The protein catalase catalyzes the reaction described by 2H202(aq) --> 2H2O(l) +

Question

The protein catalase catalyzes the reaction described by

2H202(aq) --> 2H2O(l) + O2(g)

and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s–1. The total enzyme concentration is 0.023 ?M and the initial substrate concentration is 7.27 ?M. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme.

calculate the Rmax (Vmax) for this enzyme.

calculate the initial rate (R) of this reaction.

Map eneral Chemist Donald McQuarrie Peter A. Rock Ethan Galloghy University Science Books presented by Sapling Learning The protein catalase catalyzes the reaction described by and has a Michaelis-Menten constant of KM 25 mM and a turnover number of 4.0 x 10 s-1. The total enzyme concentration is 0.023 M and the initial substrate concentration is 7.27 M. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Number mM. s Calculate the initial rate, R (often written as Vo), of this reaction. Number mM-s

Explanation / Answer

1) as we know that

Rmax = Vmax = k2 [E0] where , k2 is the turn over number , V max is maximum rate ,

E0 = concentratio of enzyme

Rmax = 4.0 × 107 s–1. * 0.023 * 10-6 M

= 0.092 * 10= 0.92 Ms-1 = 920 mM s-1

as

2) r = k2 [E0] [S] / Km + [S] where ,   [E0] = total enzyme concentration ,

[S] =initial substarte concentartion Km = Michaelis constant

r = 4.0 × 107 s–1 * 0.023 * 10-6 M * 7.27 * 10-6M / 25 * 10-3M+ 7.27 * 10-6M

= 0.668 * 10-5 M2 s-1 / 25 * 10-3 M ( we neglect  7.27 * 10-6M in comparision to 25 * 10-3M)

= 0.026 * 10-2 M   s-1 = 26 mM  s-1

conversion factor

as 1milli M = 10-3M

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