Question

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 V_max and K_m 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)

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

Step 3: Obtain the regression equations for both the plots (see image) which appears in the form of y = mx + c (or) y = (k_m/V_max).1/[S] + 1/V_max. The reciprocal of y-intercept gives V_max. Multiplying this V_max with slope m gives K_m

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 V_max and K_m in the absence of inhibitor?

V_max = 100 mM/min and K_m = 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; K_m value increased with increase in inhibitor concentration; V_max 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

K_i = K_m . [I] / (K_{m apparent} - K_m) (Here K_{m 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 (K_cat) is calculated as

K_cat = V_max /E_t (where E_t is the total enzyme concentration)

= 100/10^-6 = 10⁸ / min

Specificity constant is given as K_cat/K_m = 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