Enzyme kinetics

Topics: Lineweaver–Burk plot, Michaelis–Menten kinetics, Reaction rate Pages: 13 (1171 words) Published: October 2, 2013
BIOCHEMISTRY 304
Enzyme Kinetic Sample Problems #1
September 2004

1Given the reaction
k1 kp
E + S  ES  E + P
k-1

where k1 = 1 x 107 M-1 sec-1 k-1 = 1 x 102 sec-1, and kp = 3 x 102 sec-1

a) Calculate Ks
b) Calculate Km

(a)k-1 1 x 102 sec-1
Ks = k1 = 1 x 107 M-1 sec-1 = 1 x 10-5 M

(b)k-1 + kp (1 x 102 sec-1) + (3 x 102 sec-1) Km = k1 = 1 x 107 M-1 sec-1 = 4 x 10-5 M

2) An enzyme was assayed at an initial substrate concentration of 2 x 10-5 M. In 6 minutes half of the substrate had been consumed. Km for this enzyme’s substrate is 5 x 10 -3 M.

a) Calculate k
b) Calculate Vmax
c) Calculate the concentration of product produced in 15 minutes.

SOLUTIONS

One can by inspection come to the conclusion that this reaction is first-order. This can be seen as follows:
[S] [2 x 10-5]
Is Km < 0.01 ? [5 x 10 -3] = .004 Yes, first order

a)Since this is a first order reaction we can use the following equation to determine the rate constant, k.

0.693 0.6930.693
k = t1/2 k = 6 min k = 6 min k = 0.115 min-1

b) k = Vmax / Km

Vmax = k x Km

Vmax = 0.115 min-1 x 5 x 10 -3 M = 0.575 x 10 -3 M min-1

Vmax = 0.575 x 10 -3 mole liter-1 min-1 = 575 moles liter-1 min-1

c)We know the initial concentration, [S] o, and we have calculated the rate constant, k. We need to determine an arbitrary [S] at 15 minutes after the initial concentration for a first order reaction.

[S] o
2.303 log [S] =kt

[2 x 10-5]
2.303 log [S] = 0.115 min-1 x 15 min.

2.303 log [2 x 10-5] - 2.303 log [S] = 1.725

2.303 log [S] = 2.303 log [2 x 10-5] -1.725

2.303 log [S] = (2.303 (-4.699)) -1.725

2.303 log [S] = (-10.82) -1.725

log [S] = ((-10.82) -1.725) / 2.303

log [S] = (-12.547) / 2.303

log [S] = - 5.47

[S] = 3.56 x 10 -6 M at 15 minutes of reaction

We know that So = 2 x 10-5 M, the initial concentration
[P] = [S]o – [S] = 2 x 10-5 M - 3.56 x 10 -6 M = 1.644 x 10-5 M

3.
An enzyme catalyzes the reaction S  P. The following data has been obtained. Plot the data to determine Km and V max using both a Lineweaver-Burke plot and also an Eadie-Hofstee plot.
[S]
 
 
v
M
 
 
lit-1 min-1
8.33E-06

1.38E-08
1.00E-05

1.60E-08
1.25E-05

1.90E-08
1.67E-05

2.36E-08
2.00E-05

2.67E-08
2.50E-05

3.08E-08
3.33E-05

3.63E-08
4.00E-05

4.00E-08
5.00E-05

4.44E-08
6.00E-05

4.80E-08
8.00E-05

5.34E-08
1.00E-04

5.71E-08
2.00E-04

6.67E-08

SOLUTIONS

Manipulate the given data on the left into forms that can be plotted:

Sample Kinetics Problems 2004

Data for Problem #

ORIGINAL DATA

DATA MANIPULATIONS

[S]
 
 
v
 
 
1/S
1/v
v/[S]
[S]/v
M
 
 
lit-1 min-1
 
M-1
M -1 min
min-1
min
8.33E-06

1.38E-08

1.20E+05
7.25E+07
1.66E-03
6.04E+02
1.00E-05

1.60E-08

1.00E+05
6.25E+07
1.60E-03
6.25E+02
1.25E-05

1.90E-08

8.00E+04
5.26E+07
1.52E-03
6.58E+02
1.67E-05

2.36E-08

5.99E+04
4.24E+07
1.41E-03
7.08E+02
2.00E-05

2.67E-08

5.00E+04
3.75E+07
1.34E-03
7.49E+02
2.50E-05

3.08E-08

4.00E+04
3.25E+07
1.23E-03
8.12E+02
3.33E-05

3.63E-08

3.00E+04
2.75E+07
1.09E-03
9.17E+02
4.00E-05...
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