115:508 Proteins & Enzymes spring 2002

Kinetics: Product and Substrate Inhibition

Let us now consider inhibition in Cleland's language.  Of course I here refer to reversible inhibitors binding in a noncovalent manner - covalently reacting, irreversible inhibitors will be discussed in a later lecture.  Rever­sible inhibitors are of two basic kinds: dead-end inhibitors, molecules which bind to the enzyme but cannot undergo reaction, so that the E.I com­plex can only dissociate inhibitors, not proceed in any other way, it is up a dead-end street; and product inhibitors, compounds which are normal products of the reaction, binding where they dissociate as products.  In the experiment, one of the products of a multi-substrate, multi-product reac­tion is added to the assay and binds, backing up the reaction and causing the rate to decrease.  It causes there to be an increased amount of all pre­vious complexes to which its point of combination is reversibly connected, and a decrease in free E and in product complexes after its point of combi­nation/dissociation.

In any kind of inhibition one looks to see whether the inhibitor af­fects only the slope of the Lineweaver-Burk plot ("competitive" inhibi­tion), only the intercept ("uncompetitive"), or both ("noncompetitive".  Many authors reserve the term noncompet­itive for the special case where the L-B plots intersect on the x axis, indicating that the Ki is the same no matter what form of the enzyme the inhibitor com­bines with; when the intersection is elsewhere they call the inhibition "mixed non­competitive" or just "mixed"

The effect of an inhibitor on slope and intercept of 1/v vs 1/[S] plots may be predicted by essentially the same rules we went over last time.  Slope effects occur only when an inhibitor combines with an enzyme form reversibly connected with the form with which the varying substrate combines; an added requirement for dead-end inhibitors is that a slope effect is seen only when the inhibitor binds to an enzyme form before (or the same as) that with which the varying substrate combines, since a dead-end inhibitor, unlike a product inhibi­tor, cannot back up the reaction sequence; it merely decreases proportionally the amounts of all the other enzyme forms present by sequestering some of the en­zyme as the E.I com­plex.  This gives a powerful method of determining whether a sequential mechanism is strictly ordered or random: add a dead-end inhibitor which is a good analog of the supposed second substrate to bind (it must be com­petitive vs that substrate).  If the reaction is strictly ordered, so that sub­strate B and the inhibitor B' which is competitive with it can bind only to the EA complex, then B' will be an uncompetitive inhibitor vs substrate A.  If the reaction is random, i.e. B and B' can also bind to free enzyme, then there will be at least some slope effect due to B' binding to free E, the same form with which A binds.

Intercept effects are seen only when the inhibitor combines with an enzyme form different from that to which the varying substrate binds - obviously if they combine with the same form, saturation with the varying substrate eliminates the inhibition.  If an inhibitor gives com­petitive inhibition vs both substrates, it indicates that the inhibitor and both substrates combine with the same form of enzyme, presumably free enzyme, and therefore the reac­tion must be random, not ordered sequential; all product inhibitions being com­petitive is a typical sign of a rapid equilibrium random mechanism.  However, this is rarely seen, because of additional dead-end complexes.

Non-competitive inhibition, i.e. both slope and intercept effects, results from application of both the above rules, i.e. the inhibitor combines with a form different from that with which the varying substrate combines, but reversi­bly connected to it. It follows that most product inhibition, especially in sequential mechanisms, is non-competitive, with the first substrate on and last product off being competitive vs each other because both bind to free enzyme.

      A         B P      Q A and Q are competitive; Q is noncompetitive vs B, P is non-
      Ø         Ø ≠      ≠ competitive vs both A and B.  There is a spe cial case of the

E  ——————————  E                        ordered sequential mechanism, called the Theorell-Chance mechanism, after Hugo Theorell and Britton Chance, who suggested it in 1951 for horse liver alcohol dehydrogenase.  In it the intercon­version of central complexes, EAB € EPQ, and the dissociation of B and P are so rapid that in presence of P the complexes EA and EQ are in rapid equilibrium with each other and B and P are competitive with each other, there are two competitive and two non-competitive product inhibitions.  If in addition there is a fairly slow isomerization  of free enzyme between the release of Q and binding of A, Q may be non-competitive vs A, so that the pattern typical of an ordered sequential mechanism, three non-competitive and one competitive product inhibition, is seen, but the points of binding of the substrates and products in the mechanism are misidentified.  This is an example of when actual kinetic results can be confusing.  Binding studies would indicate that only A and Q actually bind to free enzyme.

Question, in a Ter Ter ordered sequential mech-           A                     B                     C                     P                     Q        R anism, which inhibitions will not be noncompeti tive?             Ø                     Ø                     Ø                      ≠                      ≠                    ≠

Answer: R competitive vs A; Q uncompetitive                         E  —————————————  E
vs. all three substrates.

