Michaelis Menten Plot
Velocity
[S]
abs/min (nmol/min
(mM)
)
0
0
0.000000
0.1
0.0454
3.131034
0.2
0.0795
5.482759
0.4
0.1004
6.924138
0.6
0.1155
7.965517
0.8
0.1281
8.834483
1
0.1107
7.634483
1.2
0.1241
8.558621
2
0.1237
8.531034
LineweaverBurke
1/[S]
10.00
5.00
2.50
1.67
1.25
1.00
0.83
0.50
1/v
[S]/v
0.3194
0.1824
0.1444
0.1255
0.1132
0.1310
0.1168
0.1172
0.032
0.036
0.058
0.075
0.091
0.131
0.141
0.234
Eadie Hofstee
Velocity
v/[S]
(nmol/mi
n)
31.310
27.414
17.310
13.276
11.043
7.634
7.132
4.266
10.0
0.4000
8.0
0.3000
6.0
4.0
2.0
0.0
1
2
3
[S] (mM)
Hanes-Woolfe:
Vmax= 9.24 nmol/min
Km= 0.13 mM
0.1000
3.00
6.0000
4.0000
8.00
13.00
1/[S]
Hanes-Woolfe
y = -0.1841x + 9.8797
R² = 0.86
0.250
0.200
0.150
y = 0.1084x + 0.0144
R² = 0.9917
0.100
2.0000
0.050
0.0000
0.000 10.000 20.000 30.000 40.000
0.000
v/[S]
Eadie-Hofstee:
Vmax= 9.88 nmol/min
Km= 0.18 mM
0.2000
0.0000
-2.00
-0.1000
[S]/v
Velocity (nmol/min)
8.0000
3.1310
5.4828
6.9241
7.9655
8.8345
7.6345
8.5586
8.5310
y = 0.0213x + 0.0957
R² = 0.9715
Eadie Hofstee
10.0000
Lineweaver:
Vmax= 10.45 nmol/min
Km= 0.22
Lineweaver-Burke
1/v
Velocity (nmol/min)
Michaelis Menten
0
Michealis Menton:
Vmax= 8.9 nmol/min
Km= 0.15 mM
0
1
2
3
[S] (mM)
The Hanes Woolfe experiment is more reliable because the R2 value is closest to 1 and it is also the
most accurate graph, as most of its data is evenly spaced.
[S]
(mM)
0
0.1
0.2
0.4
0.6
0.8
1
1.2
1.5
Control
Velocity
(nmol/min)
0.0
33.3
40.0
44.4
46.2
47.1
47.6
48.0
48.4
Altered
condition
Velocity
(nmol/min)
[S]
(mM)
0.0
2.4
4.5
8.3
11.5
14.3
16.7
18.8
21.4
[S]
(mM)
[S]/v
0.1
0.2
0.4
0.6
0.8
1
1.2
1.5
0.0420
0.0440
0.0480
0.0520
0.0560
0.0600
0.0640
0.0700
0.1
0.2
0.4
0.6
0.8
1
1.2
1.5
[S]/v
(control)
0.0030
0.0045
0.0090
0.0130
0.0170
0.0210
0.0250
0.0310
Michaelis-Menten
Velocity (nmol/min)
60.0
50.0
40.0
Control
30.0
Control
Vmax= 48.6 nmol/min
Km= 0.06 mM
20.0
10.0
0.0
0
0.5
1
1.5
Altered:
Vmax = 22 nmol/min
Km= 0.57 mM
2
[S] (mM)
Hanes Woolfe
0.0800
y = 0.02x + 0.04
R² = 1
0.0700
0.0600
[S]/v
0.0500
0.0400
(control)
0.0300
(altered)
y = 0.02x + 0.001
R² = 1
0.0200
0.0100
0.0000
0
0.5
1
[S]
1.5
2
Control:
Vmax= 50 nmol/min
Km= 0.05 mM
Altered:
Vmax= 50 nmol/min
Km= 2mM
a. Compare the SAP Kcat and Km with the Kcat and Km you found for standard ALP (E.coli)
performed in class. Whch enzyme, SAP or ALP, has a higher affinity for the substrate?
Which enzyme is more efficient? In each case, justify your answer.
Km for ALP= 0.13 mM (Hanes Woolfe)
Kcat for ALP= 1026.7 s-1 (refer to appendix)
Km for SAP= 0.05 mM
Kcat for SAP= 90.0 s-1 (refer to appendix)
Kcat/Km (ALP) = 7.9 *106 M-1s-1
Kcat/Km (SAP) = 1800 M-1s-1
SAP has a higher affinity because it has a lower Km.
ALP is more efficient because it has a higher value for Kcat/Km.
b.
Determine which of the five sets of conditions corresponds to your assigned data (data B).
A competitive inhibitor was added.
c. If your data set indicates an inhibitor was added, determine the type of inhibitor and
comment briefly on how the inhibitor might act.
A competitive inhibitor: the inhibitor binds to the same site on the enzyme as the substrate
and it competes with the substrate. This inhibitor resembles the substrate but it doesn’t
react like the substrate. It will eventually reach Vmax, but it will need more substrate to do
it.
d. Considering the control enzyme assay presented in the data set given, what would be the
absorbance at 420 nm of the assay at 1.5 min in the 1.5 mM assay?
Absorbance= 1.089 (working shown in appendix)
e.
What changes could you make to the enzyme assay described on the previous page to
distinguish between thermophilic (high temperature adapted), mesophilic (mid range
temperature preferring) and psychrophilic (cold adapted) alkaline phosphatases? Show
how this modified assay would distinguish the three classes of alkaline phosphatase.
In order to distinguish between thermophilic, mesophilic and psychrophilic the experiment
can be done using different temperatures like 10˚C, 30˚C and 50˚C for the enzyme assay.