enzyme kinetics increase the substrate concentration, observe the change of enzyme activity...
TRANSCRIPT
Enzyme Kinetics
Increase the substrate concentration,observe the change of enzyme activity
Substrate concentrationExam Chapters
Score
Enzym
e activity
Student A
Student B
Student C
0 1 2 3 4 0 1 2 3 4
Juang RH (2004) BCbasics
Invertase (IT)
ITSucrose
Non-reducing sugarReducing
sugars
Glucose Fructose
Reducing Power
+HOCH2
O
OH
1
23
4
5
66
54
32
1
1
2
3
4
5
6
HOCH2
O
OH
OHOCH2 HOCH2
OH
H2O
O
HOCH2HOCH2
HO
O
HOCH2
OHOCH2HOCH2
O
CHO H-C-OH HO-C-H H-C-OH H-C-OH H2-C-OH
H2C-OH
C=O HO-C-H H-C-OH H-C-OH H2-C-OH
Jua
ng
RH
(2
00
4)
BC
ba
sics
1 2
Increase Substrate C
oncentration
21 3 4 5 6 7 80
0 2 4 6 8
Substrate (mole)
Product
80
60
40
20
0
S+E
↓
P
(in a fixed period of tim
e)
Juang RH (2004) BCbasics
Essential of Enzyme Kinetics
E S+ P+
Steady State TheorySteady State Theory
In steady state, the production and consumption of the transition state proceed at the same rate. So the concentration of transition state keeps a constant.
SE E
Juang RH (2004) BCbasics
Constant ES Concentration at Steady State
S P
EES
Reaction Time
Concentration
Juang RH (2004) BCbasics
An Example for Enzyme Kinetics (Invertase)
Vmax
Km S
vo
1/S
1vo
Double reciprocal Direct plot
1)1) Use predefined amount of Enzyme → E
2)2) Add substrate in various concentrations → S (x 軸 )
3)3) Measure Product in fixed Time (P/t) → vo (y 軸 )
4)4) (x, y) plot get hyperbolic curve, estimate→ Vmax
5)5) When y = 1/2 Vmax calculate x ([S]) → Km
1Vmax
- 1 Km
1/2
Jua
ng
RH
(2
00
4)
BC
ba
sics
A Real Example for Enzyme KineticsD
ata
no
1234
0.250.501.02.0
0.420.720.800.92
Absorbance v (mole/min)[S]
0.210.360.400.46
(1) The product was measured by spectroscopy at 600 nm for 0.05 per mole(2) Reaction time was 10 min
VelocitySubstrate Product Double reciprocal
1/S 1/v421
0.5
2.081.561.351.16
→→ → →
1.0
0.5
0
v
Dire
ct p
lot
Dou
ble
reci
proc
al 2.0
1.0
0
1/v
-4 -2 0 2 41/[S]
0 1 2[S]
1.0
-3.8
Jua
ng
RH
(2
00
4)
BC
ba
sics