Lesson 8: Enzyme Kinetics Flashcards
possibilities to explain enzyme structure function problems
- sterics/size
- polarity/charge
give the 3 S and P reactions
S1 + S2 –> P (K1)
S –> P1 + P2 (k-1)
S1 + S2 –> P1 + P2
keq
[p]/[S1][S2]`
k1 =
rate of forward rxn
S1 + S2 –> P
K-1 =
rate of reverse rxn
S –> P1+P2
in the simple case of S being converted to P,
the enzyme [E] must form a complex with substrate [S] to yield an enzyme-substrate complex [ES] in order to form product [P]
Michaelis and Menten (M & M)
began investigating the effects of [S] on formation of [ES] complex
- examined the effects by measuring the initial reaction velocity (Vo)
what are held constant when investigating velocity of rxn
[E] and reaction vol. are held constant
– plot slope of [product] v. time graph over [s]
what kind of graph is substrate v. velocity
hyperbolic
trends for graphs
1 – the higher the initial [S] the higher the initial velocity
2 – although initial velocities increase as [S] increases, the increase is not as great at high [S] because reaching point of enzyme saturation
assumption of equilibrium
early in the reaction, little P has accumulated so k-2 can be ignored
steady state assumption
once reaction gets started, the [ES] remains constnat. As a result, the formation of ES must equal the Breakdown of ES
when does [ES] stay constant
during steady-state conditions
graph trends
[E] decreases as it forms ES
[ES] increases and plateaus
[S] and [P] are inversely proportional
what did they define the michaelis constnat Km to be
constant where the concentration of ES is not changing
km =( k-1 + k2) / K1
Km
a bundle of 3 rate constants, also a measure of an enzyme’s affiniyy for S
michaelis-menten equation
Vo = (Vmax [S])/(Km + [S])
how can we determine Vmax and Km experimentially
by meauring initial rates and plotting [S] vs. velocity
what does the michaleis-menten equation allow us to do
determine the velocity for any enzyme catalyzed reaction
when the velocity = 1/2 Vmax, then
Km = [S] that yields 1/2 Vmax
is K, unique for each enzyme-substrate pair
yes
an enzyme with a high Km
has a low affinity for S because [S] must be high to reach 1/2 Vmax
—– Km and affinity are inversely proportional
an enzyme with a low Km
“high affinity” for S because E only requires small amount of [S] to reach 1/2 Vmax
true/false: the K, for any given equation is independent of enzyme concentration when enzyme concentration [E] is limiting for the reaction
true
is it difficult to determine Km and Vmax form a graph
YES
double reciprocal plot
- plot the reciprocals of the Vo vs. [S] data
- accurate methods to determine Km and Vmax
equation for double reciprocol plot
1/Vo = ( Km/Vmax ) (1/[S]) + 1/Vmax
where the Y intercept is
1/Vmax
where the X intercept is
-1/Km
slope =
Km/Vmax
Km app
when we say that Km increasses or decreases in the presence or absence of inhibitor, are describing this
competitive
KI (E to EI)
no effect Vmax
increase Km
uncompetitive
KI’ (ES to ESI)
decrease Vmax
decrease Km
mixed
KI (E tp EI) and KI’ (Es to ESI)
decrease Vmax
may increase or decrease Km
competitive inhibiton
substrate and inhibitor both compete for same binding site on enzyme
- ES is formed then EI cannot form, because S and I compete for the same binding site
- if EI is formed then ES cannot form, because S and I compete for the same binding site
small KI
tight binding
big KI
loose binding
lines with difference slopes and a single Y-intercept are characteristic of ()
competitive inhibition
how can you dilute the effect of competitive inhibitor
increaseing [S]
uncompetitive inhibition
- UI can only bind to ES complex (not free E)
- often act on enzymes with multiple substrates
- ESI is catalytically incompetent due to altered active site conformation. ES needs to dissociate from I go back through ES to product product
for UI can you directly form products from ESI
no - it will take longer to go back to ES to make products
^^^ reason why Vmax decreases
for UI why dos Km decrease
because the substrate can bind in 2 places, ,there is an apparent increase for substrate affinity, this a decrease in Km
parallel lines with identical slopes is a characteristic of ()
uncompetitive inhibition
mixed inhibition
- is a mixture of aspects of competitive and incompetitive inhbitition
- Effect on Km is the same variable
- Efect on Vmax is same as uncompetitive
Mixed inhibitor can bing to Free E to form EI complex AND can also bind to ES to form ESI complex. EI can bind S to form ESI complex
for MI is there a reaction from EI and ESI
no – EI and ESI are catalytically incompetent. Both complexes need to dissociate I and go back thorough E to generate product
for mixed inhibition – when KI’ >KI
- think competitive
- Km increases
for mixed inhibition —- when KI > KI’
- think UNcompetitive
- Km decreases, increased affinity for [S]
for mixed inhibition —- when KI = KI’
- Km is unaffeced
- Vmax decreases
SPECIAL CASE –> NONcompetitive
rate determining step
K2
Vo =
= K2 [ES]
= Kcat [ES]
Kcat
(turnover number)
- represents the amount of S converted to P on a per enxyme nasis
Kcat =
Vmax/[Et]
trends
Vmax increases with [Et] but Vmax is unchanged and therefore Km unchanged
where does a’ come from
the derevation of M-M equation in presence of a competitive inhibior
a’ = 1 + [I]/Ki’
