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Michaelis constant 2026-10-14

The substrate concentration giving half the limiting reaction rate in the Michaelis-Menten equation. In the elementary mechanism

E + S ⇌ ES \to E + P

, it is

K_{M} = (k_{- 1} + k_{cat}) / k_{1}

under the quasi-steady-state approximation, whereas the binding dissociation constant is

K_{d} = k_{- 1} / k_{1}

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / b / i / Solution 2026-10-20

The intermediate balance is

\frac{d [ ONOO ]}{d t} = k_{1} [NO] [O_{2}] - k_{- 1} [ONOO] - k_{2} [NO] [ONOO] .

Setting this derivative approximately to zero under the quasi-steady-state approximation gives

[ONOO]_{ss} = \frac{k _{1} [ NO ] [ O _{2} ]}{k _{- 1} + k _{2} [ NO ]} .

Both dissociation and reaction with NO remove the intermediate. Its steady concentration is not zero and does not require the first reversible pair to be at equilibrium.

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 5 / d / Solution 2026-10-20

The quasi-steady-state approximation for the nitrate radical balances its formation against its removal:

\frac{d [ N O _{3} ]}{d t} = r_{1} - r_{2} + r_{- 2} ≃ 0.

Equivalently,

[N O_{3}]_{ss} = \frac{k _{1} [ N O _{2} ] [ O _{3} ] + k _{- 2} [ N _{2} O _{5} ]}{k _{2} [ N O _{2} ]} .

The net dinitrogen pentoxide production is

r_{2} - r_{- 2}

. Substituting the steady-state balance gives

\frac{d [ N _{2} O _{5} ]}{d t} ≃ k_{1} [N O_{2}] [O_{3}] .

The reverse rate has been retained throughout and cancels algebraically. The approximation requires sufficiently rapid radical adjustment and fails if, for example, nitrogen dioxide becomes too scarce to sustain the presumed trapping rate. It does not assume that the radical concentration is zero.

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 5 / e / Solution 2026-10-20

Nitrogen dioxide is consumed in reactions 1 and 2 and regenerated in reaction

- 2

, so

\frac{d [ N O _{2} ]}{d t} = - r_{1} - r_{2} + r_{- 2} .

Using

r_{2} - r_{- 2} ≃ r_{1}

from the nitrate-radical quasi-steady-state approximation gives

\frac{d [ N O _{2} ]}{d t} ≃ - 2 k_{1} [N O_{2}] [O_{3}] .

The factor of two is essential: an initiating event uses one nitrogen dioxide molecule, and the ensuing net radical-trapping event uses a second one. The exact nitrogen balance before making this approximation is

d ([N O_{2}] + [N O_{3}] + 2 [N_{2} O_{5}]) / d t = 0

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 5 / f / Solution 2026-10-20

Adding the initiation step to the net forward trapping step cancels the nitrate-radical reaction intermediate:

2N O_{2} + O_{3} ⟶ N_{2} O_{5} + O_{2} .

The net stoichiometry agrees with

- d [N O_{2}] / d t ≃ 2 d [N_{2} O_{5}] / d t

under the quasi-steady-state approximation.

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 5 / g / Solution 2026-10-20

The remaining fraction is $0.120 \approx 12.0%$ . Using the tabulated

300 K

value rather than replacing it by the fitted value, the constant ozone concentration makes nitrogen dioxide loss a pseudo-first-order reaction:

\frac{d [ N O _{2} ]}{d t} = - k_{obs} [N O_{2}], k_{obs} = 2 k_{1} [O_{3}] = 2 (23 \times 1 0^{6}) (1.6 \times 1 0^{- 12}) = 7.36 \times 1 0^{- 5} s^{- 1} .

Integrating with initial concentration

[N O_{2}]_{0}

and

t = 8 (3600) = 28800 s

gives

\frac{[ N O _{2} ] ( t )}{[ N O _{2} ] _{0}} = e^{- k_{obs} t} = e^{- 2.11968} = 0.12007.

This uses the nitrate-radical quasi-steady-state approximation over the interval, constant temperature and ozone concentration, and no extra nitrogen dioxide sources or sinks. The stated

c m^{3}

and

mol c m^{- 3}

units cancel directly; no factor of

1000

is needed.

Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2007 / B / 5 / iv / Solution 2026-10-31

For quasi-steady enzyme kinetics at low substrate saturation, the quasi-steady-state approximation is controlled here by

ε = e_{0} / (s_{0} + K_{m})

, rather than by

e_{0} / s_{0}

alone. Even with

e_{0} \approx s_{0}

, taking

K_{m} ≫ s_{0}

makes

ε ≪ 1

and

σ = s_{0} / K_{m} = O (ε)

After the initial fast layer

T = O (ε)

, the third equation gives

y \sim \frac{σ x}{( 1 + σ ) ( 1 + ρ ) ψ} = O (ε) .

Since

ϕ = O (ε)

, the last equation gives

(e_{i})_{T} = (ϕ / ε) y = O (ε)

. Starting from

e_{i} (0) = 0

, this implies

e_{i} = O (ε)

on bounded intervals of the slow time. The quasi-steady second equation then gives

x = s + O (ε)

. In the first equation,

s_{T} = - s + \frac{ρ}{1 + ρ} s + O (ε) = - \frac{s}{1 + ρ} + O (ε) .

Matching to the initial substrate value

s (0) = 1

therefore yields

s (T) \sim e^{- T / (1 + ρ)}

. The relation

x \sim s

is an outer approximation: it does not replace the initial layer needed to reconcile

x (0) = 0

with

s (0) = 1

. Likewise the smallness of accumulated inactivated enzyme has been justified for bounded slow times, not arbitrarily long times.

Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / B / 4 / d / Solution 2026-10-31

For an outer solution with bounded derivatives as

ϵ \to 0

, the final complex equation gives

r_{n} = 0

. Working backwards in

0 = r_{i} - r_{i + 1}

then gives every

r_{i} = 0

, including

r_{1}

; setting only their sum to zero would not suffice. This is the quasi-steady-state approximation, valid after the initial complex-formation layer with the relevant dimensionless rate parameters held fixed.

The recurrence is

C_{i} = S C_{i - 1} / K_{i}

, where

K_{i} = (k_{- i} + k_{i}^{+}) / k_{i}

. Thus

C_{j} = ϕ_{j} S^{j} C_{0}

ϕ_{j} = \prod_{i = 1}^{j} K_{i}^{- 1}

. Substitution into enzyme conservation yields

C_{0} = \frac{E _{0}}{1 + \sum _{j = 1}^{n} ϕ _{j} S ^{j}}, C_{j} = \frac{E _{0} ϕ _{j} S ^{j}}{1 + \sum _{ℓ = 1}^{n} ϕ _{ℓ} S ^{ℓ}} .

Each state

j

produces reaction product at rate

k_{j}^{+} C_{j}

, so

ρ = E_{0} \frac{\sum _{j = 1}^{n} k _{j}^{+} ϕ _{j} S ^{j}}{1 + \sum _{j = 1}^{n} ϕ _{j} S ^{j}} .

The source's uppercase

N

in the index range is read as the number

n

of binding sites. This approximation preserves the total enzyme population and the different catalytic rates of the occupied states.

Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / B / 4 / e / Solution 2026-10-31

The sequential-enzyme Hill limit uses the intended distinguished limit for

n \geq 2

holds the other rates fixed while

k_{1} \to 0

k_{n} \to \infty

, and

k_{1} k_{n}

tends to a positive finite constant. Then for every

1 \leq j < n

ϕ_{j} = \prod_{i = 1}^{j} k_{i} / (k_{- i} + k_{i}^{+}) = O (k_{1}) \to 0

, whereas

ϕ_{n}

remains finite and nonzero. Hence intermediate complexes have negligible occupancy relative to the unbound and fully bound states.

In the quasi-steady-state approximation rate, all intermediate numerator and denominator terms disappear. Multiplying by

1/ ϕ_{n} = \prod_{i = 1}^{n} K_{i}

gives

ρ ≃ E_{0} \frac{k _{n}^{+} S ^{n}}{\prod _{i = 1}^{n} K _{i} + S ^{n}} .

This is the Hill equation with maximal rate

E_{0} k_{n}^{+}

and effective half-saturation concentration

(\prod K_{i})^{1/ n}

. Here its power

n

follows from the sequential binding mechanism and the suppression of intermediate states.

The phrase “

k_{1} k_{n}

finite” needs this scaling interpretation for a nontrivial limiting rate. Merely bounding that product, while permitting unrelated rates to vary, does not ensure that all intermediate weights vanish relative to the terminal one. The derivation explicitly supplies the usual fixed-other-rates, positive-product assumptions. For

n = 1

, the rate in the box is already the exact one-state quasi-steady-state approximation rate, without this two-endpoint limiting argument.

Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / B / 4 / f / Solution 2026-10-31

Write

c = c_{1}

β = (k_{- 1} + k_{1}^{+}) / (k_{1} S_{0})

and

κ = k_{1}^{+} / (k_{1} S_{0})

. Since

c_{0} = 1 - c

, the exact equations are

ϵ c_{τ} = s - (s + β) c, s_{τ} = - s + (s + β - κ) c .

At the prescribed initial data

s = 1, c = 0

r_{1} = 1

, rather than zero. The quasi-steady-state approximation would instead demand

c = s / (s + β)

, giving

c (0^{+}) = 1/ (1 + β)

. This mismatch signals an initial layer.

Set

σ = τ / ϵ

and expand

s = s_{0} + ϵ s_{1} + \dots

c = c_{0} + \dots

. At leading order

s_{0}^{'} = 0

, so

s_{0} = 1

, while

c_{0}^{'} = 1 - (1 + β) c_{0}

and

c_{0} (0) = 0

. Thus

s_{in} = 1 + O (ϵ), c_{in} = \frac{1 - e ^{- (1 + β) σ}}{1 + β} + O (ϵ) .

For a more informative substrate approximation, integration of

s_{1}^{'} = - 1 + (1 + β - κ) c_{0}

with

s_{1} (0) = 0

gives

s_{in} = 1 - ϵ [\frac{κσ}{1 + β} + \frac{1 + β - κ}{( 1 + β ) ^{2}} (1 - e^{- (1 + β) σ})] + \dots .

On the slow scale

c_{out} = s_{out} / (s_{out} + β)

and

s_{out, τ} = - κ s_{out} / (s_{out} + β)

, with matched leading value

s_{out} (0) = 1

. Integrating gives

s_{out} + β lo g s_{out} = 1 - κ τ

. In the overlap

1 ≪ σ ≪ 1/ ϵ

, the inner complex tends to

1/ (1 + β)

and its substrate slope matches the expansion of this outer solution. A leading composite asymptotic expansion is

c_{comp} (τ) = \frac{s _{out} ( τ )}{s _{out} ( τ ) + β} - \frac{e ^{- (1 + β) τ / ϵ}}{1 + β}, s_{comp} = s_{out} .

(4)

It satisfies both initial values at leading order and smoothly joins the initial complex build-up to substrate depletion. The displayed

O (ϵ)

inner substrate term records the small binding-induced initial slip if greater accuracy is wanted.

Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 6B / Solution Created 2026-09-24 Updated 2026-10-14

Applying the law of mass action to the sequential two-site enzyme reaction network gives

\overset{s}{˙} \overset{e}{˙} \overset{c}{˙}_{1} \overset{c}{˙}_{2} \overset{p}{˙} = - k_{1} se + k_{- 1} c_{1} - k_{3} s c_{1} + k_{- 3} c_{2}, = - k_{1} se + (k_{- 1} + k_{2}) c_{1}, = k_{1} se - (k_{- 1} + k_{2}) c_{1} - k_{3} s c_{1} + (k_{- 3} + k_{4}) c_{2}, = k_{3} s c_{1} - (k_{- 3} + k_{4}) c_{2}, = k_{2} c_{1} + k_{4} c_{2} .

Adding the appropriate equations gives the conservation laws

e + c_{1} + c_{2} = e_{0}

and

s + c_{1} + 2 c_{2} + p = s_{0}

. In particular

e = e_{0} (1 - v_{1} - v_{2})

after nondimensionalization.

Define the dimensionless reaction rates

a_{1} = k_{- 1} / (k_{1} s_{0})

a_{3} = k_{- 3} / (k_{1} s_{0})

b_{2} = k_{2} / (k_{1} s_{0})

b_{4} = k_{4} / (k_{1} s_{0})

and

r_{3} = k_{3} / k_{1}

. Substitution, with primes denoting

d / d τ

, yields

u^{'} ϵ v_{1}^{'} ϵ v_{2}^{'} = - u (1 - v_{1} - v_{2}) + a_{1} v_{1} - r_{3} u v_{1} + a_{3} v_{2}, = u (1 - v_{1} - v_{2}) - (a_{1} + b_{2}) v_{1} - r_{3} u v_{1} + (a_{3} + b_{4}) v_{2}, = r_{3} u v_{1} - (a_{3} + b_{4}) v_{2} .

These right sides are

f, g_{1}, g_{2}

, respectively, with

u (0) = 1

and

v_{1} (0) = v_{2} (0) = 0

. When

ϵ ≪ 1

, the form of these ordinary differential equations displays the faster relaxation of the complexes, which motivates a subsequent quasi-steady-state approximation.

