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A mutation changes inherited genetic sequence. Under an infinite sites mutation model, every mutation occurs at a previously unmutated position, so mutations below the sample's common ancestor correspond to segregating sites.

More precisely, a mutation changes a genetic nucleotide sequence; in cellular organisms this is a DNA sequence. An altered sequence can be copied by DNA replication into descendant biological cells, changing their genotype whether or not an observable phenotype changes. Inheritance along a cell lineage does not require transmission to an offspring organism. In animals, germline mutations can enter offspring through gametes, whereas somatic mutations remain in body-cell lineages.^[ref]

A DNA lesion is damage to the physical DNA molecule, such as a damaged nucleobase or a broken strand. It need not become a lasting mutation: DNA repair can restore the original sequence. Conversely, a copied nucleotide substitution can leave an ordinary, correctly paired DNA duplex whose sequence differs from the ancestor. The distinction is between a damaged template and a changed inherited message, not simply between harmful and harmless effects.^[ref]

Small mutations include nucleotide substitutions and indels. Larger changes can remove a gene, duplicate a chromosome region or rearrange its order. Genetic recombination can bring existing alleles together without creating a new nucleobase at the exchanged positions; the mechanism and the resulting sequence change are different questions. A molecular description of a mutation, its effect on a gene product and its effect on a phenotype are likewise three separate statements.

For an original consequence ledger, choose the coding strand segment $5^{\prime }\text{-}\mathrm{ATG}\mathrm{TTT}\mathrm{GAA}\mathrm{TAA}\text{-}{3}^{\prime }$ with initiation at the first displayed codon. Use the standard genetic code, the same reading frame and suitable gene expression context throughout. Its messenger RNA has U in place of T, and the predicted amino acid chain is methionine–phenylalanine–glutamic acid residue, followed by a stop codon. This is a constructed coding segment, not an experimentally measured gene.

In the second row, position 6 changes from T to C, replacing TTT by TTC. Both corresponding RNA codons specify phenylalanine, so the predicted chain is unchanged. Changing position 5 from T to C instead gives TCT, whose RNA codon specifies serine. Changing position 7 from G to T gives TAA and therefore a stop codon before the original glutamic acid residue. Each is one nucleotide substitution, but their coding consequences differ. Even preservation of the amino acid chain does not guarantee unchanged gene expression or phenotype.

Insert GGA after ATG to add a glycine codon, or remove the complete TTT codon to remove phenylalanine. Both indels preserve the downstream reading frame because their lengths are three nucleotides. A one-nucleotide insertion at that location instead produces a frameshift mutation; the positional explanation and a worked regrouping appear under frameshift mutation. The ledger illustrates why interpreting a genetic variant requires its position and context, rather than just the label mutation.

Mutation supplies new inherited variation on which natural selection can act. The Luria-Delbrück experiment makes the distinction unusually visible: natural selection reveals resistant descendants, while the timing of a prior mutation determines how many descendants are available to reveal. This connects an invisible change in one biological cell to a visible pattern across cell cultures.

In Kingman's coalescent, the epoch with $j$ ancestral lineages lasts $T_j\sim \mathrm{Exp}((\genfrac{}{}{0ex}{}{j}{2}))$, independently across epochs. If each lineage carries an independent Poisson process of mutations with rate $\theta /2$, its epoch mutation count satisfies $Y_j\mid T_j\sim \mathrm{Poisson}(j\theta T_j/2)$. The total mutation count is $S={\sum }_{j=2}^n{Y}_j$. Under the infinite sites mutation model, this also counts segregating sites; without that assumption it still counts mutation events, which need not correspond one-to-one to observed polymorphisms.

During the epoch with $j$ ancestral lineages, Kingman's coalescent has waiting time $T_j\sim \mathrm{Exp}((\genfrac{}{}{0ex}{}{j}{2}))$. In the stated time units, each lineage accumulates mutations at rate $Nu=\theta /2$, so the mutation count on a neutral coalescent tree decomposes asThe law of total expectation and the mean of an exponential distribution give $\mathbb{E}{Y}_j=(j\theta /2)/\left(\genfrac{}{}{0ex}{}{j}{2}\right)=\theta /(j-1)$. ThereforeThis calculation counts mutation events. Equating the count with observed segregating sites additionally uses the infinite sites mutation model.

Let $\mu$ be the mutation rate per chromosomal segment per generation, and use the population-scaled parameter $\theta =4N_e\mu$. In $2N_e$-generation coalescent units, mutations occur on each ancestral lineage at rate $\theta /2$. Independent mutations on all branches of a given phylogenetic tree superpose to give a Poisson distribution with mean $\theta L/2$.

Under the infinite sites mutation model, each mutation occurs at a distinct position, and every mutation below the sample's common ancestor separates a nonempty proper subset of sampled chromosomes from the remainder. Thus the mutation count is exactly the segregating site count, givingBy the law of total expectation and the previous branch-length formula,This also gives the Watterson estimator $S/a_n$ as an unbiased moment estimator of $\theta$. Here all sample segregating sites are counted, including singletons; filtering sites by a minimum sample frequency or by external SNP discovery would introduce genetic-study ascertainment and change this calculation.

Use standard diploid coalescent time units of $2N_e$ generations, where $N_e$ is the fixed effective population size, and take $n\ge 2$. In the neutral Kingman's coalescent, each unordered pair of ancestral lineages merges at rate one. With $j$ lineages, the next coalescence time therefore has an exponential distribution with rateThe memoryless construction and the Markov property make these epoch durations independent. During epoch $j$ there are $j$ branches, each of duration $T_j$, so the total branch length of a neutral coalescent isBy linearity of expectation,The ancestral branch above the sample's most recent common ancestor is excluded: mutations there would be shared by all sampled chromosomes and would not create segregating sites.

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.

Use standard neutral coalescent time units, so each pair of ancestral lineages merges at rate 1. With $j$ lineages, there are $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ possible pairs; consequently the successive waiting times are independent withDuring this epoch there are $j$ branches, each of length $T_j$. Hence the total branch length of a neutral coalescent is $L={\sum }_{j=2}^{n}j{T}_j$, and linearity of expectation givesThe root branch above the sample's common ancestor is excluded: its mutations would be shared by every chromosome and would not be segregating sites in the sample.

In the infinite sites mutation model, mutations occur as independent Poisson processes at rate $\theta /2$ per lineage per unit of the coalescent time used in part (a). Conditional on the genealogy, superposition over branches gives a Poisson distribution with mean $\theta L/2$. Since this depends only on $L$, conditioning on total length alone givesEach mutation occurs at a new site and below the common ancestor, so it contributes exactly one segregating site. The law of iterated expectation therefore gives, with $a_{n-1}={\sum }_{i=1}^{n-1}1/i$,Here $\theta$ is fixed when taking the expectation. If it is also random under a proper prior with finite mean, the unconditional expectation is $a_{n-1}E\theta$.

Assume $n\ge 2$, $k\ge 0$, a proper prior density $\pi$ supported on nonnegative mutation rates, and independence of the mutation-rate prior from the neutral genealogy. PutBy Bayes' theorem, the joint posterior isThe normalizing constant is the prior predictive probability $P(S=k)$ and must be positive. For $k=0$, interpret the factor $(\theta L/2{)}^0$ as 1 even at zero. If $n=1$, there are no segregating sites: $S=0$ surely, so that case supplies no information about $\theta$.