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An acid hydrolase is an enzyme that catalyses hydrolysis most effectively under acidic conditions; lysosomes contain multiple kinds.

An acid phosphatase catalyses hydrolysis of phosphate esters under acidic conditions; its lysosomal enzyme latency supplied evidence for lysosomes.

Active transport moves a substance uphill in its electrochemical gradient by coupling the transfer to a downhill process. Some carrier proteins use ATP hydrolysis; others couple uphill transfer of one solute to downhill transfer of another solute, often an ion. The latter mechanism uses a stored electrochemical gradient and need not consume ATP directly at that carrier protein. Maintaining the driving electrochemical gradient still requires a source of energy. See Alberts et al.: carrier proteins and active membrane transport.

For the neutral 8:2 example, B-to-A transfer costs $RT\ln 4\simeq 1.38629RT$ per mole. As an explicitly assumed bookkeeping example, suppose a tightly coupled cycle supplies a downhill free energy change of $-2RT$ per mole of transferred solute. The total is then $\Delta G_{\mathrm{cycle}}=RT\ln 4-2RT\simeq -0.61371RT$. This makes forward cycling energetically possible under those assumptions; it specifies neither a molecular mechanism nor a rate. Merely adding a carrier protein cannot supply the missing free energy. Conversely, as the transported concentration ratio rises, the required uphill cost rises, so a fixed driving process has a finite thermodynamic limit.

ADP contains the same adenosine unit as ATP, with two phosphate groups instead of three. One ATP converted by the stated hydrolysis gives one ADP and one inorganic phosphate: $3=2+1$ accounts for the phosphate groups while adenine and ribose are retained. ADP is therefore a reusable material carrier, rather than ATP with its entire structure discarded.

Regeneration reverses the net chemical reaction:At the same stipulated reference conditions its $\Delta G^{\circ \prime }$ is $+30.5\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, the negative of the hydrolysis value. It is an endergonic reaction and requires energy input through energy coupling; a catalyst alone cannot supply that input. Recycling the material carrier does not recycle the expended driving energy. Cellular respiration and light-dependent reactions supply routes for regeneration. See Cooper, The Cell: ATP generation and energy coupling.

ADP is a nucleotide with two phosphate groups. Formation of ATP from ADP and phosphate couples to energy supply in light-dependent reactions; the Calvin cycle returns ADP after using ATP. ^[ref].

ATP is a nucleotide consisting of adenosine, itself adenine joined to ribose, carrying three phosphate groups. From the ribose outward, these groups are labelled $\alpha$, $\beta$, and $\gamma$. ATP connects energy-supplying processes to energy coupling for synthesis and active transport; the same carrier can be regenerated and reused. ^[ref].

The net hydrolysis commonly used in energy coupling isHere ${\mathrm{P}}_i$ denotes inorganic phosphate. At buffered pH $7$, this biochemical shorthand groups the appropriate protonation states into pools and leaves proton exchange with the buffer solution implicit. An explicit, charge-balanced choice is ${\mathrm{ATP}}^{4-}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{ADP}}^{3-}+{\mathrm{HP}{\mathrm{O}}_4}^{2-}+{\mathrm{H}}^{+}$, without implying these are the only chemical species present. See the IUBMB recommendations for biochemical thermodynamics.

The quoted textbook biochemical standard state value is approximatelyThis convention uses fixed pH $7$, temperature $298.15\mathrm{K}$ ($25{}^{\circ }\mathrm{C}$), reference pressure $100\mathrm{kPa}$ ($1\mathrm{bar}$), unit reference thermodynamic activities for dissolved reactants and reaction products, and water thermodynamic activity approximated by one. The prime indicates the fixed-pH biochemical convention. See OpenStax Biology 2e, 6.4: standard ATP hydrolysis value and Biology 2e, 6.2: biochemical standard conditions. This is a reference value, rather than a universal value inside a biological cell; precise tables must also specify chemical solution conditions and any magnesium ion binding convention. Different reference conventions and data treatments give different tabulated values, as discussed in Li et al., 2010: a database of thermodynamic quantities for biochemical reactions.

