Lecture 3: Factors affecting enzyme activity: [substrate] and inhibitors

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Lecture 3 Factors affecting enzyme activity
substrate and inhibitors
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How can we mathematically describe catalysis?
Lets start with a relatively simple scheme (1)
enzyme (E) binds substrate (S), forming an
enzyme-substrate complex (ES) (2) ES breaks
down to either release product (P) or substrate
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Reaction velocity is the rate of product formation

• The rate of product formation is determined by
• (1) Catalytic speed - How fast can enzyme convert
  substrate into product?
• (2) The availability of substrate - How much time
  does an enzyme molecule spend looking around for
  a substrate molecule?
• (3) Concentration of product - Rate of reverse
  reaction

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Now we can simplify the situation a bit

• (1) Lets just consider initial conditions,
  where there is essentially no product around
  (only E and S).This allows us to ignore the rate
  of the reverse reaction.
• (2) Also, lets think only about reactions in
  very high substrate concentrationswe will come
  back to consider lower S in a few slides.

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As very high S

1. The enzyme is operating at its maximal velocity
   (defined as Vmax).
2. This rate gives us the turnover number, which
   is the number of molecules of product formed per
   molecules of enzyme. Thus, Vmax is the turnover
   number times the number of enzyme molecules.
3. This Vmax is constant for a given enzyme

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Now lets consider low S

1. The enzyme is not saturated with substrate and
   must search for itthis takes time.
2. This makes the rate of the reaction slower than a
   saturating S.
3. What does a plot of the reaction rate as a
   function of S look like then???

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Rate as a function of S
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Km- a convenient term in kinetics
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Km- what is it?
Km is a substrate concentration (units must be
in concentration) Km is a property of every
enzyme moleculeit does NOT depend upon the
enzyme. Thus it is an intensive constant, in
contrast to Vmax. Km is inversely related to
the affinity of an enzyme for its substrate the
higher the affinity the lower the Km.
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Putting it together Vmax and Km
What would 0.5 units of enzyme look like?
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Time for a little bit of math
We would like to have an equation that
mathematically describes reaction velocity as a
function of S (the plots that we have been
looking at) Recall our basic reaction
scheme In this scheme, Km (k2 k3)/k1
We will not go through the complete derivation of
the equation, (and you will not be responsible
for deriving it in the future), but starting with
the definition of Km we can derive the equation
we want. This is the famous Michaelis-Menten
equation!
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Michaelis-Menten kinetics
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Lineweaver-Burke of the big M-M
Through some algebra that we will not go through,
the Michaelis-Menten equation can be rearranged
to a linear form (y mx b) called the
Lineweaver-Burke equation 1/v
(Km/Vmax)(1/S) 1/Vmax
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Enzyme inhibitors
Inhibitors are small molecules that bind to an
enzyme and reduce its catalytic ability. There
are two major classes of inhibitors Reversible
inhibitors can dissociate from the enzyme once
they are bound. Irreversible inhibitors can not
dissociate from the enzyme. We will first
consider reversible inhibitors for which there
are three major subclasses competitive,
non-competitive, and uncompetitive.
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Competitive enzyme inhibitors
Competitive inhibitors react only with the free
enzyme, often by binding to the active sitethus
the compete with substrate for binding.
We can write two equations that describes this
interaction
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Competitive inhibitor kinetics
The lineweaver-burke equation that describes
the kinetics in the presence of a competitive
inhibitor are altered by the term (1 I/KI) as
follows 1/v (1 I/KI)(Km/Vmax)(1/S)
1/Vmax
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Non-competitive enzyme inhibitors
Non-competitive inhibitors react with the free
enzyme and the enzyme substrate complex. Thus
they usually bind to a site on the enzyme surface
away from the active site.
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Non-competitive inhibitor kinetics
The lineweaver-burke equation that describes
the kinetics in the presence of a non-competitive
inhibitor are altered by the term (1 I/KI) as
follows 1/v (1 I/KI)(Km/Vmax)(1/S)
(1 I/KI)(1/Vmax)
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Uncompetitive enzyme inhibitors
Unompetitive inhibitors react only with the
substrate-bound form of the enzyme. We can
write two equations that describes this
interaction
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Uncompetitive inhibitor kinetics
The lineweaver-burke equation that describes
the kinetics in the presence of an uncompetitive
inhibitor are altered by the term (1 I/KI) as
follows 1/v (Km/Vmax)(1/S) (1
I/KI)1/Vmax)
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Summary of reversible inhibitor effects
Type of Inhibitor Reactive Form(s) Slope y
intercept of Enzyme No inhibitor Km/Vmax
1/Vmax Competitive Efree
(1I/KI) Km/Vmax 1/Vmax Non-competitive
Efree and ES (1I/KI) Km/Vmax (1I/KI)1
/Vmax Uncompetitive ES Km
/Vmax (1I/KI)1/Vmax
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Irreversible inhibitor effects
Irreversible inhibitors inactivate enzymes by
covalently binding them. The kinetics of an
irreversible inhibitor are quite easy to
interpret addition of inhibitor continually
lowers Vmax until all enzyme molecules have
reacted stoichiometrically with the inhibitor, at
which point there will be no active enzyme
molecules left. Some irreversible inhibitors,
called suicide substrates, look like the natural
substrate but covalently attach to the enzyme at
some point during binding and/or catalysishigh
concentrations of substrate can temporarily
protect the enzyme from such inhibitors.
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DFP - an example of an irreversible inhibitor
Diisopropyl-fluorophosphate (DFP), the active
agent in nerve gas, reacts with a serine at the
catalytic active site of a number of peptidases
and esterases, including acetyl choline esterase.