In: Chemistry
In the lab, you measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an unknown inhibitor (0.05 mM). The following data are obtained when the total enzyme concentration is 1 x 10-6mM:
[S] mM | 0.0001 | 0.0002 | 0.0005 | 0.001 | 0.002 | 0.005 | 0.01 | 0.02 | 0.05 | 0.1 | 0.2 | |
No Inhibitor | V0 (mM/min) | 33 | 50 | 71 | 83 | 91 | 96 | 98 | 99 | 100 | 100 | 100 |
With Inhibitor | V0 (mM/min) | 17 | 29 | 50 | 67 | 80 | 91 | 95 | 98 | 99 | 100 | 100 |
A) What are Vmax and Km in the absence of inhibitor? (2 marks)
B) What kind of inhibitor is it likely to be and why? (1 mark)
C) Calculate the inhibition constant for this inhibitor with the enzyme. (3 marks)
D) Calculate the turnover number and specificity constant for this enzyme. (2 marks)
E) When [S] = 0.0004, what will V0 be in the i) absence ii) presence of inhibitor? (2 marks)
Step 1: Calculate the reciprocals of [S] and V0 to obtain 1/[S] and 1/V0
Given Data | ||
[S] mM | V0 without inhibitor | V0 with inhibitor |
0.0001 | 33 | 17 |
0.0002 | 50 | 29 |
0.0005 | 71 | 50 |
0.001 | 83 | 67 |
0.002 | 91 | 80 |
0.005 | 96 | 91 |
0.01 | 98 | 95 |
0.02 | 99 | 98 |
0.05 | 100 | 99 |
0.1 | 100 | 100 |
0.2 | 100 | 100 |
Calculate the reciprocals of [S] and V0 | ||
1/[S] | 1/V without inhibitor | 1/V with inhibitor |
10000 | 0.030 | 0.059 |
5000 | 0.020 | 0.034 |
2000 | 0.014 | 0.020 |
1000 | 0.012 | 0.015 |
500 | 0.011 | 0.013 |
200 | 0.010 | 0.011 |
100 | 0.010 | 0.011 |
50 | 0.010 | 0.010 |
20 | 0.010 | 0.010 |
10 | 0.010 | 0.010 |
5 | 0.010 | 0.010 |
Step 2: Plot a graph between 1/[S] and 1/V0 to obtain the Lineweaver-Burk plot
Step 3: Obtain the regression equations for both the plots (see image) which appears in the form of y = mx + c (or) y = (km/Vmax).1/[S] + 1/Vmax. The reciprocal of y-intercept gives Vmax. Multiplying this Vmax with slope m gives Km
Parameter | Without inhibitor | With inhibitor |
Regression equation | y = 2E-06x + 0.01 | y = 5E-06x + 0.01 |
1/Vmax | 0.01 | 0.01 |
Vmax | 100 | 100 |
Km/Vmax | 0.00 | 5.00E-06 |
Km | 2.00E-04 | 5.00E-04 |
This is Km | This is Km apparent |
Based on the above calculations, find the answers below for the given questions
A) What are Vmax and Km in the absence of inhibitor?
Vmax = 100 mM/min and Km = 2.00E-04 (or) 2 x 10-4 mM
B) What kind of inhibitor is it likely to be and why?
This is a competitive inhibition because the plots are converging on y-axis; Km value increased with increase in inhibitor concentration; Vmax is constant irrespective of the inhibitor concentration
C) Calculate the inhibition constant for this inhibitor with the enzyme.
The inhibition constant Ki is calculated using the formula
Ki = Km . [I] / (Km apparent - Km) (Here Km apparent can be found in Step 3)
= (2 x 10-4) . 0.05 / ((5 x 10-4) - (2 x 10-4)) = 13.33 mM
D) Calculate the turnover number and specificity constant for this enzyme.
The turnover number (Kcat) is calculated as
Kcat = Vmax /Et (where Et is the total enzyme concentration)
= 100/10-6 = 108 / min
Specificity constant is given as Kcat/Km = 108/(2 x 10-4) = 5 x 1011 min/mM
E) When [S] = 0.0004, what will V0 be in the i) absence ii) presence of inhibitor?
[S] = 0.0004 or 1/[S] = 2500. Calculate 1/V0 using the above regression equations, i.e.,
In the absence of inhibitor y = 2E-06x + 0.01 (Here y = 1/V0 and x= 1/[S])
So 1/V0 = [(2 x 10-6). 2500] + 0.01 = 0.015
V0 = 1/0.015 = 66.67 mM/min
Similarly,
In the presence of inhibitor y = 5E-06x + 0.01
So 1/V0 = [(5 x 10-6). 2500] + 0.01 = 0.0225
V0 = 1/0.0225 = 44.44 mM/min