Protein kinetics
Petr Louša
16. 11. 2017
Petr Louša Protein kinetics 16. 11. 2017 1 / 28
Outline
1 Introduction
2 Enzymatic kinetics
3 Kinetics of conformational changes
4 Kinetics of oligomerisation
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Introduction
Outline
1 Introduction
2 Enzymatic kinetics
3 Kinetics of conformational changes
4 Kinetics of oligomerisation
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Introduction
General kinetics
aA + bB → c C + d D (1)
v = −
1
a
d[A]
dt
= −
1
b
d[B]
dt
= +
1
c
d[C]
dt
= +
1
d
d[D]
dt
=
dξ
dt
(2)
v = k[A]α
[B]β
(3)
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Introduction
General kinetics
aA + bB → c C + d D (1)
v = −
1
a
d[A]
dt
= −
1
b
d[B]
dt
= +
1
c
d[C]
dt
= +
1
d
d[D]
dt
=
dξ
dt
(2)
v = k[A]α
[B]β
(3)
Reaction velocity v is derivation of reaction extent ξ by time.
Sign convention – reactants decrease, products increase.
Petr Louša Protein kinetics 16. 11. 2017 4 / 28
Introduction
General kinetics
aA + bB → c C + d D (1)
v = −
1
a
d[A]
dt
= −
1
b
d[B]
dt
= +
1
c
d[C]
dt
= +
1
d
d[D]
dt
=
dξ
dt
(2)
v = k[A]α
[B]β
(3)
Reaction velocity v is derivation of reaction extent ξ by time.
Sign convention – reactants decrease, products increase.
For elemental reactions α = a, β = b, where α, β are partial
reaction orders.
This does NOT hold for more complex mechanisms.
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Introduction
Integrated rate equation
Simplest interesting case
A → B (4)
d[A]
dt
= −k[A] (5)
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Introduction
Integrated rate equation
Simplest interesting case
A → B (4)
d[A]
dt
= −k[A] (5)
Let’s integrate:
d[A]
[A]
= −kdt (6)
1
[A]
d[A] = −k dt (7)
ln[A] − ln[A]0 = −kt (8)
[A] = [A]0e−kt
(9)
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Introduction
2nd
order integrated rate equation
Slightly more difficult case
A → B (10)
d[A]
dt
= −k[A]2
(11)
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Introduction
2nd
order integrated rate equation
Slightly more difficult case
A → B (10)
d[A]
dt
= −k[A]2
(11)
After integration:
1
[A]2
d[A] = −k dt (12)
1
[A]
−
1
[A]0
= kt (13)
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Introduction
Reaction half time
First order – concentration independent:
ln
[A]0
2
− ln[A]0 = −kt1/2 (14)
t1/2 =
ln 2
k
(15)
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Introduction
Reaction half time
First order – concentration independent:
ln
[A]0
2
− ln[A]0 = −kt1/2 (14)
t1/2 =
ln 2
k
(15)
Second order – decreasing concentration prolongs the half
time:
2
[A]0
−
1
[A]0
= kt1/2 (16)
t1/2 =
1
k[A]0
(17)
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Introduction
Kinetics of equilibrium processes
Example of reversible reaction – isomerisation:
A k
k B (18)
d[A]
dt
= −k[A] + k [B] (19)
d[A]
dt
= −(k + k )[A] + k [A]0, pokud [B]0 = 0 (20)
[A] =
k + k−(k + k )t
k + k
[A]0 (21)
time
Concentration
[A]
[B]
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Introduction
Convergence to equilibrium
In equilibrium, velocities equalize.
v = v (22)
k[A] = k [B] (23)
[B]
[A]
=
k
k
= Keq (24)
Rate of relaxation to equilibrium can be studied by eg.
