202 Enzymes: Michaelis‐Menten Kinetics

Question 1

The basic Michaelis-Menten equation considers only

Answer 1

one substrate an no chemistry intermediates (I.e., S goes to P, not S goes to X and then X goes to Y and eventually Y goes to P).

Question 2

In the Michaelis-Menten equation what simplifications/assumptions do we make? What does this say about the two functional aspects of the enzyme reaction?

Answer 2

1) that the chemistry is irreversible (and this is usually assumed to be the rate-determining step)
2) there is very little product formed yet and so there is no EP complex formed by P rebinding to the enzyme.
3) these simplifications reduces the rate constants to three (three equation arrows) instead of 6

These schemes reduces the enzyme reaction to two functional aspects

1) binding of substrate and 2) the chemistry step to convert S to P

Question 3

In the michaelis-menten reaction when is initial velocity measured?

Answer 3

before 10% of the substrate has been consumed.

Question 4

Since the substrate concentration is so much higher than the enzyme concentration, what is an important assumption for the M-M kinetics?

Answer 4

that the amount of substrate bound to enzyme at any point in time is negligible compared to the total substrate concentration; hence we can assume that [S] total is constant

Question 5

The initial velocity in M-M kinetics is determined by what?

Answer 5

determined by the breakdown of the ES complex to form enzyme and product. This is affected by the concentration of ES; hence initial velocity (Vo) equals the rate constant times [ES]

Question 6

The rate constant equals what?

Answer 6

the disappearance of substrate over time or the appearance of product over time

AKA K2 * [ES] = d[P]/dt or -d[S]/dt measured in linear range

Question 7

What is the steady state assumption?

Answer 7

it describes the conditions for experiments that can assay enzyme kinetics. This assumption coupled with the assumptions in the original derivation by M-M kinetic, that chemistry is the rate-determining step) allows derivation of the parameters in the M-M equation

Question 8

What is the definition of steady-state (not the steady state assumption, just steady state)

Answer 8

the concentration of an intermediate (ES complex), is constant w/time, I.e., d [ES]/dt = 0

Question 9

The concentration of intermediates is fixed by what rates?

Answer 9

the rates of its formation minus the rates of its decay.

Question 10

What is the equation for steady state?

Answer 10

rates of formation = the rates of decay of the intermediate.

this requires that [S] total be much higher than [E] total, aka [S] total = [S] free

Question 11

What occurs during the pre-steady sate?

Answer 11

the pre-steady state is the initial period when the enzyme is first added to the excess substrate, the ES concentration does not stay constant but is increasing with time. This period is short and cannot be easily observed. The steady sate follows this period an the initial velocity is defined by this steady-state

Question 12

In a progress curve, explain how substrate concentration looks as a function of time?

Answer 12

When t=0, the substrate concentration is higher than the enzyme concentration. Then as time proceeds, substrate concentration decreases but not in a linear fashion

After a certain point, it becomes linear

this linear region is the initial velocity range used in MM kinetics

Question 13

In a progress curve, explain how the product concentration proceeds as a function of time

Answer 13

Initially, there is a non-linear increase in product concentration. This rate of increase becomes linear at the same time as the rate of disappearance of substrate becomes linear.

this product linear range is used when assaying an enzyme, I.e., the slope of the linear portion of either of those curves = the initial velocity Vo

Question 14

In a progression curve, when the substrate and product are linear, how does the concentration of the ES look?

Answer 14

the ES concentration stays constant. aka the slope in the curve in this region is zero d[ES]/dt = 0

Question 15

in the progression curve graph, what does the steady state region illustrate?

Answer 15

it illustrates that [ES] does not change significantly with time.

Question 16

in the progression curve graph, what does the pre-steady-state region illustrate?

Answer 16

it illustrates that there is a build up of the ES complex. and all four functions in this region are non linear

Question 17

What are the common methods to study pre-steady-state kinetics?

Answer 17

1) stopped-flow kinetics - in which enzyme an substrate are mixed rapidly and then the reaction almost immediately stopped (sub-second scale)
2) ESI-MS (electrospray ionization mass spectrometry) - this provides information regarding the formation of the enzyme-substrate complex and thus the mechanisms of enzyme catalysis

Question 18

steady state kinetics measures what?

Answer 18

the rates of enzyme-catalyzed reactions under the steady-state conditions as exist in the cell, and thus help us understand metabolism.

Question 19

If chemistry is the rate determining step, then velocity is determined how?

Answer 19

by how much is present of the species that precedes this slowest step (I.e. [ES]) as well as by the rate constant of chemistry, k2. I.e., v0 = k2 ∙ [ES].

Question 20

[E] total is

Answer 20

the sum of concentrations of free enzyme and enzyme bound to substrate

Question 21

What is Vmax?

