Transcript Systems_lectures_14

Systems Biology I: Bistability
Macromolecules
Jonathan Weissman
Systems Biology
How does complex biological behavior emerge from the
organization of proteins into pathways and networks?
Common properties:
1. Bistability
• convert continuous stimulus into discrete response
2. Hysteresis
• “memory” of stimulus long after it is withdrawn
Outline for this lecture:
Use simple graphical method to describe conditions that
lead to bistability
Discuss two biological examples of bistability:
Xenopus oocyte maturation - MAPK
Lac operon
Go through experiments that demonstrate system is bistable
Understand what the origin of bistability is
Describe the physiological consequences of bistability
Cellular responses to stimuli: switches
versus dimmers
Michaelian system - no feedback
stimulus


A*
A

inactivation
k 1S


A*
A 

k 1 I
Assume S, I far from
saturation
forward rate  k 1S A   k1S Atot  k 1S A *
Atot  A  A *
reverse rate  k  1 I A *
At steady state:
k  1 I A *  k 1S Atot  k 1S A *
System is monostable
Perturb [A*/Atot], return to ss
k1S
k  1I
Michaelian system - no feedback, cont.
k 1S


A*
A 

k 1 I
At steady state:
k  1 I A *  k 1S Atot  k 1S A *
S
A * 
A *  0.5
, when S  k  1 I / k 1,
Atot  k  1I / k1  S
Atot 
A *  S , EC50  S req to produce 50% response
Atot  EC50  S
Response levels off because
the larger S gets, the less
inactive A there is for the
kinase to act on, and the more
A* there is for the p’tase
increase S
Michaelian system with linear feedback
forward rate  basal rate + feedback rate
forward rate  k1S A  k 2 A *A 
forward rate  (k1S  k 2 A *)(Atot  A *)
e.g. A* increases activity of S
or phosphorylates A in trans
feedback rate = k 2 A *A 
Extremes:
With no [A*] feedback will be 0
When [A*]/[Atot]=1 and [A]=0 so
feedback also equals 0
In between:
Will be maximum when [A*]=[A]
Michaelian system with linear feedback, cont.
forward rate  basal rate + feedback rate
forward rate  k1S A  k 2 A *A 
Consider simple case where
k1S is small so basal rate
is negligible
Michaelian system with linear feedback, cont.
Two steady states:
strong feedback
k 2 Atot   k  1 I
stable ON state but
unstable OFF state
One steady state:
weak feedback
k 2 Atot   k  1 I
stable OFF state
but no ON state
How can we stabilize the OFF state?
1. Make feedback sigmoidal function of A*
• three steady-states, one unstable (threshold) and two stable
• at threshold, if decrease A*, back reaction is faster than
forward reaction and system is driven to OFF state
• at threshold, if increase A*, forward reaction is faster than
back reaction and system is driven to ON state
Making feedback sigmoidal function of A*, cont.
i. cooperativity
e.g. A* activates S and takes n molecules of A* to activate S
feedback  A*
n
ii. zero-order ultrasensitivity
Zero-order conditions
rate independent of [substrate]
First-order conditions
rate [substrate]
Making feedback sigmoidal function of A*, cont.
ii. zero-order ultrasensitivity, cont.
Opposing reactions with
the [substrate] >> Km for
the modifying enzymes
A
kinase
phosphatase
First-order
Atot  0.01 K m
A*
First-order: changing
p’tase by 1.5x has small
effect on steady state
Zero-order: changing p’tase
by 1.5x changes ss from >90%
A* to >90% A
Zero-order
Atot  100  K m
Making feedback sigmoidal function of A*, cont.
ii. zero-order ultrasensitivity, cont.
• small changes in enzyme activity lead to large changes
in modified protein, steady state
iii. inhibitor ultrasensitivity
• low abundance, high affinity inhibitor of S
• once inhibitor is saturated, S will be effective at causing
increases in [A*] (same as S-P below)
- inhibitor: 1/9 EC50 causes 10% response
90% response at 9 x EC50
81-fold change in kinase 10 to 90%
+ inhibitor: 10/9 units kinase causes 10%
response 90% response at 10 units kinase
9-fold change in kinase 10 to 90%
How can we stabilize the OFF state?
2. Saturate back reaction with A* before feedback saturates
• back reaction saturates as A* accumulates
