Home / Courses / Biochemistry / Unit 4
Biochemistry Unit 4: Enzymes and Kinetics (Michaelis-Menten)
Enzymes are protein catalysts that make the reactions of life fast enough to matter. This unit explains how they lower activation energy and how to measure and compare their activity with Michaelis-Menten kinetics.
Covers: How enzymes lower activation energy, Active sites and catalytic mechanisms, Cofactors, coenzymes and enzyme classes, Michaelis-Menten equation, Km, Vmax, kcat and kcat/Km, Lineweaver-Burk plots, Effects of temperature and pH
Study guide
An enzyme speeds a reaction by lowering its activation energy, the barrier between substrate and transition state. It does not change ΔG, the equilibrium constant, or which direction is favored; it speeds forward and reverse reactions equally. Enzymes bind substrates in an active site, a pocket lined with specific side chains. The induced-fit model says the enzyme changes shape on binding, and the active site is most complementary to the transition state. Because rate depends exponentially on Ea, lowering the barrier by about 5.7 kJ/mol at 25 °C speeds the reaction about tenfold.
Enzymes use a few main strategies. In acid-base catalysis, side chains such as His donate or accept protons. In covalent catalysis, a nucleophile like the Ser of chymotrypsin forms a temporary bond with the substrate. Metal ions such as Zn²⁺ stabilize charges or activate water. Enzymes also bring substrates together in the right orientation. Many need cofactors: metal ions or organic coenzymes, often from vitamins, such as NAD⁺ (niacin), FAD (riboflavin) and coenzyme A (pantothenate). Enzymes are grouped into six classes: oxidoreductases, transferases, hydrolases, lyases, isomerases and ligases.
Michaelis-Menten kinetics describes E + S ⇌ ES → E + P. Measuring initial rates at many substrate concentrations gives a hyperbola: v = Vmax[S]/(Km + [S]). At low [S], rate rises almost linearly; at high [S], the enzyme is saturated and v approaches Vmax. Km is the [S] giving half Vmax. It equals (k₋₁ + kcat)/k₁ and approximates the ES dissociation constant only when catalysis is slow compared with dissociation. Vmax equals kcat times total enzyme, so it depends on how much enzyme is present while Km does not.
kcat, the turnover number, is how many substrate molecules each active site converts per second at saturation. The specificity constant kcat/Km measures efficiency at low [S], where v ≈ (kcat/Km)[E][S]. It is the right way to compare substrates or enzymes, and it cannot exceed the diffusion limit of about 10⁸ to 10⁹ M⁻¹s⁻¹. The Lineweaver-Burk plot graphs 1/v against 1/[S]: the y-intercept is 1/Vmax, the x-intercept is −1/Km and the slope is Km/Vmax. It is useful for seeing inhibition patterns, but it exaggerates errors in low-[S] points.
Enzyme activity also depends on conditions. As temperature rises, rate increases until the enzyme begins to denature, giving an optimum. pH affects the protonation of catalytic side chains and substrate, so each enzyme has a pH optimum, such as about 2 for pepsin in the stomach and about 7 to 8 for trypsin in the intestine. On tests, always check units: convert µM to M before computing kcat/Km, and remember that v/Vmax = [S]/(Km + [S]) gives the fraction of active sites occupied.
Key ideas
- Enzymes lower activation energy but do not change ΔG or the equilibrium constant.
- The active site binds the transition state more tightly than the substrate.
- Main catalytic strategies are acid-base, covalent, metal-ion and proximity/orientation catalysis.
- Km is the [S] at half Vmax and does not depend on enzyme concentration.
- Vmax = kcat[E]total, so it scales with enzyme amount.
- kcat/Km measures catalytic efficiency and is limited by diffusion.
- Lineweaver-Burk intercepts give 1/Vmax and −1/Km.
Formulas and rules
Common mistakes
- Saying enzymes change ΔG or shift equilibrium.
- Thinking Km changes when enzyme concentration changes.
- Comparing enzymes by kcat alone instead of kcat/Km at low substrate.
- Mixing up Lineweaver-Burk intercepts: y gives 1/Vmax, x gives −1/Km.
- Forgetting to convert µM to M when computing kcat/Km.
Before the test, make sure you can
- Explain how enzymes lower activation energy without changing ΔG.
- Describe acid-base, covalent and metal-ion catalysis with an example.
- Calculate v, Km or Vmax from the Michaelis-Menten equation.
- Calculate kcat and kcat/Km with correct units.
- Read Km and Vmax from a Lineweaver-Burk plot.
- Compare enzymes or substrates using kcat/Km.
- Explain how temperature and pH affect enzyme activity.
- Explain when Km approximates Kd.
Words to know
- Activation energy (Ea)
- The free-energy barrier between reactants and the transition state.
- Transition state
- The highest-energy, short-lived arrangement along the reaction path.
- Active site
- The region of an enzyme where substrate binds and catalysis occurs.
- Induced fit
- The model in which substrate binding changes the enzyme's shape to fit.
- Cofactor
- A non-protein helper, metal ion or organic molecule, needed for activity.
- Coenzyme
- An organic cofactor, often vitamin-derived, such as NAD⁺, FAD or CoA.
- Holoenzyme
- An enzyme with its cofactor bound; the protein alone is the apoenzyme.
- Michaelis-Menten equation
- v = Vmax[S]/(Km + [S]), relating rate to substrate concentration.
- Km
- The substrate concentration at which v = ½Vmax.
- Vmax
- The maximal rate when enzyme is saturated; equals kcat[E]total.
- kcat
- Turnover number: reactions per active site per second at saturation.
- kcat/Km
- The specificity constant, a measure of catalytic efficiency at low [S].
- Diffusion limit
- About 10⁸–10⁹ M⁻¹s⁻¹, the upper bound on kcat/Km.
- Lineweaver-Burk plot
- A double-reciprocal plot of 1/v versus 1/[S].
- Steady state
- Condition in which [ES] stays roughly constant during the measurement.
- Catalytic triad
- Ser, His and Asp in serine proteases that work together in catalysis.
- Oxyanion hole
- Part of a serine protease active site that stabilizes the negatively charged tetrahedral intermediate.
Sample practice questions
30 more questions like these, from level 1 to 5, are in the practice set.
Enzymes speed up reactions by:
- Lowering the activation energy
- Making ΔG of the reaction more negative
- Raising the energy of the products
- Shifting the equilibrium toward products
Show the answer and steps
Answer: A) Lowering the activation energy
- A catalyst provides a lower-energy path through the transition state.
- It speeds forward and reverse reactions equally, so the equilibrium constant is unchanged.
- ΔG, which depends only on reactants and products, is also unchanged.
In the Michaelis-Menten model, Km equals:
- The maximum rate of the reaction
- The rate constant for product release
- The [S] at which v = Vmax
- The [S] at which v = ½Vmax
Show the answer and steps
Answer: D) The [S] at which v = ½Vmax
- v = Vmax[S]/(Km + [S]).
- Setting [S] = Km gives v = Vmax·Km/(2Km) = ½Vmax.
- v only approaches Vmax as [S] becomes very large; it never reaches it at a finite [S].
From a Lineweaver-Burk plot, the x-intercept is −0.25 µM⁻¹ and the y-intercept is 0.020 min/µM. What is Km?
- 4.0 µM
- 0.08 µM
- 0.25 µM
- 50 µM
Show the answer and steps
Answer: A) 4.0 µM
- The x-intercept equals −1/Km.
- Km = 1/0.25 = 4.0 µM.
- 50 is Vmax (1/0.020); 0.08 is the slope Km/Vmax.