Transcript Special Topics in Computational Biology Lecture #5: Cooperativity ¦ Bud Mishra Professor of Computer Science and Mathematics (Courant, NYU) Professor (Watson School, CSHL) 2 ¦ 19 ¦

Special Topics in Computational Biology
Lecture #5:
Cooperativity
¦
Bud Mishra
Professor of Computer Science and Mathematics (Courant, NYU)
Professor (Watson School, CSHL)
2 ¦ 19 ¦ 2002
11/7/2015
©Bud Mishra, 2002
L5-1
Metabolism
• The complex of chemical reactions that
– convert foods into cellular components
– provide the energy for synthesis,
– and get rid of used up materials.
• Metabolism is (artificially) thought to consist of
– ANABOLISM:
• Building up activities
– CATABOLISM:
• Breaking down activities.
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Anabolism
• Requirements for synthesis of macromolecules:
– Varieties of organic compounds (e.g., amino acids)
– Precursor molecules of nucleic acids
– Energy.
• External supply of precursors consists of molecules that can
enter the cells and may have to be drastically altered by
chemical reactions inside the cells.
• Energy needed for metabolism is used up or made available
in the reshuffling of chemical bonds.
• In order for these processes (construction or destruction) to
proceed stably, metabolisms need to be unidirectional.
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ATP Reaction
• ATP (adenosine triphosphate)
ATP À
-H2O
+H2O
-P-P-P
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ADP + H3PO4
-P-P iP
– The energy currency of the cell
– Placed in water ATP splits to form
ADP (adenosine diphosphate) and
phosphoric acid (iP for inorganic
phosphate)
• Hydrolysis: unidirectional
– This reaction has enormous tendency to
proceed from left to right.
– At equilibrium the ratio
ADP/ATP ¼ 1:105
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Energy Released by ATP
• Energy Potential Barrier:
Energy
Potential
Barrier
ATP
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ADP + iP
Energy
Released
– In a population of molecules of
ATP, there is a distribution of
energies…
– A small percentage of molecules
has enough energy to make the
transition over the barrier
– To undergo a chemical reaction,
a molecule must be activated
(distorted to a transition state)
from where it glides into the new
structure … ADP + H3PO4
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Catalyst/Enzyme
enzyme
or heat
DE
ATP
DE
ADP + iP
• A catalyst such as an enzyme
forms with the substrate
molecules a complex that
distorts the molecules forcing
them into a state close to the
activated transition state at the
top of the energy barrier…
– An enzyme does not change the
difference in energy between the
substrate and the product;
– It reduces effective potential
barrier
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ATP Reaction
• In the hydrolysis of ATP,
– the enzyme ATPase operates primarily on the P-O-P
bond of ATP and on the HO-H bond of the water
molecule.
– the surface groups of the enzyme recognizes the whole
ATP molecule…this makes the enzyme specific for this
reaction.
– the enzyme also recognizes the product substances, ADP
& H3PO4, otherwise the reaction could not be reversibly
catalyzed.
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Unidirectionality of a
Reaction
•
enzyme
or heat
In summary, there are two
critical parameters:
1.
The energy difference
between substrate and
product, which determines
in which direction the
reaction will proceed;
2. The potential barrier, which
controls the rate of the
reaction.
DE
ATP
DE
ADP + iP
•
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These two parameters are
unrelated (independent)
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Futile Cycles
• All living organisms require a high degree of control over
metabolic processes so as to permit orderly change without
precipitating catastrophic progress towards thermodynamic
equilibrium.
• Examples: Processes such as glycolysis and gluconeogenesis
are
– essentially reversal of each other
– but cannot occur simultaneously
– as it would simply result in continuous hydrolysis of ATP resulting
in eventual death.
• These complementary processes are either in different segregated
populations of cells or in different compartments of the same cell (e.g.
glycolysis and gluconeogenesis in liver tissues)
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Problem
• Such unidirectional processes cannot be easily
explained with the classical Michelis-Menten
model.
• Why?
