Catalysts only affect reactionrates (not equilibria)
Enzymes catalyse by stabilizingtransition states and thus loweringactivationbarriers
Steady-state studies involve conditions in which [S] >> [E], so that v can be measured over a reasonable period
(EXPLAINED) Steady-state studies involve conditions in which the concentration of the substrate is much higher than the concentration of the enzyme, so that rate can be measured over a reasonable period
The useful parameters we can derive from using the Michaelis-Menten equation: Affinity, Selectivity, Efficiency, Inhibitor characteristics - pharmacology
Measuring the rates of enzyme-catalysed reactions:
Keep pH and T constant
Continuousassays preferred (e.g. spectrophotometry), not samples taken at multiple time-points.
Conditions where [E] is low and [S] >> [E], allowing the enzyme to turn over many times
Steady-state kinetics: The rate declines as substrate is depleted so measure early while v is constant and the system is in ‘steady-state’. The enzyme is also present but in limiting (low) concentrations.
Steady-state rate (v) is proportional to [E]
With [E] constant, at a lower [S] the rate of reaction v is proportional to [S] but at higher [S] v becomes constant. The enzyme becomes saturated at high [S], giving Vmax
At high [S], all enzyme is converted to the ES further increases in [S] do not increase rate. In these conditions, the reaction rate is limited by how quickly the ES complex (Michaelis complex) can produce and liberate the product and regenerate free enzyme= Vmax
E + S <-> ES -> E + P
chymotrypsin + protein <-> chymotrypsin:protein -> chymotrypsin + peptides
K is a dissociation constant. K = [E] [S] / [ES]
ka is the forward rate constant, kd the backward rate constant
The rate of appearance of ES = ka.[E].[S]
The rate of disappearance of ES = kd.[ES]
K = kd/ka
In most cases the conversion of ES to E + P is the slowest step and is therefore rate-limiting. The rate constant for this reaction is kcat. The rate of appearance of product = kcat[ES]
EQUATION 1: v = kcat[ES]
at Vmax [ES] = [ETotal]
EQUATION 2: Vmax = kcat [ETotal]
EQUATION 3: [ES] = [E].[S]/K
Relatively little of S is tied up in the complex ES, since [Stotal] >> [E], so [S] = [Stotal] which we know
To find enzyme concentration: [E] = [Etotal] – [ES]