
mass action
Ibrahim et al 2008 PLoS ONE 3: e1555
Result of the simulations for WT cells
Ibrahim et al 2008 PLoS ONE 3: e1555
Result of the simulations for mutant and biochemically inhibited cells
Note that most, but not all parameters were known. Four unknown parameters were found using genetic
algorithm by fitting the data (to be explained in later lecture)
Ibrahim et al 2008 PLoS ONE 3: e1555
So far, we did not discuss where the reactions take place; how the molecules go where they are needed.
The following two papers examined two central requirements:
(i) capacity of single kinetochore to maintain tight inhibition of the APC–Cdc20 complex throughout the nucleus,
(ii) the rapid removal of this inhibition once the final kinetochore is attached
without assuming the exact form of reactions (this is what physics is good for)
Yeast cell: mitosis
in the nucleus
Doncic et al PNAS 102 6332 2005
Direct
Inhibition
Self-Propagating
Inhibition
Emitted
Inhibition
Doncic et al PNAS 102 6332 2005
∂C * ∂ 2C *
=D
− αC *
2
∂T
∂X
0≤ X ≤ L
X = Lx, T = t / α , C* = Cc *
∂c *
∂ 2c *
D
= D 2 − c*, D = 2 , 0 ≤ x ≤ 1
∂t
∂x
Lα
% ca - active; ci - inhibited
a = 0; b = 1; t0 = 0; t1 = 6; k = 0.1; h = 0.0005;
D = 10; D = D*h/k^2; N = (b-a)/k; M = (t1-t0)/h;
ca = ones(M+1
ones(M+1,N+1);
N+1); ci = ones(M+1,N+1);
ones(M+1 N+1); % Init Cond
%-------------------------MainLoop-------------------for i = (1:M/2)
ca(i+1,2:N) = ca(i,2:N) + h*ci(i,2:N) + ...
D*(ca(i,1:N-1) + ca(i,3:N+1) - 2*ca(i,2:N));
ci(i+1,2:N) = ci(i,2:N) - h*ci(i,2:N) + ...
D*(ci(i,1:N-1) + ci(i,3:N+1) - 2*ci(i,2:N));
ca(i+1,1) = 0;
ci(i+1,1) = ci(i,1) + D*ca(i,2) + D*(ci(i,2) - ci(i,1));
Boundary conditions: no flux
at the right; activated concentration
is zero at the left; activated ‘flux in’
i equall tto iinactivated
is
ti t d ‘fl
‘flux out’t’ att
the left.
ca(i+1,N+1) = ca(i,N+1) + h*ci(i,N+1) + D*(ca(i,N) - ca(i,N+1));
ci(i+1,N+1) = ci(i,N+1) - h*ci(i,N+1) + D*(ci(i,N) - ci(i,N+1));
end
for i = (M/2+1:M)
ca(i+1 2:N) = ca(i,2:N)
ca(i+1,2:N)
ca(i 2:N) + h*ci(i
h ci(i,2:N)
2:N) + ...
D*(ca(i,1:N-1) + ca(i,3:N+1) - 2*ca(i,2:N));
ci(i+1,2:N) = ci(i,2:N) - h*ci(i,2:N) + ...
D*(ci(i,1:N-1) + ci(i,3:N+1) - 2*ci(i,2:N));
D ~ 1μ m 2 / s
D > 10 for effective inactivation
D
1
α<
~
0.1
s
,
L
~
1
μ
m
,
T
~
> 10s
2
10 L
α
Where are 100’s of sec in the paper coming from?
ca(i+1,1) = ca(i,1) + h*ci(i,1) + D*(ca(i,2) - ca(i,1));
ci(i+1,1) = ci(i,1) - h*ci(i,1) + D*(ci(i,2) - ci(i,1));
ca(i+1,N+1) = ca(i,N+1) + h*ci(i,N+1) + D*(ca(i,N) - ca(i,N+1));
ci(i+1,N+1) = ci(i,N+1) - h*ci(i,N+1) + D*(ci(i,N) - ci(i,N+1));
end
%------------------------GraphicOutput----------------plot((0:N)/N,ca(M/2,:),'r--',(0:N)/N,ci(M/2,:),'m--',...
