1. Enzymes Pt 2:
Kinetics
Relicardo M. Coloso, Ph. D.
College of Medicine
Central Philippine University

2. Kinetics of enzyme action
Michaelis-Menten
model
An enzyme converts one chemical (the substrate), into another (the
product). A graph of product concentration vs. time follows three phases
as shown in the following graph.
At very early time points, the rate of product accumulation increases over time.
Special techniques are needed to study the early kinetics of enzyme
action, since this transient phase usually lasts less than a second (the figure
greatly exaggerates the first phase).

3. Enzyme velocity as a function of substrate concentration
If you measure enzyme velocity at many different concentrations
of substrate, the graph generally looks like this:
Enzyme velocity as a function of substrate concentration
often follows the Michaelis-Menten equation:
Where KM –Michaelis-Menten
constant
Vmax – maximum velocity of the
reaction

4. Vmax is the limiting velocity as substrate concentrations get very large.
Vmax (and V) are expressed in units of product formed per time. If you
know the molar concentration of enzyme, you can divide the observed
velocity by the concentration of enzyme sites in the assay, and express
Vmax as units of moles of product formed per second per mole of
enzyme sites. This is the turnover number, the number of molecules of
substrate converted to product by one enzyme site per second. In
defining enzyme concentration, distinguish the concentration of enzyme
molecules and concentration of enzyme sites (if the enzyme is a dimer
with two active sites, the molar concentration of sites is twice the molar
concentration of enzyme).
KM is expressed in units of concentration, usually in Molar units. KM is
the concentration of substrate that leads to half-maximal velocity. To
prove this, set [S] equal to KM in the equation above. Cancel terms and
you'll see that V=Vmax/2.

5. Note that KM is not a binding constant that measures the strength
of binding between the enzyme and substrate. Its value includes
the affinity of substrate for enzyme, but also the rate at which the
substrate bound to the enzyme is converted to product.The
Michaelis-Menten model is too simple for many purposes. The
Briggs-Haldane model has proven more useful:
Under the Briggs-Haldane model, the graph of enzyme velocity vs.
substrate looks the same as under the Michaelis-Menten model, but
KM is defined as a combination of all five of the rate constants in the
model.

6. Significance of KM of an enzyme
Example: Hexokinase – enzyme that phophorylates glucose
Glucose + ATP Glucose – 6-P + ADP + H+
Rates of phosphorylation of glucose and fructose in the brain
Properties of brain Sugar concn in Calculated rate
Sugar brain cell
hexokinase of phosphorylation
Vmax KM In vivo
Glucose 17 10-5 10-5 8.5
Fructose 25 10-3 10-6 10-2
Units: Vmax – micromol/min/g; KM – Molar; Sugar concn – Molar;
rate of phosphorylation – micromol/min/g
KM value tells us whether or not the enzyme is physiologically important.
It also gives information on the affinity of the enzyme for its substrate

7. Assumptions of enzyme kinetic analyses
Standard analyses of enzyme kinetics (the only kind discussed here) assume:
The production of product is linear with time during time interval is used
•
• •The concentration of substrate vastly exceeds the concentration of enzyme. This means
that the free concentration of substrate is very close to the concentration you added, and
that substrate concentration is constant throughout the assay.
• A single enzyme forms the product.
• There is negligible spontaneous creation of product without
enzyme
•
No cooperativity. Binding of substrate to one enzyme binding site doesn't influence
the affinity or activity of an adjacent site.
• Neither substrate nor product acts as an allosteric modulator to alter the
enzyme velocity.

8. Linear form of the Michaelis-Menten equation
Transform the curved data into a straight line, so they could be
analyzed with linear regression
One way to do this is with a Lineweaver-Burk plot.
Take the inverse of the Michaelis-Menten equation and simplify:
Ignoring experimental error, a plot of 1/V vs. 1/S will be linear, with a Y-intercept
of 1/Vmax and a slope equal to Km/Vmax. The X-intercept equals 1/Km.

9. Example of double reciprocal plot to solve for KM and Vmax

10. Enzymes can be affected by inhibitory compounds or inhibitors
•most clinical drug therapy is based on inhibiting the activity of enzymes,
•analysis of enzyme reactions using the tools described above has been
fundamental to the modern design of pharmaceuticals.
Enzyme inhibitors fall into two broad classes:
1) those causing irreversible inactivation of enzymes and
2) those whose inhibitory effects can be reversed.
Irreversible inhibitors
•These inhibitors usually cause an inactivating, covalent modification of enzyme
structure.
Examples: many poisons, such as cyanide, carbon monoxide and
polychlorinated biphenols (PCBs)
Cyanide is a classic example of an irreversible enzyme inhibitor: by covalently
binding mitochondrial cytochrome oxidase, it inhibits all the reactions associated
with electron transport.
• are usually considered to be poisons and are generally unsuitable for
therapeutic purposes.

