Question

10.What type of inhibitor is this? How do you know? (2)

11.For your assigned inhibitor 1, what are the apparent Km & Vmax? (NOTE: apparent Km& Vmax are just the Km & Vmax in presence of inhibitor, at a given concentration.) (2)

Kinetics experiments were performed on PGI. Enzyme activity (initial velocity, Vo) was measured at varying concentrations of Glucose-6-phosphate (G6P). The enzyme concentration used in all experiments was 1.5 μM.

[G6P) (mm) 0.01 0.02 0.06 0.1 V. (mm/s) + Inhibitor A 0.093 0.177 0.443 0.635

12.What will be the reaction rate with 0.500 mM [G6P], 4.0 μM PGI and your inhibitor 1 (assuming no change in inhibitor concentration)? (Use the apparent Km and Vmaxthat you calculated for inhibitor A.) (2)

Answer 1

For comparing inhibition velocity of reaction is not given without inhibitor so solving next question

Michaelis Menten equation is

v = Vmax / (1 + (Km/[S]))

Lineweaver burk plot is between 1/[S] on x-axis and 1/Vo on y-axis.

substrate(mm) velocity (mm/min) 1/ 1 0.010 0.093 100 0.020 0.177 50 0.060 0.443 16.66667 0.100 0.635 10 /VUM/min) 10.75268817 5.649717514 2.257336343 1.57480315 1/V(UM/min) y = 0.102x+0.5555 1 20 40 60 80 100 120

Equation of this graph is y = 0.102x + 0.5555

According to this equation, slope = 0.102 and intercept = 0.5555

Also, intercept = 1/Vmax = 0.5555

Vmax = 1.80 mM/min

Slope = Km/Vmax = 0.102

So, Km = 0.102 *Vmax = 0.102 *1.80 = 0.1836 mM

Answer : Km = 0.1836 mM

Vmax = 1.80 mM/min