The decomposition of hydrogen peroxide (substrate S) is catalyzed by the enzyme

Question

The decomposition of hydrogen peroxide (substrate S) is catalyzed by the enzyme catalase (E). At an initial catalase concentration [S_0] of 3.5 times 10^-9M the variation of 1/R_0 (the reciprocal of the initial rate) with 1/[s]_0. (the reciprocal of the initial concentration of the substrate) is given by (Lineweaver-Burk equation), (1/R_0) = 0.6977 s (1/[S]_0) + 26.557 s M based on this experimental result, which one of the following choices is correct? (left to right: Maximum rate R_max in M/s, Michaelis constant K_m in M and the turnover number k_2 in 1/s)

Explanation / Answer

compare 1 and 2 we get Km/Rmax= 0.6977s, 1/Rmax=26.55 s/M there fore Rmax=1/(26.55 s/M),

Rmax=0.0377 M/s substitue this value in Km/Rmax= 0.6977 gives Km=0.677*0.0377 M

Km=0.0263M then we know Rmax=K2*[So]

K2= Rmax/[So]

K2=0.0263/(3.5*10-9)

k2=1.08*1071/s

there fore option C is correct

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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