4.5.4Biomolecules

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

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1. The active site & the two binding models

Lock-and-Key (Emil Fischer, 1894)

Induced Fit (Daniel Koshland, 1958)

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

2. Michaelis–Menten kinetics — derived from scratch

Step 1 — The mechanism

E+S  k1k1  ES  k2  E+PE + S \;\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\; ES \;\overset{k_2}{\longrightarrow}\; E + P Why this step? S must bind first (forming ES), then ES converts to product P, releasing free enzyme.

Step 2 — Rate of product formation

v=k2[ES]v = k_2[ES] Why this step? Product only comes from the ES complex breaking down, so rate [ES]\propto [ES].

Step 3 — Steady-state assumption (Briggs–Haldane)

After a fast start, [ES][ES] stays roughly constant: formation = breakdown. k1[E][S]formation=(k1+k2)[ES]breakdown\underbrace{k_1[E][S]}_{\text{formation}} = \underbrace{(k_{-1}+k_2)[ES]}_{\text{breakdown}} Why this step? ES is a short-lived intermediate; its concentration barely changes during measurement, so d[ES]dt0\frac{d[ES]}{dt}\approx 0.

Step 4 — Solve for [ES][ES]

[ES]=k1[E][S]k1+k2=[E][S]Km,Kmk1+k2k1[ES] = \frac{k_1[E][S]}{k_{-1}+k_2} = \frac{[E][S]}{K_m}, \qquad K_m \equiv \frac{k_{-1}+k_2}{k_1} Why this step? We defined the Michaelis constant KmK_m to bundle the rate constants — it has units of concentration.

Step 5 — Conserve total enzyme

Free enzyme [E][E] is unknown, but total is fixed: [E]T=[E]+[ES]    [E]=[E]T[ES][E]_T = [E] + [ES] \;\Rightarrow\; [E] = [E]_T - [ES] Substitute into Step 4 and solve: [ES]=([E]T[ES])[S]Km[ES]Km=[E]T[S][ES][S][ES] = \frac{([E]_T-[ES])[S]}{K_m} \Rightarrow [ES]\,K_m = [E]_T[S]-[ES][S] [ES]=[E]T[S]Km+[S][ES] = \frac{[E]_T[S]}{K_m+[S]} Why this step? We can't measure free [E][E], but we do know total enzyme [E]T[E]_T.

Step 6 — The Michaelis–Menten equation

Put Step 5 into v=k2[ES]v=k_2[ES], and note maximum rate Vmax=k2[E]TV_{max}=k_2[E]_T (all enzyme as ES):

The three regimes


3. Worked examples


4. Lineweaver–Burk (double reciprocal) — straightening the hyperbola


Flashcards

What does an enzyme do to activation energy and to equilibrium position?
Lowers EaE_a (speeds reaction); does NOT change equilibrium position.
Lock-and-key model: shape of active site?
Rigid, pre-shaped, complementary to substrate.
Induced-fit model: what happens on binding?
Active site is flexible and changes shape to wrap around the substrate.
Why is induced fit better than lock-and-key?
It explains catalysis (binding strains/orients S) and some substrate flexibility, not just recognition.
Write the Michaelis–Menten equation.
v0=Vmax[S]Km+[S]v_0 = \dfrac{V_{max}[S]}{K_m+[S]}
Define KmK_m in terms of rate constants.
Km=k1+k2k1K_m=\dfrac{k_{-1}+k_2}{k_1}
Physical meaning of KmK_m?
The [S][S] at which v0=Vmax/2v_0=V_{max}/2; lower KmK_m = higher substrate affinity.
What is VmaxV_{max} in terms of k2k_2 and enzyme?
Vmax=k2[E]TV_{max}=k_2[E]_T; proportional to total enzyme.
At [S]Km[S]\gg K_m, what order is the reaction?
Zero order; v0Vmaxv_0\approx V_{max} (saturated).
At [S]Km[S]\ll K_m, what order?
First order; v0(Vmax/Km)[S]v_0\approx (V_{max}/K_m)[S].
What assumption gives the MM equation?
Steady state: d[ES]/dt0d[ES]/dt\approx0 (formation = breakdown of ES).
Lineweaver–Burk y-intercept and x-intercept?
y-intercept =1/Vmax=1/V_{max}; x-intercept =1/Km=-1/K_m.
Recall Feynman: explain to a 12-year-old

