michaelis menten

Question 1

Km

Answer 1

Km = [s] when v is 1/2 Vmax
steady state constant

Question 2

small Km

Answer 2

max catalytic efficiency at low S

Question 3

uncompetitive vs allosteric

Answer 3

allosteric based on a different kinetic model.
UI only binds to ES, AI could bind to E or ES

Question 4

initial rate of reaction occurs when…

Answer 4

steady state is achieved.
why? This is following the pre steady state where enzyme is encountering substrate only sometimes…
- substrate concentration remains approximately constant, equal to the initial substrate concentration, while the enzyme-substrate complex concentration builds up.

Question 5

What is the initial rate?

Answer 5

• 10% of the rxn is used up and has achieved steady state

Question 6

Chemical equation, following the assumptions of MM equation

Answer 6

E + S >1> ES >2> E + P

1 = k1, k-1
2 = kcat.

Can ignore k2 step, due to k1 and k3 being fast.

Question 7

kcat

Answer 7

kcat is essentially equal to k2, because k3 happens so fast.
(ES to EP)
It is the apparent rate constant for the rate determining conversion (k2) of substrate to product.
But, because k2 is happening so slowly, we are saying that is is k3.

(time required by an enzyme molecule to “turn over” one substrate molecule)

Question 8

Michaelis menten.
- 1) what is it
- 2) equilibrium state assumptions, what does it yield
- 3) Role of k2?

Answer 8

1. E and S associate reversible to form an ES complex E+S <-k1,-1-> ES <-k2-> E+P
2. Assume ES is in rapid equilibrium with free E (k1 time E+S = k-1 times ES… helps us determine Ks (ES dissociation constant)
3. Breakdown of ES to form P is slower than ES formation, also slower than ES forming E+S.
   k2««k-1,k1

Question 9

Vmax

Answer 9

• maximum rate
• occurs when all enzyme saturated
• all E is in [ES]

Question 10

rate determining step

Answer 10

k2 is the rate determining step, it is when [S] is high enough to convert all E to ES
(kcat) is the apparent rate constant for ES to EP.

Question 11

Why is overall rate unafected when more S is added at saturation?

Answer 11

there is no enzyme to bind to the S, all enzyme is already bound at saturation

Question 12

rate of product formation

Answer 12

v = kcat[ES]
(ES <–> E + P)

Question 13

why cant you measure E or ES but can measure Etotal?

Answer 13

We know how much enzyme is in the reaction, but we dont know how much of that enzyme is actually in ES complex or free E

Question 14

Ks

Answer 14

dissociation constant,

Question 15

when kcat &laquo_space;k-1

Answer 15

Ks = k-1/k1 = [E] [S] / [ES]
( E and S are in equilibrium with ES)
these are not truly in equilibrium, does not apply to everything, because some ES is always being converted to P, very slowly

Question 16

Briggs haldane model

Answer 16

the more ES that is present, the faster ES will dissociate either to prod (kcat) or reactant (k-1).

Therefore, when first mixing E and S, [ES] builds fast but then reaches a steady state. This state persists until all substrate is used up

Question 17

steady state

Answer 17

formation and breakdown of ES are =
k1 [E] [S] = k-1 [ES] + kcat [ES]

formation of ES complex = diss. of ES + breakdown to E+P

Question 18

burst phase

Answer 18

• presteady state
• rapid formation of ES complexes why??

Question 19

Km equation

Answer 19

Km = (k-1 + kcat) / k1
substrate concentration when the velocity is 1/2 Vmax.
NOT a measure of affinity

Question 20

Michaelis menten equation

Answer 20

v = Vmax[S] / Km + [S]

Question 21

Km three definitions

Answer 21

complex rate constant
concentration of S when v is 1/2 Vmax
steady state constant

Question 22

Km varies with

Answer 22

enzyme, substrate, temp, pH

Question 23

enzyme efficiency

Answer 23

kcat /Km
large kcat (rapid turnover) or small Km (1/2 Vmax at low [S]) will make enzyme efficiency large

Question 24

kcat equation

Answer 24

Vmax / [E] total

Question 25

Vmax equation

Answer 25

Vmax = kcat [E]total

Question 26

diffusion controlled limit

Answer 26

(kcat/Km) corresponds to the second order rate constant for enzyme substrate combination under low [S] conditions.

this rate constant has a MAXIMUM value, which is determined by frequency of collision between E and S

this maximum value of collision (10^8, 10^9… GOVERNOR) is diffusion limited

Question 27

Catalytic perfection

Answer 27

when an enzyme has the maximum possible efficiency… P formed as soon as E collides with S.
E efficiency in range of 10^8 and 10^9.