Print Calculator sal Periodic Table 4 of 9 incorrect ral Chemistry 4th Edition U

Question

Print Calculator sal Periodic Table 4 of 9 incorrect ral Chemistry 4th Edition University Science Books presented by Sapling Learning Ma We define a quantity called the turnover number to be the maximum number of substrate molecules that ca be converted active sites molecules is given by into product molecules per unit time by an enzyme molecule. The concentration of enzyme is not necessarily equal to the concentration of enzyme molecules because some enzyme have more than one active site. If the enzyme molecule has one active site, the turnover number turnover number= ET-k2 (Rmax is often written as Vmax) If the enzyme molecule has more than one active site, then we multiply [Elt 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-249 ol. L-1, s-1 and [E]t = 2.26 nmol, L-1. Carbonic anhydrase has a single active site Number Tools x 102 Previous Give Up & View SolutionCheck Answer Next

Explanation / Answer

turn over number, Kcat= Rmax/ET

where Rmax= maximum rate and ET= enzyme concentration

given Rmax= 249 umol/L.sec, 1 umole= 10-6 moles, hence Rmax in temrs of moles= 249*10-6 moles/L.sec

ET= 2.26 nmoles/L, 1 nano moles= 10-9 moles, ET= 2.26*10-9 moles/L

KCat=turn over number= 249*10-6 moles/L/sec/ 2.26*10-9 moles/L=110177/sec

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