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A disulfide bond is the single covalent bond between the two adjacent sulfur chemical atoms in a disulfide, conventionally written $\mathrm{R}-\mathrm{S}-\mathrm{S}-{\mathrm{R}}^{\prime }$. Here $R$ and $R^{\prime }$ stand for the attached organic groups; they may belong to the same molecule or to initially separate molecules. A disulfide bond is a chemical bridge, rather than a hydrogen bond or a mere contact between nearby groups. Two thiols, $\mathrm{R}-\mathrm{SH}$ and ${\mathrm{R}}^{\prime }-\mathrm{SH}$, can supply its two sulfur endpoints. ^[ref]

The stoichiometry of forming a disulfide bond is especially revealing. Written as an oxidation half-reaction, it isThe reverse reduction half-reaction consumes two Electrons and two hydrogen ions to restore two thiols. Both sides contain two sulfur chemical atoms; the two hydrogen chemical atoms have moved into the two hydrogen ions, and the net electric charge remains zero because $+2-2=0$. Thus one disulfide bond corresponds to two thiol groups and two transferred Electrons. These half-reactions describe redox bookkeeping, not a complete standalone chemical reaction: an accompanying reactant must accept or supply the Electrons, and pH affects the actual protonation of the groups. Mercaptoethanol can promote reduction through exchange of disulfide bonds, but adequate excess and appropriate conditions are needed; adding it does not by itself prove complete reduction. ^[ref]

In a protein disulfide bond, the endpoints are normally the amino acid side chains of two cysteine amino acid residues. An intrachain protein disulfide bond closes an intramolecular bridge within one polypeptide chain; an interchain protein disulfide bond connects different polypeptide chains, and can arise by intermolecular crosslinking of initially separate molecules. Forming or reducing these bridges changes covalent bond connectivity and can change protein structure, while preserving the order of amino acid residues and the peptide bonds in the protein backbone. A complete chemical account of protein primary structure includes crosslinks as well as sequence, so preservation of sequence does not imply preservation of every covalent bond. Mature insulin has two interchain protein disulfide bonds and one intrachain protein disulfide bond. ^[ref]

A bridge is useful because it constrains which parts of a polypeptide chain can move apart without breaking a covalent bond. Its placement therefore matters as well as its presence. The ribonuclease A refolding experiment provides the existing experimental example: breaking and reforming four protein disulfide bonds can accompany loss and recovery of function. Counting four bridges alone cannot tell whether their endpoints have the native pairing. This is an attractive connection between simple chemical counting and the richer problem of protein folding.

How can an experiment reveal an initially hidden disulfide bond? Ellman's reagent, also called DTNB, measures free thiols by producing a yellow absorbing species. Its reactive disulfide bond undergoes exchange with the conjugate base $\mathrm{R}-{\mathrm{S}}^{-}$ of a thiol, producing a mixed disulfide attached to $R$ and one released 2-nitro-5-thiobenzoate species, abbreviated TNB. At the assay pH, the strongly colored form is the dianion $\mathrm{TN}{\mathrm{B}}^{2-}$: one negative charge is on sulfur and the other is on the carboxylic acid-derived $\mathrm{CO}{\mathrm{O}}^{-}$ group. With excess DTNB and complete reaction, one released TNB reports one reacted thiol. George L. Ellman's 1959 paper introduced the tissue assay; it was subsequently reexamined quantitatively. ^{[ref] [ref]}

The modern apparatus can be simple: tubes, pipettes, a buffer solution, cysteine standards, DTNB, and an instrument that measures absorbance through a cuvette at wavelength $412\mathrm{nm}$. The Pierce protocol uses $50\mu \mathrm{L}$ of a $4\mathrm{mg}\mathrm{m}{\mathrm{L}}^{-1}$ DTNB stock, $2.50\mathrm{mL}$ of buffer solution, and $250\mu \mathrm{L}$ of sample. Its buffer solution is $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ sodium phosphate at pH 8.0 with a metal-binding additive. After 15 minutes at room temperature, measure absorbance against a buffer solution plus DTNB blank; run known cysteine concentrations in parallel. The observable is increased yellow color and increased blank-corrected absorbance as more accessible thiols react, rather than direct imaging of disulfide bonds. These are published protocol conditions, not measurements performed for this article. ^[ref]

