Question

30.0 uM(microMolar) solution of enzyme catalyzed breakdown of 0.250M of substrate in 2 minutes. How many molecules of substrate were converted to product each second by one enzyme molecule?

Answer 1

Ans. For simplicity of calculations, lets the reaction volume be 1.0 liters

# Given, [E] = 30.0 uM = 3.0 x 10^-5 M

Moles of Enzyme = Molarity x Volume of reaction vessel

                                    = (3.0 x 10^-5 M) x 1.0 L

                                    = (3.0 x 10^-5 mol/ L) x 1.0 L

                                    = 3.0 x 10^-5 mol

Number of enzyme molecules = Moles of enzyme x Avogadro number

                                    = 3.0 x 10^-5 mol x (N_A atoms/ mol)

                                    = (3.0 x 10^-5)N_A molecules              ; [N_A = Avogadro number]

# Given, [substrate] catalyzed = 0.250 M

Moles of substrate catalyzed = Molarity x Volume of reaction vessel

                                    = 0.250 M x 1.0 L

                                    = 0.250 mol

Number of substrate molecules catalyzed = 0.250 mol x N_A = (0.250)N_A molecules

# Now,

Number of substrate molecules catalyzed per second =

            = (no. of substrate molecules catalyzed / no. of enzyme molecules) / time in s

                                    = [(0.250)N_A molecules / (3.0 x 10^-5)N_A molecules] / (2 x 60 s)

                                    = 69.44 s^-1

                                    = 69 s^-1 (nearest whole number)

Therefore, 1 enzyme molecule catalyzes (breaks down) 69 substrate molecules per second.