• back reaction can more than keep up with first bit of feedback,
but then is overwhelmed as feedback continues to rise
How can we stabilize the OFF state?
1. Make feedback a sigmoidal function of A*
i. cooperativity
ii. zero-order ultrasensitivity
iii. inhibitor ultrasensitivity
2. Saturate back reaction with A* before feedback saturates
Systems are bistable only for a limited range of kinetic parameters
Switching between the OFF and ON states
• continuously increase feedback-independent stimulus S
• bistable system converts continuous change in S into discontinuous
change in output (A*)
forward rate  basal rate + feedback rate
basal rate  k1S A 
increase S - increase slope of basal rate
Switching between the OFF and ON states
What happens when we lower the stimulus?
OFF state reappears when S is lowered to 3, but no driving force to leave
ON state
HYSTERESIS/IRREVERSIBILITY
Hysteresis and irreversibility
Hysteresis
Path from ON to OFF is different than that from OFF to ON
Any bistable system will exhibit some degree of hysteresis
Irreversibility
System stays in ON state indefinitely after S is removed
Occurs when feedback is strong
Decreases chattering between ON and OFF states when S is near threshold
Mechanism for biochemical memory - unless positive feedback is broken
system will remain in ON state - can remember a stimulus long after it
is removed
Example 1: Xenopus oocyte maturation
immature
mature
progesterone
G2 arrested
meiosis I
GVBD
How does this system turn a
continuously varying stimulus
(progesterone) into an all-ornone response?
MAPK
cascade
metaphase arrest,
meiosis II
Xenopus oocyte maturation, cont.
Measure response of oocytes to different [progesterone] by measuring
MAPK phosphorylation (activation)
Response of population - treat oocytes with give [progesterone], make
extract, measure MAPK phosphorylation and tabulate % activation
Response is
graded - Michaelian!
Xenopus oocyte maturation, cont.
How do we interpret these observations?
OR
[progesterone]
[progesterone]
Answer: Look at the response of individuals
Xenopus oocyte maturation, cont.
Response is all-or-none - no intermediate levels of MAPK activation
BISTABLE
Where does this behavior come from?
Clue: Microinjection of activated Mos gives same result (bistability),
indicating it comes from pathway downstream of Mos (MAPK cascade)
Xenopus oocyte maturation, cont.
What are the functional consequences of bistability for this system?
Small change in stimulus near threshold throws switch
At low stimulus buffered a bit from change
No intermediate states
Xenopus oocyte maturation, cont.
Can quantitate the sensitivity of the response - for a given change in
stimulus, how much change In output (MAPK phosphorylation) results
Michaelian - requires 81-fold change in [ligand] to drive system from
10% to 90% on
Ultrasensitive - requires <81-fold change in [ligand] to drive system from
10% to 90% on
Subsensitive - requires >81-fold change in [ligand] to drive system from
10% to 90% on
Xenopus oocyte maturation, cont.
Hill coefficient - measure of sensitivity
Michaelian: nH=1
Ultrasensitive: nH>1
Subsensitive: nH<1
Relationship exists between Hill
coefficient and frequency of
occurrence of intermediate states
Higher Hill coefficient - fewer
intermediate states
Counted 190 oocytes - lower
bound on nH is 42
Xenopus oocyte maturation, cont.
What is the origin of the bistability?
REMEMBER: Bistability arises from positive feedback and
a mechanism to stabilize the off state
1. Multistep ultrasensitivity - stabilizing the off state
MAPKK
MAPK
MAPKKK
MAPKK
MAPKK-P
MAPK-P
MAPKKK
MAPKK
MAPKK-PP
MAPK-PP
Two phosphorylation events required for MAPK and MAPKK
activation
Activating kinase must dissociate in between phosphorylation
events (to load ATP)
Rate will vary as [upstream kinase]2
Two steps - effects are multiplicative
Xenopus oocyte maturation, cont.
1. Multistep ultrasensitivity - stabilizing the off state
Michaelian
Multistep ultrasensitivity