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 6C / d / Solution Created 2026-09-24 Updated 2026-09-25

On times

τ ≫ ϵ

, the quasi-steady-state approximation sets

g_{1} = g_{2} = 0.

The second equation gives

v_{2} = \frac{β}{γ} u v_{1} .

In the first equation the terms

- β u v_{1} + γ v_{2}

then cancel, leaving

u (1 - v_{1} - v_{2}) = α v_{1} .

Solving these two equations gives

v_{1} = \frac{u}{α + u + ( β / γ ) u ^{2}}, v_{2} = \frac{( β / γ ) u ^{2}}{α + u + ( β / γ ) u ^{2}} .

(4)

Since

\overset{p}{˙} = k_{3} c_{2} = k_{3} e_{0} v_{2}

\frac{d p}{d t} = A \frac{u ^{2}}{α + u + ( β / γ ) u ^{2}}, A = \frac{k _{3} e _{0} β}{γ} = k_{2} e_{0} s_{0} .

(5)

At small substrate concentration this allosteric rate is quadratic,

d p / d t \sim (A / α) u^{2}

, so its graph leaves the origin with zero slope. At large

u

it saturates at

A γ / β = k_{3} e_{0}

. A Michaelis-Menten equation also saturates, but its rate

V_{m a x} u / (K_{M} + u)

is linear near the origin. The allosteric curve is therefore more sigmoidal:

allosteric Michaelis-Menten u \to 0 O (u^{2}) O (u) u \to \infty k_{3} e_{0} V_{m a x}

(6)

This is the quasi-steady rate law for a two-substrate allosteric enzyme.

Past exam of the systems biology course of the University of Cambridge / 2024 / Paper 3 / 3 / e / Solution 2026-10-20

The quasi-steady-state approximation for

Y

sets

Y = a_{Y} h (X; K_{X Y}, n_{X Y})

. Substituting into the

Z

equation gives

\frac{d Z}{d t} = a_{Z} h (X; K_{XZ}, n_{XZ}) h (a_{Y} h (X; K_{X Y}, n_{X Y}); K_{Y Z}, n_{Y Z}) - Z .

Its equilibrium concentration is therefore

Z_{AND}^{*} (X) = a_{Z} \frac{X ^{n_{XZ}}}{K _{XZ}^{n_{XZ}} + X ^{n_{XZ}}} \frac{( a _{Y} X ^{n_{X Y}} / ( K _{X Y}^{n_{X Y}} + X ^{n_{X Y}} ) ) ^{n_{Y Z}}}{K _{Y Z}^{n_{Y Z}} + ( a _{Y} X ^{n_{X Y}} / ( K _{X Y}^{n_{X Y}} + X ^{n_{X Y}} ) ) ^{n_{Y Z}}} .

For constant input this is also the exact equilibrium of the full two-variable model: the fast-

Y

assumption is needed to replace the transient dynamics, not to calculate a fixed point.

Relative to the direct-activation result, $Z_{AND}^{*} = Z_{direct}^{*} h (Y^{*}; K_{Y Z}, n_{Y Z}) \leq Z_{direct}^{*}$ . The extra activation gate suppresses the response at low input and can sharpen its rise. The high-input limit is

a_{Z} h (a_{Y}; K_{Y Z}, n_{Y Z})

, which need not reach

a_{Z}

because

Y^{*}

saturates at

a_{Y}

For the parameters in part (d), the following Python continuation plots the expression and changes each Hill constant separately:

def z_equilibrium(X, Kxy=1, Kxz=2, Kyz=1):
    Yeq = 2 * hill(X, Kxy, 2)
    return 3 * hill(X, Kxz, 4) * hill(Yeq, Kyz, 4)

X_grid = np.linspace(0, 5, 501)
plt.figure(figsize=(7, 4))
plt.plot(X_grid, 3*hill(X_grid, 2, 4), ":", label="Direct activation")
for parameters, label in [((1, 2, 1), "Baseline"),
                          ((2, 2, 1), "Kxy=2"),
                          ((1, 3, 1), "Kxz=3"),
                          ((1, 2, 2), "Kyz=2")]:
    plt.plot(X_grid, z_equilibrium(X_grid, *parameters), label=label)
plt.xlabel("X concentration")
plt.ylabel("Equilibrium Z")
plt.legend()
plt.tight_layout()
plt.show()

Increasing

K_{X Y}

lowers

Y^{*}

at fixed finite

X

, delaying opening of the indirect gate. Increasing

K_{XZ}

delays opening of the direct gate. Neither changes the limiting response as

X \to \infty

, provided the constants remain finite. Increasing

K_{Y Z}

makes a given

Y

less effective and also lowers the high-input plateau. These changes are visible in the preceding figure. At small

X

, the response is proportional to

X^{n_{XZ} + n_{X Y} n_{Y Z}}

, while both gates are far below saturation, explaining the increased low-input suppression.