Answer 21

the maximal initial velocity, which is the initial velocity when the enzyme is fully saturated with substrate

at this point [E]total = [ES]. So Vmax = k2 ∙ [E]total. Substituting, we get v0 = (Vmax ∙ [S]total ) / (KM + [S]total), the Michaelis‐Menten equation.

Question 22

What are the MM parameters?

Answer 22

Vmax
Kcat
Km
Kcat/Km

Question 23

What is does the turnover number Kcat describee?

Answer 23

it describes the limiting rate of an enzyme-catalyzed reaction at saturation by substrate. it is equivalent to the number of substrate molecules converted to product in a given unit of time (usually sec is used) by a single enzyme molecule when fully saturated

Question 24

What information does Vmax and Kcat give?

Answer 24

information on speed (velocity)

Question 25

What information does Km give?

Answer 25

information on saturation by substrate

Question 26

What information does Kcat/Km give

Answer 26

gives info on efficiency (=the maximum rate constant, which is at very low substrate concentration)

Question 27

Does Kcat depend on [E]?

Answer 27

no, kcat does not depend on [E].

Kcat is a catalyst constant that characterizes the M-M equation at very high substrate concentration and is also called a turnover number

Question 28

Maximal velocity Vmax occurs when?

Answer 28

occurs at high substrate concentration when the enzyme is fully saturated I..e., entirely in the ES form.

Also, when more enzyme is present the greater the volume of Vmax

Question 29

[E] total is measured in what units? Why?

Answer 29

measured in units of weight not concentration

umoles or ug

The reason for this is that an experimental volume is being assumed.

Question 30

What does Km equal and what does this report?

Answer 30

Km= the concentration of S that yields 50% of the maximal velocity or rate

-this reports on saturation

Question 31

When can Km be used as a measure of the binding-affinity?

Answer 31

if k2 is the rate‐limiting step and k2 is much less than k‐1, then Km equation reduces to k‐1 / k1. Now if you remember, this is the equation for KD or the dissociation constant. In such a scenario, KM can be used as a measure of the binding‐affinity of the enzyme for its substrate. This is often done without really checking whether k2 &laquo_space;k‐1. Watch out.

Question 32

Kcat is the rate constant for what?

Answer 32

S goes to P. and is a second order rate constant that describes the MM function at very low substrate concentrations. Also called specificity constant. Since at low substrate concentration the Etotal is mostly free enzyme, this rate also reflects the efficiency of the enzyme-substrate collisions

Question 33

If there is a single rate-limiting step in a catalytic mechanisms, then Kcat =

Answer 33

the rate constant for that step

Question 34

At low concentrations the E total is mostly what?

Answer 34

free enzymes

Question 35

When look at the efficiency of enzymes, what do you want to measure?

Answer 35

Kcat/Km, since this is more reflective of physiological conditions, where substrates are typically not in a very high concentration

Question 36

Describe the type of turnover rate and affinity strength an efficient enzyme would have

Answer 36

A highly efficient enzyme will have a large Kcat (high turnover) and a low Km (strong affinity for substrate).

Question 37

When can Kcat/Km be used to compare which substrate a single enzyme prefers?

Answer 37

if and only if that enzyme can use more than one substrate (e.g., glucose vs fructose in the hexokinase rx).

Question 38

Kcat/Km take home lesson:

Answer 38

tells us how fast a substrate molecule and an enzyme molecule can collide and result in substrate binding to the enzyme followed by the reaction taht produces product bound to the enzyme and that product is “immediately” released.

Question 39

Why can’t Kcat/Km values be directly compared between different enzymes catalyzing different reactions?

Answer 39

b/c there can be two enzymes with the same Kcat/Km values and one enzyme may hav a large Kcat value (the reaction proceeds fast) and a large Km value (the enzyme does not bind the substrate very tightly); the other enzyme can have a small Kcat value (the reaction proceeds slowly) and small Km value (the enzyme binds the substrate tightly)

Question 40

What is the standard way of linearizing the MM equation?

Answer 40

Lineweaver-Burk Plot, in which the MM equation (which is a hyperbola, y = ax/[b + x]), is inverted to get a linear equation (y= ax + b)

Question 41

What is the disadvantage of linearizing the MM equation via L-B plot? And if there is a disadvantage, why convert?

Answer 41

there is an uneven error envelope.

Useful - useful for visualizing trends in data, especially when changing conditions between experiments and would like to see how those changes affects enzyme kinetics.

Question 42

When chemistry is not the rate determining step, what is?

Answer 42

sometimes product-dissociation is the rate determining step or very efficient enzymes like TIM and other perfect enzymes

Question 43

The MM equation can be used to fit data for what enzymes?

Answer 43

any enzymes that exhibit a hyperbolic dependence of velocity on substrate concentration