Question

The following data set was collected from an experiment conducted in the lab where a new enzyme is being characterized.

Use the Lineweaver-Burk method in order to determine the values of KM and Vmax. Complete your work on a separate piece of paper and upload the excel file or picture of your work. Your work must include the following:

1. Table of reciprocals

2. Reciprocal plot (straight-line graph)

3. Calculations and results for KM and Vmax

In order to get full credit for this exercise you must:

-Specify the KM and Vmax (Values with units)

Substrate Concentration (mM)	Velocity (mM. Sec^-1)
0.1	0.34
0.2	0.53
0.4	0.74
0.8	0.91
1.6	1.04

(recipricals of substrate concetration) x axis parameters: -0.3 - 0.5 mM*sec-1

(recipricals of velocity) y-axis paremeters: 0 - 50 mM

Answer 1

The reciprocal values of the substrate concentration and the velocity are mentioned below along with Km and Vmax.

Substrate concentration [S] (mM)	Velocity vo (mM.sec-)	1/[S] (1/mM)	1/vo (1/m.Msec-)
0.1	0.34	10	2.941176
0.2	0.53	5	1.886792
0.4	0.74	2.5	1.351351
0.8	0.91	1.25	1.098901
1.6	1.04	0.625	0.961538
1/Vmax	Vmax (mM.sec- )	1/Km	Km (mM)
0.8	1.25	4	0.25

| N + 1/10 (1/mM.sec.) 1/Km 1/Vmax 0 -6 -4 -2 0 6 8 10 12 0 2 4 1/[S] (1/mM)

Vmax is the maximum rate of reactio or the velocity an enzyme can achieve At maximum velocity, the enzyme is completely saturated with the substrate molecules and with any increase in the substrate concentration the rate of reaction will not increase.

KM is the substrate concentration at which the maximum velocity (VMAX) is half of the original value. It indicates the affinity of the enzyme towards the substrate, lower the KM of the enzyme, higher is its affinity towards the substrate and vice versa