202 Enzymes: Michaelis‐Menten Kinetics Flashcards
The basic Michaelis-Menten equation considers only
one substrate an no chemistry intermediates (I.e., S goes to P, not S goes to X and then X goes to Y and eventually Y goes to P).
In the Michaelis-Menten equation what simplifications/assumptions do we make? What does this say about the two functional aspects of the enzyme reaction?
1) that the chemistry is irreversible (and this is usually assumed to be the rate-determining step)
2) there is very little product formed yet and so there is no EP complex formed by P rebinding to the enzyme.
3) these simplifications reduces the rate constants to three (three equation arrows) instead of 6
These schemes reduces the enzyme reaction to two functional aspects
1) binding of substrate and 2) the chemistry step to convert S to P
In the michaelis-menten reaction when is initial velocity measured?
before 10% of the substrate has been consumed.
Since the substrate concentration is so much higher than the enzyme concentration, what is an important assumption for the M-M kinetics?
that the amount of substrate bound to enzyme at any point in time is negligible compared to the total substrate concentration; hence we can assume that [S] total is constant
The initial velocity in M-M kinetics is determined by what?
determined by the breakdown of the ES complex to form enzyme and product. This is affected by the concentration of ES; hence initial velocity (Vo) equals the rate constant times [ES]
The rate constant equals what?
the disappearance of substrate over time or the appearance of product over time
AKA K2 * [ES] = d[P]/dt or -d[S]/dt measured in linear range
What is the steady state assumption?
it describes the conditions for experiments that can assay enzyme kinetics. This assumption coupled with the assumptions in the original derivation by M-M kinetic, that chemistry is the rate-determining step) allows derivation of the parameters in the M-M equation
What is the definition of steady-state (not the steady state assumption, just steady state)
the concentration of an intermediate (ES complex), is constant w/time, I.e., d [ES]/dt = 0
The concentration of intermediates is fixed by what rates?
the rates of its formation minus the rates of its decay.
What is the equation for steady state?
rates of formation = the rates of decay of the intermediate.
this requires that [S] total be much higher than [E] total, aka [S] total = [S] free
What occurs during the pre-steady sate?
the pre-steady state is the initial period when the enzyme is first added to the excess substrate, the ES concentration does not stay constant but is increasing with time. This period is short and cannot be easily observed. The steady sate follows this period an the initial velocity is defined by this steady-state
In a progress curve, explain how substrate concentration looks as a function of time?
When t=0, the substrate concentration is higher than the enzyme concentration. Then as time proceeds, substrate concentration decreases but not in a linear fashion
After a certain point, it becomes linear
this linear region is the initial velocity range used in MM kinetics
In a progress curve, explain how the product concentration proceeds as a function of time
Initially, there is a non-linear increase in product concentration. This rate of increase becomes linear at the same time as the rate of disappearance of substrate becomes linear.
this product linear range is used when assaying an enzyme, I.e., the slope of the linear portion of either of those curves = the initial velocity Vo
In a progression curve, when the substrate and product are linear, how does the concentration of the ES look?
the ES concentration stays constant. aka the slope in the curve in this region is zero d[ES]/dt = 0
in the progression curve graph, what does the steady state region illustrate?
it illustrates that [ES] does not change significantly with time.
in the progression curve graph, what does the pre-steady-state region illustrate?
it illustrates that there is a build up of the ES complex. and all four functions in this region are non linear
What are the common methods to study pre-steady-state kinetics?
1) stopped-flow kinetics - in which enzyme an substrate are mixed rapidly and then the reaction almost immediately stopped (sub-second scale)
2) ESI-MS (electrospray ionization mass spectrometry) - this provides information regarding the formation of the enzyme-substrate complex and thus the mechanisms of enzyme catalysis
steady state kinetics measures what?
the rates of enzyme-catalyzed reactions under the steady-state conditions as exist in the cell, and thus help us understand metabolism.
If chemistry is the rate determining step, then velocity is determined how?
by how much is present of the species that precedes this slowest step (I.e. [ES]) as well as by the rate constant of chemistry, k2. I.e., v0 = k2 ∙ [ES].
[E] total is
the sum of concentrations of free enzyme and enzyme bound to substrate
What is Vmax?
the maximal initial velocity, which is the initial velocity when the enzyme is fully saturated with substrate
at this point [E]total = [ES]. So Vmax = k2 ∙ [E]total. Substituting, we get v0 = (Vmax ∙ [S]total ) / (KM + [S]total), the Michaelis‐Menten equation.
What are the MM parameters?
Vmax
Kcat
Km
Kcat/Km
What is does the turnover number Kcat describee?
it describes the limiting rate of an enzyme-catalyzed reaction at saturation by substrate. it is equivalent to the number of substrate molecules converted to product in a given unit of time (usually sec is used) by a single enzyme molecule when fully saturated
What information does Vmax and Kcat give?
information on speed (velocity)