Vmax represents the rate at which an enzyme is completely saturated with substrate. Vmax is often estimated because the value is an asymptote, which cannot be obtained under normal laboratory conditions. The Lineweaver-Burk method uses the reciprocal values of substrate concentration (1/[S]) and reaction velocity (1/V) to obtain a straight-line graph. The graph allows for the calculation of both the Vmax and Michaelis-Menten constant (Km) through linear regression of the data points. You will need data from the enzymatic assay, such as substrate concentration and the velocity of reaction. For example, the substrate concentrations are: 2.5, 5.0, 10.0, 15.0 and 20.0 (mM). The velocity values are 0.024, 0.036, 0.053, 0.060 and 0.061 (mM sec-1).

- Skill level:
- Moderately Challenging

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### Things you need

- Graphing calculator
- Graphing software

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## Instructions

- 1
Calculate the inverse of both sets of data. Divide one by each data point. For example, the inverse of the first substrate concentration value is 0.40mm, or 1/2.5 = The inverse of the substrate concentrations are 0.400, 0.200, 0.100, 0.067, and 0.50mm The inverse values of the initial velocities are 41.67, 27.78, 18.87, 16.67 and 15.63 (mm sec-1).

- 2
Use a graphing calculator or software to plot a graph with the inverse values. The 1/[s] values are plotted along the x-axis and 1/v (time values) are plotted along the y-axis.

- 3
Complete a linear regression for the data points. The linear regression yields the equation for the straight-line, y = mx + b. The m value equals the slope of the line. The b value equals the y-intercept. The equation is y = 75.46x + 11.791 based on the listed data points. The y-intercept represents the Vmax value for the enzymatic assay.

- 4
Calculate the inverse of the y-intercept by dividing one by the y-intercept (1/11.791 = 0.0847mm sec-1). Vmax is equal to the inverse of the y-intercept value.