For a before/after test, split a purified protein sample into matched aliquots. Measure accessible free thiols in one aliquot; subject the other to complete reduction of its disulfide bonds, remove the reducing reagent, then measure the resulting thiols with fresh DTNB. Process the unreduced aliquot through a matching cleanup. Normalize to recovered protein and known dilution, since cleanup can change both. To estimate all reducible protein disulfide bonds, expose buried groups using matching protein denaturation conditions in both aliquots and calibrate in that same chemical solution; a native-accessibility measurement answers a different question. A larger corrected reading after reduction supports conversion of previously unreactive groups into detectable thiols. Attribution to disulfide bonds requires excluding other reducible sulfur modifications. This is a proposed comparison based on the detection principles, not a reconstruction of Ellman's original tissue measurements. ^[ref]

Under the Beer-Lambert law, the blank-corrected absorbance is $A=\epsilon lc$, where $l$ is the cuvette path length, $c$ is the released TNB concentration, and $\epsilon$ is its molar absorption coefficient. Riddles, Blakeley and Zerner measured ${\epsilon }_{412}=14150\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ in dilute aqueous salt chemical solutions, confirming the value independently using cysteine. It must not be applied indiscriminately to other solvents or protein denaturation mixtures. ^[ref] For aliquots compared on the same final volume and protein concentration basis, the ideal estimate isThe factor of two follows from the reduction stoichiometry; ${\bar{n}}_{\mathrm{S}-\mathrm{S}}$ is the number of disulfide bonds per protein molecule, or an average for a mixed population. If the aliquots have different dilutions or recoveries, convert each reading to a common sample basis before subtracting.

As a hypothetical calculation, choose $c_{\mathrm{protein}}=10\mu \mathrm{mol}{\mathrm{L}}^{-1}$ in each final cuvette, two free thiols and two disulfide bonds per protein molecule, and $l=1\mathrm{cm}$. Before reduction, $c_{\mathrm{SH}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.283$; complete reduction adds four thiols per protein molecule, so $c_{\mathrm{SH}}=60\mu \mathrm{mol}{\mathrm{L}}^{-1}$ gives $A=0.849$. The difference $0.566$ gives $c_{\mathrm{S}-\mathrm{S}}=20\mu \mathrm{mol}{\mathrm{L}}^{-1}$ and ${\bar{n}}_{\mathrm{S}-\mathrm{S}}=2$. The published tube volumes imply a dilution factor $2.80/0.250=11.2$; reversing that dilution multiplies both thiol and protein concentrations by 11.2 and leaves the bridge count per protein molecule unchanged.

Controls determine whether that arithmetic has chemical meaning. Use fresh cysteine standards and a processed blank without protein to detect color generated by residual reducing reagent. Even a reducing reagent containing no thiols can interfere: Han and Han showed that tris(2-carboxyethyl)phosphine, abbreviated TCEP, reduces DTNB to two TNB equivalents per TCEP. ^[ref] In the hypothetical model, $5\mu \mathrm{mol}{\mathrm{L}}^{-1}$ residual TCEP in only the reduced cuvette would add $A=0.1415$ and falsely add $0.5$ bridge per protein molecule. A stable endpoint and agreement after a further reduction treatment help check completion. A sample without DTNB checks its own color; cloudy samples can also attenuate light without producing TNB. Match pH, temperature, accessibility and calibration chemical solution, and minimize delay that permits reoxidation. If two already-free but buried thiols become accessible only in the reduced aliquot, they mimic one new disulfide bond per protein molecule. Finally, bulk absorbance supplies a count under these assumptions: it does not locate the cysteine endpoints, distinguish intrachain from interchain bridges, or establish the native pairing of a protein disulfide bond.

DNA replication copies a double-stranded DNA molecule into two daughter double-stranded DNA molecules. In semiconservative replication, each daughter retains one parental strand and receives one newly synthesized complementary strand. DNA polymerases build the new strands from incoming nucleotides; they do not make RNA or proteins in this copying step. See Alberts et al., Molecular Biology of the Cell, 4th edition (2002): DNA replication mechanisms.

