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In the lab, you measure the initial rate of an enzyme reactionas a function of...

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)

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Answer #1

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

Lineweaver-Burk plot 0.070 y 5E-06x0.01 0.060 0.050 0.040 - Linear (Without Inhibitor) 0.030 0.020 y 2E-06x 0.01 -Linear (Wit

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

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