• forward rate (slope) varies as [kinase]2
• 1 unit of kinase gives 50% S-P; double kinase and rate increases 4-fold
• S-P must increase to higher level than in Michaelian case for rates
to balance and reach new steady state (80% vs. 67%)
• leads to sigmoidal stimulus-response curve
Xenopus oocyte maturation, cont.
1. Multistep ultrasensitivity, cont. - stabilizing the off state
Measure dependence of MAPK activation on Mos input in vitro in oocyte
extracts
nH ~ 5
~2.5 fold change in MAPKKK
changes MAPK from 10%
to 90% activity
Xenopus oocyte maturation, cont.
2. Zero-order ultrasensitivity - stabilizing the off state
• no clear data that reactions in MAPK cascade are operating
in zero-order conditions, but concentrations and kinetic
properties that have been measured are consistent with this
idea
• small change in activity leads to large change in ss position
3. Positive feedback
+
Mos 
 Mos* 
 MEK 
 MAPK
• protein synthesis-dependent feedback from active MAPK
to Mos
Xenopus oocyte maturation, cont.
Is the biochemistry underlying oocyte maturation irreversible?
• Is the [progesterone] required to activate MAPK and Cdc2
different from the [progesterone] required to maintain activities?
Induction: incubate with
progesterone, wait until
maturation plateaus
Maintenance: incubate with
600 nM progesterone, wait
until GVBD plateaus, wash
for 10 hr, incubate with
different [progesterone]
Maturation is irreversible - “memory” of progesterone
Xenopus oocyte maturation, cont.
• based on what we learned about bistability, postulate it is the
strength of the feedback that gives rise to irreversibility
• predict that if feedback is disrupted, system should lose
irreversibility or “memory”
+E2=MAPK pathway
activated
+E2, wash=MAPK pathway
activated then 16 hr wash
Disrupt feedback with
CHX, Mos antisense
“Memory” requires positive feedback
Xenopus oocyte maturation, cont.
• also predict that disrupting positive feedback should alter
bistability - make less ultrasensitive
• quantitate response of individual oocytes to microinjected Mos
in presence and absence of CHX
• observe oocytes with intermediate amounts of phosphorylation - loss of
bistability
Xenopus oocyte maturation, cont.
• ultrasensitivity of MAPK cascade plus positive feedback generates
bistability
• Michaelian MAPK with feedback - unstable OFF state
• ultrasensitive MAPK with feedback - stable OFF state, filters out small
stimuli
Lac operon
• beta-galactosidase is formed by bacterial cells grown in the
presence of lactose - beta-gal is necessary for metabolism
of lactose
• known that non-metabolizable galactosides (e.g. TMG;
thiomethyl-galactoside) induce beta-gal formation
• add high [TMG], see induction of beta-gal at maximal
induction rate - under these conditions cells respond
uniformly
• at lower [TMG], see dose-dependent rate of induction of
beta-gal
• cells that have been exposed to a high [TMG] and then shifted
to lower [TMG] (maintenance concentrations) continue
to synthesize beta-gal at the maximum rate
Phenotypic change following transient signal memory of inducer!
Lac operon, cont.
Is the Lac system bistable?
Cells growing in [TMG]
that gives intermediate level
of beta-gal synthesis (e.g. 30%)
Dilute to maintenance [TMG] to
give ~1 bacterium/10 tubes
• 10% of the tubes developed bacterial populations
• 30% of the cultures from single bacteria had maximal beta-gal
levels; 70% had only small amounts
Population at intermediate [TMG] consists of cells that are fully
induced and those that are uninduced - BISTABLE
Lac operon, cont.
TMG
LacI = Lac repressor
LacY = Lac permease
TMG
LacI
LacZ = beta-galactosidase
LacY
LacZ
• autocatalytic positive feedback (double negative)
• if there is non-linearity in response, might expect
bistability in TMG, LacI activity, LacY and LacZ expression
Lac operon, cont.
d TMG 
net rate 
 vin  vout
dt
vout  vbind  vcatab  kbind TMG  0  kbind TMG 
For vin , know  TMG e
But TMG also indirectly activates entry of TMG through LacI
TMG i
n
K  TMG i

