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An enzyme is a biological catalyst: it accelerates a chemical reaction and is regenerated after a successful catalytic cycle. Most enzymes are proteins, but catalytic RNA molecules, called ribozymes, also exist. Many reactions needed by a biological cell would proceed too slowly under its ordinary conditions without enzymes. See Cooper, The Cell: the central role of enzymes as biological catalysts.

The biological beauty is selective speed. A biological cell contains many possible reactants; its enzymes help determine which transformations happen on a useful timescale. In metabolism, one step's reaction product can become another step's substrate, so selective catalysts support an organized flow of material.

The ribosome is a striking biological example: the machinery that builds proteins has RNA at its catalytic centre. The complete ribosome contains both RNA and proteins, but its bond-forming active site is built from RNA. Structural evidence established this distinction in Nissen et al., The structural basis of ribosome activity in peptide bond synthesis, Science 289, 920–930 (2000).

Selection occurs at an active site, the region where substrate binding and catalytic chemistry take place. Shape and chemical interactions favour some substrates over others; binding can also reshape the enzyme, an induced fit. Recognition alone is insufficient: holding a substrate extremely tightly need not make it react quickly. Catalysis requires a favourable route through the transition state, the high-barrier configuration between reactants and reaction products. Positioning reactive groups and stabilizing this configuration relative to the bound substrate can lower that barrier. Specificity is often strong, but need not mean exactly one acceptable substrate. See Cooper, The Cell: mechanisms of enzymatic catalysis.

Read the diagram numerically. Its chosen Gibbs free energies are $0$ for the reactants and $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the reaction products. The uncatalysed peak is $60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, whereas the catalysed peak is $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The forward free-energy barriers are therefore $60$ and $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; viewed from the reaction products, the reverse barriers are $70$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Both decrease by $20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, while the endpoint difference stays $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are constructed values, not measurements of a particular enzyme.

This separates chemical kinetics from chemical equilibrium. An enzyme changes how rapidly the system approaches chemical equilibrium; it does not change the equilibrium constant for the same overall chemical reaction under the same conditions. Faster catalysis therefore supplies no extra driving energy. In a biological cell, continuing supply of reactants and removal of reaction products can nevertheless sustain a net flow: the surroundings keep that system away from chemical equilibrium. See IUBMB, Symbolism and terminology in enzyme kinetics: basic definitions.

Some enzymes require a cofactor, an additional nonprotein component such as a metal ion or a small organic molecule. Organic helpers are often called coenzymes; they can assist transfers that the protein alone cannot readily perform. The cofactor may remain bound or exchange with the surroundings, depending on the system. A coenzyme can be chemically changed in one reaction and regenerated by another: regeneration of the enzyme does not mean every helper is unchanged in that individual reaction. Some enzymes need no additional cofactor. See Alberts et al., Molecular Biology of the Cell: small molecules associated with proteins and Alberts et al., activated carriers and their regeneration.

How much faster does adding substrate make a reaction? Enzyme kinetics gives a quantitative answer when a model's assumptions fit. For an illustrative reaction obeying Michaelis-Menten kinetics, use the Michaelis-Menten equation with substrate concentration $s$, limiting reaction rate $V_{\max }$ and Michaelis constant $K_M$. At $s/K_M=0.1$, substitution gives $v/V_{\max }=0.1/1.1=1/11\approx 0.0909$. At $s/K_M=1$, it gives $1/2$; at $s/K_M=10$, it gives $10/11\approx 0.9091$. Thus a hundredfold increase in substrate between the first and last examples yields only a tenfold increase in reaction rate.

For a chosen $V_{\max }=110\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, those fractions mean reaction rates of $10$, $55$ and $100\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Doubling substrate from $10K_M$ to $20K_M$ raises the normalized rate only from $10/11$ to $20/21\approx 0.9524$, a relative increase of about $4.76\%$. In this simplest model, the catalytic machinery is already nearly fully occupied. Doubling the amount of active enzyme instead doubles $V_{\max }$ and the predicted reaction rates at fixed free substrate concentration, provided the other conditions are unchanged. This distinction helps decide whether additional substrate or additional catalytic capacity could improve throughput.

The interpretation has limits. The simplest Michaelis-Menten kinetics assumes a single substrate, fixed active enzyme amount, a suitable quasi-steady-state approximation, and initial measurements with negligible reverse reaction from accumulated reaction product. Substantial depletion, interacting binding sites, or inhibition by excess substrate can invalidate this model. Within it, $K_M$ is the substrate concentration giving half the limiting reaction rate; it is not generally the binding dissociation constant $K_d$. See IUBMB, enzyme kinetics: steady-state approximation and substrate dissociation and IUBMB, Michaelis-Menten and non-Michaelis-Menten kinetics. Measuring activity also requires specified temperature, pH and necessary cofactor availability: a weak signal can reflect unsuitable conditions rather than little enzyme. See Brooks et al., Assay Guidance Manual: basics of enzymatic assays for HTS.

Michaelis-Menten kinetics models substrate binding to an enzyme, complex dissociation and product formation. Under the relevant quasi-steady-state regime, its initial-rate law is the saturating relation given by the following equation.

The Monod equation is an empirical saturating law for the specific growth rate of a culture limited by one nutrient. Write $s\ge 0$ for its residual nutrient concentration, ${\mu }_{\max }>0$ for the limiting specific growth rate, and $K_s>0$ for the nutrient concentration at half that rate:Here $\mu$ and ${\mu }_{\max }$ have units of inverse time, whereas $K_s$ has the same units as $s$. The notation $K(C)$ in the title denotes the same growth function, with $C=s$, $K_{\max }={\mu }_{\max }$ and $K_0=K_s$. For $s\ll K_s$, $\mu (s)\approx ({\mu }_{\max }/K_s)s$; for $s\gg K_s$, $\mu (s)$ approaches ${\mu }_{\max }$. Its derivative ${\mu }^{\prime }(s)={\mu }_{\max }{K}_s/(K_s+s{)}^2$ is positive: more nutrient increases the specific growth rate, with progressively smaller increments.

For example, take ${\mu }_{\max }=1{\mathrm{h}}^{-1}$ and $K_s=1\mathrm{g}{\mathrm{L}}^{-1}$. Nutrient concentrations of $0$, $1$ and $9\mathrm{g}{\mathrm{L}}^{-1}$ give specific growth rates $0$, $0.5$ and $0.9{\mathrm{h}}^{-1}$. In a chemostat with dilution rate $D=0.5{\mathrm{h}}^{-1}$, the intersection $\mu (s)=D$ selects the residual nutrient concentration $s=1\mathrm{g}{\mathrm{L}}^{-1}$, provided the feed can supply enough nutrient for a positive biomass concentration.

The Monod equation is useful because two measurable parameters describe both nutrient scarcity and saturation. It assumes the other conditions affecting growth remain controlled and does not represent inhibition at high nutrient concentration. Although its formula resembles the Michaelis-Menten equation, a whole culture's specific growth rate combines many processes; the resemblance alone does not derive it from Michaelis-Menten kinetics or identify $K_s$ with a Michaelis constant.

In his 1949 review, Monod presents the hyperbolic rate law as an empirical fit in equation (2), pp. 383-384, and discusses the difficulty of identifying a single growth-controlling reaction. His reported division rates use doublings per unit time; the natural-exponential specific growth rate used here is that division rate multiplied by $\ln 2$. ^[ref].