• Recall that…
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Michelis-Menten’s Model (1913)
E + A À EA ! E + P
– E = Enzyme,
• e= instantaneous enzyme concentration
– A = Substrate,
• a = instantaneous substrate concentration
– EA=Enzyme-Substrate Complex,
• x = instantaneous EA concentration
– P = Product,
• p = instantaneous concentration
• Assumption:
– The reversible first step (E + A À EA) was fast enough for it to be
represented by an equilibrium constant for substrate dissociation:
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v = dp/dt = Va/(Km+a)
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Product Concentration
•
•
•
•
dp/dt = v = V a/(Km+a) = V(a0-p)/(Km+a0-p)
s V dt = s (Km+a0-p)/(a0-p) dp
s Km dp/(a0-p) + s dp = s V dt
-Km ln (a0-p) + p = Vt + a
– a = -Km ln a0
• Vt = p + Km ln a0/(a0-p)
• Vappt = p + Kmapp ln a0/(a0-p)
t/[ln a0/(a0-p)]
= (1/Vapp) { p /[ln a0/(a0-p)]} + Kmapp/Vapp
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Simulation
• Product concentration:
tpFun t_, a0_, K_, V_ :
NSolve t
Log a0
a0 p
1 V p
Log a0
a0 p
K V, p ;
Plot p . rulet1, p . rulet2, p . rulet4,
p . rulet8 , t, 0, 10 , PlotRange
All
0.8
0.6
x-axis = t
y-axis = p
0.4
0.2
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4
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Model of transcription
a = concentration of a TF
transcription
{X,k, n}
p = concentration of an mRNA
• n = Cooperativity coefficient
• k = Concentration of a at which transcription of m is
“half-maximally” activated.
• dp/dt = F(a, k, n) = V an/[kn + an]
• A graph of function F = Sigmoid Function
• If n =1 then, the transcription activation function
resembles the classical Michaelis-Menten!
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Effect of Cooperativity
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Rate of Change:
Michaelis-Menten
• In Michalis-Menten Equation:
v = Va/(K+a), and dv/da = KV/(a+K)2
• At the half way point,
(dv/da)|a=K = 1/4 (V/K)
– The rate is 0.1 V at a = K/9 and it is 0.9 V at a = 9 K.
– An enormous increase in substrate concentration (81
fold) is needed to increase the rate from 10% to 90%.
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Rate of Change:
Cooperativity
• In general, with cooperativity, (n = cooperativity
constant), the rate increases rapidly:
v = V an/(Kn + an) and dv/da = n an-1 Kn V/(an+Kn)2
• At the half way point,
(dv/da)|a=K = (n/4) (V/K).
– The rate is 0.1 V at a = K/(91/n) and it is 0.9V at a = (91/n)
K.
– The needed increase in substrate concentration reduces
to 3 fold for n = 2.
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Substrate Concentration and
Cooperativity
• Increase in substrate concentration needed to increase the rate from
10% to 90%.--As a function of the cooperativity coefficient n:
80
60
40
20
2
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3
4
n, cooperativity coefficient )
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6
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Difference in rates:
1
0.8
MichaelisMenten
0.6
0.4
Cooperative
0.2
log a a
-6
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-4
-2
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2
4
6
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Cooperativity
• A property arising from “cooperation” among
many active sites from polymeric enzymes…
• The main role in metabolic regulation:
– Property of responding with exceptional sensitivity to
change in metabolic concentrations.
– Shows a characteristic S-shaped (sigmoid) response curve
(as opposed to a rectangular hyperbola of MichaelisMenten)
– The steepest part of the curve is shifted from the origin
to a positive concentration a typically a concentration
within the physiological range for the metabolite.
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Allosteric Interaction
• To permit inhibition or activation by metabolically
appropriate effectors, many regulated enzymes have evolved
sites for effector binding that are separate from catalytic
sites.
• These sites are called “allosteric sites”
– (Greek for different shape or another solid):
– Monod, Cahngeux and Jacob: 1963.
– Enzymes possessing allosteric sites are called “allosteric enzymes.”
• Many allosteric enzymes are cooperative and vice versa.
– Haemoglobin was known to be cooperative for more than 60 years
before the allosteric effect of 1,2-bisphosphoglycerate was
understood.
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Activation & Inhibition
•
2
1
1
2
•
1.
2.
•
1 active site
•
2 regulatory site
•
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In cells, not only the amounts, but also the activity
of enzyme is regulated (via allosteric regulatory
sites).
The regulator substances fall into two categories:
Activators &
Inhibitors.
Inhibitors decrease enzyme activity, either by
competing with the substrate for the active sites, or
by changing the configuration of the enzyme…
Activators increase the activity of enzymes when they
combine with them…
These kinds of regulation occurs by the regulators
combining with the enzyme at a site other than the
active site…allosteric regulatory sites.
©Bud Mishra, 2002
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Feedback Inhibition
A
1
B
2
C
3
D
4
• In the example shown, an amino acid (say X) is
made by a pathway in which each arrow indicates
an enzyme reaction.
• The level of X in the cell controls the activity of
enzyme 1 of the pathway..
– The more X is present, the less active the enzyme is.
– When external X decreases the enzyme becomes active
again.
– Example of negative feedback accomplished by
allosteric sites and competition for the active sites.
• Unidirectionality of A !1 B reaction is very
important as a small amount of X should have
X
large effect on the rate of the reaction.
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©Bud Mishra, 2002
The Hill equation
• Hill (1910): As an empirical description of the cooperative
binding of oxygen to haemoglobin.
v = V ah/(K0.5h + ah)
• h = Hill coefficient = cooperativity coefficient.
