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

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

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

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

Now we can simplify the situation a bit… (1) Let’s 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, let’s think only about reactions in very high substrate concentrations…we will come back to consider lower [S] in a few slides.

As very high [S]… • The enzyme is operating at its maximal velocity (defined as Vmax). • 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. • This Vmax is constant for a given [enzyme]

Now let’s consider low [S] • The enzyme is not saturated with substrate and must search for it…this takes time. • This makes the rate of the reaction slower than a saturating [S]. • What does a plot of the reaction rate as a function of [S] look like then???

Rate as a function of [S]

Km is the [S] at which the reaction proceeds at 1/2 Vmax Km- a convenient term in kinetics

Km- what is it? • Km is a substrate concentration (units must be in concentration) • Km is a property of every enzyme molecule…it 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.

Putting it together: Vmax and Km What would 0.5 units of enzyme look like?

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!

The big equation is: v = Vmax[S](Km + [S]) One problem: it is hard to work with non-linear plots Michaelis-Menten kinetics

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

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.

Competitive enzyme inhibitors • Competitive inhibitors react only with the free enzyme, often by binding to the active site…thus the “compete” with substrate for binding. • We can write two equations that describes this interaction:

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

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.

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)

Uncompetitive enzyme inhibitors • Unompetitive inhibitors react only with the substrate-bound form of the enzyme. • We can write two equations that describes this interaction:

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)

Summary of reversible inhibitor effects Type of Inhibitor Reactive Form(s) Slope y intercept of Enzyme No inhibitor Km/Vmax 1/Vmax Competitive Efree (1+[I/KI]) Km/Vmax 1/Vmax Non-competitive Efree and ES (1+[I/KI]) Km/Vmax (1+[I/KI])1/Vmax Uncompetitive ES Km /Vmax (1+[I/KI])1/Vmax

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 catalysis…high concentrations of substrate can temporarily protect the enzyme from such inhibitors.

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.

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