- Describe formation and breakdown of the enzyme–substrate complex and identify the roles of k1, k−1, and k2.
- Explain the relative-concentration, steady-state, and early-reaction conditions presented for the Michaelis–Menten model.
- Use the Michaelis–Menten equation and interpret Km, Vmax, saturation, kcat, and the specificity constant.
- Identify the equations, slopes, and intercepts associated with the presented linearized kinetic plots.
Reaction Model and Rate Constants
An enzyme binds substrate to form an enzyme–substrate complex. Complex formation is reversible: the complex can return to free enzyme and substrate, or it can break down into free enzyme and product.
| Constant | Assigned step |
|---|---|
k1 | Free enzyme and substrate form the enzyme–substrate complex. |
k−1 | The enzyme–substrate complex dissociates into free enzyme and substrate. |
k2 | The enzyme–substrate complex converts into free enzyme and product. |
A complex separates into free enzyme and substrate. Which rate constant is assigned to this step?
Model Assumptions and Initial Velocity
- Substrate concentration is much greater than enzyme concentration under the relative-concentration assumption.
- At steady state, complex formation and complex breakdown occur at equal rates.
- At steady state, there is no net change in enzyme–substrate complex concentration.
At steady state, what two statements describe the formation, breakdown, and net concentration change of the enzyme–substrate complex?
Derivation of the Michaelis–Menten Equation
In the mass-action relationship used here, enzyme–substrate complex formation is directly proportional to enzyme concentration and substrate concentration. Introducing k1 gives the forward formation rate shown below.
- Begin with the reversible steady-state balance: k1[E][S] + k−2[P][E] = k−1[ES] + k2[ES].
- During the early reaction stage, product concentration is very low, and the product-dependent reverse term k−2[P] is omitted.
- Use Km = (k−1 + k2) / k1.
- Use total enzyme concentration = free enzyme concentration + enzyme–substrate complex concentration.
In the early reaction stage, which term is omitted from the full reversible steady-state balance, and what stated concentration condition supports that omission?
Km, Saturation, and Substrate Affinity
For the saturation kinetics described, plotting velocity against substrate concentration produces a rectangular hyperbola. Velocity stops increasing beyond the saturation limit even when additional substrate is supplied.
Substitute v0 = Vmax/2 into the Michaelis–Menten equation. What relationship between Km and [S] applies at that velocity?
Concentration Effects and Catalytic Performance
- At low substrate concentration, most enzyme molecules are free.
- Before saturation, increasing substrate concentration increases initial velocity.
- At the stated saturation point, all enzyme molecules are present as enzyme–substrate complexes.
- After all enzyme molecules are occupied, further substrate increases do not increase velocity.
All enzyme molecules are occupied. Predict what happens to velocity when substrate concentration is increased further.
Linearized Kinetic Plots
| Plot | Equation | Slope |
|---|---|---|
| Lineweaver–Burk | 1/v0 = (Km/Vmax)(1/[S]) + 1/Vmax | Km/Vmax |
| Hanes–Woolf | [S]/v0 = Km/Vmax + [S]/Vmax | 1/Vmax |
| Eadie–Hofstee | v0 = Vmax − Km(v0/[S]) | −Km |
For a Lineweaver–Burk plot, the x-intercept is −1/Km and the y-intercept is 1/Vmax.
A linearized plot has slope −Km. Identify the presented plot and state its equation.
Common misconceptions
Watch out: At steady state, the enzyme–substrate complex does not form or break down.
Instead: At steady state, complex formation and breakdown continue at equal rates, with no net change in complex concentration.
Watch out: A high Km indicates high enzyme affinity for substrate.
Instead: A high Km indicates low affinity; a low Km indicates high affinity.
Watch out: Adding more substrate always increases reaction velocity.
Instead: Before saturation, increasing substrate increases initial velocity. After all enzyme molecules are occupied, further substrate increases do not increase velocity.
Lesson recap
- Enzyme and substrate form a reversible enzyme–substrate complex; k1, k−1, and k2 identify the presented reaction steps.
- At steady state, complex formation equals complex breakdown, so complex concentration has no net change.
- The Michaelis–Menten equation is v0 = Vmax[S] / (Km + [S]).
- Km equals [S] at one-half Vmax; lower Km indicates higher substrate affinity.
- Before saturation, more substrate increases initial velocity; after enzyme occupancy is complete, it does not.
- The presented linearized forms are Lineweaver–Burk, Hanes–Woolf, and Eadie–Hofstee.