– Based on a limiting physical model of substrate binding, h takes an
integral value.
– Experimentally determined h coefficient is often non-integral.
• K0.5 = Value of the substrate concentration at which the rate
of reaction is half of the maximum achievable.
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Hill Plot
• Note that
v/(V-v) = (a/K0.5)h, and
log[v/(V-v)] = h log a – h log K0.5
• h =Hill coefficient can be
determined from a plot of
log a vs. log[v/(V-v)] .
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¢ Break ¢
10 minutes
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Adair Equation
• An enzyme E has two active sites that bind
substrate A independently:
A
+
A+E
A
+
EA
Ks1
Ks2
A+EA
Ks2
Ks1
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E AA
– At equilibrium the dissociation constants are
Ks1 and Ks2
– The rate constants at the two sites are
independent and equal k1 and k2
• Applying Michaelis-Menten, we have:
v = k1 e0 a/(Ks1 +a) + k2 e0 a/(Ks2+a)
• Assuming that V/2 = k1 e0 = k2 e0, we have
(2v/V) = 1/(1+ Ks1/a) + 1/(1+Ks2/a)
= (1 + K’/a)/(1+K’/a + K2/a2)
= (a2 +K’a)/(K2 + K’a + a2)
©Bud Mishra, 2002
L5-27
Adair Equation
• General Formula with two active
sites:
(k1+k2) e [a2 + (Ks1 k2 +Ks2 k2)/(k1 + k2) a]
[a2 + (Ks1+Ks2) a + Ks1 Ks2]
• The formula is approximated as
¼ V a2/(K2 + a2),
• where
V = (k1+k2)e and K = p(Ks1 Ks2)
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Adair Equation for
Haemolobin
• An equation of the same general form with four
binding sites, as haemoglobin can bind upto four
molecules of oxygen:
y = (v/V) =
[a/K1 + 3a2/K1K2 + 3a3/K1K2K3 + a4/K1K2K3K4]
[1+4a/K1 + 6a2/K1K2 + 4a3/K1K2K3 + a4/K1K2K3K4]
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The Monod-WymanChangeux Model
• The earliest mechanistic model proposed to account for
cooperative effects in terms of the enzyme’s conformation:
(1965).
• Assumptions:
– Cooperative proteins are composed of several identical reacting units,
called protomers, that occupy equivalent positions within the protein.
– Each protomer contains one binding site per ligand.
– The binding sites within each protein are equivalent.
– If the binding of a ligand to one protomer induces a conformational
change in that protomer, an identical change is induced in all
protomers.
– The protein has two conformational states, usually denoted by R and T,
which differ in their ability to bind ligands.
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Example: A protein with two
binding sites…
R2
T2
sk3 2k-3
sk1 2k-1
R1
2sk1
T1
2sk3 k-3
k-1
R0
k2
k-2
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T0
• Consider a protein with two binding
sites:
• The protein exist in six states:
– Ri, i=0,1,2; or Ti, i=0,1,2,
– where i à the number of bound ligands.
• Assume that R1 cannot convert
directly to T1 or viceversa.
• Similarly, Assume that R2 cannot
convert directly to T2 or viceversa.
• Let s denote the concentration of the
substrate.
©Bud Mishra, 2002
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Saturation Function
R2
T2
sk3 2k-3
sk1 2k-1
R1
2sk1
T1
2sk3 k-3
k-1
R0
k2
k-2
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T0
• Saturation function = Fraction Y of
occupied sites:
Y=
[r1 + 2 r2 + t1 +2t2]/2(r0+r1+r2+t0+t1+t2)
• Let Ki = k-i/ki…Then
–
–
–
–
–
r1 = 2sK1-1r0;
r2 = s2K1-2r0;
t1 = 2sK1-1t0;
t2 = s2K1-2t0
and r0/t0 = K2.
©Bud Mishra, 2002
L5-32
Final Result
• Substituting into the saturation function:
Y=
s K1-1(1+sK1-1)+k2-1[sK3-1(1+ sK3-1)]
(1+sK1-1)2 + k2-1 (1+ sK3-1)2
• More generally, with n binding sites:
Y=
s K1-1(1+sK1-1)n-1+k2-1[sK3-1(1+ sK3-1)n-1]
(1+sK1-1)n + k2-1 (1+ sK3-1)n
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Calculation
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Calculation
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Regulatory Networks
• All cells in an organism have the same genomic
data, but the proteins synthesized in each vary
according to cell type, time and environmental
factors
• There are network of interactions among various
biochemical entities in a cell (DNA RNA, protein,
small moleules)
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Gene Regulation
DNA
transcription
Transport to
cytosol
mRNA
Nonphosphorylated
protein
Transport to
nucleus
Nonphosphorylated
protein
Post-translational modifications
Nonphosphorylated
protein
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Transcriptional Regulation:
Example: The lac Operon
Regions coding for proteins
Regulatory Regions
Diffusable regulatory proteins
RNA
polymerase
lacI
P
P
lacZ
O
lacY
mRNA +
ribosomes
I
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lacA
mRNA +
ribosomes
Z
Y
©Bud Mishra, 2002
A
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The lac Operon
• Regulates utilization of lactose by the bacterium E.
coli.
