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Enzyme kinetics

Biochemistry (B.Sc) · Biomolecules and enzymes · free study notes for exams, placements and interviews

In short

Enzyme kinetics is the quantitative study of how fast enzymes convert substrate into product and how that rate changes with substrate concentration, enzyme concentration, pH, temperature, inhibitors and activators. It gives the two numbers that define a catalyst — Km (substrate concentration at half-maximal velocity, a measure of apparent affinity) and Vmax (saturation velocity, from which kcat is derived) — and it is the physical basis of a large share of modern drugs — roughly a quarter to a third of marketed drug targets are enzymes, acted on by inhibitors. It is a guaranteed scoring section in B.Sc/M.Sc Biochemistry papers, CSIR-UGC NET Life Sciences, GAT-B/DBT-JRF, ICMR-JRF, GATE Biotechnology (BT) and Life Sciences (XL) papers. In jobs, it is tested in interviews for analytical and QC roles in Indian pharma and enzyme companies (Biocon, Sun Pharma, Dr Reddy's, Advanced Enzyme Technologies, Novozymes South Asia), in clinical biochemistry labs (NABL-accredited, running ALT, AST, ALP, CK, LDH assays) and in biofuel, detergent and food-enzyme process development.

Explained

Basic concepts and rate measurement

Michaelis–Menten model and its derivation

Linearised plots and parameter estimation

PlotLinear formx-axisy-axisSlopeIntercepts
Lineweaver–Burk (double reciprocal)1/v = (Km/Vmax)(1/[S]) + 1/Vmax1/[S]1/vKm/Vmaxy: 1/Vmax; x: −1/Km
Eadie–Hofsteev = −Km(v/[S]) + Vmaxv/[S]v−Kmy: Vmax; x: Vmax/Km
Hanes–Woolf[S]/v = (1/Vmax)[S] + Km/Vmax[S][S]/v1/Vmaxy: Km/Vmax; x: −Km
Direct linear (Eisenthal–Cornish-Bowden)lines from (−[S], 0) to (0, v) intersect at (Km, Vmax)———median intersection used

Inhibition: types, equations and diagnosis

Allosteric and cooperative kinetics

Multi-substrate reactions and bisubstrate kinetics

Effect of pH, temperature and other factors

Key points

Formulas and examples

v₀ = Vmax[S]/(Km + [S]) — v₀ = initial velocity (µmol min⁻¹ or M s⁻¹); [S] = substrate concentration (M); Km = Michaelis constant (M); Vmax = maximal velocity (same units as v₀)
Km = (k₋₁ + k₂)/k₁ ; Ks = k₋₁/k₁ (rapid-equilibrium dissociation constant, M)
Vmax = kcat[E]ₜ ; kcat = Vmax/[E]ₜ (s⁻¹) ; catalytic efficiency = kcat/Km (M⁻¹ s⁻¹)
1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax (Lineweaver–Burk) ; v₀ = −Km(v₀/[S]) + Vmax (Eadie–Hofstee) ; [S]/v₀ = [S]/Vmax + Km/Vmax (Hanes–Woolf)
v = Vmax[S]/(αKm + α′[S]) where α = 1 + [I]/Ki and α′ = 1 + [I]/Ki′ ([I] = inhibitor concentration, Ki = inhibition constant, M)
Ki = IC₅₀/(1 + [S]/Km) (Cheng–Prusoff, competitive inhibition)
v = Vmax[S]ⁿ/(K₀.₅ⁿ + [S]ⁿ) ; log[v/(Vmax − v)] = n log[S] − n log K₀.₅ (Hill equation and plot; n = Hill coefficient, dimensionless)
Activity (IU/mL) = (ΔA/Δt × V_total) / (ε × l × V_sample) with ΔA/min, ε in M⁻¹cm⁻¹ (×10⁶ for µmol basis), l = path length (cm), volumes in mL
k = A e−Ea/RT ; ln k = ln A − Ea/RT (R = 8.314 J mol⁻¹K⁻¹, T in K, Ea in J mol⁻¹) ; Q₁₀ = k(T+10)/k(T)
ΔG‡ = −RT ln(kcat h / k_B T) (Eyring); % inhibition = (1 − v_i/v₀) × 100

Example 1 — Km and Vmax from the Michaelis–Menten equation. An enzyme has Vmax = 60 µmol min⁻¹ mg⁻¹ and Km = 2.0 mM. Find v₀ at [S] = 0.5, 2.0 and 20 mM, and the [S] needed for 90% of Vmax.

Example 2 — kcat and catalytic efficiency. A purified enzyme of molecular mass 50,000 g/mol (one active site) is assayed at 0.20 µg per 1.0 mL reaction; Vmax = 0.015 µmol min⁻¹ per mL and Km = 0.40 mM.

Example 3 — identifying the inhibition type from data. With Km = 5 mM and Vmax = 100 units, adding 1 mM of an inhibitor gives apparent Km = 20 mM and apparent Vmax = 100 units.

Example 4 — clinical assay: activity in IU/L from absorbance. An ALT (SGPT) assay follows NADH oxidation at 340 nm (ε = 6,220 M⁻¹cm⁻¹, l = 1 cm). Total reaction volume 1.0 mL contains 0.100 mL serum; ΔA/min = −0.040.

Example 5 — Lineweaver–Burk from experimental points. Data: [S] = 0.2 mM, v = 20; [S] = 1.0 mM, v = 60 (arbitrary units).

Inhibition typeBindsApparent KmVmaxLineweaver–Burk lines intersectReversed by excess SExample
CompetitiveFree E onlyIncreases (αKm)UnchangedOn y-axisYesMethotrexate / DHFR
Non-competitive (pure)E and ES equallyUnchangedDecreasesOn x-axisNoSome metal-chelate inhibitors
UncompetitiveES onlyDecreasesDecreases (same factor)Parallel (no intersection)NoLi⁺ / inositol monophosphatase
MixedBoth, unequallyIncreases or decreasesDecreases2nd quadrant, off-axisPartlyMany drug inhibitors
IrreversibleCovalent at active siteTime-dependent fall in active [E]DecreasesBehaves like loss of enzymeNoAspirin / COX
EnzymeSubstrateKm (approx.)kcat (s⁻¹, approx.)kcat/Km (M⁻¹s⁻¹)
CatalaseH₂O₂25 mM4 × 10⁷~1.6 × 10⁹
Carbonic anhydraseCO₂12 mM1 × 10⁶~8.3 × 10⁷
ChymotrypsinAcetyl-Phe amide15 mM0.14~9
Hexokinase (I)Glucose0.05 mM~ 10²~10⁶
Glucokinase (hexokinase IV)Glucose~10 mM~ 10²~10⁴

Where it is used

Interview questions

Common mistakes

Related topics

Biochemistry: biomolecules and enzymes · Bioenergetics and oxidative phosphorylation · Enzyme mechanisms and catalytic strategies · Protein purification and assay techniques · Metabolic regulation and glycolysis · Pharmacology of enzyme-inhibitor drugs

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