Transcript Document

Deriving
Kinetic Parameters &
Rate Equations for
Multi-Substrate Systems
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.
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
Secondary Plot
1/Vapp for different
fixed [B]’s
-1/KA, app for
different
fixed [B]’s
-
,true
* 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.
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-firstorder constant by including the
concentration of A: k+1a
A master pattern is drawn representing the skeleton of the scheme; here a square:
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:
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-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 enzyme
concentration e0 present as the species in question. So for all four species we have:
The denominator S 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 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.
“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)
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-M
equation
v
= dp/dt
p = 0 due to
initial velocities
= 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)
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!
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:
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.
Ordered Sequential BiBi
Max velocities for
forward & reverse rxns
Substrate Michaelis
constants for
forward & reverse rxns
Inhibition constants for
forward & reverse rxns
Ping-Pong BiBi
For Ordered Sequential BiBi, Can Calculate Individual Rate Constants by Rearranging
Definitions 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 are
both very large, v = V.
When b is very large:
When b is very small (but not zero):
KmA is the limiting Michaelis constant
for A when B is saturating. Similarly
KmB is limiting when A is saturating.
KiA is the limiting Michaelis constant
for A when B approaches zero, and is also
the 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-Menten
equation--
Plots of Vapp or Vapp/Kapp vs. b
give rectangular hyperbolas
Analysis same as single substrate M-M
Primary Plot Using Hanes Plot:
Increasing b
Secondary Plots
Ping-Pong
BiBi
Initial velocities (p = q = 0):
No constant term in denominator!
Increasing b
a = variable, b = fixed (normal)
Only one secondary
plot is necessary-b/Vapp vs. b (Hanes)
like sequential.
Product Inhibition--
Forward
component
Reverse
component
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 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
Applying Principles 1-5 to Rate Equation for Ordered Sequential BiBi:
Forward
component
Reverse
component
If a = variable substrate, b = fixed substrate, p = added product inhibitor, & q = 0, then:
The constant part of the denominator is
1 + KmAb/KiAKmB + KmQp/KmPKiQ
And the variable part of the denominator is
(a/KiA)(1 + b/KmB + KmQp/KmPKiQ + bp/KmBKiP)
Both expressions contain p, so inhibition is MIXED.
Similar analyses => P & Q behave as mixed inhibitors when B is variable substrate.
Applying Principles 1-5 to Rate Equation for Ordered Sequential BiBi:
Forward
component
Reverse
component
If a = variable substrate, b = fixed substrate, p = 0, & q = added product inhibitor, then:
The constant part of the denominator is
1 + KmAb/KiAKmB + q/KiQ + KmAbq/KiAKmBKiQ
And the variable part of the denominator is
(a/KiA)(1 + b/KmB)
Only constant expression contains q, so inhibition is COMPETITIVE.
SUMMARY: Product Inhibition Patterns for Ordered Sequential BiBi
Inhibition at
Inhibition at
Variable
Product
Normal Levels
Saturating Levels
Substrate
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/KiAKmB + KmQp/KmPKiQ
DOMINATES => constant part of denominator
becomes effectively independent of p
And the variable part of the denominator is
(a/KiA)(1 + b/KmB + KmQp/KmPKiQ + bp/KmBKiP)
DOMINATE => but variable part of denominator
remains dependent on p
Effectively, only variable portion of denominator is dependent on p when fixed substrate b
approaches saturation. So, inhibition is UNCOMPETITIVE.
Siesta
Time!