For a persistence detector, retain finite

Y

dynamics and use, for example,

a_{Y} = a_{Z} = 1, K_{X Y} = K_{XZ} = 1, n_{X Y} = n_{XZ} = 4, K_{Y Z} = 0.7, n_{Y Z} = 8,

with unit loss rates,

X_{off} = 0

X_{on} = 2

, and initially

Y = Z = 0

. On switching the input on,

Y_{\infty} = 16/17

and

Y (t) = \frac{16}{17} (1 - e^{- t}) .

(4)

The time to reach the half-activation threshold of the indirect gate is

t_{gate} = - lo g (1 - \frac{0.7}{16/17}) = lo g (160/41) \approx 1.362 .

(5)

An isolated input pulse of duration

0.25

ends well before this threshold; a duration-four pulse persists long enough to open the gate. The large but finite Hill coefficient makes this a graded filter, not a perfectly sharp cutoff. Once

X

vanishes, its direct gate closes immediately and stops production of

Z

, which then decays with time constant one. The protein output itself does not drop discontinuously. This delayed-on, prompt-production-off mechanism is the characteristic persistence detector interpretation of an AND-type coherent feed-forward loop.^[ref]

This final Python continuation simulates both pulse durations. Each discontinuity is placed on the time grid and the input is held constant within each integration step:

def persistence_rhs(state, X):
    Y, Z = state
    return np.array([hill(X, 1, 4) - Y,
                     hill(X, 1, 4)*hill(Y, 0.7, 8) - Z])

def pulse_trajectory(duration, dt=0.005):
    times = np.arange(round(8/dt) + 1) * dt
    out = np.zeros((len(times), 2))
    for j in range(len(times)-1):
        X = 2.0 if 1 <= times[j] + dt/2 < 1 + duration else 0.0
        # Each step is integrated with its constant input.
        out[j+1] = integrate(lambda s: persistence_rhs(s, X),
                             out[j], [0, dt])[-1]
    return times, out

fig, axes = plt.subplots(1, 2, figsize=(10, 4))
for duration in [0.25, 4]:
    times, out = pulse_trajectory(duration)
    axes[0].plot(times, out[:, 0], label=f"Y, duration {duration}")
    axes[1].plot(times, out[:, 1], label=f"Z, duration {duration}")
    print(duration, "maximum output", out[:, 1].max())
axes[0].axhline(0.7, color="black", linestyle=":")
for ax in axes:
    ax.set_xlabel("Time")
    ax.set_ylabel("Concentration")
    ax.legend()
plt.tight_layout()
plt.show()

The chosen example produces peak

Z

of about

1.72 \times 1 0^{- 6}

for the short pulse and

0.766

for the long pulse. These are original illustrative simulations, not fitted experimental measurements.

Figure 1.
A coherent feed-forward loop distinguishes short and persistent input pulses
. Both inputs switch from zero to X=2 at time one. One ends after 0.25 time units and the other after four. The intermediate Y crosses its activation threshold only for the sustained input, producing a much larger target response. **Details**.

Finally, the strict instantaneous-

Y

reduction used for the equilibrium calculation has no separate intermediate accumulation delay. Its remaining one-variable dynamics can still attenuate short pulses through ordinary output relaxation, but it cannot demonstrate the additional delay caused by finite

Y

buildup. The pulse test therefore uses the full two-variable model.

Persistence detector Created 2026-10-20 Updated 2026-10-25

A persistence detector responds strongly to a sustained input and weakly to short isolated inputs. In an activating coherent feed-forward loop, an intermediate protein accumulates toward a threshold while the target also requires the direct input. Eliminating that intermediate by an instantaneous quasi-steady-state approximation removes its accumulation delay.