(0:N)/N,ca(M/2+1000,:),'r',(0:N)/N,ci(M/2+1000:),'m')
2
∂c *
∂ 2c *
= D 2 − c * + rcc *
∂t
∂x
∂c
∂ 2c
= D 2 + c * −rcc *
∂t
∂x
1.8
Simplest model
Red – activated, blue – inactivated
Solid – after, dashed – before
D=10
1.6
1.4
1.2
1
08
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D ~ 1, r >> 1: decent spatial inhibition;
alpha does not have to be small
anymore, so switch can be fast.
But: ‘Auto-lock’ in an inhibited state;
do not need KT anymore:
2
18
1.8
1.6
c * − rcc* = c * (1 − rc ) = 0, c + c* = 1
Simplest model
Red – activated, blue – inactivated
Solid – after, dashed – before
D=3
1.4
1.2
1) c* = 0, c = 1
1
1
2) c = , c* = 1 −
r
r
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
1.8
1.6
1.4
Self-propagating inhibition
Red – activated, blue – inactivated
Solid – after, dashed – before
D=1, r=4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Emitted inhibition:
α c * −γ ce* = 0
α c * +λ e* = 0
λ e * −γ ce* = 0
c* = 0
c + c* = 1
e + e * +c* = 1
e =1
Inhibition range:
Switching time:
c =1
e* = 0
D / λ or
D /α
1/ λ or 1/ α
So, benefit is not obvious… just twice better?
Small yeast; big animal cells
nonautocatalytic amplification
Doncic’s scheme does not catalytically amplify the inhibitory
g
One e* molecule can interact with only
y one c
signal.
molecule. In Sear’s scheme, a single e* molecule can
convert many molecules
into the inhibiting form, thereby producing amplification.
Sear and Howard PNAS 103 16763 2006
Can we figure out the type of certain reaction from a general requirement?
Doncic et al Molecular Systems Biology 2 1 2006
There is a considerable noise in protein expression
Doncic et al examined the capacity of the mitotic spindle checkpoint to buffer
temporal fluctuations in Cdc20 production rate. Their results suggest that inhibiting
Cdc20 through a sequestering mechanism allows for a significant buffering of
protein production noise.
Noise
N
i iin th
the Cd
Cdc20
20 production
d ti iis th
thus b
buffered
ff d
by its tethering to the activated complexes
Doncic et al Molecular Systems Biology 2 1 2006
o
k1
k2
k2
c
k 4 k3
ddc
= k1 − ( k2 + k4 ) c + k3 m
dt
dm
= k 4 c − ( k 2 + k3 ) m
dt
m
Degradation: perturbation decays with rate k2
k2.
Sequestering: perturbation decays with rates k2
and k2+k3+k4. We can keep k2 low, if k4 is high,
and k3 is medium – then most of c is in the form
of m,
m and average c is low
low. Noise is ‘filtered
filtered out’
out
then, because very small in amplitude perturbation
of c decays rapidly, with rate k2+k3+k4, while
greater in amplitude perturbation of m decays with
slow rate k2.
k2
C =c+m
dc
= ( k1 + k3C ) − ( k2 + k3 + k4 ) c
dt
dC
= k1 − k2C
dt
k1
k1 + ε
k1 = 1;k2 = 0.1;k3 = 1;k4 = 10;eps=0.1;
h=0.01; t = (0:h:100);
c = ones(size(t)); m = 9*ones(size(t));
for n = (1:length(t)-1)
c(n+1) = c(n) + h*(k1+3*randn-(k2+k4)*c(n)+k3*m(n));
m(n+1) = m(n) + h*(k4*c(n)-(k2+k3)*m(n));
end
plot(t,c), axis([0 100 0 2])
1.12
1.1
1.1
1.08
τ ~1
1.08
1.06
1.06
τ ~ 10
1.04
δ c ≈ 0.09
1.04
k1 = 1;k2 = 1;k3 = 0;k4 = 0;eps=0.1;
1 02
1.02
1.02
δ c ≈ 0.1
1
1
0 98
0.98
0
1
2
3
4
5
6
7
8
9
10
0 98
0.98
k1 = 1;k2 = 0.1;k3 = 1;k4 = 10;eps=0.1;
0
5
10
15
20
25
30
35
40
45
50
2
2
1.8
1.8
k2 = 0.1;k3 = 1;k4 = 10;
k2 = 1;k3 = 0;k4 = 0;
1.6
1.6
1.4
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
10
20
30
40
50
60
70
80
90
100
0
0
10
20
30
40
50
60
70
80
90
100