Explore BrainMass

Explore BrainMass

    Enzymatic rate laws

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    We are given:
    k1 [ES]= [E][S] / Km
    E+S---->ES
    <---- with Km=(k1+K3)/K1
    k1'

    k2 material balance gives:
    E+I------>EI
    <------ [Eo] = [E][S]/Km + [E][I]Ki
    k2'
    where Ki=[EI]/ [E][I] is the equilibrium constant
    k3 for step 2 on the left.

    ES--------->E+P

    Use this to show that the initial reaction rate is given by:

    k3[E]o[S]
    v=d[P]/dt=----------------------------- approximately=to k3[E]o[S]o/Km+[S]o
    Km+[S]+KmKi[I]

    © BrainMass Inc. brainmass.com March 4, 2021, 6:22 pm ad1c9bdddf
    https://brainmass.com/chemistry/enzymatic-rate-laws-42200

    Solution Preview

    The first thing to realize is what we are looking for...

    d[P]/dt = k3[ES] "from last reaction"

    substitute in for [ES]
    d[P]/dt = k3*([E][S]/km) "from first relationship [ES]= [E][S] / Km"

    Notice that [E] is not in the intital reaction rate wanted. Thus we need to get rid ...

    Solution Summary

    This solution is provided in 335 words. It uses step-by-step problem to simplify the expression.

    $2.49

    ADVERTISEMENT