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CHEM 430 > Lesson 8: Enzyme Kinetics > Flashcards

Lesson 8: Enzyme Kinetics Flashcards

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1
Q

possibilities to explain enzyme structure function problems

A
  • sterics/size
  • polarity/charge
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2
Q

give the 3 S and P reactions

A

S1 + S2 –> P (K1)

S –> P1 + P2 (k-1)

S1 + S2 –> P1 + P2

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3
Q

keq

A

[p]/[S1][S2]`

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4
Q

k1 =

A

rate of forward rxn

S1 + S2 –> P

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5
Q

K-1 =

A

rate of reverse rxn

S –> P1+P2

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6
Q

in the simple case of S being converted to P,

A

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

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7
Q

Michaelis and Menten (M & M)

A

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

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8
Q

what are held constant when investigating velocity of rxn

A

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

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9
Q

what kind of graph is substrate v. velocity

A

hyperbolic

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10
Q

trends for graphs

A

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

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11
Q

assumption of equilibrium

A

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

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12
Q

steady state assumption

A

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

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13
Q

when does [ES] stay constant

A

during steady-state conditions

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14
Q

graph trends

A

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

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15
Q

what did they define the michaelis constnat Km to be

A

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

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16
Q

Km

A

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

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17
Q

michaelis-menten equation

A

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

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18
Q

how can we determine Vmax and Km experimentially

A

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

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19
Q

what does the michaleis-menten equation allow us to do

A

determine the velocity for any enzyme catalyzed reaction

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20
Q

when the velocity = 1/2 Vmax, then

A

Km = [S] that yields 1/2 Vmax

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21
Q

is K, unique for each enzyme-substrate pair

A

yes

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22
Q

an enzyme with a high Km

A

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

23
Q

an enzyme with a low Km

A

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

24
Q

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

A

true

25
Q

is it difficult to determine Km and Vmax form a graph

A

YES

26
Q

double reciprocal plot

A
  • plot the reciprocals of the Vo vs. [S] data
  • accurate methods to determine Km and Vmax
27
Q

equation for double reciprocol plot

A

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

28
Q

where the Y intercept is

A

1/Vmax

29
Q

where the X intercept is

A

-1/Km

30
Q

slope =

A

Km/Vmax

31
Q

Km app

A

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

32
Q

competitive

A

KI (E to EI)
no effect Vmax
increase Km

33
Q

uncompetitive

A

KI’ (ES to ESI)
decrease Vmax
decrease Km

34
Q

mixed

A

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

35
Q

competitive inhibiton

A

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

36
Q

small KI

A

tight binding

37
Q

big KI

A

loose binding

38
Q

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

A

competitive inhibition

39
Q

how can you dilute the effect of competitive inhibitor

A

increaseing [S]

40
Q

uncompetitive inhibition

A
  • 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
41
Q

for UI can you directly form products from ESI

A

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

42
Q

for UI why dos Km decrease

A

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

43
Q

parallel lines with identical slopes is a characteristic of ()

A

uncompetitive inhibition

44
Q

mixed inhibition

A
  • 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

45
Q

for MI is there a reaction from EI and ESI

A

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

46
Q

for mixed inhibition – when KI’ >KI

A
  • think competitive
  • Km increases
47
Q

for mixed inhibition —- when KI > KI’

A
  • think UNcompetitive
  • Km decreases, increased affinity for [S]
48
Q

for mixed inhibition —- when KI = KI’

A
  • Km is unaffeced
  • Vmax decreases
    SPECIAL CASE –> NONcompetitive
49
Q

rate determining step

A

K2

50
Q

Vo =

A

= K2 [ES]
= Kcat [ES]

51
Q

Kcat

A

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

52
Q

Kcat =

A

Vmax/[Et]

53
Q

trends

A

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

54
Q

where does a’ come from

A

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

CHEM 430 (23 decks)

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