contestada

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

Respuesta :

usamakld99

Answer:

[tex]K_{cat} = 1.30\times 10^{-5}[/tex]

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‡ )

         [tex]K_{cat} =\frac{K_{b} T}{h} e^{\frac{\Delta G}{RT} }[/tex]

      

where

               ΔG‡ = -RTlnKeq‡

     [tex]K_{cat} =\frac{K_{b} T}{h} e^{\frac{-(-RTlnK_{eq}) }{RT} }[/tex]

     [tex]K_{cat} =\frac{K_{b} T}[/tex]K‡eq

     

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

     [tex]K_{cat} = \frac{3.297\times 10^{-27}\frac{Cal}{K}\times(273+27)K }{1.583\times 10^{-34}\frac{Cal}{K}} \times\frac{500}{60} \times\frac{2.5\frac{KCal}{mol} }{1\frac{KCal}{mol} }[/tex]

    [tex]K_{cat} = 1.30\times 10^{-5}[/tex]

     

Otras preguntas