A simple model for the eukaryotic cell cycle · - In Xenopus early development, with large mass,...

A simple model for the eukaryoticcell cycle

Andrea Ciliberto

The cell division cycle

Start

S(DNA Replication)

cell divis

ion

G1

Fini

sh

G2G2/MM(mitosis)

Kohn, Mol. Biol. Cell., 1999

How did we get to this mess??

Murray and Kirschner, Science, 1989

Xenopus and the clock paradigm

Cell mass decreases during early divisions

Alberts et a., Molecular Biology of the Cell.2002

In Xenopus oscillations progress independentlyof DNA presence and cell cycle events

Autonomous oscillations!

Alberts et a., Molecular Biology of the Cell.2002

MPF, the mitosis promoting factor

Murray and Kirschner, Science, 1989

MPF is a heterodimer

CDK

Cyc

cyclin dependent kinase

cyclin (regulatory subunit)

Only cyclin synthesis and degradation are required for Xenopus early cycles.

Alberts et a., Molecular Biology of the Cell.2002

CDK

CycCyc

CDK

APCa APCi

CDK is activated by cyclin binding and onceactivated it induces cyclin degradation

‘X’

CDK

CycCyc

CDK

APCa APCi

But something else must be at work...

‘X’

Cyclin threshold

Solomon et al, Cell, 1990

Yeast and the domino paradigm

S

G1

G2Metaphase

Anaphase

Balanced growth and division

Size control

TDTC

Cell cycleengine

Cytoplasmicgrowth

Cyt

opla

smic

mas

s

exponentialbalancedgrowth

0 1 2 3 40

1

2

3

4

TC > TD

TC < TD

TC = TD

Cell division cycle (cdc) mutants are temperature sensitive

Alberts et a., Molecular Biology of the Cell.2002

Hartwell, Genetics, 1991

wee1wild type cdc25

Wee1 controls a rate limiting step in the cell cycle

Cell division and cell growth are coupled

unreplicated DNA

Nurse, Noble lecture, 2000

Basic cell cycle properties

- Cell physiology-

- Coupling of mass growth and cell division.

- Once the cell enters the cycle,it is commited to finish it: irreversibility.

- The cell halts during cell cycle progression if something has gone wrongly.

-Molecular network-

-Oscillations of MPF drive cells into and out of mitosis.

- Cdc28 activity is controlled by Wee1 (negative) and Cdc25 (positive).

Dominoes and clocks: Cdc28 is thebudding yeast homologous of MPF’s

catalytic subunit

MPF Cdc28

Clb2= Cdc2

Cdc13= = CDK1

CycB

Cdc2

Cyc

P

Cdc25P

Wee1

Cdc2

CycCyc

Cdc2

APCaAPCi

Phosphorylation as well as cyclin binding controls MPF activity

‘X’

mass

Cdc25

Cdc2

Cyc

P

Cdc25P

Wee1 Wee1P

Cdc2

CycCyc

Cdc2

APCaAPCi

‘X’

mass

Phosphorylation as well as cyclin binding controls MPF activity

Cdc2

Cyc

P

Wee1 Wee1P

Isolation and analysis of a positive feedback: the network...

G2

M

Cdc2

Cyc

Notice, here no cyclin synthesis, no cyclin degradation!!

...and the physiology

Solomon et al, Cell, 1990

Part IIStandard laws of biochemical kinetics

applied to molecular networks

pMPF MPF

�

dMPF

dt= ka ! pMPF

pMPF =MPFtot -MPF

dMPF

dt= ka ! (MPFtot -MPF)

ka

dMPF

dt= 0

MPFSS=MPF

tot

Steady State solution (MPFSS)

Law of Mass Action: forward reaction

Notice: no dimer, only MPF. Cdk is supposed to be present in excess throughoutthe cycle. Increasing MPF total mimics an increase in cyclin total.

