Lesson 8: Enzyme Kinetics

Question 1

possibilities to explain enzyme structure function problems

Answer 1

• sterics/size
• polarity/charge

Question 2

give the 3 S and P reactions

Answer 2

S1 + S2 –> P (K1)

S –> P1 + P2 (k-1)

S1 + S2 –> P1 + P2

Question 3

keq

Answer 3

[p]/[S1][S2]`

Question 4

k1 =

Answer 4

rate of forward rxn

S1 + S2 –> P

Question 5

K-1 =

Answer 5

rate of reverse rxn

S –> P1+P2

Question 6

in the simple case of S being converted to P,

Answer 6

the enzyme [E] must form a complex with substrate [S] to yield an enzyme-substrate complex [ES] in order to form product [P]

Question 7

Michaelis and Menten (M & M)

Answer 7

began investigating the effects of [S] on formation of [ES] complex
- examined the effects by measuring the initial reaction velocity (Vo)

Question 8

what are held constant when investigating velocity of rxn

Answer 8

[E] and reaction vol. are held constant
– plot slope of [product] v. time graph over [s]

Question 9

what kind of graph is substrate v. velocity

Answer 9

hyperbolic

Question 10

trends for graphs

Answer 10

1 – the higher the initial [S] the higher the initial velocity
2 – although initial velocities increase as [S] increases, the increase is not as great at high [S] because reaching point of enzyme saturation

Question 11

assumption of equilibrium

Answer 11

early in the reaction, little P has accumulated so k-2 can be ignored

Question 12

steady state assumption

Answer 12

once reaction gets started, the [ES] remains constnat. As a result, the formation of ES must equal the Breakdown of ES

Question 13

when does [ES] stay constant

Answer 13

during steady-state conditions

Question 14

graph trends

Answer 14

[E] decreases as it forms ES
[ES] increases and plateaus
[S] and [P] are inversely proportional

Question 15

what did they define the michaelis constnat Km to be

Answer 15

constant where the concentration of ES is not changing
km =( k-1 + k2) / K1

Question 16

Km

Answer 16

a bundle of 3 rate constants, also a measure of an enzyme’s affiniyy for S

Question 17

michaelis-menten equation

Answer 17

Vo = (Vmax [S])/(Km + [S])

Question 18

how can we determine Vmax and Km experimentially

Answer 18

by meauring initial rates and plotting [S] vs. velocity

Question 19

what does the michaleis-menten equation allow us to do

Answer 19

determine the velocity for any enzyme catalyzed reaction

Question 20

when the velocity = 1/2 Vmax, then

Answer 20

Km = [S] that yields 1/2 Vmax

Question 21

is K, unique for each enzyme-substrate pair

Answer 21

yes

Question 22

an enzyme with a high Km

Answer 22

has a low affinity for S because [S] must be high to reach 1/2 Vmax
—– Km and affinity are inversely proportional

Question 23

an enzyme with a low Km

Answer 23

“high affinity” for S because E only requires small amount of [S] to reach 1/2 Vmax

Question 24

true/false: the K, for any given equation is independent of enzyme concentration when enzyme concentration [E] is limiting for the reaction

Answer 24

true

Question 25

is it difficult to determine Km and Vmax form a graph

Answer 25

YES

Question 26

double reciprocal plot

Answer 26

- plot the reciprocals of the Vo vs. [S] data - accurate methods to determine Km and Vmax

Question 27

equation for double reciprocol plot

Answer 27

1/Vo = ( Km/Vmax ) (1/[S]) + 1/Vmax

Question 28

where the Y intercept is

Answer 28

1/Vmax

Question 29

where the X intercept is

Answer 29

-1/Km

Question 30

slope =

Answer 30

Km/Vmax

Question 31

Km app

Answer 31

when we say that Km increasses or decreases in the presence or absence of inhibitor, are describing this

Question 32

competitive

Answer 32

KI (E to EI) no effect Vmax increase Km

Question 33

uncompetitive

Answer 33

KI' (ES to ESI) decrease Vmax decrease Km

Question 34

mixed

Answer 34

KI (E tp EI) and KI' (Es to ESI) decrease Vmax may increase or decrease Km

Question 35

competitive inhibiton

Answer 35

substrate and inhibitor both compete for same binding site on enzyme - ES is formed then EI cannot form, because S and I compete for the same binding site - if EI is formed then ES cannot form, because S and I compete for the same binding site

Question 36

small KI

Answer 36

tight binding

Question 37

big KI

Answer 37

loose binding

Question 38

lines with difference slopes and a single Y-intercept are characteristic of ()

Answer 38

competitive inhibition

Question 39

how can you dilute the effect of competitive inhibitor

Answer 39

increaseing [S]

Question 40

uncompetitive inhibition

Answer 40

- UI can only bind to ES complex (not free E) - often act on enzymes with multiple substrates - ESI is catalytically incompetent due to altered active site conformation. ES needs to dissociate from I go back through ES to product product

Question 41

for UI can you directly form products from ESI

Answer 41

no - it will take longer to go back to ES to make products ^^^ reason why Vmax decreases

Question 42

for UI why dos Km decrease

Answer 42

because the substrate can bind in 2 places, ,there is an apparent increase for substrate affinity, this a decrease in Km

Question 43

parallel lines with identical slopes is a characteristic of ()

Answer 43

uncompetitive inhibition

Question 44

mixed inhibition

Answer 44

- is a mixture of aspects of competitive and incompetitive inhbitition - Effect on Km is the same variable - Efect on Vmax is same as uncompetitive Mixed inhibitor can bing to Free E to form EI complex AND can also bind to ES to form ESI complex. EI can bind S to form ESI complex

Question 45

for MI is there a reaction from EI and ESI

Answer 45

no -- EI and ESI are catalytically incompetent. Both complexes need to dissociate I and go back thorough E to generate product

Question 46

for mixed inhibition -- when KI' >KI

Answer 46

- think competitive - Km increases

Question 47

for mixed inhibition ---- when KI > KI'

Answer 47

- think UNcompetitive - Km decreases, increased affinity for [S]

Question 48

for mixed inhibition ---- when KI = KI'

Answer 48

- Km is unaffeced - Vmax decreases SPECIAL CASE --> NONcompetitive

Question 49

rate determining step

Answer 49

K2

Question 50

Vo =

Answer 50

= K2 [ES] = Kcat [ES]

Question 51

Kcat

Answer 51

(turnover number) - represents the amount of S converted to P on a per enxyme nasis

Question 52

Kcat =

Answer 52

Vmax/[Et]

Question 53

trends

Answer 53

Vmax increases with [Et] but Vmax is unchanged and therefore Km unchanged

Question 54

where does a' come from

Answer 54

the derevation of M-M equation in presence of a competitive inhibior a' = 1 + [I]/Ki'