Enzyme kinetics

Question 1

what enzyme phosphorylates glucose

Answer 1

hexokinase

Question 2

what bonds do hydrolases cleave

Answer 2

CN
CO
CC

Question 3

lyases

Answer 3

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.

Question 4

isomerases

Answer 4

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

Question 5

which enzyme group requires ATP

Answer 5

ligases

Question 6

Vo

Answer 6

initial rate of catalysis
moles of product formed at start

Question 7

equation for initial rate

Answer 7

k2[ES]

Question 8

rate of formation of ES

Answer 8

k1[S][E]

Question 9

rate pf breakdown of ES

Answer 9

k-1 + k2 [ES]

Question 10

pre steady state

Answer 10

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

Question 11

steady state

Answer 11

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

Question 12

applying steady state to the equations

Answer 12

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

Question 13

Km

Answer 13

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

Question 14

insert Km into equation

Answer 14

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

Question 15

conc free enzyme

Answer 15

[E]T-[ES]

Question 16

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

Answer 16

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

Question 17

subsitute euqation into Vo

Answer 17

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

Question 18

Vmax

Answer 18

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

Question 19

when [S] < Km

Answer 19

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

Question 20

when [S] > Km

Answer 20

Vo=Vmax
initial rate independent of [S]

Question 21

when [S]=Km

Answer 21

Vo=Vmax/2

Question 22

what is Km?

Answer 22

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

Question 23

Km values for different enzymes

Answer 23

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

Question 24

what does a lower Km mean

Answer 24

the better the enzyme can process the substrate as binds stronger

Question 25

what do Km values of enzymes depend on

Answer 25

substrate, pH, temperature

Question 26

dissociation constant of ES complex

Answer 26

Km~k-1/k1

Question 27

equilibrium constant for ES

Answer 27

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

Question 28

turnover number

Answer 28

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

Question 29

kcat

Answer 29

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

Question 30

catalytic efficiency of an enzyme for a particular substrate

Answer 30

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

Question 31

most enzymes are not saturated [S]<

Answer 31

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

Question 32

what does the specificity constant consider

Answer 32

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

Question 33

modern method for measuring Km and Vmax

Answer 33

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

Question 34

old method for measuring Km and Vmax

Answer 34

take reciprocal: 1/Vo=Km/Vmax x 1/S + 1/Vmax 1/Vo on y axis and 1/[S] on x axis

Question 35

slope of curve

Answer 35

Km/Vmax

Question 36

x intercept

Answer 36

-1/Km

Question 37

y intercept

Answer 37

1/Vmax