Problem #2: checkpoint mechanism

Li et al Trends Cell Biol 13 553 2003

Shah and Cleveland Cell 103 997 2000

F t i l tt h d KT t h id ti f i l d t i diff i i thFacts: a single unattached KT prevents anaphase; rapid reactions of involved proteins diffusing in thecytoplasm; 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

[ ] [ ] [ ] [ ]8 8

::

d APC Cdck APC Cdc k APC Cdc= − + ×

Ibrahim et al 2008 PLoS ONE 3: e1555

[ ] [ ] [ ]8 8:k APC Cdc k APC Cdcdt − + ×

mass action

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 geneticalgorithm 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(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)

Doncic et al PNAS 102 6332 2005

Yeast cell: mitosisin the nucleus

Direct Self-Propagating EmittedInhibition Inhibition Inhibition

Doncic et al PNAS 102 6332 2005

2

2

* * *C CD CT X

α∂ ∂= −

∂ ∂

0 X L≤ ≤

% ca - active; ci - inhibiteda = 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 N+1); ci = ones(M+1 N+1); % Init Cond0 X L≤ ≤

, / , * *X Lx T t C Ccα= = =

ca = ones(M+1,N+1); ci = 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));

2* *c c D∂ ∂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));

2 2*, ,0 1c c DD c D xt x Lα

∂ ∂= − = ≤ ≤

∂ ∂

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) + h*ci(i 2:N) +

Boundary conditions: no fluxat the right; activated concentrationis zero at the left; activated ‘flux in’i l t i ti t d ‘fl t’ t 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));

is equal to inactivated ‘flux out’ atthe left.

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));

2~ 1 /D m sμ

D > 10 for effective inactivation1D 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

1~ 0.1 , ~ 1 , ~ 1010

D s L m T sL

α μα

< >

Where are 100’s of sec in the paper coming from?

1.6

1.8

2

Simplest modelRed – activated, blue – inactivated

2

2

2

* * * *c cD c rcct x

∂ ∂= − +

∂ ∂∂ ∂

0 8

1

1.2

1.4Red activated, blue inactivatedSolid – after, dashed – beforeD=10

2

2 * *c cD c rcct x

∂ ∂= + −

∂ ∂

0.2

0.4

0.6

0.8

D ~ 1, r >> 1: decent spatial inhibition;alpha does not have to be smallanymore, so switch can be fast.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1 8

2

anymore, so switch can be fast.But: ‘Auto-lock’ in an inhibited state;do not need KT anymore:

1.2

1.4

1.6

1.8

Simplest modelRed – activated, blue – inactivatedSolid – after, dashed – beforeD=3

( )* * * 1 0, * 1c rcc c rc c c− = − = + =

1) * 0, 1c c= =

0.6

0.8

1 1 12) , * 1c cr r

= = −

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

2

1.4

1.6

1.8

Self-propagating inhibition

0.8

1

1.2 Red – activated, blue – inactivatedSolid – after, dashed – beforeD=1, r=4

0.2

0.4

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

Emitted inhibition:Emitted inhibition:

* * 0* * 0

c cec e

α γα λ

− =+ =

* 01

cc==

* * 0* 1* * 1

e cec ce e c

λ γ− =+ =+ + =

* 01

ee==

1e e c+ + =

Inhibition range: / or /D Dλ α

Switching time: 1/ or 1/λ α

So, benefit is not obvious… just twice better?

Small yeast; big animal cells

nonautocatalytic amplification

Doncic’s scheme does not catalytically amplify the inhibitorysignal. One e* molecule can interact with only one cg ymolecule. 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 bufferDoncic et al examined the capacity of the mitotic spindle checkpoint to buffer temporal fluctuations in Cdc20 production rate. Their results suggest that inhibitingCdc20 through a sequestering mechanism allows for a significant buffering of protein production noise.

N i i th Cd 20 d ti i th b ff dNoise in the Cdc20 production is thus buffered by its tethering to the activated complexes

Doncic et al Molecular Systems Biology 2 1 2006

co 1kk dco

2k

2k 4k 3k( )

( )

1 2 4 3

4 2 3

dc k k k c k mdtdm k c k k m

= − + +

= − +C c m= +

m

2 ( )4 2 3k c k k mdt

+

( ) ( )1 3 2 3 4dc k k C k k k cdt

= + − + +m1 2

dtdC k k Cdt

= −

Degradation: perturbation decays with rate k2

1k 1k ε+Degradation: perturbation decays with rate k2.Sequestering: perturbation decays with rates k2and k2+k3+k4. We can keep k2 low, if k4 is high,and k3 is medium – then most of c is in the formof m and average c is low Noise is ‘filtered out’of m, and average c is low. Noise is filtered outthen, because very small in amplitude perturbationof c decays rapidly, with rate k2+k3+k4, whilegreater in amplitude perturbation of m decays withslow rate k2slow rate k2.

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));

endplot(t,c), axis([0 100 0 2])

1.08

1.1

1.08

1.1

1.12

1τ

1 02

1.04

1.06

1.04

1.06

k1 = 1;k2 = 1;k3 = 0;k4 = 0;eps=0.1;

~ 10τ

~ 1τ

0.09cδ ≈

0 98

1

1.02

k1 = 1;k2 = 0.1;k3 = 1;k4 = 10;eps=0.1;

0 98

1

1.02

0.1cδ ≈

0 5 10 15 20 25 30 35 40 45 500.98

0 1 2 3 4 5 6 7 8 9 100.98

2 2

1.4

1.6

1.8k2 = 0.1;k3 = 1;k4 = 10;

1.4

1.6

1.8k2 = 1;k3 = 0;k4 = 0;

0.8

1

1.2

0.8

1

1.2

0.2

0.4

0.6

0.2

0.4

0.6

0 10 20 30 40 50 60 70 80 90 1000

0 10 20 30 40 50 60 70 80 90 1000