Enzymes

Enzymes are biological catalysts. They speed reactions up enormously — not by changing the thermodynamics (they do not), but by providing a lower-energy pathway. Everything the MCAT tests about enzymes flows from that one insight.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Mechanism: How Enzymes Actually Work

The Thermodynamic Picture

Must know

An enzyme does not change $\Delta G$. The free-energy gap between substrates and products is fixed by chemistry. What an enzyme does is lower the activation energy ($E_a$) by stabilizing the transition state, making the reaction far faster at body temperature.

$\Delta G^\ddagger_{\text{uncatalyzed}} \gg \Delta G^\ddagger_{\text{catalyzed}}$

Because rate depends exponentially on $E_a$, even a modest drop in $E_a$ gives a huge rate acceleration.

Molecular Strategies for Catalysis

Know the logic

Understand these mechanisms; don't memorize every example enzyme.

• Proximity/orientation: holding substrates together in the right geometry.
• Transition-state stabilization: the active site is complementary to the transition state, not the ground-state substrate — a subtle but critical point.
• Covalent catalysis: an active-site nucleophile transiently bonds the substrate. Serine proteases (chymotrypsin, trypsin) are the canonical example, using a Ser–His–Asp catalytic triad as a charge relay.
• Acid–base catalysis: active-site residues donate/accept protons.
• Metal-ion catalysis: a metal (e.g., $\ce{Zn^2+}$ in carbonic anhydrase) acts as a Lewis acid.

Substrate Specificity and the Binding Models (Brief)

Know the logic

To frame catalysis: an enzyme acts on a specific substrate because the active site's shape, charge, and H-bonding are complementary to it. The lock-and-key model pictures a rigid, preformed pocket; the accepted induced-fit model has a flexible site that closes around the correct substrate to align catalytic residues. Either way, the site is ultimately complementary to the transition state, not the ground-state substrate — the basis for the transition-state stabilization above. (Enzyme classification, binding-model detail, and cofactors/coenzymes are covered in the 1A Enzyme Structure and Function guide.)

Quick check: Hexokinase undergoes a "clam-shell" closure around glucose on binding. Which model does this illustrate, and what's the advantage?

Answer: Induced-fit. The closure excludes water, preventing wasteful ATP hydrolysis — the conformational change creates the right environment for catalysis, not just binding.

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Kinetics

General Concepts

Must know

Two intuitions:

1. An enzyme lowers $E_a$ for both directions equally — it speeds the approach to equilibrium but does not change $K_{eq}$.
2. Enzyme catalysis is saturable: finite active sites mean that at high $[S]$, all sites are full and adding substrate does nothing. This is what distinguishes enzyme kinetics from simple chemical kinetics.

Turnover number ($k_{cat}$) = substrate molecules converted per enzyme per second at saturation. Catalytic efficiency $k_{cat}/K_m$ is the best single measure of how good an enzyme is (fast and tight-binding); the most efficient enzymes approach the diffusion-controlled limit.

Michaelis-Menten Kinetics

Must know

The model (single-substrate, steady-state assumption that $[\ce{ES}]$ stays roughly constant):

$\ce{E + S <=>[\mathit{k_1}][\mathit{k_{-1}}] ES ->[\mathit{k_2}] E + P}$

$\boxed{v = \frac{V_{\max}[S]}{K_m + [S]}}$

• $v$ = velocity; $V_{\max} = k_{cat}[E]_{\text{total}}$ = max velocity; $[S]$ = substrate conc.
• $K_m$ = Michaelis constant = $[S]$ at which $v = V_{\max}/2$.

Low $K_m$ = high affinity (half-saturation reached at low $[S]$); high $K_m$ = low affinity. $K_m$ approximates $K_d$ of the ES complex when $k_2 \ll k_{-1}$.

The $v$ vs. $[S]$ plot is a rectangular hyperbola: first-order at low $[S]$, approaching zero-order ($v \approx V_{\max}$) at high $[S]$.

