enzyme kinetics vo=vo= v max [s] k m + [s] kmkm v max & e1 e2 e3 1st order zero order...
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![Page 1: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/1.jpg)
Enzyme Kinetics
vo=Vmax [S]Km + [S]
Km Vmax &
E1E2E3
1st order
zero order
Competitive
Non-competitive
Uncompetitive
Direct plot
Double reciprocal
Bi-substrate reaction alsofollows M-M equation, butone of the substrate shouldbe saturated when estimate the other
Affinity withsubstrate
Maximumvelocity Inh
ibition
Activity
Observe vo change under various [S], resulted plotsyield Vmax and Km
k3 [Et]
kcatTurn overnumber
kcat / Km
Activity Unit
1 molemin
Specific Activity
unitmg
Significance
Juang RH (2004) BCbasics
![Page 2: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/2.jpg)
Significance of Enzyme Kinetics
vo =Vmax [S]Km + [S]
Obtain Vmax and Km
[S] = Low → High [S] = Fixed concentration
zero order
1st order
E3
E2E1
Proportional to
enzyme concentration
v0 = Vmax × K = k3 [Et] × K
Juang RH (2004) BCbasics
![Page 3: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/3.jpg)
Km = [S]
Km + [S] = 2 [S]
Vmax
2=
Vmax [S]Km + [S]
Km: Affinity with Substrate
If vo =Vmax
2
S2S1 S3
S1 S2 S3
Vmax
1/2
When using different substrate
Affinity changesKm
vo =Vmax [S]Km + [S]
Jua
ng
RH
(2
00
4)
BC
ba
sics
![Page 4: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/4.jpg)
Km: Hexokinase Example
Glucose + ATP → Glc-6-P + ADP
1
2
3
4
5
6
Glucose Allose Mannose Substratenumber
Km = 8 8,000 5 M
CHO H-C-OH HO-C-H H-C-OH H-C-OH H2-C-OH
CHO H-C-OH H-C-OH H-C-OH H-C-OH H2-C-OH
CHO HO-C-H HO-C-H H-C-OH H-C-OH H2-C-OH
Juang RH (2004) BCbasics
![Page 5: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/5.jpg)
Turn Over Number, kcat
vo =Vmax [S]Km + [S]
(vo)E + S ES E + P
k2
k1 k3
When substrate excess, k3 = kcat, turn over number (t.o.n)
When [substrate] is low
=k3 [E][S]
Km + [S]=
Km
k3 [E][S]
Omit the [substrate] Substrate specificityStart from M-M eq.
Second order(IV)
Juang RH (2004) BCbasics
![Page 6: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/6.jpg)
Turn Over Numbers of Enzymes
Catalase H2O2
Carbonic anhydrase HCO3-
Acetylcholinesterase Acetylcholine
40,000,000
400,000
140,000
-Lactamase Benzylpenicillin 2,000
Fumarase Fumarate 800
RecA protein (ATPase) ATP 0.4
Enzymes Substrate kcat (s-1)
The number of product transformed from substrate by one enzyme molecule in one second
Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.263
![Page 7: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/7.jpg)
Chymotrypsin Has Distinct kcat / Km to Different Substrates
O R O
H3C–C–N–C–C–O–CH3
H H
= – =––
–HGlycine
kcat / Km
1.3 ╳ 10-1
–CH2–CH2–CH3Norvaline 3.6 ╳ 102
–CH2–CH2–CH2–CH3Norleucine 3.0 ╳ 103
–CH2–Phenylalanine 1.0 ╳ 105
(M-1 s-1)
R =
Adapted from Mathews et al (2000) Biochemistry (3e) p.379
![Page 8: Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal](https://reader035.vdocuments.net/reader035/viewer/2022072016/56649ef45503460f94c06b38/html5/thumbnails/8.jpg)
Enzyme Activity Unit
Reaction time (min)P
rodu
ct [P
]0 10 20 30 40
Slopetan
S → P mole
vo = [P] / min
Unit =
Activity Units
Protein (mg)
t
mole/min
yx
yx = tan
Jua
ng
RH
(2
00
4)
BC
ba
sics
Specific Activity =