7.29.10 enzymes (kinetics) coloso

Enzymes Pt 2: Kinetics

Relicardo M. Coloso, Ph. D.College of MedicineCentral Philippine University

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).

Kinetics of enzyme action

Michaelis-Menten model

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 constantVmax – maximum velocity of the reaction

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.

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.

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

SugarProperties of brain hexokinase

Sugar concn in brain cell

Calculated rateof phosphorylationIn vivoVmax KM

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

Assumptions of enzyme kinetic analysesStandard 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.

One way to do this is with a Lineweaver-Burk plot. Take the inverse of the Michaelis-Menten equation and simplify:

Linear form of the Michaelis-Menten equationTransform the curved data into a straight line, so they could be analyzed with linear regression

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.

Example of double reciprocal plot to solve for KM and Vmax

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.

•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.

Irreversible inhibitors

• are usually considered to be poisons and are generally unsuitable for therapeutic purposes.

Reversible inhibitors can be divided into two main categories; competitive inhibitors and noncompetitive inhibitors, with a third category, uncompetitive inhibitors, rarely encountered.

Inhibitor Type Binding Site on Enzyme Kinetic effect

1) Competitive Inhibitor

Specifically at the catalytic site, where it competes with substrate for binding in a dynamic equilibrium- like process. Inhibition is reversible by substrate.

Vmax is unchanged; Km, as defined

by [S] required for ½ maximal activity, is increased.

2)Noncompetitive Inhibitor

Binds E or ES complex other than at the catalytic site. Substrate binding unaltered, but ESI complex cannot form products. Inhibition cannot be reversed by substrate.

Km appears unaltered; Vmax is

decreased proportionately to inhibitor concentration.

3) Uncompetitive Inhibitor

Binds only to ES complexes at locations other than the catalytic site. Substrate binding modifies enzyme structure, making inhibitor- binding site available. Inhibition cannot be reversed by substrate.

Apparent Vmax decreased; Km, as

defined by [S] required for ½ maximal activity, is decreased.

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.

Michaelis–Menten curves for enzyme with or without inhibitor

A ---------> B Enz (Q)

  Tube A Tube B Tube C Tube D

[S], orConc of A

4.8 mM 1.2 mM 0.6 mM 0.3 mM

1/[S] 0.21 0.83 1.67 3.33

Δ OD540

(Vi) or

Rate of reaction

0.081 0.048 0.035 0.020

1/Vi 12.3 20.8 31.7 50.0

Example: Given the enzyme Q which Converts substrate A to product B

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.)

Competitive inhibitor

WithNon competitive

inhibitor

Lineweaver-Burk Plot

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

(Vi)0.060 0.032 0.019 0.011

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.)

Competitive inhibitor

WithNon competitive

inhibitor

Lineweaver-Burk Plot

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

(Vi)0.040 0.024 0.016 0.010

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.

Competitive inhibitor

WithNon competitive

inhibitor

Lineweaver-Burk Plot

Random bi bi mechanism

Ordered bi bi mechanism

Enzyme reaction mechanisms

Intersecting LB plots

Ordered bi bi mechanism in acyl homoserine lactone synthase

ACP,acyl carrier protein;HSL,homoserine lactone;

SAM,S-adenosylmethionine;MTA,5′-methylthioadenosine

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 randombi bi mechanism

Ping pong mechanism

Parallel LB plots

A friendly animated ping pong game

Ping pong mechanism for uridylyltransferase

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 bindbefore substrate B).

2. In ping-pong reactions, one or more products are released from theenzyme before all the substrates have added.

Keep your eye on the ball!