Question

The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?

Answer 1

use the data (1) From the given a data in the ecuation Vman [s Km+ [s] tere reaction velocity in the unit at NM (min M Now 6.25x10-6 15 nM/min - 15 x 10 g Vsec 1 2.5X10-10 M/sec So cauation 5 Uman can be simplified Км [s] Nman l + Vonax Inax 6.258106 25*10-10 1x 10-4 ['s 2, a= 60 nm/ min = N GOX torg M/sec 1x 10 9 m/see km 1 f 1810-9 Umor 1810-4 2) Now , we get. 1 1 KM Clou lo-g 6.25x10 104 2.5x10² => X 3x109 Unan * 1.5x 105 Iman km a KM 20000 vnan

Now put the value in canation we get. 1 + 20000 * ido 10-4 log 2 ) 8 x to 3 V man 109 - 2x 10 Imax Umar = 1.25x109 uman Now put the value of Vare in equation 3 we get 2oooo &max Yst M 20000 x 1.25 x 10 9 - 2.5x10-5 km = 2.5x 105 M and &max= 1.25x og m/sec = 7 5 nM /min equation [5] = 2.58 10 5 M Now putting the value of kar and Iman we get the value of we from Vonan [s the KMT [s] 1.25 X 10-9x2.5x10-5 2.5x105 +2.5x10-5 = 6.2 5x10-10 M/sec = 37.5 nm nor/min So, I would be 37.5 nM/min (Ans).