An exergonic reaction has a negative change in Gibbs free energy, so the overall ATP hydrolysis is favourable under those reference conditions. Breaking a chemical bond by itself costs energy; forming new chemical bonds can release energy. The net change compares all reactants with all reaction products, including their interactions with water and the entropy contribution. Calling the phosphate connection a "high-energy bond" describes a large favourable change in Gibbs free energy on hydrolysis, rather than energy released by bond breaking alone. See 7.5 Strengths of Ionic and Covalent Bonds and Cooper, The Cell: ATP and metabolic energy.

Composition changes the driving force. At the same fixed pH, temperature, pressure, and binding convention, write the molar reaction change asHere $R$ is the molar gas constant, $T$ the absolute temperature, and $a$ a dimensionless thermodynamic activity for a biochemical pool. Constant water thermodynamic activity is absorbed into the reference term; fixed pH accounts for buffered protons. The shorthand $\Delta G$ means transformed molar reaction Gibbs free energy change, ${\Delta }_r{G}^{\prime }$, rather than a finite sample's total change. The natural logarithm acts on a dimensionless reaction quotient, never on concentrations with units. See the IUBMB biochemical thermodynamics recommendations and the general equilibrium constant relation.

For a stipulated ideal dilute chemical solution, let $a_i=[i]/c^{\circ }$, with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. Choose $[\mathrm{ATP}]=5\mathrm{mmol}{\mathrm{L}}^{-1}$, $[\mathrm{ADP}]=0.5\mathrm{mmol}{\mathrm{L}}^{-1}$, and $[{\mathrm{P}}_i]=1\mathrm{mmol}{\mathrm{L}}^{-1}$, ignoring activity coefficient corrections and magnesium binding. These are illustrative inputs, not a claim about a particular biological cell. ThenandIncreasing $Q$ tenfold raises $\Delta G$ by $RT\ln 10\approx 5.708\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at this temperature. Thus the ATP to ADP ratio and available inorganic phosphate matter, even with the same standard state value. These are local molar predictions at the stipulated composition, rather than a constant total yield as an isolated sample reacts.

An allylic organotitanium species that can add its allyl group to a carbonyl compound, furnishing a homoallylic alcohol after hydrolysis.

Hexokinase provides a concrete connection between sugar chemistry and energy coupling: it transfers a terminal phosphate group from ATP to glucose, forming a phosphorylated sugar and adenosine diphosphate. The enzyme accelerates the transfer, while consuming ATP changes the net chemical reaction being carried out. The donated group reaches the substrate through the catalytic route.^[ref]

For a symbolic teaching model, write $S$ for an unphosphorylated sugar, $S\text{-}P$ for its phosphorylated form, and $P_i$ for inorganic phosphate. These are labels for chemical states, not complete elemental formulas. Using one consistent reference convention at fixed pH and temperature, stipulate the following molar changes in Gibbs free energy:All three numbers are chosen illustrative values, not measured properties of glucose or a particular enzyme. The prime marks a biochemical reference convention with pH held fixed; explicit proton and metal-binding bookkeeping is suppressed consistently. The first two lines are a thermodynamic decomposition: cancelling water and inorganic phosphate gives the net transfer. They do not assert that the reaction mechanism first releases free inorganic phosphate and then recaptures it.

Because Gibbs free energy is a state function, consistently defined reaction changes add. A working reaction mechanism must also connect the desired conversion to ATP consumption; independently allowing ATP hydrolysis beside the sugar does not establish energy coupling. The general distinction is developed under energy coupling. ATP-dependent phosphorylation of a membrane protein can likewise connect a chemical change to active transport.^[ref]

Actual Gibbs free energy changes depend on thermodynamic activities. Even a favourable reference sum can be overcome by accumulation of reaction products or depletion of ATP; it supplies no prediction of the reaction rate. A living biological cell also regenerates ATP. Sustained operation therefore requires a connected metabolism, not an inexhaustible stock of a special molecule.