Imagine a vending machine (enzyme) with one slot shaped for one coin (substrate). Lock-and-key: the slot is hard plastic — only the exact coin fits. Induced fit: the slot is soft rubber that squeezes around the coin to grab it tightly. Now, how fast does it give snacks? With few coins it's slow; as you feed more coins it speeds up — but there's only one machine, so once it's working flat-out, more coins just wait in line. That top speed is VmaxV_{max}, and the number of coins needed to run at half-speed is KmK_m. A machine that runs at half-speed with very few coins is a "greedy/grabby" machine — low KmK_m, high affinity.

Connections

  • Proteins — enzymes are globular proteins; structure → active site.
  • Catalysis — enzymes vs inorganic catalysts; activation energy.
  • Chemical Kinetics — rate laws, order of reaction, rate constants.
  • Equilibrium — why catalysts don't shift KeqK_{eq}.
  • Enzyme Inhibition — competitive (↑KmK_m) vs non-competitive (↓VmaxV_{max}).
  • Vitamins and Coenzymes — cofactors that complete active sites.

Concept Map

works by

does not change

contains

binds

model 1

model 2

special case of

forms with enzyme

breaks down to

uses

predicts

Enzymes / biological catalysts

Lower activation energy Ea

Equilibrium unchanged

Active site

Substrate S

Lock-and-Key rigid

Induced Fit flexible

Enzyme-substrate complex ES

Michaelis-Menten kinetics

Steady-state assumption

Rate v0 vs S, saturates

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, enzyme ek biological catalyst hota hai — yeh reaction ko fast karta hai by activation energy ko kam karke, lekin equilibrium ko shift nahi karta. Har enzyme ke paas ek special pocket hoti hai jise active site kehte hain, jahan substrate (reactant) aa ke fit hota hai. Ab sawaal: substrate fit kaise hota hai? Do model hain. Lock-and-key kehta hai active site bilkul rigid hai, jaise lock mein sirf sahi chaabi ghusti hai — isliye specificity samajh aati hai. Induced fit isse behtar hai: active site flexible hai, substrate aate hi enzyme apna shape badal ke uske around mould ho jaata hai, jaise haath ball ke around band hota hai. Yeh moulding hi catalysis mein madad karti hai.

Ab speed ki baat — Michaelis–Menten kinetics. Agar tum v0v_0 (rate) ko [S][S] ke against plot karo, toh ek hyperbola milta hai: pehle tezi se badhta, phir saturate ho jaata hai (flat). Reason simple hai — enzyme limited hai, ek baar saare enzyme busy ho gaye toh aur substrate daalo bhi toh fayda nahi. Equation hai v0=Vmax[S]Km+[S]v_0=\frac{V_{max}[S]}{K_m+[S]}.

Isme do star players: Vmax=k2[E]TV_{max}=k_2[E]_T matlab top speed, jo enzyme ki amount par depend karti hai. Aur KmK_m — yeh wo [S][S] hai jahan rate aadha VmaxV_{max} ho jaata hai. Yaad rakho: chhota KmK_m = high affinity (enzyme substrate ko strongly grab karta hai). Exam ka 80/20 yahi hai — "Km = half Vmax", "low Km = high affinity", aur teen regimes (low SS → first order, high SS → zero order). Itna pakka kar lo toh zyादातर questions ban jaate hain.

Go deeper — visual, from zero

Test yourself — Biomolecules

Connections