Question

In: Chemistry

1. Carbonic anhydrase catalyzes the reverse reaction (of course!): HCO3 - + H+  CO2 +...

1. 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).

Expert Solution

Ans. 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

#I. 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.

#II. 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

#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

= 0.12 s^-1

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