0 0.2 0.4 0.6 0.8 1-0.1

0

0.1

0.2

MPF

�

dMPF

dt

�

dMPF

dt> 0

�

dMPF

dt= 0

0 4 8 12 16 200

0.2

0.4

0.6

0.8

1

timeM

PF

MPFtot

dMPF

dt= ka ! (MPF

tot-MPF)

�

dMPF

dt= ka ! pMPF - ki !MPF

ki

dMPF

dt= 0

MPFSS=ka !MPF

tot

ka + ki

Steady State solution

Law of Mass Action: reversible reaction

pMPF MPFka

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

MPF

�

dMPF

dt> 0

�

dMPF

dt< 0

�

dMPF

dt= ka ! (MPFtot "MPF) - ki !MPF

production+

elimination-

�

dMPF

dt= 0

ki

pMPF MPFka

�

dMPF

dt

0 4 8 12 16 200

0.2

0.4

0.6

0.8

1

time

MPF

t

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

1

2

34

5

pMPF MPFka

ki

Wee1

MPF

rate

12345

Law of Mass Action: catalyzed reversible reaction

�

dMPF

dt= ka ! (MPFtot "MPF) - ki !MPF !Wee1

production+

elimination-

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

1

2

34

5

MPF

rate

12345 0 2.5 50

0.5

1MPFSS

Wee11 2 3 4

�

dMPF

dt= 0

�

dMPF

dt< 0

�

dMPF

dt> 0

Nullclines

pMPF MPFka

ki

Wee1

�

dMPF

dt= ka ! (MPFtot "MPF) - ki !MPF !Wee1

production+

elimination-

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

1

2

34

5

MPF

rate

12345 0 2.5 50

0.5

1MPFSS

Wee11 2 3 4

�

dMPF

dt= 0

�

dMPF

dt< 0

�

dMPF

dt> 0

What happens if MPF total increases?

Wee1P Wee1

�

dWee1

dt=kwa !Wee1P

J +Wee1P

dWee1

dt=kwa ! (Wee1tot -Wee1)

J + (Wee1tot -Wee1)

kwa

dWee1

dt= 0

Wee1SS= Wee1

tot

Steady State solution

Michaelis-Menten: forward reaction

0 4 8 12 16 200

0.2

0.4

0.6

0.8

1

�

Wee1tot

time

b

0 0.2 0.4 0.6 0.8 1-0.1

0

0.1

0.2

Wee1

�

dWee1

dt

�

dWee1

dt= 0�

dWee1

dt> 0

Michaelis-Menten: reversible reaction

Wee1P Wee1kwa

kwi

Wee1P Enzyme1:Wee1Pk1

Enzyme1 Enzyme1

Wee1k1r

k2

Wee1 Enzyme2:Wee1k3

Enzyme2 Enzyme2

Wee1Pk3r

k4

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1

J +Wee1

production+

elimination-

if [enzym1TOT], [enzyme2TOT] << [Wee1TOT]

