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Biochemistry > Enzyme kinetics > Flashcards

Enzyme kinetics Flashcards

1
Q

what enzyme phosphorylates glucose

A

hexokinase

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2
Q

what bonds do hydrolases cleave

A

CN
CO
CC

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3
Q

lyases

A

catalyze the breaking of C-C (carbon-carbon), C-O (carbon-oxygen), and C-N (carbon-nitrogen) bonds by elimination, which is different from hydrolysis (breaking bonds with water)
can also catalyze the addition of groups to double bonds.

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4
Q

isomerases

A

Catalyse geometric or structural changes within one molecule (isomerization)
Conversely, catalyse the addition of groups to double bonds

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5
Q

which enzyme group requires ATP

A

ligases

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6
Q

Vo

A

initial rate of catalysis
moles of product formed at start

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7
Q

equation for initial rate

A

k2[ES]

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8
Q

rate of formation of ES

A

k1[S][E]

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9
Q

rate pf breakdown of ES

A

k-1 + k2 [ES]

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10
Q

pre steady state

A

enzyme first added to excess of substrate so concentration of ES complexes slowly building up

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11
Q

steady state

A

[ES] remains approximately constant
reflects most of reaction
rate of ES formation = rate of ES breakdown

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12
Q

applying steady state to the equations

A

k1[E][S]=k-1+k2[ES]
rearrange:
[E][S]/[ES]=k-1+k2/k1

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13
Q

Km

A

michaelis constant
k-1+k2/k1
units of concentration
independent of [E] and [S]

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14
Q

insert Km into equation

A

Km=[E][S]/[ES]
rearrange:
[ES]=[E][S]/Km

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15
Q

conc free enzyme

A

[E]T-[ES]

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16
Q

insert equation for [E] into [ES]=[E][S]/Km

A

[ES]=([E]T-[ES]) [S] / Km
[ES]=[E]T x [S]/[S]+Km

17
Q

subsitute euqation into Vo

A

Vo=k2[E]T x [S]/[S]+Km

18
Q

Vmax

A

k2[E]T
enzyme saturated with substrate
[ES]=E]T

19
Q

when [S] < Km

A

Vo=Vmax[S]/Km
initial rate proportional to [S]

20
Q

when [S] > Km

A

Vo=Vmax
initial rate independent of [S]

21
Q

when [S]=Km

A

Vo=Vmax/2

22
Q

what is Km?

A

[S] at which reaction rate is 1/2 the maximal rate
half the active sites are saturated

23
Q

Km values for different enzymes

A

enzymes have different Km values for different substrates and different enzymes will have different Km values for the same substrate

24
Q

what does a lower Km mean

A

the better the enzyme can process the substrate as binds stronger

25
Q

what do Km values of enzymes depend on

A

substrate, pH, temperature

26
Q

dissociation constant of ES complex

A

Km~k-1/k1

27
Q

equilibrium constant for ES

A

[E][S]/[ES]=k-1/k1=Km

28
Q

turnover number

A

the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate.
indicated by Vmax

29
Q

kcat

A

turnover number
Vmax=kx[E]T=kcat[E]T
kcat=Vmax/[E]T

30
Q

catalytic efficiency of an enzyme for a particular substrate

A

Kcat/Km
it is a rate constant (specificity constant)

31
Q

most enzymes are not saturated
[S]«Km

A

Vo=kcat/Km x [E][S]
where [E] is actually [E]T due to so few enzyme substrate complexes/saturation

32
Q

what does the specificity constant consider

A

rate of catalysis with a particular substrate, kcat
strength of ES interaction (Km)

33
Q

modern method for measuring Km and Vmax

A

measure Vo for several different concentrations
plot on graph
algorithm used to form line and predict Vmax
Km is 1/2 Vmax

34
Q

old method for measuring Km and Vmax

A

take reciprocal: 1/Vo=Km/Vmax x 1/S + 1/Vmax

1/Vo on y axis and 1/[S] on x axis

35
Q

slope of curve

A

Km/Vmax

36
Q

x intercept

A

-1/Km

37
Q

y intercept

A

1/Vmax

Biochemistry (14 decks)

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