TMG i 

vin  TMG e  k0 
n
K

TMG
 i 

Response 
Could come from TMG binding
LacI, relationship between active
LacI and LacY production…
Need k0 parameter because if there is no permease the first
molecule of TMG cannot enter the cell, independent of the external
concentration - either operon is not totally repressed or TMG can
slowly diffuse
Lac operon, cont.
vout  kbind TMG 

TMG i 

vin  TMG e  k0 
n
K

TMG
 i 

[TMG]
[TMG]
Three steady-states:
ss1, 3 stable
ss2 unstable
Bistable
Lac operon, cont.
Can see bistability and hysteresis in single cell experiments
Take cells expressing Lac promoter-GFP and grow in
different [TMG] - either initially uninduced or induced
[maintenance]
start
induced
Uninduced then grown
20 hr in 18 mM TMG
start
uninduced
Bistability and hysteresis
Lac operon, cont.
When inducer is added, permease synthesis will be initiated at
a rate determined by inducer in medium
At low [inducer] rate of permease synthesis may be so low that
probability of making a permease molecule during lifetime of bacterium
is small
Once bacterium has permease, [inducer] inside cell will increase, which
will increase probability that second molecule of permease will be formed
Two permeases further increases rate of permease - autocatalytic rise to
maximal permease
Once have maximal permease, bacterium and progeny will be induced
indefinitely since [inducer] inside is above maintenance concentration
Maintenance can be explained if the number of permease molecules is large
relative to threshold, there is high probability that each daughter cell will
receive sufficient permease to insure induction
Transfer to less than maintenance - observe exponential decrease in activity,
equivalent to constant chance of becoming uninduced - partitioning between
daughter cells - some will not get enough to insure induction
Flipping the switch: induction at intermediate lac concentration occurs by
complete dissociation of the tetrameric lac repressor
(A) A high concentration of intracellular inducer can force dissociation of the repressor from its operators, (B) At low or
intermediate concentrations of intracellular inducer, partial dissociation from one operator by the tetrameric LacI repressor is
followed by a fast rebinding. Consequently, no more than one transcript is generated during such a brief dissociation event.
However, the tetrameric repressor can dissociate from both operators stochastically and then be sequestered by the inducer so
that it cannot rebind, leading to a large burst of expression. (C) A time-lapse sequence captures a phenotype-switching event. In
the presence of low inducer, one such cell switches phenotype to express many LacY-YFP molecules (yellow fluorescence
overlay) whereas the other daughter cell does not
Bistability Biobliography
Reviews & C omments on Research Articles:
Ferrell, J. E., Jr. (1996). Tri pping the switch fantastic: how a protein kinase
cascade can convert graded inputs into switch-like outputs. T rends Biochem Sci
21, 460-6.
Koshland, D. E., J r. (1998). The era of pathway quantifi cation. Science
280, 852-3.
Koshland, D. E., Jr., Goldbeter, A., and Stock, J. B. (1982). Amplification
and adaptation in regulatory and sensory systems. Science 217, 220-5.
Laurent, M. and Kellershohn , N. (1999) Multi stability: a major means of
differentiation and evo lution in biological systems. Trends Biochem. Sci. 24, 418422.
Research articles:
Novick, A. and Wei ner, M. (1957). Enzyme induction as an all-or-none
phenomenon. Proc. Natl. Acad. Sci. U S A 43, 553-566.
Cohn, M. and Horibata, K. (1959). Analysis of the differentiation and of the
heterogeneity within a populati on of Escherichia coli undergoing induced betagalactosidase syn thesis. J. Bacteriology 78, 613-633.
Ferrell, J. E., Jr., and Machleder, E. M. (1998). The biochemical basis of
an all -or-none cell fate switch in Xenopus oocytes [see comments]. Science 280,
895-8.
Goldbeter, A., and Koshland , D. E., Jr. (1981). An amplified sensitivity
arising from covalent modification in biological systems. Proc Natl Acad Sci U S
A 78, 6840-4.
Goldbeter, A., and Koshland , D. E., Jr. (1984). Ultrasensitivity in
biochemical syst ems controlled by covalent modification. J. Biol. Chem. 259,
14441-14447.
Huang , C. Y., and Ferrell, J. E., Jr. (1996). Ultrasensitivity in the mitogenactivated protein kinase cascade. Proc Natl Acad Sci U S A 93, 10078-83.
Xiong, W. a nd Ferrell, J. E., Jr. (2003). A positive-feedback-based
bistable 'memory module' that governs a cell fate decision. Nature 426, 460-465.
Ozbudak, E. M., Thattai , M., Lim, H. H., Shraiman, B. and van
Oudenaarden, A. Multistability in the lactose utilization network of Eschericia
coli. Nat ure 427, 737-740.
Tanaka M, Collins S, Toyama B, Weissman JS. (2006). The physical basis
of how prion conformations determine strain phenoty pes. Nature 442, 585-589.
Choi PJ, Cai l,Freda K, Xie XS (2008) A stochastic single gene event
triggers switching of a bacterial celll. Science 332. 442-446.
Bayes Theorem
Example: Drug testing
•Pr(D), or the probability that the employee is a drug user, regardless of any other information. This is 0.005, since 0.5% of
the employees are drug users.
•Pr(N), or the probability that the employee is not a drug user. This is 1-Pr(D), or 0.995.
•Pr(+|D), or the probability that the test is positive, given that the employee is a drug user. This is 0.99, since the test is
99% accurate.
•Pr(+|N), or the probability that the test is positive, given that the employee is not a drug user. This is 0.01, since the test
will produce a false positive for 1% of non-users.
•Pr(+), or the probability of a positive test event, regardless of other information. This is 0.015 or 1.5%, which found by
adding the probability that the test will produce a true positive result in the event of drug use (= 99% x 0.5% = 0.495%)
plus the probability that the test will produce a false positive in the event of non-drug use (= 1% x 99.5% = 0.995%).
•Given this information, we can compute the probability that an employee who tested positive is actually a drug user:
Receiver Operator Correspondance (ROC)
curves