enzyme kinetics + inhibition

The michaelis menten equation

The initial velocity is equal to Vmax times the substrate concentration over the Michaelis constant plus the substrate concentration. Michaelis Menten equation applies to enzyme catalyzed reactions that display this hyperbolic behavior, which is first order in substrate at low substrate concentration and zero order at saturation. When the concentration of substrate is smaller than Km the bottom substrate concentration is not counted and because Vmax and Km are both constants, then Vo is directly proportional to the concentration of substrate. All enzymes that have this hyperbolic graph follow Michaelis Menten kinetics. A​

Allosteric (regulatory) enzymes with cooperative substrate binding give sigmoidal curves and do not follow Michaelis Menten kinetics.​

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The Michaelis-Menten equation is a rectangular hyperbola with a = Vmax, b = Km, and x = [S], the same form as the oxygen-binding curve of myoglobin, because V0/Vmax is the fraction of active sites holding substrate.​

When [S] ≪ Km, the [S] in the denominator becomes negligible, leaving V0 = (Vmax/Km)[S], so V0 is directly proportional to [S] and the reaction is first order in substrate.​

Substrate concentrations far above Km make Km negligible in the denominator, so [S] cancels and V0 = Vmax, with the reaction zero order in substrate; the approach is gradual, since V0 reaches 91% of Vmax at 10 Km and 99% at 100 Km.​

Enzymes with hyperbolic curves follow this equation, whereas allosteric enzymes with cooperative substrate binding give sigmoidal curves resembling the oxygen-binding curve of hemoglobin and fall outside Michaelis-Menten kinetics.​

what is kcat

kcat, the turnover number, is the number of substrate molecules one active site converts to product per second at saturation, and its reciprocal gives the duration of one catalytic cycle.​

kcat = Vmax/[E]T, where [E]T counts total active sites, so kcat for an enzyme with several active sites is expressed per site.​

The table spans more than a millionfold, from carbonic anhydrase at 600,000 s−1 (one cycle every 1.7 μs) to lysozyme at 0.5 s−1 (one cycle every 2 s).​

kcat describes speed only at saturation and gives no information about performance when substrate is limiting, so the measure of catalytic efficiency is kcat/Km, which combines turnover with binding.​

what does kcat/Km represent?

Substituting Vmax = kcat[E]T into each limiting form of the Michaelis-Menten equation shows which constant sets the rate in each region of the curve.​

Most of the enzyme is free far below Km, so each turnover depends on E and S meeting and on ES converting to product, and kcat/Km combines both steps.​

Nearly all of the enzyme is ES at saturation, so binding no longer limits the rate, and the turnover number alone sets the rate.​

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Substituting Vmax = kcat[E]T into the two limiting forms shows which constant controls the rate in each region of the curve.​

The low-[S] form, V0 = (kcat/Km)[E]T[S], applies when most of the enzyme exists as free E, so each catalytic event requires an encounter between E and S followed by conversion of ES to product, and kcat/Km is the second-order rate constant (M−1s−1) for that whole sequence.​

kcat/Km equals k1k2/(k−1 + k2), so it can never exceed k1, the rate constant for binding, which diffusion caps near 108 to 109 M−1s−1; the three enzymes in the pre-lecture table all sit near that ceiling.​

The saturating form, V0 = kcat[E]T, applies when nearly all of the enzyme is ES, so binding no longer limits the rate and the catalytic step alone determines the speed; kcat is a first-order rate constant (s−1) for that step.​

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Enzyme kinetics is ligand binding plus a catalytic step​

Kₘ versus Kd: Both constants are the concentration that fills half the sites, both carry concentration units, and both fall as binding tightens. The difference is that a bound ligand can leave only the way it came, whereas ES has a second exit, forward to product. Kₘ = (k₋₁ + k₂)/k₁ counts both exits. Splitting the fraction gives Kₘ = Kd + k₂/k₁, where Kd = k₋₁/k₁ is the true dissociation constant of ES. Keeping half the sites full takes more substrate when ES has two exits than when it has one, so Kₘ can never be smaller than Kd. The two values match when ES falls apart far faster than it reacts (k₂ ≪ k₋₁). They diverge when catalysis is fast, which is why Kₘ is best described as an apparent affinity. Kd also comes from binding measurements at equilibrium, whereas Kₘ comes from rate measurements at steady state. Kₘ is therefore a kinetic constant even when its value happens to equal Kd.​

kcat

how fast does a filled site work?

