The turnover number for an enzyme is known to be 500 min-1 . a mutant of the enzyme was made that resulted in the destabilization of the es complex of 1 kcal/mole, and a destabilization of transition state (es‡ ) of 2.5 kcal/mol. calculate the kcat of the mutant enzyme (assuming a temperature of 27°c). kcat = (kbt/h).e-(δg‡/rt) t = temperature in kelvin kb = 3.297 ×10−24 cal/k h = 1.583 x 10-34 cal. s r = 2 cal/k. mol

Explanation:

turnover number for an enzyme = 500/min

destabilization of the ES complex = 1Kcal/mole

destabilization of transition state (ES‡ ) = 2.5 kcal/mol

kB = 3.297×10−24 cal/K

h = 1.583 x 10-34 cal.s  

R = 2 cal/K.mol

we Know that

kcat = (kBT/h).e-(ΔG‡/RT)

T = temperature in Kelvin

By considering the following reaction

                                         K1                K2        

                             E  +  S  ⇆   (ES‡ )     ⇒         E + P

                                         K-1  

                               Ф  = K2(ES‡ )

where

               ΔG‡ = -RTlnKeq‡

     K‡eq

     K(cat)= Kb T/h   x    K2x ES‡/ES Now by putting values we get

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