One may carry out further analysis by carrying out product inhibition studies at saturating levels of one substrate - best done by carrying out studies at several levels of one substrate and extrapolating to infinite levels of it.  For instance, one could assay the enzyme at five levels of A, five levels of B and five levels of P - 125 assays; plot the results as 1/v vs 1/[B] for each combination of [A] and [P], and determine the y intercepts - 25 of them; then plot them vs 1/[A] along five lines corre­sponding to the five levels of P, i.e. P is the changing com­pound.  Since the intercepts of 1/v vs 1/[B] plots were used, B is saturating, and an addi­tional irreversible step is introduced.  If the mechanism is simple ordered sequential, P will now be an uncompetitive inhibitor vs A.  A slope effect would suggest that P can also bind to free E in competition with A.  It can­not be binding to EA, because at saturating B any EA is immediately converted to EAB; and binding to EAB would still yield uncompetitive inhibition.

One can then plot slope and intercept values, whichever vary with inhibitor concentra­tion, vs the inhibitor concentration, to see if this plot is linear.  All dead-end inhibitions give linear replots, because dead-end inhibitors do not affect the relative amounts of other enzyme forms present (unless they create alternate pathways, as in the example I gave where the EIA complex could go ahead to EI + P).  Product inhibitors may give linear plots, if they combine only at one point, or if the points of com­bination are not reversibly connected.  How­ever, if combination at two points has opposing effects - for instance, combina­tion with free E de­creases the amount of EQ complex with which P would norm­ally bind - hyperbolic replots will be observed, i.e. a replot of slopes or inter­cepts vs [P] does not increase linearly, but levels off.  If combination of product at one point increases the amount of complex to which it binds at another point, as can happen with P as product and dead-end inhibitor vs B, parabolic replots are seen, i.e. the slope or intercept increases more than linearly with P concentration.  This will also be the case if any EP2 complex is formed.

Linear product inhibition (a linear replot vs [P]) requires that 1) there is only one reaction sequence, 2) all products are different (normal except in a few reactions such as myokinase), 3) no dead-end inhibition by the product occurs.  Observed linear product inhibition thus indicates that these conditions do hold (except for the minor case where irreversible steps intervene between the two points of combination).

Unfortunately, the greatest problem in the use of product inhibitors is that they may also act as dead-end inhibitors, binding at another point in the reac­tion sequence, and you may have to unscramble the effects of two or three dif­ferent combinations with enzyme.  In addition, high levels of substrates often cause inhibition.  With single-susbtrate enzymes this can happen only by parts of two substrate molecules binding, neither able to undergo reaction, but more usually this occurs in multi-substrate reac­tions by binding where a product should bind.  These so-called abortive ternary complexes are very common among dehydrogenases, either E. NAD+.carbonyl or E.NADH.alcohol. being formed, depending on which way the reaction  is being run.  The reaction is completely inhibited if coenzyme cannot dissociate from this complex, or slowed if it dissoci­ates more slowly than from the binary complex; in some cases it actually dissoci­ates faster than from the binary complex, and substrate activation is seen.

Substrate inhibition normally gives a L-B plot which is a sort of hyperbola, one asymptote being the straight line which would be the plot in absence of in­hibition, and the other being the y axis.  The plot diverges from the uninhibited line, passes through a minimum, and swings up along the y axis.  To determine the normal kinetic constants, Km and Vmax, one needs the true asymptote, which is not quite the same as the apparently linear part of the plot.  For simple linear substrate inhibition v = ;  the minimum point of the curve is at [A] = , and a line connecting it to the y intercept of the asymptote has slope = 2Ka/Vmax or double that of the asymptote; thus if the minimum of the plot is well defined, the y int­ercept of the asymptote can be determined by adjusting it until this relationship is fulfilled.   But the best method is to fit the data to the equation by non-linear least squares to get the statistically most accurate values of Km and Ki.

Cleland says the proper way to study substrate inhibition is to plot 1/v vs. 1/a non-in­hibitory substrate at different inhibitory levels of the inhibiting sub­strate and determine whether the slopes increase - competi­tive substrate inhib­ition - or the intercepts increase – uncompet­itive sub­strate inhibition - or both - non-competitive.  Competi­tive substrate inhibition is typical of ping-pong mech­anisms - due to sub­strate B combining with E, the form with which A com­bines, as well as F, the form with which B normally combines.  Uncompetitive substrate inhib­ition is typical of sequential mechanisms where EBQ is formed and cannot dissociate Q.  The intercepts go through a minimum and then rise again.

With simple substrate inhibition the replot of slopes or intercepts should eventually become linear with increasing substrate concentration - a plot of 1/v vs. [B] rather than 1/[B], but in some cases it is hyperbolic, approaching a maxi­mum value.  This is what happens with dehydrogen­ases which form EBQ com­plexes from which Q can dissociate, but at a slow­er rate than from the EQ com­plex; 1/v approaches a maximum value, v a minimum, corresponding to the lim­iting rate of Q dissociation from the EBQ complex.