Km

how much substrate fills half the sites?

kcat/Km

How well does free enzyme capture and convert scarce substrate?​

what does Kcat/Km report

A rate depends on how far the transition state sits above wherever the enzyme spends most of its time. A saturated enzyme sits as ES, so kcat reports the climb from ES to the transition state. An enzyme far below Kₘ sits mostly as free E, so kcat/Kₘ reports the climb from E + S to the same transition state.

X-axis: (1/[S])

The reciprocal of substrate concentration. As substrate concentration increases, \(1/[S]\) decreases.

Y-axis: (1/V_0)

The reciprocal of initial reaction velocity. Lower values correspond to faster reaction rates.

Slope: Km/Vmax

The ratio of the Michaelis constant to the maximum reaction velocity.

Y-intercept: 1/Vmax

The reciprocal of maximum reaction velocity. Vmax = 1/y intercept

X-intercept: -1/Km

The negative reciprocal of the Michaelis constant. Therefore, Km=-1/x-intercept

competitive inhibition LBP

• Vmax: unchanged

• Km: increases

• Lines intersect on the y-axis.

noncompetitive inhibition LBP

• Vmax: decreases

• Km: unchanged

• Lines intersect on the x-axis.

uncompetitive inhibition

• Vmax: decreases

• Km: decreases

• Lines are parallel.

competitive inhibiton

• Inhibitor competes with the substrate for the active site.

• Often resembles the substrate structurally.

• Adding more substrate can outcompete the inhibitor.

noncompetitive inhibition

• Inhibitor binds at a site other than the active site (an allosteric site).

• Can bind to either the free enzyme or the enzyme–substrate complex with equal affinity in pure noncompetitive inhibition.

• Adding more substrate cannot fully overcome inhibition.

uncompetitive inhibition

• Inhibitor binds only after the substrate has bound to the enzyme.

• The inhibitor traps the enzyme–substrate complex, preventing product formation.

• Adding more substrate does not overcome inhibition.

KI

inhibitor binding to the free enzyme (E).

KI’

• inhibitor binding to the enzyme–substrate complex (ES).

A lower \(K_I\) or \(K_I'\) means stronger inhibitor binding.

Inhibition type	\(K_I\) vs. \(K_I'\)
Competitive	KI' to infinity (inhibitor does not bind ES)
Pure noncompetitive	KI = KI' (equal binding affinity)
Uncompetitive	KI to infty (inhibitor does not bind free E)
Mixed	KI does not equal KI (different binding affinities)

Vmax

Rate with every active site holding substrate, no

inhibitor present

a (alpha)

All substrate-free enzyme, E plus EI, divided by E

a’ (alpha prime)

All substrate-bound enzyme, ES plus ESI, divided

by ES

apparent Vmax

Ceiling measured with the inhibitor present

Apparent Km

[S] that gives half of that ceiling

alpha vs alpha prime for the different inhibitors

Inhibition type	\(\alpha\)	\(\alpha'\)	\(V_{\max}\)	\(K_m\)
Competitive	\(>1\)	\(1\)	Unchanged	Increases
Pure noncompetitive	\(>1\)	\(=\alpha\)	Decreases	Unchanged
Uncompetitive	\(1\)	\(>1\)	Decreases	Decreases
Mixed	\(>1\)	\(>1\), unequal to \(\alpha\)	Decreases	Increases or decreases

• Competitive: only \(\alpha\) changes because the inhibitor binds free enzyme.

• Uncompetitive: only \(\alpha'\) changes because the inhibitor binds only ES.

• Pure noncompetitive: both change equally.

• Mixed: both change, but unequally.

competitive inhibition

Binds E only

α′ = 1

1/α left at low [S]

All left at saturation

uncompetitive inhibition

Binds ES only

α = 1

All left at low [S]

1/α′ left at saturation

noncompetitive inhibition

Binds E and ES equally

α = α′

1/α left at every [S]

Mixed, binds tighter to E

Binds E more tightly

α > α′

1/α left at low [S]

1/α′ left at saturation

Mixed, tighter to ES

Binds ES more tightly

α′ > α

1/α left at low [S]

1/α′ left at saturation