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Deriving Kinetic Parameters & Rate Equations for Multi-Substrate Systems.

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Presentation on theme: "Deriving Kinetic Parameters & Rate Equations for Multi-Substrate Systems."— Presentation transcript:

1 Deriving Kinetic Parameters & Rate Equations for Multi-Substrate Systems

2 Determination of Kinetic Parameters in Multi-Substrate Systems * With one substrate variable and the other(s) fixed: -- generate kinetics plots that can be analyzed using any of the graphical or computational 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 used for 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.

3 Use of Lineweaver-Burk Plot to Estimate “True” Multi-Substrate Kinetic Parameters Primary LB plot for sequential enzyme A + B P + Q A is variable, B is fixed 1/V app for different fixed [B]’s -1/K A, app for different fixed [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/K A,true from the horizontal intercept; (2) also gives 1/V app from the vertical intercept, which should equal the value above.,true

4 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 the concentration of A:k +1 a

5 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:

6 Next, for each enzyme species, draw arrows on each 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 the rate constants corresponding to the arrows. So, from the patterns leading to E, the sum of products is: k -1 k -2 k -3 p + k -1 k -2 k +4 + k -1 k +3 k +4 + k +2 k +3 k +4 b This sum is then the numerator of an expression representing the fraction of the total enzyme concentration e 0 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.

7 The rate of the reaction is then the sum of the rates of steps that generate one particular product, 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 the reverse rxn, and all rate constants for a complete cycle in the reverse direction.

8 “For most purposes it is more important to know the form of the steady-state rate equation than to know its detailed expression in terms of rate constants.” -- A. Cornish-Bowden COEFFICIENT FORM(Ordered Sequential BiBi)

9 Modifications to the King-Altman Method [E]/e 0 = (k -1 + k +2 )/(k -1 + k +2 + k +1 s + k -2 p) [ES]/e 0 = (k +1 s + k -2 p)/(k -1 + k +2 + k +1 s + k -2 p) M-M equation

10 v = dp/dt = k +2 [ES] - k -2 [E]p = k +2 e 0 (k +1 s + k -2 p)/(k -1 + k +2 + k +1 s + k -2 p) = k +2 e 0 s/((k -1 + k +2 )/k +1 + s) = Vs/(Km + s) p = 0 due to initial velocities

11 This can greatly simplify the derivation for more complicated mechanisms such as random sequential: As shown, this master pattern requires 12 patterns, but if the parallel paths between E and ES and between EX and EXS are added, the master pattern becomes a square, which requires only 4 patterns!

12 Recall Coefficient Form of Rate Equation for Ordered Sequential BiBi 13 coefficients defined in terms of only 8 rate constants -------- Must be inter-related! Cleland devised a system for defining these coefficients in terms of measurable kinetic parameters. Thereby the rate equation for this mechanism becomes:

13 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.

14 Max velocities for forward & reverse rxns Substrate Michaelis constants for forward & reverse rxns Inhibition constants for forward & reverse rxns Ordered Sequential BiBiPing-Pong BiBi

15 For Ordered Sequential BiBi, Can Calculate Individual Rate Constants by Rearranging Definitions of Kinetic Parameters: Doesn’t work for Ping-Pong!

16 INITIAL VELOCITIES (p = q = 0) reduce steady-state rate equation for Ordered Sequential BiBi… … to this form: In limiting case where a & b are both very large, v = V. When b is very large: K m A is the limiting Michaelis constant for A when B is saturating. Similarly K m B is limiting when A is saturating. When b is very small (but not zero): K i A is the limiting Michaelis constant for A when B approaches zero, and is also the true dissociation constant for EA.

17 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-Menten equation-- Plots of V app or V app /K app vs. b give rectangular hyperbolas Analysis same as single substrate M-M

18 Primary Plot Using Hanes Plot: Increasing b

19 Secondary Plots

20 Ping-Pong BiBi Initial velocities (p = q = 0): No constant term in denominator! a = variable, b = fixed (normal) Only one secondary plot is necessary-- b/V app vs. b (Hanes) like sequential. Increasing b

21 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 the forward reaction. (4) Knowing which substrate is variable and which is fixed, the denominator of any rate equation can be separated into variable and constant terms depending on whether they contain the variable substrate concentration, or not. -- expression for V app depends on variable terms. -- expression for V app /K app depends on constant terms. (5) An inhibitor is classified according to whether it affects V app /K app (competitive), V app (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 Forward component Reverse component

22 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 + K m A b/K i A K m B + K m Q p/K m P K i Q And the variable part of the denominator is (a/K i A )(1 + b/K m B + K m Q p/K m P K i Q + bp/K m B K i P ) Both expressions contain p, so inhibition is MIXED. Similar analyses => P & Q behave as mixed inhibitors when B is variable substrate. Forward component Reverse component

23 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 + K m A b/K i A K m B + q/K i Q + K m A bq/K i A K m B K i Q And the variable part of the denominator is (a/K i A )(1 + b/K m B ) Only constant expression contains q, so inhibition is COMPETITIVE. Forward component Reverse component

24 SUMMARY: Product Inhibition Patterns for Ordered Sequential BiBiInhibition at VariableProductNormal LevelsSaturating Levels SubstrateInhibitorof Fixed Substrateof Fixed Substrate ---------------------------------------------------------------------------------------------------------- APNon-competitiveUncompetitive AQCompetitiveCompetitive BPNon-competitiveNon-competitive BQNon-competitiveNone

25 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 + K m A b/K i A K m B + K m Q p/K m P K i Q And the variable part of the denominator is (a/K i A )(1 + b/K m B + K m Q p/K m P K i Q + bp/K m B K i P ) Effectively, only variable portion of denominator is dependent on p when fixed substrate b approaches 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

26 Siesta Time!


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