Kinetics Presentation

8/3/2019 Kinetics Presentation

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College of the RedwoodsMath 55, Differential Equations

Michaelis-Menten Enzyme Kinetics

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Introduction

Figure 1: Triophosphate Enzyme

What is Enzyme Kinetics?

Kinetics is the study of rates of chemical reactions

Enzymes are little molecular machines that carry outreactions in cells

Enzyme kinetics is the study of rates of chemical reactions that

involve enzymes
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Michaelis-Menten Equation

The Michaelis-Menten Equation is a differential equation used to

model the rate at which enzymatic reactions occur

This model allows scientist to predict how fast a reaction will take

place based on the concentrations of the chemicals being reacted.

Figure 2: A Model of an Enzyme


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Typical Enzymatic Reactions

E0 + Sk1

k1E1

E1k2 E0 + P

S the concentration of the substrate

(the unreacted molecules)

P the concentration of product

(the reacted molecules)

E0 the concentration of the unoccupied enzymes

E1 the concentration of occupied enzymes.

k1, k1, k2 the rate constants


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Conditions for Michaelis-Menten Modelling

In order to model an enzymatic reaction, some conditions must be

maintained:

Temperature, ionic strength, pH, and other physical

conditions that might affect the rate must remain constant


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maintained:



Each enzyme can act on only one other molecule at a time


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maintained:




The enzyme must remain unchanged during the course of the

reaction.


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maintained:




The enzyme must remain unchanged during the course of the

reaction. The concentration of substrate must be much higher than the

concentration of enzyme


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Rate Equations

E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)

The rate at which reaction (1) occurs is derived as follows:

The number of possible contacts between S and E0 is directly

proportional to SE0.

The number of successful contacts over a certain amount of time

is proportional to the number of possible contacts.

Thus, the rate of reaction is directly proportional to SE0:

Rate1 = k1SE0.

where k1 is the rate constant.


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E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)

The rate at which the reverse of reaction (1) occurs is derived as

follows:

A certain proportion of E1 will release S over a certain amount

of time before the reaction is carried out.

The rate of the reverse reaction is directly proportional to E1:

Rate1 = k1E1

where k1 is the rate constant.


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Specific Rates of Reactions for each Compound

E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)

Rate1 = k1SE0, Rate1 = k1E1, and Rate2 = k2E1The rate equations associated with each reaction determines the rate

of change ofS, E0, E1, and P. Realizing this, we can write the follow-

ing:

dS

dt= Rate1 + Rate1

= k1SE0 + k1E1


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E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)



ing:

dE0dt

= Rate1 + Rate1 + Rate2

= k1SE0 + k1E1 + k2E1


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E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)



ing:

dE1dt

= Rate1 Rate1 Rate2

= k1SE0 k1E1 k2E1


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E0 + Sk1

k1E1 (1)

E1k2 E0 + P (2)

Rate1 = k1SE0, Rate1 = k1E1, and Rate2 = k2E1

The rate equations associated with each reaction determines the rate


ing:

dP

dt= Rate2

= k2E1


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Thus the system of differential equations modelling the process is:

dS

dt= k1SE0 + k1E1

dE0

dt =k1SE0 + k1E1 + k2E1

dE1dt

= k1SE0 k1E1 k2E1

dP

dt

= k2E1

The rate constants can be difficult or impossible to determine. For the

purpose of seeing the behavior of the system, we give them the the

values k1 = 10, k1 = 1, and k2 = 5, with initial conditions S = 1.0

and an E0 = 0.08.


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0 2 4 6 80

0.2

0.4

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1Substrate/ Enzyme Reaction Model

Time

Co

ncentration

Substrate, SUnoccupied Enzyme, E

0Occupied Enzyme, E

1Product, P


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Reducing the Four Equations to Two

By adding equations and doing some algebraic manipulation, we find

that

dSdt

= k1SET + (k1 + k1S)E1

dE1dt

= k1SET (k1S+ k1 + k2)E1

Figure 3: A model of an enzyme


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The Quasi-Steady-State Assumption

As long as ET S then we can assume that dE1/dt 0.

dSdt

= k1SET + (k1 + k1S)E1

0 k1SET (k1S+ k1 + k2)E1


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dSdt

= k1SET + (k1 + k1S)E1

0 k1SET (k1S+ k1 + k2)E1

Solve dE1/dt for E1

E1 =k1SET

(k1 + k2 + k1S)

Th Q i S d S A i


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dSdt

= k1SET + (k1 + k1S)E1

0 k1SET (k1S+ k1 + k2)E1

Solve dE1/dt for E1

E1 =k1SET

(k1 + k2 + k1S)

Plug this into dS/dt and evaluate various steps to obtain

dSdt

= VmaxS

KM + Swhere Vmax = k2ET and KM =

k1 + k2k1

.


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dS

dt=

VmaxS

KM + Swhere Vmax = k2ET and KM =

k1 + k2k1

.

This is the Michealis-Menten enzyme equation.

This one equation replaces the system for the modelling the substrate

rate equation.

There are only two parameters, and they can both be determined

experimentally.


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0 2 4 6 80

0.2

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1Comparing the Approximation

Time, t

ConcentrationofSubstrate,

S

MichaelisMenten ApproximationOriginal Rate Equation

Figure 4: The solutions for the Michaelis-Menten Equation and the

Original dS/dt.


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Summary

The overall process of converting a substrate to a product is given

by the following two reactions:

E0 + Sk1

k1E1 , and

E1k2 E0 + P.

These reactions give rise to a system of differential equations:

dS

dt= k1SE0 + k1E1

dE0

dt= k1SE0 + k1E1 + k2E1

dE1dt

= k1SE0 k1E1 k2E1

dP

dt= k2E1.

By manipulating these equations we can derive the Michaelis Menten


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By manipulating these equations we can derive the Michaelis-Menten

equation for dS/dt.

dS

dt=

VmaxS

KM

+ Swhere Vmax = k2ET and KM =

k1 + k2k1

.

Figure 5: A model of an enzyme


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References

[1] Edelstein-Keshet, Leah. Mathematical Modelling in Biology,

McGraw-Hill,1988.

[2] Garret and Grisham. BioChemistry,

Saunders College Publishing, 1988.

[3] More Mechanism: A Model of Enzyme Kinetics

Department of Physics, University of Pennsylvania,

http://dept.physics.upenn.edu/courses/gladney

/mathphys/subsection4 1 6.html

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