”T-jump”techniques.
fast change of temperature changes Keq
system starts to relax – measurable signal changes
even very fast processes can be analyzed – orders of µs
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Introduction
What influences reaction velocity?
concentration of all components included in rate equation
usually reactants, also products for reversible
reactions
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Introduction
What influences reaction velocity?
concentration of all components included in rate equation
usually reactants, also products for reversible
reactions
catalyzer often changes rate constant k itself
Petr Louša Protein kinetics 16. 11. 2017 10 / 28
Introduction
What influences reaction velocity?
concentration of all components included in rate equation
usually reactants, also products for reversible
reactions
catalyzer often changes rate constant k itself
temperature very important factor
empirically 10 ◦
C → 2 − 4× acceleration
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Introduction
Temperature dependence
Arrhenius equation – empiric
Petr Louša Protein kinetics 16. 11. 2017 11 / 28
Introduction
Temperature dependence
Arrhenius equation – empiric
k = A · exp −
EA
RT
(25)
ln k = −
EA
R
·
1
T
+ ln A (26)
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Introduction
Temperature dependence
Arrhenius equation – empiric
k = A · exp −
EA
RT
(25)
ln k = −
EA
R
·
1
T
+ ln A (26)
Eyring equation – derived from statistical thermodynamics
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Introduction
Temperature dependence
Arrhenius equation – empiric
k = A · exp −
EA
RT
(25)
ln k = −
EA
R
·
1
T
+ ln A (26)
Eyring equation – derived from statistical thermodynamics
k =
kBT
h
· exp −
∆G‡
RT
(27)
ln k = −
∆G‡
R
·
1
T
+ ln T + ln
kB
h
(28)
ln k = −
∆H‡
R
·
1
T
+ ln T +
∆S‡
R
+ ln
kB
h
(29)
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Enzymatic kinetics
Outline
1 Introduction
2 Enzymatic kinetics
3 Kinetics of conformational changes
4 Kinetics of oligomerisation
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Enzymatic kinetics
Energetic barrier
Conversion of substrate S to product P
S → P
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Enzymatic kinetics
Energetic barrier
Conversion of substrate S to product P
S → P
Enzyme accelerates the reaction by „decrease“ of activation
energy EA
In reality, it takes the reaction through different reaction
coordinate
S
E
−→ P
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Enzymatic kinetics
Michaelis–Menten kinetics
Assumptions
S + E
k1
−−−−
k−1
ES
k2
−→ P + E (30)
Negligible amount of product – otherwise we must include
reverse reaction and the analysis gets complex.
Substrate exceeds the enzyme: [S]0 [E]0
Stationary state:
d
dt
[ES] = 0
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Enzymatic kinetics
Michaelis–Menten kinetics
Derivation
v0 = k2[ES] (31)
d[ES]
dt
= k1[E][S] − k−1[ES] − k2[ES] (32)
d[ES]
dt
= k1[E]0[S] − [ES] (k1[S] + k−1 + k2) = 0 (33)
[ES] =
k1[E]0[S]
k1[S] + k−1 + k2
(34)
v0 =
k2[E]0[S]
k−1+k2
k1
+ [S]
(35)
Vlim = k2[E]0, KM =
k−1 + k2
k1
(36)
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Enzymatic kinetics
Michaelis–Menten kinetics
Analysis
Initial velocity is proportional to enzyme concentration.
Dependence of v0 on [S] is hyperbolic, approaching limit
velocity vlim.
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Enzymatic kinetics
Michaelis–Menten kinetics
Analysis
Initial velocity is proportional to enzyme concentration.
Dependence of v0 on [S] is hyperbolic, approaching limit
velocity vlim.
Michaelis constant KM has units of concentration
(mol.dm−3
).
KM matches [S]0 at half limit velocity.
KM is independent of enzyme concentration [E]0.
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Enzymatic kinetics
Michaelis–Menten kinetics
Analysis
Initial velocity is proportional to enzyme concentration.
Dependence of v0 on [S] is hyperbolic, approaching limit
velocity vlim.
