DerivingKinetic Parameters &

Rate Equations forMulti-Substrate Systems

References for Multi-Substrate Enzyme Kinetics

On-line:

Dr. Peter Birch, University of Paisleyhttp://www-biol.paisley.ac.uk/kinetics/contents.html

Dr. John E. Wampler, University of Georgiahttp://bmbiris.bmb.uga.edu/wampler/8010/lectures/kinetics/steady/

Off-line:

A. Cornish-Bowden (1979) Fundamentals of Enzyme Kinetics, Butterworths, London.

Determination of Kinetic Parameters in Multi-Substrate Systems

* With one substrate variable and the other(s) fixed:-- generate plots kinetics plots that can be analyzed used any of the graphical orcomputational tools that we discussed with single substrate enzymes.

* BUT, V & Km parameters derived are apparent values only. What do they mean?-- change in conc. of fixed substrate(s) will change apparent V & Km for variable one.

* To get true values, we have to extend our definitions of V & Km a bit from those we usedfor single substrate systems.

FOR MULTIPLE SUBSTRATES:

(1) Maximal velocity, V, is defined as the reaction velocity which occurs when all substrates are at saturation levels.

(2) Each substrate will have its own Michaelis constant which is defined as the concentration of that substrate which gives a velocity of half the maximal velocity when all other substrates are present at saturation levels.

* Analysis methods are also extensions of methods used for single substrate systems.

Use of Lineweaver-Burk Plot to Estimate “True” Multi-Substrate Kinetic Parameters

Primary LB plot for sequential enzymeA + B <=> P + QA is variable, B is fixed

1/Vapp for differentfixed [B]’s

-1/KA, app for differentfixed [B]’s

Secondary Plot

-

* Likewise, for primary plot with B variable and A fixed, a secondary plot of 1/Vapp vs 1/[A]:(1) gives -1/KA,true from the horizontal intercept;(2) also gives 1/Vapp from the vertical intercept, which should equal the value above.

,true

Deriving Rate Equations:King-Altman Method

Ordered Sequential BiBi Mechanism:

All rate constants must be first-order;e.g. the second-order rate constant k+1

must be represented by a pseudo-first-order constant by including theconcentration of A: k+1a

Now find every pattern that: (1) consists only of lines from the master pattern;(2) connects every enzyme species; and(3) contains no closed loops.

YES:

NO:

A master pattern is drawn representing the skeleton of the scheme; here a square:

Next, for each enzyme species, draw arrows oneach pattern, leading to the species considered,regardless of starting point. Thus for E:

Then a sum of products of rate constants is written, such that each product contains therate constants corresponding to the arrows. So, from the patterns leading to E, the sumof products is:

k-1k-2k-3p + k-1k-2k+4 + k-1k+3k+4 + k+2k+3k+4b

This sum is then the numerator of an expression representing the fraction of the total enzymeconcentration e0 present as the species in question. So for all four species we have:

The denominator is the sum of all 4 numerators, i.e. the sum of all 16 products obtained from the pattern.

The rate of the reaction is then the sum of the rates of steps that generate one particularproduct, minus the sum of the rates of steps that consume the same product.

In this example, there is only one step that generates P: (EAB + EAQ) --> EQ + P,and only one step that consumes P: EQ + P --> (EAB + EPQ), so we have

GENERAL RULE FOR NUMERATOR: -- Positive term is the product of total enzyme conc., all substrate concentrations for the forward rxn, and all rate constants for a complete cycle in the forward direction. -- Negative term is the product of total enzyme conc., all substrate concentrations for thereverse rxn, and all rate constants for a complete cycle in the reverse direction.

“For most purposes it is more important to know the form of the steady-state rate equation than to know its detailed expression interms of rate constants.” -- A. Cornish-Bowden

COEFFICIENT FORM (Ordered Sequential BiBi)

Modifications to the King-Altman Method

[E]/e0 = (k-1 + k+2)/(k-1 + k+2 + k+1s + k-2p)

[ES]/e0 = (k+1s + k-2p)/(k-1 + k+2 + k+1s + k-2p)

M-Mequation

v = dp/dt

= k+2[ES] - k-2[E]p

= k+2e0(k+1s + k-2p)/(k-1 + k+2 + k+1s + k-2p)

= k+2e0s/((k-1 + k+2)/k+1 + s)

= Vs/(Km + s)

p = 0 due toinitial velocities

This can greatly simplify the derivation for more complicatedmechanisms such as random sequential:

As shown, this master pattern requires 12 patterns, but if the parallel paths betweenE and ES and between EX and EXS are added, the master pattern becomes a square, which requires only 4 patterns!

Recall Coefficient Form of Rate Equation for Ordered Sequential BiBi

13 coefficients defined interms of only 8 rate constants--------Must be inter-related!

Cleland devised a system for defining these coefficients in terms of measurablekinetic parameters. Thereby the rate equation for this mechanism becomes:

From King-Altman Coefficient Form, Can Write Rate Equations for Other Kinetic Mechanisms in Terms of Kinetic Parameters, Too

Ping-Pong BiBi

Random Sequential BiBi

-- simpler because it assumes all steps except (EAB <=> EPQ) are at equilibrium.

