Question

The data to the below were collected for the bovine carbonic anhydrase-catalyzed reaction described by

CO2 (aq) + H2O (l) -->H+ (aq) + HCO3- (aq)

[CO2] (mmol * L^-1)	R (ݵmol * L^-1 * s^-1)
1.25	28.74
2.50	48.61
5.00	80.28
20.00	155.59

Use these values to determine the value of KM, the Michaelis-Menten constant. Answer the four parts below.

a) If you were to plot this data to graphically determine Rmax and KM using a Lineweaver-Burk plot, what would you plot for the following data? If you would plot the original data, enter the original value in the corresponding blank.

[CO2] (mmol * L^-1) graph as R (ݵmol * L^-1 * s^-1) graph as

1.25--> ________ 28.74 --> ________

20.00--> ________ 155.59 --> ________

b) Calculate the slope using the data points from part (a). Then determine the y-intercept using the second point.

Slope= _______ s y-intercept= ___________ L * s * mmol^-1

c) Use the y-intercept to calculate Rmax.

Rmax=___________ mmol * L^-1 * s^-1

d) Use Rmax and the slope to calculate Km.

Km=___________ mmol * L^-1

Answer 1

Ans. #a. Note that the question seeks rate of reaction and other variables in terms of mmol whereas the question gives the rate of reaction in terms of umol. We can take following two approaches-

            I. Convert the umol unit in given table to mmol, OR

            II. Plot the graph using original data table, and later convert umol to mmol as needed.

# For simplicity, we take the first approach so that we won’t be needing the unit conversion again and again for b, c and d.

# 1 mmol = 1000 umol

#b. The trendline equation for LB plot is y = 37.748x + 4.8788.

Slope, m = 37.748 mmol L^-1 (mmol^-1 L s) = 37.748 s

Intercept, c = 4.8788 mmol^-1 L s

#c. Rmax = Vmax = 0.20 mmol L^-1 s^-1 (see graph for calculation)

#d. Km = 7.74 mmol L^-1 (see graph for calculation)