Enzymes — lock-and-key vs induced fit; Michaelis-Menten kinetics: Question bank

Before we start, one reminder of the symbols so nothing here is used unexplained. Everything below traces back to the same one-line mechanism:

$E + S k_{- 1} ⇌ k_{1} E S ⟶ k_{2} E + P$

Definition Symbol refresher

$E$ = free enzyme, $S$ = substrate (the reactant), $E S$ = the bound enzyme–substrate complex, $P$ = product.
$k_{1}$ = rate constant for binding — the "on" speed at which free $E$ and $S$ collide and stick to form $E S$ (units of $1/ (conc \cdot time)$ , since two things must meet).
$k_{- 1}$ = rate constant for release — the "off" speed at which $E S$ falls back apart into $E + S$ without reacting (units of $1/ time$ ).
$k_{2}$ = catalytic rate constant — the speed at which a bound $E S$ actually converts to product $E + P$ (units of $1/ time$ ). A rate constant is just the fixed number that tells you how fast a given step runs at unit concentration (see Chemical Kinetics).
$v_{0}$ = initial rate (how fast product forms right at the start).
$V_{ma x} = k_{2} [E]_{T}$ = the top speed when every enzyme molecule is busy ( $[E]_{T}$ = total enzyme).
$K_{m} = \frac{k _{- 1} + k _{2}}{k _{1}}$ = the substrate concentration $[S]$ at which $v_{0} = \frac{1}{2} V_{ma x}$ .
$E_{a}$ = activation energy — the energy hill a reaction must climb.
$Δ G$ = Gibbs free-energy change — the net downhill (or uphill) energy difference between reactants and products; its sign fixes the equilibrium position, and it is untouched by a catalyst (see Equilibrium).

To keep this bank visual-first, two sketches anchor the two ideas most items lean on — a reaction-coordinate picture for $E_{a}$ vs $Δ G$ , and the Michaelis–Menten saturation curve. Glance at them before answering.

Figure — Enzymes — lock-and-key vs induced fit; Michaelis-Menten kinetics

True or false — justify

An enzyme shifts the equilibrium toward more product.

False. A catalyst lowers

E_{a}

for both forward and reverse directions equally (look at figure 1: it shaves the same hill from either side), so it speeds arrival at equilibrium but leaves the equilibrium position — fixed by

Δ G

— untouched. See Equilibrium.

Adding more enzyme raises

V_{ma x}

True.

V_{ma x} = k_{2} [E]_{T}

is proportional to total enzyme, so doubling

[E]_{T}

doubles the ceiling.

Adding more enzyme lowers

K_{m}

False.

K_{m} = \frac{k _{- 1} + k _{2}}{k _{1}}

contains only the rate constants

k_{1}, k_{- 1}, k_{2}

— it is intrinsic to the enzyme–substrate pair and independent of how much enzyme you add.

Lock-and-key is simply wrong and has been discarded.

False. It is a limiting case of induced fit (near-zero shape change); it still correctly captures specificity — induced fit only adds flexibility and catalytic strain.

At very high

[S]

the reaction is first order in substrate.

False. At

[S] ≫ K_{m}

every enzyme is saturated (the flat right side of figure 2), so

v_{0} \approx V_{ma x}

is constant — that is zero order in

[S]

A small

K_{m}

means the enzyme binds its substrate tightly.

True. Small

K_{m}

means half-speed is reached at low

[S]

, i.e. high affinity for the substrate.

The Michaelis–Menten equation assumes

[E S]

never changes at all.

False. It assumes

[E S]

is approximately steady (

d [E S] / d t \approx 0

) after a brief build-up — formation equals breakdown, not that

[E S]

is literally frozen forever.

An enzyme is consumed in the reaction it catalyses.

False. The final step

E S \to E + P

(governed by

k_{2}

) regenerates free

E

, so the enzyme emerges unchanged and can cycle again — the defining trait of a catalyst (Catalysis).

Induced fit means the substrate changes shape to fit the enzyme.

False (main effect reversed). In induced fit the enzyme's active site moulds around the substrate; the strain this imposes may then help distort the substrate's bonds.

Spot the error

V_{ma x}

is a fixed, unchangeable property of the enzyme."

The error:

V_{ma x}

depends on

[E]_{T}

(through

V_{ma x} = k_{2} [E]_{T}

), so it changes with enzyme amount. Only

K_{m}

is intrinsic.

"Since the enzyme lowers

E_{a}

, it makes the reaction more exothermic."

The error:

E_{a}

(the kinetic barrier) and

Δ H

Δ G

(the thermodynamics) are separate — in figure 1 the barrier height and the reactant–product drop are two different vertical distances. Lowering the barrier changes speed, not the energy released or the equilibrium.

"At

[S] = K_{m}

the enzyme is running at maximum speed."

The error: at

[S] = K_{m}

the rate is exactly

V_{ma x} /2

— half speed, not maximum. Maximum is only approached as

[S] \to \infty

"Because a hyperbola saturates,

80%

V_{ma x}

needs only slightly more

S

than

50%

The error: saturation is slow. Reaching

80%

V_{ma x}

needs

[S] = 4 K_{m}

, whereas

50%

needs

[S] = K_{m}

— a fourfold increase for that last stretch.