Define side $P$ as the higher proton electrochemical potential reservoir and side $N$ as the synthesis-side reservoir. For transfer $P\to N$ at common temperature $T$, define the positive downhill proton-motive force magnitude $\Delta p_{\mathrm{drive}}$ byHere $a$ is thermodynamic activity, $\psi$ is electric potential, $R$ is the molar gas constant and $F$ is the Faraday constant. Thus both a higher proton thermodynamic activity and a more positive electric potential on side $P$ contribute to the driving magnitude. This definition fixes the sign; conventions for a signed proton-motive force differ. The calculation under chemiosmosis separates these two contributions numerically.

Assume fixed reservoirs, tight mechanical energy coupling, one transported proton per $c$ subunit per revolution, and three ATP produced per revolution. Neglect leaks, slipping and membrane transport of the chemical reagents. Let $g_{\mathrm{ATP}}>0$ be the local molar Gibbs free energy cost of ATP formation at the stipulated composition, pH and magnesium-binding convention. It is not necessarily the standard state value. Adding changes in the state function gives, per mole of complete turns,For this model net synthesis is thermodynamically favoured when $\Delta G_{\mathrm{turn}}<0$. Equality is a reversible balance, not a promise of a useful reaction rate. A negative value establishes an available driving margin; it does not specify how fast the enzyme runs.

Choose $n=10$, $g_{\mathrm{ATP}}=50.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$ and $\Delta p_{\mathrm{drive}}=0.180\mathrm{V}$. These are illustrative inputs, not measurements of the preparation in the experiment below. Thenand $\Delta G_{\mathrm{turn}}=-23.673\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Per ATP, the ideal input is $57.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, leaving a driving margin of $7.891\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The balance point isAt $0.140\mathrm{V}$ the same synthesis cycle instead has $\Delta G_{\mathrm{turn}}=+14.921\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$: the chosen reservoirs favour its reverse.

The machine is reversible: sufficient ATP hydrolysis can drive opposite rotation and pump protons uphill. Actual direction also depends on inhibition and available substrates; a weak proton-motive force need not make every intact enzyme immediately run backward. Respiratory uncoupling supplies a different failure mode: a parallel proton leak dissipates the gradient without productive shaft rotation.^[ref]

The motor budget $10/3$ is not the whole cost of ATP delivered outside a mitochondrion. Import of ADP and inorganic phosphate, and export of ATP, also draw on the electrochemical gradient.^[ref] For a deliberately simplified system budget, assign these transports together one additional downhill-proton equivalent per delivered ATP, valued at the same $F\Delta p_{\mathrm{drive}}$. This is a chosen accounting convention, not a claim of one extra literal proton through $F_0$, nor a universal transport stoichiometry. The total becomes $10/3+1=13/3$ equivalents and $75.258\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ of gradient expenditure per delivered ATP at $0.180\mathrm{V}$. The ratio of stored chemical free energy to that expenditure is $50/75.258\approx 66.4\%$, compared with $50/57.891\approx 86.4\%$ for the motor alone. This system budget changes the resource expenditure, not the motor's balance point $\Delta p_{\ast }$. Real transport terms require their own electrochemical potentials, and leaks add further cost.

Borazine is the alternating six-membered ${\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6$ ring. Planar boron and nitrogen centers permit N-to-B pi donation, but unequal orbital energies strongly polarize the bonding. It undergoes addition and hydrolysis much more readily than benzene; the structural analogy does not give benzene-like chemical inertness.

Energy coupling links an endergonic reaction to an exergonic reaction through a reaction mechanism, so that the combined transformation has a favourable change in Gibbs free energy. ATP hydrolysis is a common driving chemical reaction. A shared reaction intermediate, produced in one step and consumed in another, can physically connect the transformations; merely placing two independent chemical reactions in the same chemical solution does not establish energy coupling. See Alberts et al., Molecular Biology of the Cell: coupled reactions and activated intermediates.

For a computed example, stipulate an endergonic reaction $\mathrm{A}⟶\mathrm{B}$ with $\Delta G_1^{\circ \prime }=+20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ under precisely the same biochemical standard state conditions as the quoted ATP hydrolysis value. Assume a reaction mechanism couples one ATP hydrolysis to each A-to-B conversion. The formal sum isThe sum $+20+(-30.5)=-10.5$ gives an exergonic reaction. The $+20$ is a chosen teaching value, not a measured property of an identified substance. Outside these reference conditions, add the actual $\Delta G_1$ and $\Delta G_2$, each evaluated using consistent thermodynamic activities; mixing a standard value for one chemical reaction with an unrelated cellular value for the other gives no justified prediction.

