An enzyme binds a substrate reversibly to form a complex

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Molecular machines, transport and bioenergetics · Molecular machines, transport and bioenergetics

Forces, Machines and Signals, Ch Enzymes and molecular motors

Problem

An enzyme

E

binds a substrate

S

reversibly to form a complex

ES

, which then converts irreversibly to product

P

and releases free enzyme:

E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \xrightarrow{k_2} E + P

Enzyme is present at a fixed total concentration

[E]_0=[E]+[ES]

, and substrate is in large enough excess that

[S]

can be treated as constant during the early, initial-rate period over which

[ES]

has reached a quasi-steady state,

d[ES]/dt\approx0

. ((a)) Write the rate equation for

[ES]

(formation minus loss), impose the steady-state condition, and eliminate

[E]

in favour of

[E]_0

[ES]

and

[S]

. Show that the initial rate

v=d[P]/dt=k_2[ES]

takes the Michaelis-Menten form

v=\dfrac{k_{\rm cat}[E]_0[S]}{K_M+[S]}

, and give

k_{\rm cat}

and

K_M

in terms of

k_1

k_{-1}

and

k_2

. ((b)) For an enzyme with

k_1=2.0\times10^7\,\mathrm{M^{-1}\,s^{-1}}

k_{-1}=300\,\mathrm{s^{-1}}

k_2=100\,\mathrm{s^{-1}}

and

[E]_0=1.0\times10^{-8}\,\mathrm{M}

, compute

K_M

and the initial rate

v

[S]=5.0\times10^{-5}\,\mathrm{M}

. ((c)) Compute the dissociation constant

K_d=k_{-1}/k_1

of the

ES

complex for this enzyme, compare it with

K_M

from part (b), state the condition on

k_2

relative to

k_{-1}

under which

K_M\to K_d

, and explain why that condition does not hold here.

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