Question

1.   The following data were obtained from 3 separate enzyme kinetic experiments using 3 different substrates S1, S2 and S3 forming products P1, P2 and P3 respectively. The amount of enzyme in each reaction is 1 µM. Find out the rate and graph the data using a Michaelis-Menton and Lineweaver-Burk plots and determine the values for Km, Vmax, Kcat, and Kcat/Km. Which among the 3 substrate is best substrate for this enzyme and why? (4 points)

[S] (mM)   [P1] (mM) in 60 min   [P2] (mM) in 240 min   [P3] (mM) in 30 min
2.5   0.170   0.126   0.194
2   0.201   0.135   0.213
1.5   0.167   0.129   0.206
1   0.162   0.131   0.203
0.6   0.159   0.135   0.174
0.3   0.148   0.131   0.163
0.15   0.122   0.111   0.122
0.2   0.095   0.117   0.114
0.1   0.075   0.072   0.104
0.05   0.049   0.040   0.071

2.   Use the Michaelis-Menton Equation to calculate the missing values of [S] given below if Vmax = 5 mmol/min. Plot [S] versus V (NOT the reciprocals!). Draw line parallel to the x-axis at Vmax and extend your plotted line to show its approach to Vmax. (2 points)

[S] (mM)   V0 (mmol/min)
10   1.2
[S]1       1.7
[S]2       2.1
[S]3       2.2
[S]4       2.5

3.   Plot the below data and determine the type of inhibition of an enzymatic reaction by inspecting the graph (give an explanation) (2 points)

[S] (mM)   V0 (mM/min)   V0 with Inhibitor present (mM/min)
1   1.3   0.8
2   2.0   1.2
4   2.8   1.7
8   3.6   2.2
12   4.0   2.4

4.   Plot the below data with and without inhibitor (I) and determine the type of inhibition of an enzymatic reaction by inspecting the graph (2 points)

[S] µM   V0 (µmol /min); [I] = 0 nM    V0 (µmol /min); [I] = 25 nM    V0 (µmol /min); [I] = 50 nM
0.4   0.22   0.21   0.20
0.67   0.29   0.26   0.24
1.00   0.32   0.30   0.28
2.00   0.40   0.36   0.32

Answer 1

Answer 3 It is an example of noncompetitive inhibition as K_M is same but V_max is different.

From the graph, Vmax of inhibited enzyme is 2.4 mM/min and of uninhibited enzyme is 4 mM/min. Intercept on x-axis for both the enymes give the K_M value of 2 mM.

Answer 4

This is an example of Uncompetitive inhibition as V_max in presence of inhibitor is smaller then V_max of uninhibited reaction. Also, K_M has decreased.

V = Vmax [S] / K_M + [S]

• Calculation of K_M of uninhibited reaction -

0.32 = 0.40*1 / K_M + 1

K_M + 1 = 0.40 / 0.32

?K_M = 1.25 - 1

?K_M = 0.25 M

• Calculation of K_M of inhibited reaction -

0.3 = 0.36*1 / K_M + 1

K_M + 1 = 0.36 / 0.3

K_M = 1.2 - 1

K_M = 0.2 M