Question

If vmax = 100 mmol/l*sec, and v= 50 mmol/l*sec at [s] = 10 nm, what is km?

Answer 1

Answer:

Please read the following text. For more details see the attached file.

You have conducted the experiment with only two substrate concentrations. In order to get accurate values of Km and Vmax you should run the experiment with at least 4 or 5 subdtrate concentrations (in the attached file, you will find a figure example of 1/V vs. 1/[S] for estimating the values of Km and Vmax. The intercept of the line is 1/Vmax. So from the intercept you find Vmax. The slop of the line is Km/Vmax; by substituting the value you got for Vmax you can calculate the value of Km).

Determining KM and Vmax experimentally

To characterize an enzyme-catalyzed reaction KM and Vmax need to be determined. The way this is done experimentally is to measure the rate of catalysis (reaction velocity) for different substrate concentrations. In other words, determine V at different values of [S]. Then plotting 1/V vs. 1/[S] we should obtain a straight line described by equation (18). From the y-intercept and the slope, the values of KM and Vmax can be determined. For example, use EXCEL to plot the data shown below. Fit the data to a straight line, and from the equation of the straight line