2020 Enzyme MM Kinetics Part II

Question 1

The MM equation can be used to describe MOST situations BUT…

Answer 1

• the Km may contain additional microscopic rates
• Kcat may not equal K2
• Thus these parameters may not be easily interpreted

Question 2

If K2» K-1 the Kcat/Km is equal to what?

Answer 2

K1

S binds and has no

Question 3

A perfect enzyme is one that?

what does this mean about the ES complex?

Answer 3

catalyzes reactions where chemistry is not rate-determining; I.e., K2 is much greater than K-1. and have a Kcat/Km value around 10^8 or 10^9 M-1 sec-1

-This means that the ES complex proceeds very quickly to EP complex and then on to free enzyme + product.

Dissociation of the ES complex back to the enzyme and substrate is negligible.

Question 4

What is the theoretical diffusion control limit for non-reacting spheres as they move through solutions?

Answer 4

10^9 M-1 Sec-1

Question 5

Why do some rare enzymes have a Kcat/Km values as high as ~10^12 sec-1?

Answer 5

b/c electrostatic interactions, which are common in enzymatic reactions can complement the binding

Question 6

In a transition free energy as a function of the reaction coordinate, what is the hallmark of a perfect enzyme?

Answer 6

the hallmark of a perfect enzyme is that all three peaks are of similar height

Question 7

In a transition free energy as a function of the reaction coordinate, why wouldn’t enzymes evolve to reduce the middle peak? (ES to EP, determines rate of chemistry)

Answer 7

-chemistry is not the rate determining step, these enzymes are limited by diffusion. Evolution does not concern itself with diffusion of substrate and enzyme. one result is that the only control of the rate of the catalyzed reaction is diffusion of substrate and/or product. and that is no way to control life processes

Question 8

What is a classic example of a perfect enzyme?

Answer 8

TIM (trisephosphate isomerase), which has a Kcat/Km of 2*10^8 M-1 sec-1

Question 9

How can altering viscosity help determine if an enzyme is a perfect enzyme or not?

Answer 9

if chemistry is the rate determining step, then altering viscosity should not affect the rate. But if binding of substrate to enzyme or release of product is rate-determining, then an increase in viscosity reduces diffusion and thus the rate decreases

Question 10

What are the 3 general types of bi bi reactions?

Answer 10

1) sequential random (e.g., kinases, some dehydrogenase)
2) Sequential ordered (eg., NAD-dep. dehydrogenase)
3) Ping-pong (e.g., transferases)

Question 11

What does the term bi bi refer to?

Answer 11

2 substrates and 2 products

Question 12

With both sequential types of bi bi reactions, what time of complex is formed? what about for ping-pong?

Answer 12

Sequential random and sequential ordered: a ternary complex (EAB) is formed

For Ping-pong: the substrates are not simultaneously bound to E, so no ternary complex is formed; instead there are 2 distinct enzymes formed

Question 13

What is a sequential random reaction?

Answer 13

• a reaction in which it does not matter which substrate binds first, or which product dissociates first - the rate and affinity are the same whatever order the reactants bind

Question 14

To simplify looking at initial velocities in sequential random reaction, we assume…

Answer 14

• that there is no product rebinding, so we can ignore many of the rate constants.
• we can also assume a rapid equilibrium; I.e., chemistry is the rate determining step

Question 15

What is a sequential ordered reaction?

Answer 15

one in which there is a fixed order for substrate binding and almost always the products are also released in a specific order

in these reactions a binary complex is first formed with E and A, B then binds to EA to form the ternary complex, EAB. In the chemistry step, products are also released in a specific order, with C coming off first and then D

Question 16

To simplify looking at initial velocities in sequential ordered reaction, we assume…

Answer 16

1) initial velocity (no product re-binding)

2) rapid equilibrium (chemistry rate determining)

Question 17

What is a ping pong reaction/mechanism,

Answer 17

in a ping-pong mechanism, one substrate is bound, one product is released, then a second substrate comes in and a second product is released. The enzyme E gets modified in some way after the binding of the first substrate. There is no ternary complex formed

Question 18

What are the steps of a ping pong reaction ?

Answer 18

The first substrate A, binds to enzyme E< which is altered from an EA complex to a FC complex.

When the first product C is released, the enzyme is now in the F form.

Free F then binds to the second substrate B, to form the complex FB.

The enzyme is modified back to its original form and B becomes D, which then comes off leaving behind the original enzyme E

Question 19

What is the hallmark of the ping pong mechanism/reaction

Answer 19

the 2 form enzyme, where the enzyme modifications are generally covalent in nature. so when studying this reaction you can break it up into 2 half-reactions and simplify rate constants (I.e., ignore some of them) to look at initial velocities

Question 20

What are the two types of enzyme inhibition?

Answer 20

1) reversible inhibition

2) irreversible inhibition

Question 21

What are the types of reversible inhibitors?

Answer 21

1) competitive inhibition
2) uncompetitive inhibition
3) mixed or non-competitive inhibition

Question 22

What does reversible inhibition involve?

Answer 22

involves non-covalent binding, with the ability of the inhibitor to bind as well as dissociate from the enzyme.

Question 23

What does irreversible inhibition involve? and when do you see irreversible inhibition?

Answer 23

the inhibitor is covalently linked to the enzyme or forms a strong stable non-covalent association, inactivating the enzyme permanently

often seen in the case of toxins and poisons where death results from toxins incapacitating the key enzymes.

Question 24

What is α = 1 + [I]/Ki?

Answer 24

[I] is the conc. of inhibitor and Ki the dissociation constant of the enzyme‐inhibitor complex.

Question 25

An α‐factor alters:

Answer 25

1) KM in competitive inhibition

2) KM and Vmax in both uncompetitive and mixed inhibitions

Question 26

Competitive inhibitors bind to what part of the enzyme?

Answer 26

the active site.

this creates a subtle but crucial change in protein conformation that occurs when a ligand (such as a substrate or inhibitor) bind s to the protein

Question 27

How can competitive inhibition be overcome?

Answer 27

by increasing the concentration of substrate so there are so many substrate molecules (compared to the inhibitor molecules) that they “outcompete” the inhibitor for the substrate-binding site of the enzyme

Question 28

How do competitive inhibitors “fool” the enzyme active site into binding them?

Answer 28

competitive inhibitors typically have structures similar enough to the structure of the substrate ‘fooling’ the enzyme active site into binding them.

HOWEVER, the inhibitor structure does not allow the reaction to proceed to produce product

Question 29

In competitive inhibitors how does Vmax and KM look?

Answer 29

there is no change in Vmax but there is an apparent increase in Km

“Apparaent” KM in presence of a competitive inhibitor = α∙ KM