Michaelis constant KM has units of concentration
(mol.dm−3
).
KM matches [S]0 at half limit velocity.
KM is independent of enzyme concentration [E]0.
Turnover number – kcat = k2 = vlim
[E]0
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Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
Petr Louša Protein kinetics 16. 11. 2017 17 / 28
Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
competitive – inhibitor and substrate compete about one active site –
KM increases, vlim constant
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Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
competitive – inhibitor and substrate compete about one active site –
KM increases, vlim constant
uncompetitive – inhibitor binds to complex ES that cannot convert to
product
– KM and vlim decrease in the same ratio
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Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
competitive – inhibitor and substrate compete about one active site –
KM increases, vlim constant
uncompetitive – inhibitor binds to complex ES that cannot convert to
product
– KM and vlim decrease in the same ratio
non-competitive – Inhibitor binds to different site than substrate
– typically allosteric inhibitors
– KM constant, vlim decreases
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Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
competitive – inhibitor and substrate compete about one active site –
KM increases, vlim constant
uncompetitive – inhibitor binds to complex ES that cannot convert to
product
– KM and vlim decrease in the same ratio
non-competitive – Inhibitor binds to different site than substrate
– typically allosteric inhibitors
– KM constant, vlim decreases
mixed – non-ideal conditions – e.g. uncompetitively inhibited
complex ES can convert to product, however slowly
Petr Louša Protein kinetics 16. 11. 2017 17 / 28
Enzymatic kinetics
Inhibition
by substrate – more molecules in active site – reaction stops
competitive – inhibitor and substrate compete about one active site –
KM increases, vlim constant
uncompetitive – inhibitor binds to complex ES that cannot convert to
product
– KM and vlim decrease in the same ratio
non-competitive – Inhibitor binds to different site than substrate
– typically allosteric inhibitors
– KM constant, vlim decreases
mixed – non-ideal conditions – e.g. uncompetitively inhibited
complex ES can convert to product, however slowly
irreversible – permanent deactivation of enzyme – e.g. by covalent
bond
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Kinetics of conformational changes
Outline
1 Introduction
2 Enzymatic kinetics
3 Kinetics of conformational changes
4 Kinetics of oligomerisation
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Kinetics of conformational changes
Kinetics of denaturation and renaturation
Usual assumption of simple two-state process:
D
k
−−
k
N (37)
For kinetics of folding and unfolding:
At − AR = (AN − AR) e−(k+k )t
(38)
AR − At = (AR − AD) e−(k+k )t
(39)
where A denotes values of e.g. absorbation AN for native,
AD for denatured state, AR in equilibrium and At in time t
Classical first order kinetics.
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Kinetics of conformational changes
Isomerisation of proline
Often cause of folding problems – isomerisation of proline
peptidic bond
In oligopeptides, ca 10–30 % bonds of X-Pro in cis state
In proteins, only ca 7 % in cis
Isomerisation slow – tens of seconds.
Helper enzyme – prolylisomerase
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Kinetics of oligomerisation
Outline
1 Introduction
2 Enzymatic kinetics
3 Kinetics of conformational changes
4 Kinetics of oligomerisation
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Kinetics of oligomerisation
Oligomerisation
Simplest and very often case – homodimerisation:
M + M
kon
−−−−
koff
D (40)
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Kinetics of oligomerisation
Oligomerisation
Simplest and very often case – homodimerisation:
M + M
kon
−−−−
koff
D (40)
v = −
1
2
d[M]
dt
= +
d[D]
dt
(41)
d[D]
dt
= kon[M]2
− koff[D] (42)
d[M]
dt
= 2koff[D] − 2kon[M]2
(43)
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Kinetics – exercise
Petr Louša
16. 11. 2017
Petr Louša Kinetics – exercise 16. 11. 2017 23 / 28
Alcoholic
Grown man (80 kg) got 1.5 ‰ of alcohol in blood after
drinking vodka.