"A competitive inhibitor lowers

V_{ma x}

The error: a competitor merely fights for the active site, so enough substrate can still saturate the enzyme —

V_{ma x}

is unchanged while apparent

K_{m}

rises (see Enzyme Inhibition).

"The Lineweaver–Burk x-intercept gives

+ 1/ K_{m}

The error: setting

1/ v_{0} = 0

\frac{1}{v _{0}} = \frac{K _{m}}{V _{ma x}} \frac{1}{[ S ]} + \frac{1}{V _{ma x}}

gives

1/ [S] = - 1/ K_{m}

, so the x-intercept is negative,

- 1/ K_{m}

"All enzymes are proteins, so no non-protein molecule can catalyse biologically."

The error: most enzymes are proteins (Proteins), but catalytic RNA (ribozymes) exists, so "enzyme = protein" is a strong tendency, not an absolute law.

Why questions

Why does the

v_{0}

-vs-

[S]

curve level off instead of rising forever?

Because there is a fixed amount of enzyme; once every active site is occupied (saturated), extra substrate must wait, so the rate cannot exceed

V_{ma x}

— the plateau in figure 2.

Why is the steady-state assumption justified during a rate measurement?

There is a timescale separation:

E S

is consumed (via

k_{- 1}

and

k_{2}

) far faster than the bulk

[S]

is depleted, so after a very brief build-up phase the small

[E S]

pool settles into a near-constant value where its rate of formation (

k_{1} [E] [S]

) balances its rate of loss (

(k_{- 1} + k_{2}) [E S]

). Over the measurement window

d [E S] / d t \approx 0

because

E S

is being replenished as fast as it disappears.

Why do we plot the initial rate

v_{0}

rather than a later rate?

Early on

[S]

is essentially unchanged and product

P

is negligible, so the reverse reaction and substrate depletion don't yet distort the measurement — the mechanism runs strictly left-to-right through

k_{1}

and

k_{2}

Why does induced fit explain catalysis better than lock-and-key?

Because the shape change on binding strains and correctly orients the substrate's bonds, actively helping them break — rigid lock-and-key only explains recognition, not bond-breaking.

Why bother taking reciprocals to get the Lineweaver–Burk plot?

A rectangular hyperbola never truly reaches

V_{ma x}

, making it hard to read the ceiling by eye; the double-reciprocal turns it into a straight line whose intercepts give

V_{ma x}

and

K_{m}

precisely.

Why is

K_{m}

measured in units of concentration?

Because it equals a specific value of

[S]

(the half-saturation point); dimensionally

\frac{k _{- 1} + k _{2}}{k _{1}}

works out to concentration since

k_{1}

is second-order (units

1/ (conc \cdot time)

) while

k_{- 1}, k_{2}

are first-order (units

1/ time

), leaving concentration.

Why can some coenzymes/vitamins be essential for enzyme activity?

Many enzymes need a non-protein helper (coenzyme, often vitamin-derived) to complete a functional active site — without it the protein alone cannot catalyse (Vitamins and Coenzymes).

Edge cases

What is

v_{0}

when

[S] = 0

Zero — with no substrate no

E S

forms, so

v_{0} = \frac{V _{ma x} \cdot 0}{K _{m} + 0} = 0

; the curve passes through the origin in figure 2.

What happens to

v_{0}

[S] \to \infty

It approaches

V_{ma x}

asymptotically but never exactly reaches it, because the fraction

\frac{[ S ]}{K _{m} + [ S ]} \to 1

only in the limit.

k_{2} ≫ k_{- 1}

(product forms much faster than

E S

falls apart), what does

K_{m}

approach?

K_{m} \to k_{2} / k_{1}

; it no longer reflects true binding affinity because breakdown is dominated by catalysis (

k_{2}

), not by release of substrate (

k_{- 1}

Is lock-and-key ever the correct picture?

Yes — for a rigid enzyme with negligible conformational change on binding, lock-and-key is the accurate limiting form of induced fit.

At the exact half-saturation point

[S] = K_{m}

, what fraction of enzyme is bound as

E S

Exactly half — since

[E S] = \frac{[ E ] _{T} [ S ]}{K _{m} + [ S ]}

gives

[E S] = [E]_{T} /2

when

[S] = K_{m}

, matching

v_{0} = V_{ma x} /2

What would a hypothetical

K_{m} = 0

imply?

Infinitely tight binding — the enzyme would sit at half-speed even at vanishingly small

[S]

, effectively saturating for any

[S] > 0

; a physical idealisation, not a real value.

Recall One-line self-test before you close this page

"Which two quantities can I read straight off a Michaelis–Menten graph, and which one depends on how much enzyme I added?" Answer ::: $K_{m}$ (x-value at half-height) and $V_{ma x}$ (the ceiling) are both readable; only $V_{ma x} = k_{2} [E]_{T}$ depends on enzyme amount.