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Biology · Phylogeny, Classification & the Tree of Life

Molecular and morphological data in phylogenetic inference

A computer can return a tree from a table, but the result depends on what the table contains and what the calculation assumes. Follow one small example by hand before treating a long sequence alignment as a history with no uncertainty.

Heredity, Variation & Evolution introduces inherited differences and common ancestry; Evolutionary Processes & Adaptation distinguishes observations from evolutionary mechanisms. Species, Speciation & Macroevolution provides useful context for species concepts and reticulate histories, though its detailed cases are optional here. You need to read small tables, add branch values and compare fractions. Each tree and probability rule is explained before it is used; no programming or previous taxonomic vocabulary is assumed.

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By the end, you can

Understand how scored characters, change models and support measures contribute to a proposed tree.

01

Make corresponding observations comparable

Morphological analyses score defined structures or other traits in living and fossil samples. Molecular analyses compare corresponding sequence positions. Alignment proposes which positions correspond despite insertions, deletions and differing lengths. A column of an alignment is therefore a comparison claim, not just text placed in a neat row.

For a species-level question, investigators also examine which gene copies are being compared. Mistaking duplicated copies for the same inherited locus can mislead the analysis even if each sequence was read correctly. Contamination, missing samples and uncertain alignments likewise matter before tree calculation begins.

More data can add useful evidence, but repeating a systematic error does not remove it. Linked sites and genes may not be independent; different regions may have different histories. Combining regions is a modelling choice to assess, rather than an automatic guarantee of a better species tree.

02

Parsimony asks for the smallest change count

Under an equal-cost binary parsimony rule, each change between 0 and 1 counts one step. For each candidate topology, find the smallest number of steps that can explain the observed tip states. This is an inference criterion, not a rule requiring evolution to choose the simplest actual history.

Our unrooted four-tip example has A = 0, B = 0, C = 1 and D = 1. On the A/B versus C/D split, assigning 0 to the A/B internal node and 1 to the C/D node gives one change on the joining edge. On A/C versus B/D, each pair contains both states. At least two changes are needed. The third split, A/D versus B/C, also requires two.

This one character therefore favours the first split under the supplied rule. It does not determine a root or a divergence date, and it cannot exclude repeated change in the actual history. Parsimony can be misled when homoplasy and unequal evolutionary patterns do not fit its assumptions.

Compare minimum change counts

Compare minimum change countsThree unrooted quartets show fixed observed tip states and one explicitly selected minimum internal-state assignment; a second-character score table exposes a tie.A/B versus C/D splitInternal nodeU0Internal nodeV1Taxon A0Taxon B0Taxon C1Taxon D1A/C versus B/D splitInternal nodeU0Internal nodeV0Taxon A0Taxon C1Taxon B0Taxon D1A/D versus B/C splitInternal nodeU0Internal nodeV0Taxon A0Taxon D1Taxon B0Taxon C1Combined totals: 3, 3, 4A/B versus C/D split1 + 2 = 3A/C versus B/D split2 + 1 = 3A/D versus B/C split2 + 2 = 4Observed tip stateOne minimum internal-state assignmentEach unequal-state edge: one stepCombined totals: 3, 3, 4Two original binary characters, equal one-step cost in either direction,no fixed ancestral state and no rooting/date inference. Selectedmappings may not be unique or historically true.
These unrooted edges have no direction of ancestry. A change mark identifies unequal endpoint states without specifying a uniquely directed gain. Observed tip states and selected internal-state assignments provide different kinds of information.
Read the diagram in words

X needs 1/2/2 steps on AB-CD/AC-BD/AD-BC; Y needs 2/1/2. Equal-weight totals are 3/3/4.

Observed tip state
DescriptionPattern X: 0,0,1,1Pattern Y: 0,1,0,1
A00
B01
C10
D11
A/B versus C/D split
DescriptionRelationship
Internal node U ↔ Taxon AConnection
Internal node U ↔ Taxon BConnection
Internal node U ↔ Internal node VInternal split
Internal node V ↔ Taxon CConnection
Internal node V ↔ Taxon DConnection
One minimum internal-state assignment
DescriptionOne minimum internal-state assignment
Internal node U0
Internal node V1
Each unequal-state edge: one step
DescriptionOutcome
Internal node U ↔ Internal node VOne state change
A/C versus B/D split
DescriptionRelationship
Internal node U ↔ Taxon AConnection
Internal node U ↔ Taxon CConnection
Internal node U ↔ Internal node VInternal split
Internal node V ↔ Taxon BConnection
Internal node V ↔ Taxon DConnection
One minimum internal-state assignment
DescriptionOne minimum internal-state assignment
Internal node U0
Internal node V0
Each unequal-state edge: one step
DescriptionOutcome
Internal node U ↔ Taxon COne state change
Internal node V ↔ Taxon DOne state change
A/D versus B/C split
DescriptionRelationship
Internal node U ↔ Taxon AConnection
Internal node U ↔ Taxon DConnection
Internal node U ↔ Internal node VInternal split
Internal node V ↔ Taxon BConnection
Internal node V ↔ Taxon CConnection
One minimum internal-state assignment
DescriptionOne minimum internal-state assignment
Internal node U0
Internal node V0
Each unequal-state edge: one step
DescriptionOutcome
Internal node U ↔ Taxon DOne state change
Internal node V ↔ Taxon COne state change
Combined totals: 3, 3, 4
DescriptionPattern X: 0,0,1,1Pattern Y: 0,1,0,1Total
A/B versus C/D split123
A/C versus B/D split213
A/D versus B/C split224

