Question

Carbonic anhydrase catalyzes the reverse reaction (of course!): HCO3 - + H+  CO2 + H2O Initial velocities were measured at the following concentrations of HCO3 - (The enzyme concentration was 0.0025 mg/ml; the enzyme Mr = 29,100 Da.): [HCO3 -] mM 1.3 2.6 6.5 13.0 26.0

vo (µM/sec) 2.5 4.00 6.30 7.60 9.00

vo (µM/sec) with acetazolamide 1.17 2.10 4.00 5.70 7.20

a. Do the required calculations and graph the results as a Lineweaver-Burk plot. Use the X- and Y-axis scales provided or an excel graph; add labels and units to each axis. b. Calculate (i) KM (ii) Vmax and (iii) the enzyme turnover number (kcat).

c) Add a line to the graph corresponding to reactions with the acetzolamide inhibitor. Draw the line with the acetazolamide as either a (i) competitive inhibitor or (ii) noncompetitive inhibitor. What kind of affinity would azetazolamide have for ES or S?

Answer 1

Ans. #a. Determination of Vmax and Km using LB Plot”

Lineweaver-Burk plot gives an equation in from of y = mx + c

where, y = 1/ Vo, x = 1/ [S],

Intercept, c = 1/ Vmax ,

Slope, m = Km/ Vmax

#b. Enzyme kinetics without inhibitor [y = 0.3922x + 0.0987]:

Vmax = 1 / c = 1 / 0.0987 = 10.13

Hence, Vmax = 10.13uM/s

Now,

            Km= m x Vmax = 0.3922 x 10.13 = 3.97

            Hence, Km= 3.97 mM

Note: There is NO need of unit conversion in LB plot. Just calculate the numerical values of Vmax and Km and put the respective units as mentioned in the table.

#b’. Enzyme kinetics with inhibitor acetazolamide [y = 0.9804x + 0.1]:

Calculations made as above-

Vmax = 1/ 0.1 = 10.0 uM/s

Km= 10.0 x 0.9804 = 9.80 mM

#b. III. Turnover number:

Given, Molar mass of enzyme = 29100 Da = 29100 g/ mol

Given, [E] = 0.0025 mg/ mL

                        = (0.0025 x 10^-3) g/ mL

                        = (0.0025 x 10^-3 g/ 29100 g mol^-1) / mL               ; moles = Mass / MW]

                        = 8.5911 x 10^-8 mol/ mL                                          ; [1 mL = 0.001 L]

                        = 8.5911 x 10^-8 mol/ 0.001 L

                        = 8.5911 x 10^-5 mol/ L

                        = 8.5911 x 10^-5 M                                                     ; [1 M = 10⁶ uM]

                        = 85.91 uM

# Kcat (un-inhibited enzyme) = Vmax / [E]

                                    = 10.13 uM s^-1 / 85.91 uM

                                    = 0.122 s^-1

# Kcat (inhibited enzyme) = Vmax,app / [E]

                                    = 10.00 uM s^-1 / 85.91 uM

                                    = 0.116 s^-1

#c. See graph.

Type of Inhibitor: The Vmax remains almost the same but Km is increased in presence of inhibitor. It is the characteristic of a competitive inhibitor. Therefore, acetazolamide is a completive inhibitor.

# Acetazolamide’s Interaction with ES or S: A competitive inhibitor competes with the substrate for the active site. That is, acetazolamide interacts with free enzyme ([E]).

No inhibitor directly interacts with free substrate.