8 Enzyme kinetics Flashcards
Enzyme kinetics background
Many important chemical reactions in biological systems are made possible by enzymes that act as catalysts: they participate in the intermediate stages of the reaction, but are not themselves converted into product.
S substrate
P product
S →(k₀) P (slow reaction
S +E S →(k₁) SE →(k₂) P + E
←(k₋₁)
faster reaction
E Enzyme
SE substrate-enzyme complex
reversible with different rates
reaction forming the product is not reversible
ENZYME KINETICS
ODEs
The hypothesis that the rate of reaction between two species is proportional to the product
of their population sizes is something we have already seen. In chemistry, it is known as the
“law of mass action
ds(t)/dt= −k₁e(t)s(t) + k₋₁c(t)
de(t)/dt = −k₁e(t)s(t) + (k₋₁ + k₂)c(t)
dc(t)/dt= k₁e(t)s(t) − (k₋₁ + k₂)c(t)
dp(t)dt= k₂c(t).
s(t) :concentration of substrate,
e(t) : concentration of enzyme,
c(t) : concentration of complexes
p(t):concentration of product.
Enzyme kinetics dimensions
ds(t)/dt= −k₁e(t)s(t) + k₋₁c(t)
de(t)/dt = −k₁e(t)s(t) + (k₋₁ + k₂)c(t)
dc(t)/dt= k₁e(t)s(t) − (k₋₁ + k₂)c(t)
dp(t)dt= k₂c(t).
time-dependent quantities are concentrations, measured in units such as per ml, so
[s] = [e] = [c] = [p] = L⁻³
[k₂]= [k₋₁ ]= T⁻¹
[k₁] = L³T ⁻¹
Enzyme kinetics
Typical initial conditions
looking at how the values are related
s(0) > 0 and E₀= e(0) > 0
but c(0) = 0 and p(0) = 0.
Find p(t): not on RHS on any so obtained from c(t)
de(t)/dt + dc(t)/dt = 0
so e(t) +c(t) constant = e(0) E₀
THE ENZYME IS NOT DESTROYED BY REACTION only stored in complexes
so e(t)= E₀ -c(t) and we only need to consider ds/dt and dc(t)/dt using E₀ and subbing in using de/dt
Enzyme kinetics what eq do we consider now
enzyme not destroyed only stored
New eq to consider:
ds(t)/dt= −k₁E₀s(t) +( k₁s(t)+k₋₁)c(t)
dc(t)/dt= k₁E₀s(t) − (k₁S(t)+k₋₁ + k₂)c(t)
Enzyme Kinetics Michaelis and Menten
New eq to consider:
ds(t)/dt= −k₁E₀s(t) +( k₁s(t)+k₋₁)c(t)
dc(t)/dt= k₁E₀s(t) − (k₁S(t)+k₋₁ + k₂)c(t)
Approx method for solving these.
ASSUMPTION process of formation and breakup of complex SE is so fast compared to formation of the product that it can be considered to be in equilibrium dc/dt =0 assumed at any t
So
c(t) = E₀[s(t)]/[s(t)+Kₘ]
where Kₘ = (k₋₁+k₂)/k₁
using this
ds/dt = -k₂ E₀ (s/(s+Kₘ))
dp/dt =k₂ E₀ (s/(s+Kₘ))
Enzyme kinetics: Michaelis-Menten constant Kₘ
A concentration
[Kₘ]= L⁻³
concentration of substrate such that the rate of formation of product is (1/2)Vₘₐₓ
Enzyme kinetics
under Michaelis and Menten model assumptions
Vₘₐₓ
dp/dt
Let V =dp/dt rate of formation of product
k₂ E₀ is called Vₘₐₓ as its the max rate of formation of product found when susbtrate concentration is high:
We can write V= Vₘₐₓ [s/[s+Kₘ]]
Enzyme kinetics plots
S against V
