Question

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

Answer 1

`K_(cat) = 1.30* 10^(-5)`

Step-by-step 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‡ )

`K_(cat) =(K_(b) T)/(h) e^{(\Delta G)/(RT) }`

where

ΔG‡ = -RTlnKeq‡

`K_(cat) =(K_(b) T)/(h) e^{(-(-RTlnK_(eq)) )/(RT) }`

`K_(cat) =\frac{K_(b) T}`K‡eq

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

`K_(cat) = (3.297* 10^(-27)(Cal)/(K)*(273+27)K )/(1.583* 10^(-34)(Cal)/(K)) *(500)/(60) *(2.5(KCal)/(mol) )/(1(KCal)/(mol) )`

`K_(cat) = 1.30* 10^(-5)`