The Lineweaver-Burk Plot

Must know

Taking the reciprocal linearizes Michaelis-Menten:

$\frac{1}{v} = \frac{K_m}{V_{\max}} \cdot \frac{1}{[S]} + \frac{1}{V_{\max}}$

• y-intercept = $1/V_{\max}$; slope = $K_m/V_{\max}$; x-intercept = $-1/K_m$.

Must know these intercepts cold — they're how you read off $V_{\max}$ and $K_m$ and how you distinguish inhibition types.

Worked Example

Know the logic

Given data where $v \to 100\,\mu\text{M/min}$ as $[S]\to\infty$, and $v = 50$ at $[S] = 2.0$ mM:

• $V_{\max} = 100\,\mu\text{M/min}$ (the plateau).
• $K_m = 2.0$ mM (the $[S]$ giving half $V_{\max}$).
• Check at $[S] = 1.0$: $v = \dfrac{100 \times 1.0}{2.0 + 1.0} = 33.3\,\mu\text{M/min}$ ✓
• LB intercepts: y-int $= 1/100 = 0.01$; x-int $= -1/2.0 = -0.5\,\text{mM}^{-1}$; slope $= 2.0/100 = 0.02$.

(Cooperative, multi-subunit enzymes give a sigmoidal $v$ vs. $[S]$ curve rather than a hyperbola; cooperativity is covered with biological regulation in the 1A Control of Enzyme Activity guide.)

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Effects of Local Conditions on Enzyme Activity

Must know

Each enzyme has an optimum for temperature and pH; extremes denature it.

• Temperature: rate rises with temperature until the optimum (~$37°C$ for human enzymes), then drops sharply as the protein denatures — a bell-shaped curve.
• pH: activity depends on the protonation state of active-site residues; each enzyme has a pH optimum (e.g., pepsin ~2, most intracellular enzymes ~7.4).
• Passage-level Salt/substrate/product: high ionic strength can disrupt salt bridges; product accumulation slows the forward rate (product inhibition).

Quick check: A lab measures serum enzyme activity at 25°C instead of 37°C. How will measured activity compare to true physiological activity?

Answer: Lower — slower molecular motion and fewer collisions exceeding $E_a$, so $v$ underestimates true activity.

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Inhibition

Must know

Inhibition types and their effects on $K_m$, $V_{\max}$, and the Lineweaver-Burk plot — heavily tested.

Competitive

Must know

Inhibitor resembles substrate and binds the active site; the two compete.

• $V_{\max}$ unchanged (high $[S]$ outcompetes inhibitor); $K_m$ increases.
• LB: same y-intercept, steeper slope, x-intercept shifts toward zero.
• Optional Examples: statins (HMG-CoA reductase), methotrexate (DHFR).

Uncompetitive

Must know

Inhibitor binds only the ES complex.

• $V_{\max}$ decreases; $K_m$ decreases (same factor) → slope unchanged.
• LB: parallel lines, higher y-intercept.

Mixed / Pure Noncompetitive

Must know

Inhibitor binds both E and ES at a site other than the active site (allosteric).

• Pure noncompetitive (equal affinity for E and ES): $V_{\max}$ decreases, $K_m$ unchanged.
• LB: lines intersect on the x-axis (x-intercept unchanged, higher y-intercept).

Type	$K_m$	$V_{\max}$	LB lines
Competitive	↑	—	same y-int, steeper
Uncompetitive	↓	↓	parallel
Pure noncompetitive	—	↓	cross on x-axis

Irreversible

Know the logic

Some inhibitors form covalent bonds, permanently inactivating the enzyme — not described by standard $K_m/V_{\max}$ analysis. Optional Examples: aspirin (COX), organophosphates (acetylcholinesterase), penicillin (transpeptidase).

Quick check: With an inhibitor, the LB y-intercept is unchanged but the slope is steeper. What are $K_m$ and $V_{\max}$ doing, and why is the y-intercept unchanged?

