Question

An enzyme that degrades pheromone is isolated from the sensory hairs of the silk moth, Antheraea polyphemus. The kinetics of the reaction are studied at pH 7.2 using a fixed enzyme concentration and temperatures of 10 C to 40 C. The following table lists results for reaction velocity as a function of substrate concentration. (a) Determine vmax and km at Pheromone Concentration the four reaction temperatures. (umol/L) Initial Reaction Velocity (umol/L.sec) T=10C T=20C T=30C T=40C (b) Determine the Arrhenius 3.00E 5.50E 5.00E 7.70E- parameters (E and A) for this 0.5 06 06 06 06 enzyme reaction. 4.20E 9.70E 8.90E 1.40E- 06 06 06 05 6.10E 1.10E 1.30E 1.80E- 1.5 06 05 05 05 7.10E 1.60E 1.50E 2.10E- 2.2 06 05 OS 05 1.10E 1.90E 2.00E 3.50E- 5.5 05 05 05 05 1.20E 2.20E 2.90E 3.90E- 11 05 05 05 05 1

Answer 1

intial reaction velocity T = 10 C y = ax + b 1/ Vo = Km/Vmax x 1/[S] + 1/Vmax 1.40E-05 y=8E-07x + 4E-06 R2 = 0.8222 1.20E-05 1.00E-05 8.00E-06 Pheromone intial reaction concentration velocity T= 10 C 0.5 3.00E-06 1 4.20E-06 1.5 6.10E-06 2.2 7.10E-06 5.5 1.10E-05 11 1.20E-05 intial reaction velocity intial reaction velocity T= 10 C 6.00E-06 4.00E-06 - Linear (intial reaction velocity T = 10 C) 2.00E-06 Here the obtained equation is y = 8E-07x + 4E-06 applying the equation, 1/Vmax = 4E-06 so, Vmax = 2.50E+05 Km/Vmax = 8E-07 Km = 8E-07 x 2.50E+05 = 2.00E-01 0.00E+00 0 15 10 substrate concentration intial reaction velocity T = 20C Pheromone intial reaction concentration velocity 3.00E-05 y = 1E-06x + 9E-06 RP=0.7708 2.50E-05 0.5 1 2.00E-05 T= 20C 5.50E-06 9.70E-06 1.10E-05 1.60E-05 1.90E-05 2.20E-05 1.5 Here the obtained equation is y = 1E-06x + 9E-06 applying th equation, 1/Vmax = 9E-06 Vmax = 1.11E+05 Km/Vmax = 1E-06 Km = 1E-06 x 1.11E+05 = 1.11E-01 intial reaction velocity 1.50E-05 -intial reaction velocity T= 20C 1.00E-05 2.2 5.5 11 - Linear (intial reaction velocity T = 200) 5.00E-06 0.00E+00 0 5 10 15 substrate concentration intial reaction velocity T = 30C 3.50E-05 3.00E-05 y = 2E-06x + 8E-06 R= 0.9175 2.50E-05 Pheromone intial reaction concentration velocity T = 30C 0.5 5.00E-06 1 8.90E-06 1.5 1.30E-05 2.2 1.50E-05 5.5 2.00E-05 11 2.90E-05 2.00E-05 intial reaction velocity Here the obtained equation is y = 2E-06x + 8E-06 applying th equation, 1/Vmax = 8E-06 Vmax = 1.25E+05 Km/Vmax = 2E-06 Km = 2E-06 x 1.25E+05 = 2.50E-01 1.50E-05 -intial reaction velocity T= 30C — Linear (intial reaction velocity T = 30C) 1.00E-05 5.00E-06 0.00E+00 0 5 10 15 substrate concentration intial reaction velocity T = 40C y = 3E-06x + 1E-05 R= 0.8438 Pheromone intial reaction concentration velocity T = 40C 0.5 7.70E-06 1 1.40E-05 1.5 1.80E-05 2.2 2.10E-05 5.5 3.50E-05 11 3.90E-05 ntial reaction velocity 5.00E-05 4.50E-05 4.00E-05 3.50E-05 3.00E-05 2.50E-05 2.00E-05 1.50E-05 1.00E-05 5.00E-06 0.00E+00 the obtained equation is y = 3E-06x + 1E-05 1/Vmax = 1E-05 Vmax = 1.00E+05 Km/Vmax = 3E-06 Km = 3E-06 x Vmax = 3.00E-01 intial reaction velocity T= 40C Linear (intial reaction velocity T = 40C) 0 15 5 10 substrate concentration In K y=-1748.4x + 4.2759 R= 0.3477 0 3.15E-03.20E-03.25E-03.30E-03.35E-03.40E-03.45E-03.50E-03.55E-03.60E-03 -0.5 Temperat Km values Temp(kelvin) 1.00E+01 2.00E-01 2.83E+02 2.00E+01 1.11E-01 2.93E+02 3.00E+01 2.50E-01 3.03E+02 4.00E+01 3.00E-01 3.13E+02 1/T In K 3.53E-03 -1.60944 3.41E-03 -2.19823 3.30E-03 -1.38629 3.19E-03 -1.20397 -1 • Ink - Linear (In K) -1.5 *Temperature in k is obtained by adding 273 to the value of Temperature in celsius -2 -2.5 1/T *R=8.314 (a constant) In K = -Ea/Rx 1/T + In A y = ax + b -Ea/R = -1748.4 -Ea / 8.314 = -1748.4 Ea = 1748.4 x 8.314 = 14536.2 J/mol In A = 4.2759 A = EA 4.2759 = 71.944