Question

por favor mostrar el proceso

idroliza un sustrato heptapeptidico con un ca de 1000 s. del sustrato es 0.1 M, la vo es 5.31 x 10' Ms. blisis cuando la concentración de la proteasa es de 0.2 619. Una proteasa (M = 21,500) hidroliza un si Cuando la concentración inicial del sustrato e a) Calcule Vmax y de esta hidrólisis cuando la com mg/mL de esta proteasa. b) Calcule Km con los datos anteriores. c) En presencia de un inhibidor.se inhibición está ocurriendo? or, se encontró que Vmax = 9.3 x 10 M/s. ¿Qué tipo de

A protease (Mr = 21,500) hydrolyzes a heptapeptide substrate with a 1000 s kcat. -3 -1 When the initial substrate concentration is 0.1 M, the v0 is 5.31 x 10 Ms-1

a) Calculate vmax and this hydrolysis when the protease concentration is 0.2 mg / mL of this protease.

b) Calculate Km with the previous data.

c) In the presence of an inhibitor, it was found that vmax = 9.3 x 10-3 M / s. What kind of inhibition is happening?

Answer 1

Keat : 1000 57 [s= 0.1m V o = 5-31 X10-3 M57 [E]. = 0.2 mg ImL = 0.2 g/L * 21500 fot - 9.3 x10 6 M 21500 g/mol concentrahon enzyme (a Vmax = Keat [Elo = 1000 5 x9.3 X10 M 9.3 X10-3 MS-17 1. (b) Vo = Unax [5] km(s) 5.31 X205 AST = 9.3 x103 A5T 0.1 M 1. KM + 0.1M KM + 0.1M = 0.93 M 5.31 KM = 0.075 M km Vinax increases uno Hected (C) Typeof inhibition Competitive Non-competitive Un competitive reduced reduced Reduced unaffected As in presence and absence of inhibitor, Unax = 9.3x10-3 M51. The type of inhibition is competitive inhibition !