Question

Suppose that you have isolated the enzyme sucrase (able to hydrolyze sucrose into glucose and fructose), and you wish to determine the nature of inhibitor A for this enzyme. You have prepared five different concentrations of substrate (sucrose), and five different concentrations of inhibitor A (plus the control, with zero mM of inhibitor A). The following Table lists the inhibitor A concentrations [I], substrate concentrations [S], and resulting enzyme velocities (Vo) for all six of these experiments:

[I]	[S]	Vo	1/[S]	1/ Vo
0 mM	0.1 mM	0.333333333333 mM per minute
0 mM	0.2 mM	0.50
0 mM	0.3 mM	0.60
0 mM	0.4 mM	0.666666666667
0 mM	0.5 mM	0.714285714286
0.1 mM	0.1 mM	0.20
0.1 mM	0.2 mM	0.333333333333
0.1 mM	0.3 mM	0.428571428571
0.1 mM	0.4 mM	0.50
0.1 mM	0.5 mM	0.555555555556
0.20 mM	0.1 mM	0.142857142857
0.20 mM	0.2 mM	0.25
0.20 mM	0.3 mM	0.333333333333
0.20 mM	0.4 mM	0.40
0.20 mM	0.5 mM	0.454545454545
0.3 mM	0.1 mM	0.111111111111
0.3 mM	0.2 mM	0.20
0.3 mM	0.3 mM	0.272727272727
0.3 mM	0.4 mM	0.333333333333
0.3 mM	0.5 mM	0.384615384615
0.40 mM	0.1 mM	0.090909090909
0.40 mM	0.2 mM	0.166666666667
0.40 mM	0.3 mM	0.230769230769
0.40 mM	0.4 mM	0.285714285714
0.40 mM	0.5 mM	0.333333333333
0.5 mM	0.1 mM	0.076923076923
0.5 mM	0.2 mM	0.142857142857
0.5 mM	0.3 mM	0.20
0.5 mM	0.4 mM	0.25
0.5 mM	0.5 mM	0.294117647059

Construct a Michaelis-Menten plot, and a Lineweaver-Burk plot, for all six of these experiments on the same graph (for each plot). Calculate the Vmax, the Km, and the slope (Vmax/Km ) for the control (with [I] = 0 mM) and for each non-zero concentration of inhibitor A ([I] = 0.1, 0.2, 0.3, 0.4, and 0.5 mM). Which type of reversible enzyme inhibition is illustrated by inhibitor A?

Since many of you do not have scanners or cameras to send me an e-mail image of your graphs, the grading of this assignment will be based entirely on your calculations. I will be looking for your six Vmax calculations (6 points), your six Km calculations (6 points), your six Vmax/Km or slope of the line calculations (6 points), and the type of reversible enzyme inhibition that you identify for inhibitor A, along with reasons to justify your identification (2 points).

Answer 1

The question pertains to enzyme activity and the influence of an inhibitor on the activity of this enzyme.

In order to determine the V_max and K_M for the enzyme and the effect of the inhibitor on these parameters, different concentrations of substrates were added to different concentrations of inhibitors and the initial velocity or V0 was calculated. The results are presented in the question. They can be represented as:

[S]	Vo_0 mM	Vo_0.1 mM	Vo_0.2 mM	Vo_0.3 mM	Vo_0.4 mM	Vo_0.5 mM
0.1	0.3333333333	0.2	0.1428571429	0.1111111111	0.09090909091	0.07692307692
0.2	0.5	0.3333333333	0.25	0.2	0.1666666667	0.1428571429
0.3	0.6	0.4285714286	0.3333333333	0.2727272727	0.2307692308	0.2
0.4	0.6666666667	0.5	0.4	0.3333333333	0.2857142857	0.25
0.5	0.7142857143	0.5555555556	0.4545454545	0.3846153846	0.3333333333	0.2941176471

The Michaelis-Menten equation relates the parameters v_0, v_max, [S] and K_M by the formula:

In order to get a Michaelis-Menten plot for each of the different inhibitor concentrations, we can simply plot the above table with [S] on the X-axis and V_0 on the y-axis. Doing so, we get:

The Line-Weaver Burk equation is a modified version of the Michaelis-Menten equation in a linear format. The equation for the same is:

Hence, as can be seen, this is equivalent to a linear equation of y = mx + c, where,

Calculating the ratios, we get:

1 / [S]	1/ Vo_0	1/ Vo_0.1	1/ Vo_0.2	1/ Vo_0.3	1/ Vo_0.4	1/ Vo_0.5
10	3	5	7	9	11	13
5	2	3	4	5	6	7
3.3333	1.6667	2.3333	3	3.6667	4.3333	5
2.5	1.5	2	2.5	3	3.5	4
2	1.4	1.8	2.2	2.6	3	3.4

Plotting these plots and getting a trendline, we get:

The equations on the right point to the Line-weaver Burk equation, comparing to the y = mx + c, as mentioned above, we get:

Inhibitor Concentration	m = K_M/v_max	c = 1/v_max	v_max	K_M
0 mM	0.2	1	1	0.2
0.1 mM	0.4	1	1	0.4
0.2 mM	0.6	1	1	0.6
0.3 mM	0.8	1	1	0.8
0.4 mM	1	1	1	1
0.5 mM	1.2	1	1	1.2

v_max is calculated as the reciprocal of c and K_M is calculated by multiplying v_max with m.

As can be seen, the K_M increases with an increase in inhibitor concentration. However, v_max is not altered. This is an instance of competitive inhibition.