kwa=[enzyme1TOT]k2

kwi=[enzyme2TOT]k4

Michaelis-Menten: reversible reaction

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1

J +Wee1

production+

elimination-

Wee1P Wee1kwa

kwi

0.0 0.2 0.4 0.6 0.8 1.00.0

0.1

0.2

0.3

0.4

0.5

Wee1*

rate

�

dWee1

dt< 0

�

dWee1

dt> 0

�

dWee1

dt= 0

0.0 0.2 0.4 0.6 0.8 1.00.0

0.1

0.2

0.3

0.4

0.5

Wee1SS

rate

�

dWee1

dt< 0

�

dWee1

dt> 0

�

dWee1

dt= 0

Nullclines

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1

J +Wee1

production+

elimination-

Wee1P Wee1kwa

kwi

0 2.5 50

0.5

1Wee1SS

MPF1 2 3 4

�

dWee1

dt< 0

�

dWee1

dt> 0

Wee1P Wee1kwi

MPF pMPFka

ki

Phase plane analysis

�

dMPF

dt= ka ! (MPFtot "MPF) - ki !MPF !Wee1

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1

J +Wee1

0 0.5 1

Wee1SS

MPF

02.5

5

�

dWee1

dt< 0

�

dWee1

dt> 0

0 2.5 50

0.5

1MPFSS

Wee11 2 3 4

�

dMPF

dt= 0

�

dMPF

dt< 0

�

dMPF

dt> 0

0 2.5 50

0.5

1

Wee1

MP

F

0 0.5 1

Wee1SS

MP

F

02.5

5

�

dWee1

dt< 0

�

dWee1

dt> 0

0 2.5 50

0.5

1

MP

FSS

Wee1

�

dMPF

dt= 0

�

dMPF

dt< 0

�

dMPF

dt> 0

How does MPF increases with Cyclin total?

0

MPFtot

MPF

MPFSS=ka !MPF

tot

ki !Wee1+ka

0 2.5 50

0.5

Wee1

MP

F

Solomon et al, Cell, 1990

Not quite the same!

Wee1P Wee1kwi

MPF pMPFka

ki

Michaelis-Menten: catalyzed reversible reaction

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1 !MPF

J +Wee1

production+

elimination-

Wee1P Wee1kwa

kwi

MPF

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

Wee1SS

MPF

�

dWee1

dt= 0

�

dWee1

dt< 0

�

dWee1

dt> 0

Nullclines

0.0 0.2 0.4 0.6 0.8 1.00.0

0.1

0.2

0.3

0.4

0.5

1

2

3

4

5

.1

Wee1SS

rate

Wee1P Wee1kwi

MPF

dWee1

dt=kwa ! (Wee1

tot"Wee1)

J +Wee1tot-Wee1

-kwi !Wee1 !MPF

J +Wee1

production+

elimination-

0 2.5 50

0.5

1MPFSS

Wee10 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1Wee1SS

MPF

�

dWee1

dt< 0

�

dMPF

dt< 0

�

dMPF

dt> 0

�

dWee1

dt> 0

�

dWee1

dt= 0

�

dMPF

dt= 0

Phase plane analysis

0 0.5 10

2.5

5Wee1

MPFSS0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1Wee1SS

MPF

�

dWee1

dt< 0

�

dMPF

dt< 0

�

dMPF

dt> 0

�

dWee1

dt> 0

�

dWee1

dt= 0

�

dMPF

dt= 0

Phase plane analysis

0 0.4 0.8 1.2 1.60

0.2

0.4

0.6

0.8

1

Wee1

MPF

First solution, MPF wins, Wee1 loses

MPFtot=1.5

0 0.4 0.8 1.2 1.60

0.2

0.4

0.6

0.8

1

Wee1

MPF

Second solution, Wee1 wins, MPF loses

MPFtot=0.5

0 0.4 0.8 1.2 1.60

0.2

0.4

0.6

0.8

1

Wee1

MPF

Third solution, both can win: hysteresis

MPFtot=1

How does MPF increases with Cyclin total?

Wee1Wee1

MPFMPF

Wee1

MPF

MPFtot=0.5MPFtot=1.5MPFtot=1

Hysteresis in the Xenopus early cycles: simulationof an experimental result

From Sha et al, PNAS, 2003

What happens if cyclin total increases with cell mass?

Cdc25

Cdc2

Cyc

P

Cdc25P

Wee1 Wee1P

Cdc2

CycCyc

Cdc2

APCaAPCi

‘X’

mass

Conclusion

-Same wiring in different organisms, combination of positive and negative

feedbacks.

- In Xenopus early development, with large mass, the cell cycle is a limit

cycle oscillator, the negative feedback plays the key role.

- Artificially, an additional mechanism of control emerges, based on a

positive feedback loop.

- Both positive and negative feedbacks are at work in yeast. In these

organisms, mass growth drives the cell cycle.

- Positive feedbacks introduce checkpoints and irreversibility in the cycle.

The negative feedback the capability to start a new process.