11. Reversible inhibitors can be divided into two main categories;
competitive inhibitors and noncompetitive inhibitors, with a third
category, uncompetitive inhibitors, rarely encountered.
Binding Site on Enzyme Kinetic effect
Inhibitor Type
Specifically at the catalytic site, where it
Vmax is unchanged; Km, as defined
1) Competitive competes with substrate for binding in a
by [S] required for ½ maximal
Inhibitor dynamic equilibrium- like process. Inhibition
activity, is increased.
is reversible by substrate.
Binds E or ES complex other than at the
Km appears unaltered; Vmax is
2)Noncompetitive catalytic site. Substrate binding unaltered, but
decreased proportionately to
Inhibitor ESI complex cannot form products. Inhibition
inhibitor concentration.
cannot be reversed by substrate.
Binds only to ES complexes at locations other
than the catalytic site. Substrate binding Apparent Vmax decreased; Km, as
3) Uncompetitive
modifies enzyme structure, making inhibitor- defined by [S] required for ½
Inhibitor
binding site available. Inhibition cannot be maximal activity, is decreased.
reversed by substrate.
The hallmark of all the reversible inhibitors is that when the inhibitor concentration
drops, enzyme activity is regenerated. Usually these inhibitors bind to enzymes by
non-covalent forces and the inhibitor maintains a reversible equilibrium with the
enzyme.

12. Michaelis–Menten curves for enzyme with or without inhibitor

16. Example: Given the enzyme Q which A ---------> B
Converts substrate A to product B
Enz (Q)
Tube A Tube B Tube C Tube D
[S], or
4.8 mM 1.2 mM 0.6 mM 0.3 mM
Conc of A
1/[S] 0.21 0.83 1.67 3.33
Δ OD540
(Vi) or
0.081 0.048 0.035 0.020
Rate of
reaction
1/Vi 12.3 20.8 31.7 50.0
Making a Lineweaver-Burk plot of these results shows (red line in graph) that
1/Vmax = 10, so Vmax = 0.10
−1/Km = − 0.8, so Km = 1.25 mM
(In other words, when [S] is 1.25 mM, 1/Vi = 20, and Vi = 0.05 or one-half of Vmax.)

17. With
Non competitive
inhibitor
Competitive inhibitor
Lineweaver-Burk Plot

18. The table below summarizes the results with competitive inhibitor
Tube A Tube B Tube C Tube D
[S] 4.8 mM 1.2 mM 0.6 mM 0.3 mM
1/[S] 0.21 0.83 1.67 3.33
ΔOD540
0.060 0.032 0.019 0.011
(Vi)
1/Vi 16.7 31.3 52.6 90.9
The Lineweaver-Burk plot of these results is shown above (green line in graph).
1/Vmax = 10, so Vmax remains 0.10.
Now, however, −1/Km = − 0.4, so Km = 2.50 mM
(In other words, it now takes a substrate concentration [S] of 2.50 mM, to achieve
one-half of Vmax.)

19. With
Non competitive
inhibitor
Competitive inhibitor
Lineweaver-Burk Plot

20. The table below summarizes the results with non competitive inhibitor
Tube A Tube B Tube C Tube D
[S] 4.8 mM 1.2 mM 0.6 mM 0.3 mM
1/[S] 0.21 0.83 1.67 3.33
ΔOD540
0.040 0.024 0.016 0.010
(Vi)
1/Vi 25 41 62 102
The Lineweaver-Burk plot of these results is shown above( blue line in graph).
Now 1/Vmax = 20, so Vmax = 0.05.
But −1/Km = − 0.8, so Km = 1.25 mM as it was in the first experiment.
So once again it only takes a substrate concentration,[S], of 1.25 mM to achieve one-
half of Vmax.

21. With
Non competitive
inhibitor
Competitive inhibitor
Lineweaver-Burk Plot

22. Random bi bi mechanism
Ordered bi bi mechanism
Enzyme reaction Intersecting LB plots
mechanisms

23. ACP,acyl carrier protein;
HSL,homoserine lactone;
SAM,S-adenosylmethionine;
MTA,5′-methylthioadenosine
Ordered bi bi mechanism in
acyl homoserine lactone synthase

24. Adenylate kinase (myokinase) is a phosphotransferase enzyme that catalyzes
the interconversion of adenine nucleotides, and plays an important role in cellular
energy homeostasis. The reaction catalyzed is:
2 ADP ATP + AMP
The reaction is random
bi bi mechanism

25. Ping pong mechanism
Parallel LB plots

26. A friendly animated ping pong game

27. Ping pong mechanism for
uridylyltransferase

28. Thus,
1. Substrates may add in a random order
(either substrate may combine
first with the enzyme) or in a compulsory
order (substrate A must bind
before substrate B).
2. In ping-pong reactions, one or more
products are released from the
enzyme before all the substrates have
added.

29. Keep your
eye
on the
ball!