After a fast start, [ES] stays roughly constant: formation = breakdown.
formationk1[E][S]=breakdown(k−1+k2)[ES]Why this step? ES is a short-lived intermediate; its concentration barely changes during measurement, so dtd[ES]≈0.
[ES]=k−1+k2k1[E][S]=Km[E][S],Km≡k1k−1+k2Why this step? We defined the Michaelis constantKm to bundle the rate constants — it has units of concentration.
Free enzyme [E] is unknown, but total is fixed:
[E]T=[E]+[ES]⇒[E]=[E]T−[ES]
Substitute into Step 4 and solve:
[ES]=Km([E]T−[ES])[S]⇒[ES]Km=[E]T[S]−[ES][S][ES]=Km+[S][E]T[S]Why this step? We can't measure free [E], but we do know total enzyme [E]T.
What does an enzyme do to activation energy and to equilibrium position?
Lowers Ea (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=Km+[S]Vmax[S]
Define Km in terms of rate constants.
Km=k1k−1+k2
Physical meaning of Km?
The [S] at which v0=Vmax/2; lower Km = higher substrate affinity.
What is Vmax in terms of k2 and enzyme?
Vmax=k2[E]T; proportional to total enzyme.
At [S]≫Km, what order is the reaction?
Zero order; v0≈Vmax (saturated).
At [S]≪Km, what order?
First order; v0≈(Vmax/Km)[S].
What assumption gives the MM equation?
Steady state: d[ES]/dt≈0 (formation = breakdown of ES).
Lineweaver–Burk y-intercept and x-intercept?
y-intercept =1/Vmax; x-intercept =−1/Km.
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 Vmax, and the number of coins needed to run at half-speed is Km. A machine that runs at half-speed with very few coins is a "greedy/grabby" machine — low Km, high affinity.
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 v0 (rate) ko [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=Km+[S]Vmax[S].
Isme do star players: Vmax=k2[E]T matlab top speed, jo enzyme ki amount par depend karti hai. Aur Km — yeh wo [S] hai jahan rate aadha Vmax ho jaata hai. Yaad rakho: chhota Km = 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 S → first order, high S → zero order). Itna pakka kar lo toh zyादातर questions ban jaate hain.