Answer: $V_{\max}$ unchanged (same y-intercept $= 1/V_{\max}$), $K_m$ increased (steeper slope) — a competitive inhibitor. The y-intercept holds because high $[S]$ still outcompetes the inhibitor to reach $V_{\max}$.

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Common Confusions & Tricks

1. "Enzymes shift equilibrium" — WRONG. They speed the approach to equilibrium but don't change $K_{eq}$ or $\Delta G°'$. A passage claiming an enzyme makes a thermodynamically unfavorable reaction favorable is wrong.

2. Competitive ↑ $K_m$; uncompetitive decreases $K_m$. Uncompetitive inhibitors bind ES preferentially, pulling equilibrium toward ES — you need less substrate for half-saturation. (Here decreased $K_m$ is a kinetic artifact, not increased affinity.)

3. LB x-intercept with competitive inhibition. $K_m$ ↑ → $1/K_m$ ↓ → the x-intercept $(-1/K_m)$ moves toward zero (right). Remember you're plotting $-1/K_m$, so the sign flips the intuition.

4. $K_m$ ≠ affinity (always). $K_m \approx K_d$ only when $k_{cat} \ll k_{-1}$. Safe rule: lower $K_m$ = half-saturation at lower $[S]$.

5. Coenzyme vs. cofactor vs. prosthetic group. Cofactor = any non-protein helper. Coenzyme = organic, vitamin-derived. Prosthetic group = tightly/covalently bound (FAD, heme); if it floats away and returns it's a cosubstrate (NAD⁺, CoA).

6. Allosteric vs. active site. Allosteric effectors bind a separate site; competitive inhibitors bind the active site. Test: does high substrate overcome it? Yes → competitive; no → allosteric.

7. Irreversible inhibition is not kinetics. Standard $K_m/V_{\max}$ analysis doesn't apply; adding substrate can't reverse a covalent inhibitor.

8. "ATCase = cooperative AND allosteric." Cooperativity (subunit communication) and allostery (regulation by a distinct effector) often coexist but are conceptually separate.

9. Zymogen activation is irreversible. Cleaving pepsinogen to pepsin can't be undone — unlike reversible allostery and phosphorylation.

10. Temperature bell curve. Activity rises with temperature to the optimum, then drops from denaturation. Don't confuse this with temperature's effect on $K_{eq}$ (van't Hoff thermodynamics).

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Key Equations

Equation	Variables & When to Use
$v = \dfrac{V_{\max}[S]}{K_m + [S]}$	Michaelis-Menten: $v$ = initial velocity, $V_{\max}$ = max velocity, $[S]$ = substrate conc., $K_m$ = $[S]$ at half-maximal velocity
$\dfrac{1}{v} = \dfrac{K_m}{V_{\max}} \cdot \dfrac{1}{[S]} + \dfrac{1}{V_{\max}}$	Lineweaver-Burk: linearizes Michaelis-Menten; y-intercept = $1/V_{\max}$, x-intercept = $-1/K_m$, slope = $K_m/V_{\max}$
$V_{\max} = k_{cat}[E]_T$	Relates $V_{\max}$ to turnover number $k_{cat}$ and total enzyme $[E]_T$
$\text{catalytic efficiency} = \dfrac{k_{cat}}{K_m}$	Overall measure of enzyme performance (speed × binding); diffusion-limited ceiling $\sim 10^8$–$10^9\,\text{M}^{-1}\text{s}^{-1}$
$K_m^{\text{app}} = K_m \left(1 + \dfrac{[I]}{K_i}\right)$	Competitive inhibition: apparent $K_m$ increases; $V_{\max}$ unchanged
$V_{\max}^{\text{app}} = \dfrac{V_{\max}}{1 + [I]/K_i}$	Uncompetitive/noncompetitive: apparent $V_{\max}$ decreases
$\ce{E + S <=>[\mathit{k_1}][\mathit{k_{-1}}] ES ->[\mathit{k_{cat}}] E + P}$	Standard Michaelis-Menten reaction scheme