Question

The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = - -=k2 (Rmax is often written as Vmax) Et If the enzyme molecule has more than one active site, then [E], is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that Rmax for carbonic anhydrase equals 249 umol · L-1.5-1 and (E]. = 2.21 nmol · L-1. Carbonic anhydrase has a single active site. turnover number =

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Answer 1

turnover number = Vmax [Elt Vmax = maximum velocity [E] = Total engyme concentration Vmax= 249 umolt's [Ex] = 2.21 nmol i 14 mol = 10nmol 249 umol = 249 Xlo nmol = 112.78103 st Turn over number = 2490103 nay5 2.21 oombly Turnover number = 113x10 s Answer 2 Vmer = 222 umol List [E) = 3.7 pmole lu=100p 222u= 222x10°p Turn over number= numbes - 6x167 st 2227 222 X10 pment - 37P MOTE Turnover number = 6x10 st Answes.