Question is from Biochemistry, pls explain. 8. An enzyme was obtained in Dr. Cai

Question

Question is from Biochemistry, pls explain.

8. An enzyme was obtained in Dr. Cai's lab. To study the kinetics of this enzyme, a graduate student measured the velocity of the reaction at a substrate concentration of 0.1 mm substrate. He found that after 1 mm competitive inhibitor was added, the initial rate decreased to 1/4 of the original rate. Km, the Michaelis constant, is 0.5 mm. What is the K, the dissociation constant for the inhibitor binding to the enzyme? (Hint: in the presence of the competitive inhibitor, A. 0.33 mM B. 0.45 mm C. 0.14 MM D. 0.28 mM E. 0.9 mm ymax (Si Vo = *)* 189

Explanation / Answer

when there is no inhibition, V= VmaxS(/KM+S) (1)

when there is competitive inhibition, Vmax remain the same and V becomes V/4. The equation for competitive inhibition is V= VmaxS/{(KM*(1+I/KI)+S}

V/4= VmaxS/(KM*(1+I/Ki)+S (2), I is the concentration of inhibitor.

taking the ratio Eq.2/Eq,1

KM*(1+I/KI)= 4KM+3S

= 4*0.5+3*0.1= 2.3

1+I/KI= 2.3/KM= 2.3/0.5= 4.6

I/Ki= 3.6

KI= I/KI= 1/3.6= 0.28mM (D is correct)

KI= I/1.3= 1/1,3 mM

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.