Why do the changes in Gibbs free energy add? Gibbs free energy is a state function. Adding consistent reaction changes cancels any shared reaction intermediate and leaves the difference between the combined reactants and combined reaction products. This explains the thermodynamic accounting; it does not construct the reaction mechanism. In many ATP-dependent routes, phosphorylation transfers a phosphate group to a substrate or an enzyme, and subsequent transformation consumes that modified material. The shared material is what connects the steps. Other reaction mechanisms use ATP differently. See Alberts et al., Molecular Biology of the Cell: ATP-driven biosynthesis.

The useful achievement is that a common regenerated nucleotide can connect many different processes in metabolism. The net negative change in Gibbs free energy permits the coupled conversion, but supplies no reaction rate: enzymes can lower activation energy barriers without changing the endpoint difference. Energy coupling changes which overall chemical reaction is being considered; catalysis accelerates an available route. See OpenStax Biology 2e, 6.2: free-energy changes and reaction barriers.

A historical landmark is Fritz Lipmann's ^[ref]. Its sections on the phosphate cycle, generation and utilization of phosphate-bond energy, and group transfer organized these processes around a reusable chemical carrier. The historical terminology should be read with the modern distinction between net hydrolysis and the cost of breaking a chemical bond.

An ether with oxygen directly attached to an alkene carbon. Electrophilic activation and hydrolysis can convert an appropriate enol ether back to a carbonyl compound.

An ester consumed in the conductometric saponification assay, where hydrolysis under basic conditions produces ethanol and an acetate ion.

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.

In the payoff stage of glycolysis, oxidation of glyceraldehyde 3-phosphate is coupled to reduction of NAD+. Each NADH carries a two-electron reducing equivalent. The formal half-chemical equationmakes the electron count explicit: two NADH per glucose carry four electrons. This is redox bookkeeping, not a claim that free electrons travel through the chemical solution or that NADH is itself ATP.

Fermentation can return NADH to NAD+ by reducing an organic substrate; reduction of pyruvate to lactate is one example. In aerobic cellular respiration, reducing equivalents reach an electron transport chain, reducing molecular oxygen. Other forms of cellular respiration use other electron acceptors, so absence of oxygen does not uniquely imply fermentation.^[ref]

The need for NAD+ regeneration follows from a finite-pool argument. In a hypothetical closed preparation with $12\mu \mathrm{mol}$ of initially oxidized NAD+ and no regeneration, at most $6\mu \mathrm{mol}$ of glucose can complete the two NAD+-consuming conversions per glucose. This is an upper bound even with excess glucose, not a predicted stopping time: other limitations can stop the preparation earlier. With an intact fixed nicotinamide adenine dinucleotide pool and no side reactions, converting NAD+ into NADH preserves their sum but depletes the usable oxidized form.

ATP recycling imposes a second condition. Glycolysis needs some initial ATP for investment and consumes two net ADP and two inorganic phosphate per completed glucose. In a closed ideal preparation with only $8\mu \mathrm{mol}$ of initial ADP, no other source of ADP, and enough initial ATP and inorganic phosphate, at most $4\mu \mathrm{mol}$ of glucose can finish. This bound follows from the net ADP consumption, even though investment temporarily produces additional ADP. Initial investment also means that adequate final inventories alone cannot guarantee the route can start.

For continued operation at fixed carrier inventories, other chemical reactions must match the two net ATP formed per glucose with ATP use that restores ADP and inorganic phosphate, or exchange must supply those reactants. ATP hydrolysis coupled to cellular work is one route. Thus glycolysis can sustain useful energy coupling while neither its ATP inventory nor its NADH inventory grows indefinitely; matter cycles while glucose supplies fresh chemical free energy.

Hydrolysis is a chemical reaction involving water that breaks a chemical bond. The free energy released by suitably coupled ATP hydrolysis can drive active transport; the available free energy depends on the actual composition and conditions.