• Lactose is not generally available to E. coli as a food
substrate, so the bacterium does not usually
synthesize the enzymes necessary for its metabolic
use.
• There is an operon, called the lac operaon, normally
turned off, that codes for three enzymes:
– b-galactoside permease, b-galactosidase and bthiogalactoside acetyl transferase.
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Activation of the lac operon
• If the bacterium is exposed to lactose, these enzymes work
together to
– transport lactose into the cell and
– isomerizes lactose into allolactose (an allosteric isomer of lactose.
• The allolactose binds with a repressor molecule to keep it
from repressing the production of mRNA.
• Production of allolactose turns on the production of
mRNA, which then leads to production of more enzyme,
enabling production of more lactose to allolactose…
An autoctalyitic reaction..
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Transcriptional Regulation:
Example: The lac Operon
Regions coding for proteins
Binds but
cannot move to
transcribe
Regulatory Regions
Diffusable regulatory proteins
RNA
polymerase
lacI
I
P
P
I
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mRNA +
ribosomes
O
lacZ lacY lacA
No mRNA
When lactose is absent, the protein encoded
by lacI represses transcription of the lac
operon
©Bud Mishra, 2002
L5-41
Transcriptional Regulation:
Example: The lac Operon
Regions coding for proteins
Regulatory Regions
Diffusable regulatory proteins
RNA
polymerase
lacI
P
P
O
mRNA +
ribosomes
I
lacZ
Z
lacY
lacA
Y
mRNA +
ribosomes
A
Lactose
Confirmational
change
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Blocked
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L5-42
Mathematical Model
G + mP Àk-1k1 X
• Production of enzyme is turned on by m molecules
of the prouct allolactose P…
• G=Inactive state of the gene
• X=Active state of the gene
• In a large population of genes, the percentage of
active genes is given by the chemical equilibrium:
p = [P]m/(keqm + [P]m)
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L5-43
Production of mRNA
• The differential equation governing the (average)
production of mRNA
dM/dt =M0 + k1 [P]m/(keqm + [P]m) – k2 M,
• where M is the concentration of mRNA that codes
for the enzyme.
• Production of the enzymes (responsible for
tarnsforming into allolactose substrate):
dE1/dt = c1 M – d1 E1;
dE2/dt = c2 M – d2 E2.
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Lactose states
• S0 = Concentration of the lactose that is exterior to
the cell.
• S = Concentration of the lactose that is interior to
the cell.
• [P] = Concentration of allolactose.
dS0/dt = -s0 E1 S0/(k0 + S0)
dS/dt = s0 E1 S0/(k0 + S0) - s1 E2 S/(ks + S)
d[P]/dt = s1 E2 S/(ks + S) - s2 E2 [P]/(kp +
[P])
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Simplification
• Assume: mRNA is in quasi-steady state:
M = (k1/k2)[P]m/(keqm+[P]m) + M0/k2;
• Assume: d1 = d2. Degradation is slow compared to
cell growth. Also, E1 = E2.
dE1/dt = c1M0/k2 + (c1k1/k2)[P]m/(keqm+[P]m) – d1 E1;
• Assume: No delay in conversion of the lactose into
allolactose:
d[P]/dt = s0 E1 S0/(k0+S0) - s2 E1 [P]/(kp + [P]).
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L5-46
Dimensionless Form
• Dimensionless variables: S0 = k0 s, [P] = kp p, E1 = e0
e, and t = t0 t…
de/dt = m0 + pm/(km + pm) - e e,
dp/dt = m e[s/(s+1) - l p/(p+1)],
ds/dt = -e s/(s+1),
• where e02 = c1k0k1/(s0k2), t0 = k+0/(e0s0),
l = s2/s0, m = k0/kp, k = k/kp, m0 = M0/k1,
and e = t0 d1…
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Numerical Simulation:
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NDSolve
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Time Evolution
external lactose, s
b-galactosidase, e
allolactose, p
Time a
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The lac operon
external lactose, s
b-galactosidase, e
allolactose, p
Time a
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• If the amount of lactose is too small,
then the lactose is gradually depleted,
although there is no increase in
enzyme concentration.
• However, if the lactose dose is
sufficiently large, then there is an
autocatalytic response, as the lac
operon is turned on and enzyme is
produced.
• The production of enzyme shuts
down when the lactose stimuls is
consumed, and the enzyme
concentration gradually declines…
©Bud Mishra, 2002
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