Lineweaver-Burk
Hanes-woolf plot
S against V
starting at S=0 V=0 increases then rate of increase slows
reaches V= Vₘₐₓ/2 at S= Kₘ
and continues to increase
Lineweaver-Burk: displays experimental data on a plot of 1/s against 1/V
rearranged
(1/V) = [Kₘ/Vₘₐₓ] (1/s) + (1/Vₘₐₓ)
Hanes-woolf plot S against S/V
or rearrange
s/V= (s/Vₘₐₓ) + [Kₘ/Vₘₐₓ)
both linear plots starting at some positive value
Enzyme kinetics
new eq:
ds(t)/dt= −k₁E₀s(t) +( k₁s(t)+k₋₁)c(t)
dc(t)/dt= k₁E₀s(t) − (k₁S(t)+k₋₁ + k₂)c(t)
scaled variables
u(t)=s(t)/s(0)
v(t)=c(t)/E₀
𝜏=k_1 E₀t
u(0)=1
v(0)=0
du/𝜏=-u + (u +K-α)v
ε(dv/d𝜏) = u - (u+K)v
constants
α=k₂/(k₁s(0))
K=(k₋₁+k₂)/k₁s(0)
ε= E₀/s(0)
K-α = (k₋₁)/k₁s(0) >0
near 𝜏=0 approx u=1 and v=0
du/d𝜏<0 and dv/d𝜏>0
V reaches maximum when:
dv/d𝜏 =0 when v= u/(u+K)
Enzyme kinetics
new eq: STEADY STATES
du/𝜏=-u + (u +K-α)v
ε(dv/d𝜏) = u - (u+K)v
Using the jacobian
A(u,v) =
[-1+v u+K -α ]
[ (1-v)/ε -(u+K)/ε ]
(0,0)
trace = -1-(K/ε) <0
det =α/ ε >0
STABLE and as τ → ∞, both u and v approach 0
reaction happens unless both values 0
enzyme kinetics typical behaviour trajectories for scaled new model
u and v against 𝜏:
v starts at 0, increases and reaches peak then decreases to 0
u starts positive and decreases with increasing slope then reduced slope continues to decrease close to 0
above v always
Both decrease to 0 S runs out so complexes also and will give a particular amount of product and E
ENZYME KINETICS
Steady-state analysis (MATH5566)
(0,0) IS STABLE
NOW determine how fast trajectories approach (0,0) and from what direction
Eigenvalues:
ελ² + (K + ε)λ + α = 0
2ελ = - (K + ε) ± (( (K + ε)² - 4αε) ^0.5
=- (K+ε) ± (K + ε)(1 - (4αε/(K + ε)²))^0.5
USING (1+x)^0.5 = 1 + 0.5x + O(x²)
two cases
(2/K)ελ = -2 +O(ε)
(2/K)ελ = -(2α/K²) ε+O(ε)
Thus
ελ₁=-K +O(ε) (<0)
λ₂= -α/K +O(ε) (<0)
( when ε is small …. O(ε) approx 0)
both eigenvalue negative and real
Can write the general solution of linearised ODEs eigenvectors (a_1, 1) and (a_2, 1)
(u(τ)) =c₁(a₁)exp(λ₁τ)+c₂(a₂)exp(λ₂τ)
(v(τ)) (1) (1)
ENZYME KINETICS
small-ε and steady-state analyses
λ₁ and λ₂ seem to be negative
so exp(λ₁t) and exp(λ₂t) converge to 0
one exponential will dominate the dynamics: whichever is larger
As τ → ∞
exp(λ₁τ)→exp(-(k/ε)τ)
exp(λ₂τ) →exp(-(α/k)τ) will dominate as small ε
(in the lecture we subbed in constants to see this and to see E_0 is smaller than a function of the k’s)
ENZYME KINETICS
small-ε and steady-state analyses
We can approx dynamics by
NOTE:
J. v₂ =λ₂v₂
we find a₂ = K + ελ₂
(u(τ)) =c₁(a₁)exp(λ₁τ)+c₂(a₂)exp(λ₂τ)
(v(τ)) (1) (1)
using instead as τ → ∞
(u(τ))
(v(τ))
≈ c₂(a₂)exp(-(α/k)τ)
(1)
Thus the small-ε and steady-state analyses consistent:
1) rate at which u and v approach 0 proportional to exp(-(α/k)τ)
2) the approach to (0,0) along v=u/K
