Question

1 Graphic depictions of enzyme kinetic data can and in evaluation of reversible enzyme inhibition. Please use the data below to determine type of inhibition for using two different concentrations (in blue and purple). Be sure to include graphical representation (Michaelis-Menton. I Lineweaver-Burk plots), type of inhibition, appropriate kinetic schemes and rate equations. The enzyme concentration at its initial state E, was kept constant. All data points are given in arbitrary units. Completely justify your answer. Use Excel or graphing paper provided (or you may use your own graph paper) (15 points) Extra Credit: Discuss in details how K, and Vmax are affected by inhibition How can changing the substrate concentration help to achieve Vmax? No Inhibition Inhibitor at concentration A 1/v Inhibitor at concentration B 1/ 15] 1/(S) v 1/ 1.0 1.0 12 0.083 7.8 0.128 43 0.233 0.50 2.0 20 0.050 13.7 8.0 0.073 0.125 4.0 29 0.25 24.0 21.8 0.034 0.046 0.071 0.125 35 8.0 0.029 30.3 21.0 0.033 0.048 40 12.0 0.083 0.038 26.0 0.029 34.5 0.025 Focus 1006 ENG IN 425 PM 7/30/2020 e to search o

Answer 1

Answer-

According to the given question-

Vmax - is defined as the maximum velocity or maximum rate of reaction of an enzyme. At the Vmax, the enzyme totally binds with the substrate molecules or saturated with the substrate molecules; and if we increase the concentration of substrate, then there will be no increase in the rate of reaction.

K_m -- is defined as the concentration of substrate at which the maximum rate of reaction or velocity i.e. Vmax will be half of the original value. the enzyme which has Lower Km value, they have a high affinity for the substrate. It is also known as Michaelis constant.

Here the values are given in the table, then we have calculated the reciprocal values of both the substrate concentration [S] as well as initial velocity Vo to calculate and prepare a Lineweaver Burk Plot. this plot is derived when we take reciprocal of Michelis Menten Equation.

Michelis Menten Equation. = Vo= Vmax. [S]/ Km +[S]

Then reciprocal of above equation-

1/Vo= (Km/Vmax).1/[S]+ 1/Vmax

When we compare the Lineweaver Burk equation with straight line equation we get-

y = mx + c

y = 1 / Vo

m = Km / Vmax

x = 1 / [S]

c =1 / Vmax

So, on y-axis: 1/Vo, and on the x-axis: 1/[S]

here the y-intercept= 1/Vmax

while the x-intercept= -1/Km

[S]	V (No inhibition)	V ( inhibitor at concentration A )	V ( inhibitor at concentration B )
0	0	0	0
1	12	7.8	4.3
2	20	13.7	8
4	29	21.8	14
8	35	30.3	21
12	40	34.5	26

45 40 35 30 25 20 v(no inhibition) -v (inhibitor at conc. A) -V (inhibitor at conc. B) 15 10 5 0 0 5 10 15 [S]

[S]	v (no inhibition)	v (inhibitor at conc. A)	V (inhibitor at conc. B)	1/[S]	1/v (no inhibitor)	1/v (Inhibitor at conc. A)	1/v (inhibitor at conc. B)
1	12	7.8	4.3	1	0.083333	0.128205	0.232558
2	20	13.7	8	0.5	0.05	0.072993	0.125
4	29	21.8	14	0.25	0.034483	0.045872	0.071429
8	35	30.3	21	0.125	0.028571	0.033003	0.047619
12	40	34.5	26	0.083333	0.025	0.028986	0.038462

0.25 0.2 0.15 1/v 0.1 -1/v(no inhibitor) 1/v(Inhibitor at conc. A) 1/v (inhibitor at conc. B) 0.05 -0.6 -0.2 0.2 0.4 0.6 0.8 1 1.2 -0.05 -0.1 1/[S]

• The inhibition observed is competitive inhibition. Competitive inhibition is the type of reversible inhibition, where the inhibitor competes with the substrate for the active site and prevents the substrate from binding at the active site, thus decreasing the affinity of the enzyme towards the substrate, increasing the Km.
• From the Lineweaver Burk Plots, it is clearly visible that the Km (x-intercept) of the enzyme is increasing with an increase in the concentration of the inhibitor, whereas, the Vmax (y-intercept) of the enzyme remains constant in absence as well as in presence of the inhibitor.
• When substrate concentration is increased, the higher substrate concentration will cause the binding of the substrate to the active site of the enzyme instead of the competitive inhibitor, thus it will reverse the inhibition.