After several hours following concentrations were
measured:
Time [h] 2 3.5 5 6
Alcohol concentration [‰] 1.24 1.05 0.86 0.73
Petr Louša Kinetics – exercise 16. 11. 2017 24 / 28
Alcoholic
Grown man (80 kg) got 1.5 ‰ of alcohol in blood after
drinking vodka.
After several hours following concentrations were
measured:
Time [h] 2 3.5 5 6
Alcohol concentration [‰] 1.24 1.05 0.86 0.73
1. How much vodka did the man drink?
2. Calculate the order of reaction for alcohol degradation in
human body and its rate constant.
3. How long after drinking will the man be able to drive a car
without losing his driving license?
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Alcoholic – solution
1. About 5 large shots
80 kg – ca 60 % of water = 48 kg of water – 1.50 ‰ = 72 g
of alcohol – 40% vodka – ca 180 g of vodka. Be careful,
alcohol is less dense than water (ρ = 0.8 g.cm−3
), therefore
the volume of vodka was about 225 ml.
Petr Louša Kinetics – exercise 16. 11. 2017 25 / 28
Alcoholic – solution
1. About 5 large shots
80 kg – ca 60 % of water = 48 kg of water – 1.50 ‰ = 72 g
of alcohol – 40% vodka – ca 180 g of vodka. Be careful,
alcohol is less dense than water (ρ = 0.8 g.cm−3
), therefore
the volume of vodka was about 225 ml.
2. Zeroth order of reaction – same amount gets degraded
during each hour
Rate constant k = 0.13 ‰.h−1
Petr Louša Kinetics – exercise 16. 11. 2017 25 / 28
Alcoholic – solution
1. About 5 large shots
80 kg – ca 60 % of water = 48 kg of water – 1.50 ‰ = 72 g
of alcohol – 40% vodka – ca 180 g of vodka. Be careful,
alcohol is less dense than water (ρ = 0.8 g.cm−3
), therefore
the volume of vodka was about 225 ml.
2. Zeroth order of reaction – same amount gets degraded
during each hour
Rate constant k = 0.13 ‰.h−1
3. About 11 hours after finishing vodka drinking – the blood
concentration of alcohol drops below 0.1 ‰.
Petr Louša Kinetics – exercise 16. 11. 2017 25 / 28
Enzymatic activity
Initial substrate concentration – 10 µmol.dm−3
Michaelis constant – KM = 2 mmol.dm−3
After 1 minute – 2 % of substrate converted to product.
Petr Louša Kinetics – exercise 16. 11. 2017 26 / 28
Enzymatic activity
Initial substrate concentration – 10 µmol.dm−3
Michaelis constant – KM = 2 mmol.dm−3
After 1 minute – 2 % of substrate converted to product.
1. How much substrate was converted after 3 minutes?
2. What is the limiting velocity?
3. The limiting velocity will be achieved at [S]0 = 0.2 mol.dm−3
.
How much substrate will convert in 3 minutes?
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Enzymatic activity – solution
1. 5.6 %, first order kinetics ([S] KM), k = 0.02 min−1
2. vlim = 40.2 µmol.dm−3
.min−1
3. Concentration of product will be 120 µmol.dm−3
, being
0.06 % of substrate.
Petr Louša Kinetics – exercise 16. 11. 2017 27 / 28
References
Zuckerman, Daniel M. Statistical Physics of Biomolecules.
An Introduction
Atkins, Peter; de Paula, Julio. Physical Chemistry
Kodíček, Milan; Karpenko, Vladimír. Biofysikální chemie
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References
Zuckerman, Daniel M. Statistical Physics of Biomolecules.
An Introduction
Atkins, Peter; de Paula, Julio. Physical Chemistry
Kodíček, Milan; Karpenko, Vladimír. Biofysikální chemie
Wikipedia
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