Quasi-steady enzyme kinetics at low substrate saturation 2026-10-31

A quasi-steady-state approximation for enzyme-substrate complexes can remain useful when total enzyme and initial substrate concentrations are comparable. The relevant parameter can involve the sum

s_{0} + K_{m}

, rather than

s_{0}

alone. In a mechanism-based inhibition model with

s_{0} / K_{m} = O (ε)

, bounded intermediate decay rates and an inactivation rate of order

ε

, the fast complexes adjust in a layer of scaled width

O (ε)

. On bounded slow-time intervals the main complex follows the substrate, the later complex and accumulated inactivated enzyme remain

O (ε)

, and the substrate has leading exponential depletion. An outer algebraic complex concentration does not satisfy the unbound initial condition inside that initial layer.

Reversible enzyme inhibition Created 2026-10-18 Updated 2026-10-20

Construct a minimal model with one substrate and no appreciable accumulated reaction product. Only the enzyme-substrate complex

ES

produces product, with catalytic rate constant

k_{cat}

;

E I

and

ES I

are inactive. Write

s = [S]

and

I = [I]

for free molar concentrations, and

E_{T}

for total enzyme molar concentration. Assume a suitable quasi-steady-state approximation, rapid inhibitor binding and release relative to turnover, and negligible depletion of free inhibitor. In this construction the population relations are

[ES] = [E] \frac{s}{K _{M}}, [E I] = [E] \frac{I}{K _{i}}, [ES I] = [ES] \frac{I}{K _{i}^{'}}, α = 1 + \frac{I}{K _{i}}, α^{'} = 1 + \frac{I}{K _{i}^{'}} .

Here

K_{M}

is the uninhibited Michaelis constant, and

K_{i}, K_{i}^{'}

are the inhibition constants for the two indicated binding branches. The symbol

I

denotes a free concentration, not an amount added to the tube. Define

V_{m a x} = k_{cat} E_{T}

. Conservation of enzyme gives an original short derivation:

E_{T} v = [E] + [E I] + [ES] + [ES I] = [E] (α + α^{'} \frac{s}{K _{M}}), = k_{cat} [ES] = \frac{V _{m a x} s}{α K _{M} + α ^{'} s} = \frac{( V _{m a x} / α ^{'} ) s}{( α / α ^{'} ) K _{M} + s} .

Thus fitting a Michaelis-Menten equation at a fixed

I

would return the apparent parameters

V_{m a x, app} = \frac{V _{m a x}}{α ^{'}}, K_{M, app} = \frac{α}{α ^{'}} K_{M} .

“Apparent” means conditional on that inhibitor concentration. The uninhibited Michaelis constant has not been redefined. Dimensional analysis checks the denominator: both

α K_{M}

and

α^{'} s

have units of molar concentration; the resulting

v

has units of reaction rate.

Figure 1.
Chosen competitive inhibition, pure noncompetitive inhibition and mixed inhibition curves
. Original calculations, not measured rates or fits to trypsin. Every panel uses $V_{m a x} = 100 μ M mi n^{- 1}$ , $K_{M} = 20 μ M$ and $K_{i} = 10 μ M$ , with free $I = 0, 10, 30 μ M$ . Competitive inhibition uses $α^{'} = 1$ ; pure noncompetitive inhibition uses $K_{i}^{'} = 10 μ M$ ; the unequal-constant mixed inhibition example uses $K_{i}^{'} = 30 μ M$ . Dots mark $s = 20 μ M$ . Dotted horizontal lines are limiting rates, approached rather than reached at finite $s$ . **Details**.

The shared numerical choices make the mechanisms comparable. At

I = 0

and

s = 20 μ M

, all three models give

v = 50 μ M mi n^{- 1}

. At

I = 10 μ M

, the respective reaction rates are

100/3

25

, and

30 μ M mi n^{- 1}

; at

I = 30 μ M

they are

20

12.5

, and

50/3 μ M mi n^{- 1}

. These are independently chosen teaching models, not measured inhibition constants for a named enzyme.

Sequential-enzyme Hill limit 2026-10-31

For sequential enzyme binding, a quasi-steady-state approximation gives occupied-state weights

ϕ_{j} S^{j}

ϕ_{j} = \prod_{i = 1}^{j} K_{i}^{- 1}

, relative to free enzyme. If all intermediate weights tend to zero while the terminal weight remains nonzero, enzyme conservation leaves only unbound and fully bound states, and the production rate becomes the displayed Hill equation. One realization holds other rates fixed while the first binding rate tends to zero and the last tends to infinity with positive finite product. A statement merely bounding that product does not by itself specify this nontrivial distinguished limit.

Sequential two-site enzyme reaction network Created 2026-10-14 Updated 2026-10-20