10. An enzyme catalyzing the reaction shown was mixed with 4mM substrate) (compound A). The initial rate of product formation was 25% of E+AEAE+P Vmax. The Km for the enzyme is which one of the following? (A) 2mM (B) 4mM (C) 9mM (D) 12mM (E) 25mM The answer is D. The Michaelis-Menten equation is v=(Vmax[S])/(Km+[S]). In this case, the velocity (v) is,Vmax, and [S]=4mM. Thus, the Michaelis-Menten equation becomes . Vmax=(Vmax4)/(Km+4). When one solves that equation for Km,Km=12mM. 11. The liver enzyme glucokinase catalyzes the phosphorylation of glucose to glucose 6-phosphate. The value of Km for glucose is about 7mM. Blood glucose is 5mM under fasting conditions, and can rise in the liver to 20mM after a high-carbohydrate meal. Therefore, if a person who is fasting eats a high-carbohydrate meal, the velocity of the glucokinase reaction will change in which one of the following ways? (A) Remain at 50%Vmax (D) Decrease from >50%Vmax to