For a five-base pair example, separate $5^{\prime }\text{-}\mathrm{ATGCC}\text{-}{3}^{\prime }$ from its partner $3^{\prime }\text{-}\mathrm{TACGG}\text{-}{5}^{\prime }$. Base pairing specifies a new TACGG partner for the old ATGCC strand and a new ATGCC partner for the old TACGG strand. With error-free copying, both daughter DNA molecules have the original nucleotide sequence, but different physical strand histories. One initial duplex contains ten nucleotides; one complete round produces two duplexes containing twenty nucleotides, ten retained and ten newly incorporated.

This is useful for inheritance: both daughter biological cells can receive a copy of the DNA record. The elegant feature is that base pairing lets each strand specify its missing partner. Direction matters too: a replicating DNA polymerase reads its template $3^{\prime }\to 5^{\prime }$ and extends the new strand at its $3^{\prime }$ end, so synthesis proceeds $5^{\prime }\to 3^{\prime }$. These are opposite directions on antiparallel nucleic acid strands, not two alternative directions of synthesis. The enzyme needs a pre-existing paired strand end to extend; the diagram omits the machinery that starts and coordinates copying. See Alberts et al.: template-directed synthesis and the requirement for a primer.

The strand accounting predicts observable differences between copying mechanisms. In 1958, Meselson and Stahl grew Escherichia coli with the heavier nitrogen isotope ${}^{15}\mathrm{N}$, then supplied ${}^{14}\mathrm{N}$ and measured extracted DNA by density-gradient centrifugation. After one generation their samples contained intermediate-density DNA; after two, intermediate-density and light DNA occurred in equal amounts. This supported semiconservative replication in the tested Escherichia coli culture. It was an isotope assay of sampled DNA, not simultaneous testing of every kind of DNA or direct colored-strand photography. See The replication of DNA in Escherichia coli by Matthew Meselson and Franklin W. Stahl.

Endocytosis takes material into a biological cell by inward budding and separation of part of its cell membrane. At a chemical synapse, it helps recover membrane and proteins after synaptic vesicle exocytosis. This recovery is distinct from uptake of individual neurotransmitter molecules through membrane transport proteins.

Endosymbiosis is an ecological association in which one organism, the endosymbiont, lives inside another. The definition specifies location, not that both partners benefit, nor that the internal partner has become an organelle. An intracellular association can remain an association between distinct organisms. Endosymbiotic theory addresses a different question: how particular ancestral bacteria became inherited, functionally integrated organelles during evolution.

The striking consequence is that a eukaryotic biological cell can combine histories: its nuclear lineage and its mitochondria need not have the same ancestry. Mitochondria have bacterial ancestry related to Alphaproteobacteria; the plastids of plants and many algae, including chloroplasts, have ancestry within cyanobacteria. These are ancestral lineages, not a claim that a named living bacterial species is the literal ancestor. Present-day bacteria and organelles have both evolved since their shared ancestors. ^[ref]

Endosymbiotic gene transfer moves genetic material from an endosymbiont or its descendant organelle into the host genome, notably the nuclear genome. A transferred DNA segment can remain silent. For a transferred gene to replace an organelle-encoded function, nuclear gene expression must work and its protein must reach the right compartment through organelle protein import. Transfer of DNA and transport of proteins are distinct processes in opposite directions. ^[ref]

Gene loss and endosymbiotic gene transfer are different: absence from an organelle genome does not by itself show that a functional nuclear copy exists. Conversely, an imported protein may derive from a host gene rather than a transferred gene. The surviving organelle genome is therefore a partial record, not a complete inventory of either ancestral bacteria or the modern organelle's machinery. ^[ref]

Endosymbiotic theory explains the bacterial origins of mitochondria and plastids through enduring internal associations and subsequent integration. Survival inside a biological cell is only part of the problem: exchange through membrane transport, inherited division, gene expression and delivery of required proteins must work together. The elegance is evolutionary reuse of an existing cellular capability, followed by changes that make the partners operate as one system. ^[ref]

The historical On the origin of mitosing cells by Lynn Sagan (1967, abstract and Introduction, pp. 225–226) synthesized a symbiotic account of eukaryotic origins and proposed tests using molecular evidence. Its Introduction explicitly acknowledged earlier proposals. It also hypothesized origins for other structures; the modern evidence for mitochondria and plastids should be assessed separately from the rest of that historical scheme.

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.

Epidermal growth factor is a protein ligand for the epidermal growth factor receptor.

