Kinetics Quiz – Chem 391
Name
1.
The reaction of the ubiquitous substrate, S, to product, P, is accomplished by the enzyme E.
The following kinetic constants apply to the reaction.
kuncat = 1.0 x 10-4 s-1
kcat = 10 s-1
Km = 100 µM
A small molecule catalyst (SM) also promotes the reaction with a rate constant kSM = 0.10 M-1s-1.
a.
By how many kcal/mol does the enzyme stabilize the transition state of this reaction vs. the
substrate? Show your reasoning.
kcat/kuncat= 1 x 105
∆G˚ = -7 kcal/mol, noting that -1.4*5 = -7
b.
How many fold faster is the enzyme than the small molecule catalyst? Show your reasoning.
Compare kcat/Km (105 M-1s-1) to kSM (0.1 M-1s-1) and we get 106 fold.
c.
An competitive inhibitor of the enzyme binds with a Ki of 1 µM. Which binds more tightly,
the inhibitor or the substrate? Show your reasoning.
Assuming that kcat << k-1, then Km = Kd. Ki < Km so likely the inhibitor binds more tightly.
d.
A glutamate residue is present in the active site and the side chain binds the substrate and
transition state equally well. It is mutated to alanine and the contact is lost. Will the mutant enzyme
be better, worse or equally good as the wild-type? By what measure? Give brief reasoning.
The Km will likely increase as the substrate binds less tightly, but kcat won’t be altered since both S
and X≠ are bound by the side chain.
2.
Still the same enzyme… with kcat = 10 s-1 and Km = 100 µM. Assume that the total enzyme
concentration is 2 µM.
a.
What will the rate of reaction be at [S] = 100 µM? Show your reasoning.
v = ½ Vmax when [S] = Km.
so… v = 0.5(10 s-1)(2 µM) = 10 µM/s
b.
What will the rate of reaction be at [S] = 900 µM? Show your reasoning.
v = Vmax[S]/(Km + [S]) = (20 µM/s)(900 µM)/(1000 µM) = 18 µM
3.
Consider the following two reaction diagrams, which correlate to the activity of two different
enzymes, A and B. The vertical axis is free energy and the x-axis is the reaction coordinate.
a.
Label all the horizontal lines on the trace for enzyme “A”.
b.
Which enzyme has a larger kcat? Explain your reasoning.
A, since the barrier between E•S and E•X≠ is smaller
c.
Which enzyme has a faster overall rate constant. Again, briefly give your reasoning.
B, since the barrier between E + S and E•X≠ is smaller.