2020 Enzyme MM Kinetics Part II Flashcards Preview

MBIOS 513 Unit 2 > 2020 Enzyme MM Kinetics Part II > Flashcards

Flashcards in 2020 Enzyme MM Kinetics Part II Deck (29)
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1
Q

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

A
  • the Km may contain additional microscopic rates
  • Kcat may not equal K2
  • Thus these parameters may not be easily interpreted
2
Q

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

A

K1

S binds and has no

3
Q

A perfect enzyme is one that?

what does this mean about the ES complex?

A

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.

4
Q

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

A

10^9 M-1 Sec-1

5
Q

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

A

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

6
Q

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

A

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

7
Q

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)

A

-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

8
Q

What is a classic example of a perfect enzyme?

A

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

9
Q

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

A

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

10
Q

What are the 3 general types of bi bi reactions?

A

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

11
Q

What does the term bi bi refer to?

A

2 substrates and 2 products

12
Q

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

A

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

13
Q

What is a sequential random reaction?

A
  • 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
14
Q

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

A
  • 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
15
Q

What is a sequential ordered reaction?

A

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

16
Q

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

A

1) initial velocity (no product re-binding)

2) rapid equilibrium (chemistry rate determining)

17
Q

What is a ping pong reaction/mechanism,

A

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

18
Q

What are the steps of a ping pong reaction ?

A

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

19
Q

What is the hallmark of the ping pong mechanism/reaction

A

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

20
Q

What are the two types of enzyme inhibition?

A

1) reversible inhibition

2) irreversible inhibition

21
Q

What are the types of reversible inhibitors?

A

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

22
Q

What does reversible inhibition involve?

A

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

23
Q

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

A

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.

24
Q

What is α = 1 + [I]/Ki?

A

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

25
Q

An α‐factor alters:

A

1) KM in competitive inhibition

2) KM and Vmax in both uncompetitive and mixed inhibitions

26
Q

Competitive inhibitors bind to what part of the enzyme?

A

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

27
Q

How can competitive inhibition be overcome?

A

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

28
Q

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

A

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

29
Q

In competitive inhibitors how does Vmax and KM look?

A

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

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