Segal goes on to discuss various further types of inhibition.  For in­stance, the inhibitor may only reduce  the binding of substrate to enzyme - increase Ks by some factor a greater than 1, without  affecting Vmax. This causes the slope of L-B plots to increase without affecting the intercept, so this is a competitive inhibition, but the slope, instead of increasing indefi­nitely as [I] increases, approaches a maximum = aKs/Vmax.  A replot of the slopes vs [I] is a hyperbola, as is a Dixon plot of 1/v vs. [I].  The slope and the apparent Km are increased over the uninhibited state by a factor ; as [I] becomes very large this approaches and a in the denominator of the denominator becomes a in the numerator, the slope approaches aKs/Vmax.

Ki and a can be determined from a manipulation of the slopes as follows: subtract the uninhibited slope from the inhibited slope . To see the reason for this, multiply the uninhibited slope by .  Then you have  - , or = .  Multiply top and bottom of the bracketed expression by aKi and put [I] outside at the top: .  For convenience, call this ∆.  Now invert this:
= = + .  Thus a plot of 1/∆  vs. 1/[I] will have intercept , allowing a to be calculated since is known from the uninhibited case.  The slope is , allowing Ki to be calculated once a is known.

An equivalent case for non-competitive inhibition is when [I] does not affect binding of substrate, but decreases the forward velocity by some factor b.  As I increases Vm decreases, approacing a limiting value bVmax.  The equation is equivalent, the inter­cept and slope both being multiplied by a factor .  b and Ki are evalu­ated by plots of 1/∆int vs 1/[I] as suggested for the previous case, ∆int being (inter­cept in presence of inhibitor) - (intercept in absence of inhibitor).  This type of inhibition looks like ordinary pure non-competitive inhibition except for the hyperbolic dependence of Vmax on [I]; all L-B plots will intersect on the x axis.

The next simple case is that suggested earlier, where EIS does not yield product but S does change the dissociation constant for I, Ki' = aKi (and Ks' = aKs).  In this case the slope is increased by the factor 1 + [I]/Ki and the intercept by the factor 1 + [I]/aKi.  If a is greater than 1, the plots will intersect above the x axis; I and S mutually increase each other's disso­ciation constant.  It is perfectly possible for I and S to decrease each other's dissociation constant, a<1; then the L-B plots will intersect below the x axis.  The x coordinate of the int­ersection point is -1/aKs, and the y coordinate is .  The x intercepts of replots of slope or intercept vs [I] are -Ki and -aKi respectively.  The Dixon plot looks like that of competitive inhibition - intersection above the x axis - but the intersection is below 1/Vmax.

Haldanes

One other thing which can be done with Km and Vmax values derived from variation of v in each direction with each substrate and product concentration: one can determine whether the Haldane relationships come out right. Haldane relationships are expressions of the equilibrium constant in terms of Vms and Kms.  For a one-substrate, one product reaction the Haldane is simply Keq = ; note that any one term can vary without constraint from the equilibrium constant, allowing catalysis to be to some extent unidirectional (if Kp is above the steady state concentration of P), but the overall relationship must be preserved.

For multisubstrate situations different mechanisms give different Haldane relationships and consequently they can be used as tests of the mechanism sup­posed to exist; for instance, sequential 2-substrate mechanisms have the Haldane relationship Keq = øpq/øab, where øpq, øab are the coefficients of the terms in the reciprocal equation which are sensitive to both substrates or products - the slope of a replot of the slopes of primary plots.  In a ping-pong mechanism Keq = , the øpq, øab terms being lacking.  In 3-substrate mechanisms there are sev­eral Haldane relationships, using both the Kms and the inhibition constants Kia, Kib, ….

Similar but more powerful relationships were derived by Dalziel; they con­cern the relationship of the maximum rate (Vmax with both substrates saturat­ing) to the ø parameters.  In the general sequential model øab[E]t is greater than Vr times øaøb, while for the extreme cases, the rapid equilibrium random and Theorell-Chance mechanisms, øab = , or to put it another way = 1.  A value of 1 is considered by Dalziel to indi­cate an essent­ially Theorell-Chance mechan­ism, coenzyme dissociation (off rate of product Q) being rate limiting, while a value less than 1 indicates that other steps play a role in deter­mining Vmax, i.e. the mechanism is ordered but not Theorell-Chance, and a value greater than 1 indicates that something else kinetically significant is going on, such as isomerization of free enzyme.

Failure of experimental results to accord with a predicted Haldane rela­tionship is usually a signal that some kinetically important step not influenced by substrate, such as isomerization of a stable enzyme form or a transitory complex, has been left out of the expression.  I won't try to give an example, but you can read about this in Cleland's Biochim. Biophys. Acta papers, or in a paper from Dr. Pietruszko's lab on an isozyme of alcohol dehydrogenase.