Protein Kinetics Flashcards
Advantages for using initial rate
- The reverse reaction can be ignored
2) little substrate has been converted
rate of the uncatalyzed reaction
v = k[A][B]
v is proportional to both [A] and [B]
Species involved in an enzymatic reaction
S + E –> P + E
Steady state assumption of ES complex? Initial rate of reaction?
A protein E first forms a complex with a substrate S , then convert S to P.
Initial rate: Vo= k2[ES]
Rate of association = k1[E][S]
Rate of dissociation = (k-1 + k2)[ES]
Steady state is when these two rates are equal
k1[E][S] = (k-1 + k2)/[ES]
What is the Michaelis constant
Km = [E][S]/[ES] = (k-1 +k2)/k1
Michaelis Menten Equation
V0 = k2[Et][S]/ ([S] + Km) or
V0 = kcat[Et][S]/ ([S] +Km)
What is the Km relation to k1, k2, k-1? What are the assumptions for a very slow and very fast enzyme? What is Km
Km = (k2+k-1)/k1
But for a very slow enzyme k2 «k-1
Km ~ k-1/k1
For a very fast enzyme k2»_space; k-1
Km~ k2/k1
Km is the the substrate concentration when Half Vmax is reached
what is kcat
kcat is the turnover number
kcat = k2
indicates how fast ES turns into E and P
How does Km affect v
The higher Km, the lower the rate will be at that substrate concentration
What is the specificity constant
kcat/Km is the specificity constant
how would [S] «_space;Km affect Michaelis Menten equation
Km + [S] ~ Km
An enzyme close to 10^8 M-1s-1, is considered what?
Diffusion-controlled
has reached catalytic perfection
What value does an enzyme need to reach to be diffusion-controlled and reach catalytic perfection
10^8M-1s-1
Slope, x-axis, y-axis, x and y intercepts of Lineweaver-Burk double-reciprocal plot
How does increasing S affect line-weaver plot
competitive inhibition plot
Uncompetitive inhibition plot and what is uncompetitive inhibition
Inhibitor binds to ES complex preventing it from releasing P
