Acetaldehyde + NADH + H+ → Ethanol + NAD+ How is the inhibition constant for this gotten? Determine the inhibition constant(s) of AMP?

Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
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Chapter1: Biochemistry: An Evolving Science
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Acetaldehyde + NADH + H+ → Ethanol + NAD+

How is the inhibition constant for this gotten? Determine the inhibition constant(s) of AMP?

[AMP]=0
[AMP]=1 mM [AMP]=2 mM [AMP]=3 mM [AMP]=4 mM [AMP]=5 mM
[CH;CHO]
Vo
Vo
Vo
Vo
Vo
Vo
1.00
0.096
0.049
0.033
0.024
0.020
0.016
1.50
0.128
0.065
0.043
0.033
0.026
0.022
2.50
0.175
0.089
0.059
0.045
0.036
0.030
5.00
0.238
0.122
0.082
0.061
0.049
0.041
50
y = 47.5837x + 14.4554
R? = 0.9989
[AMP] = 5 mM
y = 37.2228x + 13.0690
R? = 0.9991
[AMP] = 5 mM
[AMP] = 4 mM
40
y = 31.5867x + 9.7472
R? = 0.9986
[AMP] = 3 mM
[AMP] = 4 mM
y = 22.6680x + 7.8306
R? = 0.9991
[AMP] = 2 mM
30
[AMP] = 3 mM
y = 15.2885x + 5.1429
[AMP] = 1 mM
R? = 1.0000
20
[AMP] = 2 mM
y = 7.7837x + 2.6255
R? = 1.0000
[AMP] = 0 mM
[AMP] = 1 mM
10
(AMP] = 0 mM
-0.500
-0.300
-0.100
0.100
0.300
0.500
0.700
1/[S]
Transcribed Image Text:[AMP]=0 [AMP]=1 mM [AMP]=2 mM [AMP]=3 mM [AMP]=4 mM [AMP]=5 mM [CH;CHO] Vo Vo Vo Vo Vo Vo 1.00 0.096 0.049 0.033 0.024 0.020 0.016 1.50 0.128 0.065 0.043 0.033 0.026 0.022 2.50 0.175 0.089 0.059 0.045 0.036 0.030 5.00 0.238 0.122 0.082 0.061 0.049 0.041 50 y = 47.5837x + 14.4554 R? = 0.9989 [AMP] = 5 mM y = 37.2228x + 13.0690 R? = 0.9991 [AMP] = 5 mM [AMP] = 4 mM 40 y = 31.5867x + 9.7472 R? = 0.9986 [AMP] = 3 mM [AMP] = 4 mM y = 22.6680x + 7.8306 R? = 0.9991 [AMP] = 2 mM 30 [AMP] = 3 mM y = 15.2885x + 5.1429 [AMP] = 1 mM R? = 1.0000 20 [AMP] = 2 mM y = 7.7837x + 2.6255 R? = 1.0000 [AMP] = 0 mM [AMP] = 1 mM 10 (AMP] = 0 mM -0.500 -0.300 -0.100 0.100 0.300 0.500 0.700 1/[S]
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