enzyme kinetics and inhibition Flashcards
kinetics
study of the rate at which some process progresses
equilibrium assumption
assuming P is produced more slowly that ES is dissociated, so that k2«k1, the rate of ES formation reaches equilibrium relatively rapidly
steady-state assumption
assuming the probability the P->S is very low, k2 can be ignored. therefore, the reaction reaches a steady state in which the rate of ES formation equals the state of ES loss, either to reverse reaction or product formation
what is the equation for enzyme kinetics?
E+S <->ES<-> E+P
at low [S], most of the enzyme is in the unbound form, E; what does the initial reaction velocity do?
increases linearly with [S], since the formation of [ES] depends on [S]
at high [S], most of the enzyme is in the ES complex; the initial velocity no longer increases linearly with substrate concentrations and …
approaches a maximum Vmax, since [ES] no longer depends on [S]
total enzyme concentration
[Et]=[E]+[ES]
michaelis constant
Km= (k-1+k2)/k1
what is kcat?
limiting rate constant for a general catalysis reaction so for this simple reaction, kcat=k2
what does Vmax equal?
Vmax=kcat[Et]
what is the michaelis-menten equation?
Vo=Vmax[S]/km+[S]
Km
substrate concentration that gives a rate, Vo, equal to Vmax/2
what is Kd?
k-1/k1
what is Vmax?
maximal or saturating rate that an enzyme-catalyzed reaction approaches at high substrate concentrations
when [S]»Km, Vo=
Vmax[S]/[S]=Vmax
