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Five consequences of the ancestral recombination graph are:

Shared ancestry explains correlation; recombination makes ancestry a mosaic; mutation on the resulting branches converts genealogy into observed variation. These explanations use the neutral ARG, with a variable recombination map where stated. Selection, population subdivision or changing population size can also be studied genealogically, but require extending the event model rather than being consequences of the constant-size panmictic neutral process alone.

The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.

Saccharomyces cerevisiae is a particular species of yeast, used in baking, brewing and laboratory research. During ordinary vegetative growth it reproduces by budding: a smaller daughter grows from the mother before separation. Laboratory strains can be maintained as either haploid or diploid biological cells. This makes it possible to investigate genes in a single eukaryote, then use controlled crosses and meiosis to connect a genotype to a phenotype. See Burgess, Powers and Mell, Budding Yeast Saccharomyces cerevisiae as a Model Genetic Organism (2017), pp. 1–2.

The practical value of Saccharomyces cerevisiae as a model organism is the combination of easy laboratory cultivation, accessible genetics, and cellular functions shared with other eukaryotes. A haploid strain has one set of chromosomes, so a changed allele is not accompanied by a second copy at that genetic locus. But a mutation need not produce an observable phenotype under every condition. The yeast is also a whole organism, not a miniature human: transferring a result requires evidence that the relevant function is shared in the target system. See Duina, Miller and Keeney, Budding Yeast for Budding Geneticists (2014): cultivation, genetic tools and conserved functions.

For a concrete example, Hartwell, Culotti and Reid's 1970 work followed temperature-sensitive Saccharomyces cerevisiae strains with successive microscope photographs. Failure to complete a particular part of the cell cycle helped identify required gene functions. Their distinction between when a function is required and where a cell finally stops is especially instructive: the visible endpoint need not be the time of the underlying defect. See Hartwell, Culotti and Reid, Genetic control of the cell-division cycle in yeast. I. Detection of mutants (1970): primary abstract. This is a useful way to turn an observable change in a biological cell into a question about genes and causal order.

In alcoholic fermentation, Saccharomyces cerevisiae converts sugars into ethanol and carbon dioxide. The carbon dioxide helps dough rise; the ethanol is useful in alcoholic drinks. The metabolism also supports production of ATP, a usable cellular energy carrier. This is one metabolic ability of one species, not the definition of every yeast. See Clark, Douglas and Choi, Biology 2e, 7.5: alcohol fermentation and the Burgess, Powers and Mell account above.

For glucose, the ideal net stoichiometry isOne molecule of glucose supplies six carbon chemical atoms: four end up in the two ethanol molecules and two in the two carbon dioxide molecules. The twelve hydrogen and six oxygen chemical atoms also balance. This is a net material balance across several enzyme-catalyzed steps, not a single chemical reaction step or the complete bookkeeping for cellular energy carriers. The standard pathway accounting yields a net two ATP per glucose: the initial breakdown produces the ATP, while the later conversion to ethanol regenerates a coenzyme needed to keep the breakdown going. See Clark, Douglas and Choi, Biology 2e, 7.2: net ATP accounting and the section 7.5 cited above.

At complete conversion, this toy gives $20\mathrm{mmol}$ of ethanol and $20\mathrm{mmol}$ of carbon dioxide. The carbon balance is $6\times 10=2\times 20+1\times 20=60\mathrm{mmol}$ of carbon chemical atoms. A real culture also builds cellular material, can use other metabolic routes, and may retain dissolved carbon dioxide; the toy is a reference balance rather than a prediction of collected gas or beverage yield.

Although alcoholic fermentation does not require molecular oxygen as a reactant, the presence of molecular oxygen does not force Saccharomyces cerevisiae to stop fermenting. De Deken's 1966 experiments reported predominantly fermentative glucose use during exponential growth in air. Consequently, bubbles do not establish either absence of molecular oxygen or an exclusively fermentative metabolism. See De Deken, The Crabtree Effect: A Regulatory System in Yeast (1966): primary abstract.

Wee1 is a protein kinase that inhibits CDK1 by adding a phosphate group at an inhibitory site. Purified human Wee1 phosphorylated Cdc2 at Tyr15 in the reported in vitro assay, with no Thr14 modification detected. A catalytic-site mutation abolished both the measured protein kinase activity and the ability to inhibit division. This identifies a catalytic brake rather than merely a correlated protein amount.^[ref]

Do not attribute all inhibitory CDK1 phosphorylation to Wee1: the Xenopus laevis Myt1 experiment separately identified activity toward both Thr14 and Tyr15. Nor does every phosphorylation inhibit its target: activating Thr161 modification in the studied Xenopus laevis Cdc2 has the opposite role. In cell-cycle activation feedback, suppression of Wee1 removes a brake on CDK1, reinforcing the transition without implying that the total amount of either protein instantly disappears.