Max velocities forforward & reverse rxns

Substrate Michaelisconstants forforward & reverse rxns

Inhibition constants forforward & reverse rxns

Ordered Sequential BiBi Ping-Pong BiBi

For Ordered Sequential BiBi, Can Calculate Individual Rate Constants by RearrangingDefinitions of Kinetic Parameters:

Doesn’t work for Ping-Pong!

INITIAL VELOCITIES (p = q = 0) reduce steady-state rate equation for Ordered Sequential BiBi…

… to this form:

In limiting case where a & b areboth very large, v = V.

When b is very large:

KmA is the limiting Michaelis constant

for A when B is saturating. SimilarlyKm

B is limiting when A is saturating.

When b is very small (but not zero):

KiA is the limiting Michaelis constant

for A when B approaches zero, and is alsothe true dissociation constant for EA.

Ordered Sequential BiBi -- Initial Velocities:

IN GENERAL: for a = variable and b = fixed (normal conc’s, b not very high or very low):

Terms that do not contain a are constant!

Same form as Michaelis-Mentenequation--

Plots of Vapp or Vapp/Kapp vs. bgive rectangular hyperbolas

Analysis same as single substrate M-M

Primary Plot Using Hanes Plot:

Increasing b

Secondary Plots

Ping-PongBiBi

Initial velocities (p = q = 0):No constant term in denominator!

a = variable, b = fixed (normal)

Only one secondaryplot is necessary--b/Vapp vs. b (Hanes)like sequential.

Increasing b

Product Inhibition-- Ex: Ordered Sequential BiBi

If only one product added, then:

(1) Negative (reverse) term in numerator drops out.

(2) All terms containing missing product drop out of denominator.

(3) The only effect of adding product is to increase the denominator, that is, to inhibit theforward reaction.

(4) Knowing which substrate is variable and which is fixed, the denominator of any rateequation can be separated into variable and constant terms depending on whether theycontain the variable substrate concentration, or not.

-- expression for Vapp depends on variable terms.-- expression for Vapp/Kapp depends on constant terms.

(5) An inhibitor is classified according to whether it affects Vapp/Kapp (competitive), Vapp

(uncompetitive), or both (mixed).-- competitive: product conc appears only in constant terms-- uncompetitive: product conc appears only in variable terms-- mixed: product conc appears in both constant and variable terms

Forwardcomponent

Reversecomponent

Applying Principles 1-5 to Rate Equation for Ordered Sequential BiBi:

If a = variable substrate, b = fixed substrate, p = added product inhibitor, & q = 0, then:

The constant part of the denominator is

1 + KmAb/Ki

AKmB + Km

Qp/KmPKi

Q

And the variable part of the denominator is

(a/KiA)(1 + b/Km

B + KmQp/Km

PKiQ + bp/Km

BKiP)

Both expressions contain p, so inhibition is MIXED.Similar analyses => P & Q behave as mixed inhibitors when B is variable substrate.

Forwardcomponent

Reversecomponent

Applying Principles 1-5 to Rate Equation for Ordered Sequential BiBi:

If a = variable substrate, b = fixed substrate, p = 0, & q = added product inhibitor, then:

The constant part of the denominator is

1 + KmAb/Ki

AKmB + q/Ki

Q + KmAbq/Ki

AKmBKi

Q

And the variable part of the denominator is

(a/KiA)(1 + b/Km

B)

Only constant expression contains q, so inhibition is COMPETITIVE.

Forwardcomponent

Reversecomponent

SUMMARY: Product Inhibition Patterns for Ordered Sequential BiBi

Inhibition at Inhibition atVariable Product Normal Levels Saturating LevelsSubstrate Inhibitor of Fixed Substrate of Fixed Substrate----------------------------------------------------------------------------------------------------------

A P Non-competitive Uncompetitive

A Q Competitive Competitive

B P Non-competitive Non-competitive

B Q Non-competitive None

Ordered Sequential BiBi

For a = variable substrate, b = fixed substrate (SATURATING), p = added product inhibitor, & q = 0:

The constant part of the denominator is

1 + KmAb/Ki

AKmB + Km

Qp/KmPKi

Q

And the variable part of the denominator is

(a/KiA)(1 + b/Km

B + KmQp/Km

PKiQ + bp/Km

BKiP)

Effectively, only variable portion of denominator is dependent on p when fixed substrate bapproaches saturation. So, inhibition is UNCOMPETITIVE.

DOMINATES => constant part of denominator becomes effectively independent of p

DOMINATE => but variable part of denominator remains dependent on p

Topic Assignments for In-Class Presentations:

E. coli Threonine Deaminase St. John Oct. 17

Mammalian Glycogen Phosphorylase Guth Oct. 17

E. coli Aspartate Receptor Chapin Oct. 22

E. coli Aspartate Transcarbamoylase Eckenroth Oct. 22

E. coli Phosphofructokinase Petrova Oct. 24

E. coli Phosphoenolpyruvate Carboxylase Nausch Oct. 24

E. coli Type II Citrate Synthase Guo Oct. 29

GroEL/GroES Knapp Oct. 29

Mammalian Ribonucleotide Reductase Roberts Oct. 29

Take-Home Exam:

To be handed out by the middle of this week (probably Wed. 10/1); will be due on Wed. 10/8 by 4:30 PM (there is no lecture that afternoon)

SiestaTime!