Two original binary characters, equal one-step cost in either direction, no fixed ancestral state and no rooting/date inference. Selected mappings may not be unique or historically true.

03

Likelihood asks how well a model could generate the data

A likelihood calculation asks for the probability of the observed data conditional on a proposed tree and a change model. It is not, by itself, the probability that the tree is true. Maximum-likelihood analysis compares candidate trees and ordinarily estimates model parameters and branch lengths; our small illustration keeps them fixed.

The illustration uses two character states, 0 and 1. At one chosen internal node each state has probability one half. On each of the five unrooted-tree edges, the end states match with probability 0.9 and differ with probability 0.1. These transition probabilities are assigned; 0.1 is not a measured mutation rate or a count of all substitutions along an edge.

For each possible pair of internal states, multiply the initial-state probability and all five edge transition probabilities. Then add the four terms because those internal states are unobserved alternatives. For the 0,0,1,1 tip pattern, this gives likelihood 0.0401 on the A/B versus C/D split and 0.0081 on A/C versus B/D. The first is about 4.95 times the second under exactly these assumptions.

A real sequence model generally has more states and parameters and must accommodate relevant evolutionary differences. Changing the model, samples or alignment can change the comparison. Model fit and sensitivity checks complement a high score; a single favourable calculation is not a universal validation.

Add probabilities over unobserved internal states

Add probabilities over unobserved internal statesTwo unrooted quartet graphs are paired with four internal-state alternatives and the product terms contributing to the observed 0,0,1,1 pattern.A/B versus C/D splitInternal nodeUInternal nodeVTaxon A0Taxon B0Taxon C1Taxon D1Multiply five transition factors and initial probabilityU/V: 0/00.5 × 0.9 × 0.9 × 0.9 × 0.1 ×0.1 = 0.003645U/V: 0/10.5 × 0.9 × 0.9 × 0.1 × 0.9 ×0.9 = 0.032805U/V: 1/00.5 × 0.1 × 0.1 × 0.1 × 0.1 ×0.1 = 0.000005U/V: 1/10.5 × 0.1 × 0.1 × 0.9 × 0.9 ×0.9 = 0.003645Add four alternative-state termsA/B versus C/D split0.0401A/C versus B/D splitInternal nodeUInternal nodeVTaxon A0Taxon C1Taxon B0Taxon D1Multiply five transition factors and initial probabilityU/V: 0/00.5 × 0.9 × 0.1 × 0.9 × 0.9 ×0.1 = 0.003645U/V: 0/10.5 × 0.9 × 0.1 × 0.1 × 0.1 ×0.9 = 0.000405U/V: 1/00.5 × 0.1 × 0.9 × 0.1 × 0.9 ×0.1 = 0.000405U/V: 1/10.5 × 0.1 × 0.9 × 0.9 × 0.1 ×0.9 = 0.003645Add four alternative-state termsA/C versus B/D split0.0081Probability of observed pattern given fixed tree/model0.0401 / 0.00814.950617283950617Probability of U stateUnobserved U/V statesMultiply five transition factors and initial probabilityAdd four alternative-state termsProbability of observed pattern given fixed tree/modelNot probability that the tree is trueOriginal fixed two-state transition model. At U, probability of 0/1 is 0.5each. On every edge, match probability0.9 and differ probability0.1. Theassigned reversible model permits orientation from U for calculationwithout inferring an evolutionary root. Not a full maximum-likelihoodoptimisation.
Probabilities multiply along the edges within each assignment and add across alternative internal-state assignments. The computational starting node is not a claimed phylogenetic root.
Read the diagram in words

AB-CD terms for U/V=00,01,10,11 are 0.003645,0.032805,0.000005,0.003645, sum0.0401. AC-BD terms are0.003645,0.000405,0.000405,0.003645, sum0.0081; ratio about4.95.