An epitope is the molecular feature recognized by an antibody. Protein denaturation can expose an epitope or destroy the spatial arrangement required for its recognition; an antibody that works on a folded protein need not work in SDS-PAGE-based Western blot antibody detection.

Evolutionary game theory studies how a population's composition changes when the success of a pure strategy depends on which other pure strategies it encounters. The central feedback is a frequency-dependent payoff: becoming common changes the opponents a strategy meets, which can change whether it does well. In a coordination game, common behaviour can reinforce itself; in the coexistence example below, being rare is an advantage. This makes evolutionary game theory useful for asking which behaviours can persist and which initial compositions lead to different outcomes, without assuming that each participant solves an optimization problem.

In the basic model, a large, well-mixed population has finitely many inherited pure strategies; interactions sample opponents according to their population frequencies. Assume that a strategy's expected payoff determines its per capita net reproductive growth rate, with a common baseline allowed. These are modelling assumptions: reproduction must transmit the strategy, encounters must match the proposed sampling rule, and the conversion from payoff to growth must be justified for an application. The replicator equation below describes the resulting continuous-time frequencies. Taylor and Jonker's ^[ref] explicitly derive this growth model.

For learned behaviour, reproduction can instead be replaced by an assumed copying rule: an individual samples another individual uniformly and copies a better-performing pure strategy at a rate proportional to the positive payoff difference. At frequencies $x_i,x_j$, the net flow from $j$ to $i$ is $x_i{x}_j(f_i-f_j)$, because the two directional positive parts subtract to $f_i-f_j$. Summing over $j$ gives $x_i(f_i-\bar{f})$, the replicator equation. This derivation states a particular mechanism; an arbitrary rule for learning or imitation need not give the same dynamical system.

Three questions must be distinguished. A symmetric Nash equilibrium tests whether an individual can improve its expected payoff against the resident composition. An evolutionarily stable strategy tests whether each distinct rare mutant is outperformed after it is introduced. Dynamical stability tests the response of an actual dynamical system to small changes of composition, and asymptotic stability additionally requires nearby trajectories to converge. For a finite symmetric bimatrix game under the continuous-time replicator equation, every evolutionarily stable strategy is a symmetric Nash equilibrium and a locally asymptotically stable rest point; the converse from asymptotic stability to an evolutionarily stable strategy fails in general. See Ross Cressman and Yi Tao, The replicator equation and other game dynamics (2014), Theorems 1–2 and Example 1. These statements refer to stability on the full probability simplex, including perturbations introducing currently absent pure strategies.

A concrete experiment makes the frequency feedback visible. In 2009, Gore, Youk and van Oudenaarden competed two yeast strains: a producer of the enzyme invertase, which enables sucrose hydrolysis, and a mutant lacking the corresponding SUC2 gene. Released sugars can feed neighbours, so avoiding production can be advantageous. They used shaken 5 mL cultures with 5% sucrose at $30{}^{\circ }\mathrm{C}$, diluted daily after 23 hours of growth. Different labelled proteins allowed the two strains to be counted separately by flow cytometry; selective plating periodically checked the counts. The cooperator required externally supplied histidine, so restricting that amino acid imposed an adjustable growth disadvantage. See Snowdrift game dynamics and facultative cheating in yeast, pp. 253 and 255–256, Methods Summary.

Starting with rare producers and, separately, rare nonproducers tested both invasion directions. Under the reported coexistence conditions each rare strain increased; changing histidine or added glucose shifted the eventual composition. Their explanation used nonlinear growth benefits, rather than a fixed pairwise payoff matrix. This motivates testing both rarity regimes before inferring an evolutionarily stable strategy or fitting a replicator equation; it does not identify the illustrative tables below with measured payoffs. See Gore, Youk and van Oudenaarden, Figures 1–3.

An early motivation was understanding why animal contests can involve restraint even when escalation might win an individual encounter. Maynard Smith and Price's ^[ref] introduced the evolutionarily stable strategy in this setting. Taylor and Jonker's ^[ref] then studied explicit continuous and discrete dynamics. Static resistance to invasion and the paths populations follow are related questions, but they require separate definitions.

A segment retained in the mature RNA after RNA splicing, or the corresponding segment of a gene. An exon need not encode a protein: the ribosomal RNA exons in the Tetrahymena ribozyme experiment are an example.