Inorganic phosphate, abbreviated ${\mathrm{P}}_i$, denotes the free phosphate pool in biochemical chemical equations, including ATP hydrolysis. Its constituent chemical species depend on pH and protonation; ${\mathrm{P}}_i$ does not specify a single electric charge or proton count. See the IUBMB recommendations on biochemical reactant pools.

Phospholipase C catalyzes hydrolysis of membrane PIP2, producing soluble IP3 and membrane-associated diacylglycerol. This consumes a phospholipid substrate, rather than manufacturing calcium. G protein-regulated and receptor tyrosine kinase-regulated forms of PLC provide different routes to this reaction.^[ref]

Lysosomal enzyme latency is the difference between the enzyme activity accessible to an added substrate before disruption of the surrounding membrane and the activity revealed after disruption. Low accessible activity can reflect separation of enzyme from substrate, rather than absence of enzyme. This is an assay-accessibility statement, not a claim that the intact lysosome is inactive in the living biological cell.^[ref]

Let $F$ be the blank-corrected initial rate measured with a membrane-preserving preparation and $T>0$ the rate after a treatment that exposes all assayable enzyme. Use matched amounts of starting material, pH, substrate and measurement time. Define the apparent latent fraction byThis definition requires more than a large treated signal. Disruption must preserve enzyme activity, and the treatment must not independently alter the readout. A soluble-enzyme comparison with and without treatment helps test direct activation or inhibition; an appropriate blank tests signal unrelated to enzyme-catalysed hydrolysis. These are controls for interpreting the constructed assay, rather than additional historical observations.

For a deliberately chosen example, take $F=10$ and $T=100$ arbitrary rate units. Then $\ell =0.90$: ninety percent of the assayable activity was initially inaccessible. After hypothetical partial damage, $F=25$ with unchanged $T=100$ gives $\ell =0.75$. The preparation has lost latency without gaining total assayable activity. These percentages describe activity accessibility; they do not count intact lysosomes. Unequal enzyme contents, partial leakage and mixed particle populations prevent that identification. If $F>T$, the formula is negative, flagging failure of this simple accessibility interpretation rather than a negative population of hidden enzymes.

To check the logic, divide the same uniformly mixed preparation into two equal portions. If dilution halves both measured rates, $(F/2)/(T/2)=F/T$, so $\ell$ is unchanged. In contrast, comparing an undiluted untreated portion with a diluted treated portion changes the ratio artificially. Normalization to the same amount of original material is therefore part of the definition, not an optional numerical correction.

The lysosomal lumen is the aqueous interior of a lysosome, separated from cytosol by its surrounding lipid bilayer. A membrane pump couples ATP hydrolysis to inward transport of hydrogen ions, maintaining an acidic environment for acid hydrolases. Membrane transport proteins export digestion products, including amino acids and sugars, for reuse in metabolism. Compartmentalization combines restricted access with controlled exchange.^[ref]

A lysosome is an acidic digestive and recycling organelle: one surrounding lipid bilayer encloses a lysosomal lumen containing acid hydrolases. These enzymes break down proteins, nucleic acids, carbohydrates and lipids by hydrolysis. Cargo can originate outside the biological cell, entering through endocytosis, or within its own cytoplasm, delivered through autophagy. Digestion therefore serves both disposal and reuse.^[ref]

An endosome sorts endocytosed cargo: some returns to the cell membrane, while some proceeds toward lysosomes. An autophagosome instead encloses material already in the cytoplasm. Both routes bring cargo into contact with digestive enzymes; neither implies that every internalized molecule is destroyed.^[ref]

Enzyme supply is a different itinerary from cargo disposal. Secretory protein routing and protein targeting deliver newly made digestive enzymes through the endoplasmic reticulum, Golgi apparatus and trans-Golgi network toward lysosomes.^[ref] The existing membrane transport and differential centrifugation articles explain the boundary-crossing and separation principles used below.

A proteasome is a different digestive machine: a protein complex without a surrounding lipid bilayer, acting on individual proteins in the cytosol and cell nucleus. Many targets carry ubiquitin chains. A lysosome accepts a much broader range of cargo, including material delivered inside an autophagosome. Thus an unwanted soluble protein and an unwanted mitochondrion pose different disposal problems; a proteasome does not swallow an intact mitochondrion.^[ref]