Enzyme kinetics Flashcards
what enzyme phosphorylates glucose
hexokinase
what bonds do hydrolases cleave
CN
CO
CC
lyases
catalyze the breaking of C-C (carbon-carbon), C-O (carbon-oxygen), and C-N (carbon-nitrogen) bonds by elimination, which is different from hydrolysis (breaking bonds with water)
can also catalyze the addition of groups to double bonds.
isomerases
Catalyse geometric or structural changes within one molecule (isomerization)
Conversely, catalyse the addition of groups to double bonds
which enzyme group requires ATP
ligases
Vo
initial rate of catalysis
moles of product formed at start
equation for initial rate
k2[ES]
rate of formation of ES
k1[S][E]
rate pf breakdown of ES
k-1 + k2 [ES]
pre steady state
enzyme first added to excess of substrate so concentration of ES complexes slowly building up
steady state
[ES] remains approximately constant
reflects most of reaction
rate of ES formation = rate of ES breakdown
applying steady state to the equations
k1[E][S]=k-1+k2[ES]
rearrange:
[E][S]/[ES]=k-1+k2/k1
Km
michaelis constant
k-1+k2/k1
units of concentration
independent of [E] and [S]
insert Km into equation
Km=[E][S]/[ES]
rearrange:
[ES]=[E][S]/Km
conc free enzyme
[E]T-[ES]
insert equation for [E] into [ES]=[E][S]/Km
[ES]=([E]T-[ES]) [S] / Km
[ES]=[E]T x [S]/[S]+Km
subsitute euqation into Vo
Vo=k2[E]T x [S]/[S]+Km
Vmax
k2[E]T
enzyme saturated with substrate
[ES]=E]T
when [S] < Km
Vo=Vmax[S]/Km
initial rate proportional to [S]
when [S] > Km
Vo=Vmax
initial rate independent of [S]
when [S]=Km
Vo=Vmax/2
what is Km?
[S] at which reaction rate is 1/2 the maximal rate
half the active sites are saturated
Km values for different enzymes
enzymes have different Km values for different substrates and different enzymes will have different Km values for the same substrate
what does a lower Km mean
the better the enzyme can process the substrate as binds stronger
what do Km values of enzymes depend on
substrate, pH, temperature
dissociation constant of ES complex
Km~k-1/k1
equilibrium constant for ES
[E][S]/[ES]=k-1/k1=Km
turnover number
the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate.
indicated by Vmax
kcat
turnover number
Vmax=kx[E]T=kcat[E]T
kcat=Vmax/[E]T
catalytic efficiency of an enzyme for a particular substrate
Kcat/Km
it is a rate constant (specificity constant)
most enzymes are not saturated
[S]«Km
Vo=kcat/Km x [E][S]
where [E] is actually [E]T due to so few enzyme substrate complexes/saturation
what does the specificity constant consider
rate of catalysis with a particular substrate, kcat
strength of ES interaction (Km)
modern method for measuring Km and Vmax
measure Vo for several different concentrations
plot on graph
algorithm used to form line and predict Vmax
Km is 1/2 Vmax
old method for measuring Km and Vmax
take reciprocal: 1/Vo=Km/Vmax x 1/S + 1/Vmax
1/Vo on y axis and 1/[S] on x axis
slope of curve
Km/Vmax
x intercept
-1/Km
y intercept
1/Vmax
