michaelis menten Flashcards
(27 cards)
Km
Km = [s] when v is 1/2 Vmax
steady state constant
small Km
max catalytic efficiency at low S
uncompetitive vs allosteric
allosteric based on a different kinetic model.
UI only binds to ES, AI could bind to E or ES
initial rate of reaction occurs when…
steady state is achieved.
why? This is following the pre steady state where enzyme is encountering substrate only sometimes…
- substrate concentration remains approximately constant, equal to the initial substrate concentration, while the enzyme-substrate complex concentration builds up.
What is the initial rate?
- 10% of the rxn is used up and has achieved steady state
Chemical equation, following the assumptions of MM equation
E + S >1> ES >2> E + P
1 = k1, k-1
2 = kcat.
Can ignore k2 step, due to k1 and k3 being fast.
kcat
kcat is essentially equal to k2, because k3 happens so fast.
(ES to EP)
It is the apparent rate constant for the rate determining conversion (k2) of substrate to product.
But, because k2 is happening so slowly, we are saying that is is k3.
(time required by an enzyme molecule to “turn over” one substrate molecule)
Michaelis menten.
- 1) what is it
- 2) equilibrium state assumptions, what does it yield
- 3) Role of k2?
- E and S associate reversible to form an ES complex E+S <-k1,-1-> ES <-k2-> E+P
- Assume ES is in rapid equilibrium with free E (k1 time E+S = k-1 times ES… helps us determine Ks (ES dissociation constant)
- Breakdown of ES to form P is slower than ES formation, also slower than ES forming E+S.
k2««k-1,k1
Vmax
- maximum rate
- occurs when all enzyme saturated
- all E is in [ES]
rate determining step
k2 is the rate determining step, it is when [S] is high enough to convert all E to ES
(kcat) is the apparent rate constant for ES to EP.
Why is overall rate unafected when more S is added at saturation?
there is no enzyme to bind to the S, all enzyme is already bound at saturation
rate of product formation
v = kcat[ES]
(ES <–> E + P)
why cant you measure E or ES but can measure Etotal?
We know how much enzyme is in the reaction, but we dont know how much of that enzyme is actually in ES complex or free E
Ks
dissociation constant,
when kcat «_space;k-1
Ks = k-1/k1 = [E] [S] / [ES]
( E and S are in equilibrium with ES)
these are not truly in equilibrium, does not apply to everything, because some ES is always being converted to P, very slowly
Briggs haldane model
the more ES that is present, the faster ES will dissociate either to prod (kcat) or reactant (k-1).
Therefore, when first mixing E and S, [ES] builds fast but then reaches a steady state. This state persists until all substrate is used up
steady state
formation and breakdown of ES are =
k1 [E] [S] = k-1 [ES] + kcat [ES]
formation of ES complex = diss. of ES + breakdown to E+P
burst phase
- presteady state
- rapid formation of ES complexes why??
Km equation
Km = (k-1 + kcat) / k1
substrate concentration when the velocity is 1/2 Vmax.
NOT a measure of affinity
Michaelis menten equation
v = Vmax[S] / Km + [S]
Km three definitions
complex rate constant
concentration of S when v is 1/2 Vmax
steady state constant
Km varies with
enzyme, substrate, temp, pH
enzyme efficiency
kcat /Km
large kcat (rapid turnover) or small Km (1/2 Vmax at low [S]) will make enzyme efficiency large
kcat equation
Vmax / [E] total