Probability of U state
DescriptionProbability
U: 00.5
U: 10.5
Multiply five transition factors and initial probability
DescriptionProbability
Matching edge-end states0.9
Differing edge-end states0.1
A/B versus C/D split
DescriptionRelationship
Internal node U ↔ Taxon AConnection
Internal node U ↔ Taxon BConnection
Internal node U ↔ Internal node VInternal split
Internal node V ↔ Taxon CConnection
Internal node V ↔ Taxon DConnection
Observed tip state
DescriptionObserved tip state
Taxon A0
Taxon B0
Taxon C1
Taxon D1
Multiply five transition factors and initial probability
Unobserved U/V statesProbability of U stateU ↔ AU ↔ BU ↔ VV ↔ CV ↔ DOutcome
U/V: 0/00.50.90.90.90.10.10.003645
U/V: 0/10.50.90.90.10.90.90.032805
U/V: 1/00.50.10.10.10.10.10.000005
U/V: 1/10.50.10.10.90.90.90.003645
Add four alternative-state terms
DescriptionProbability of observed pattern given fixed tree/model
A/B versus C/D split0.0401
A/C versus B/D split
DescriptionRelationship
Internal node U ↔ Taxon AConnection
Internal node U ↔ Taxon CConnection
Internal node U ↔ Internal node VInternal split
Internal node V ↔ Taxon BConnection
Internal node V ↔ Taxon DConnection
Observed tip state
DescriptionObserved tip state
Taxon A0
Taxon B0
Taxon C1
Taxon D1
Multiply five transition factors and initial probability
Unobserved U/V statesProbability of U stateU ↔ AU ↔ CU ↔ VV ↔ BV ↔ DOutcome
U/V: 0/00.50.90.10.90.90.10.003645
U/V: 0/10.50.90.10.10.10.90.000405
U/V: 1/00.50.10.90.10.90.10.000405
U/V: 1/10.50.10.90.90.10.90.003645
Add four alternative-state terms
DescriptionProbability of observed pattern given fixed tree/model
A/C versus B/D split0.0081
Not probability that the tree is true
DescriptionOutcome
0.0401 / 0.00814.950617283950617

Original fixed two-state transition model. At U, probability of 0/1 is 0.5 each. On every edge, match probability0.9 and differ probability0.1. The assigned reversible model permits orientation from U for calculation without inferring an evolutionary root. Not a full maximum-likelihood optimisation.

04

Support describes an analysis and its uncertainty

A conventional character bootstrap draws alignment columns or other character units with replacement, producing a replicate dataset of the same size. The sampled taxa remain in place. Repeating the analysis on many replicates gives the fraction in which a specified branch or clade is recovered.

That frequency is tied to the dataset, resampling scheme and analysis. It is not automatically a posterior probability that the clade is true. Bayesian posterior support is a different quantity, conditional on its model, priors and analysis. Neither label removes model error, misleading sampling or gene-history conflict.

Low support can justify leaving a connection unresolved. High support can still accompany a biased result. Examine which relationship is scored, how evidence was resampled and whether reasonable alternatives change the conclusion. Presenting uncertainty helps readers use a tree responsibly.

Resample columns, then score a specified split

Resample columns, then score a specified splitA four-taxon/four-column state grid is redrawn using a specified with-replacement column-index list; an explicitly separate assigned ten-output ledger shows recovery frequency.Original four columnsColumn 1Column 2Column 3Column 4A0000B0100C1010D1110One four-column resampleColumn 1Column 1Column 3Column 4A0000B0000C1110D1110Separate assigned output ledger1✓2✓3✓4×5✓6✓7✓8×9✓10✓8 of 10: 80% recoveryOriginal four columnsOne four-column resampleColumn 1 drawn twiceColumn 2 not drawnAll four taxa retainedSeparate assigned output ledger8 of 10: 80% recoveryOriginal illustrative matrix and one resample. The ten output flags areindependently assigned teaching results, not outputs computed fromthis matrix or the preceding two-character parsimony example.Conventional character bootstrap assumes suitable characterindependence; no posterior truth probability is calculated.
The repeated and omitted column indices show sampling with replacement while keeping each taxon's row. The separate ten-outcome ledger is assigned, rather than computed from this tiny alignment. Its recovery frequency is not a posterior probability that a tree is true.
Read the diagram in words

One replicate draws columns1,1,3,4 with all four taxa; column1 repeats and2 is omitted. Separate assigned flags recover the specified split8/10 times, 80%.

Original four columns
DescriptionColumn 1Column 2Column 3Column 4
A0000
B0100
C1010
D1110
One four-column resample
DescriptionColumn 1Column 1Column 3Column 4
A0000
B0000
C1110
D1110
Column 1 drawn twice
Position in resampleOriginal column
1Column 1
2Column 1
3Column 3
4Column 4
Separate assigned output ledger
Assigned analysis outputSpecified A/B versus C/D split
1✓
2✓
3✓
4×
5✓
6✓
7✓
8×
9✓
10✓
8 of 10: 80% recovery
DescriptionCount
Recovered outputs8
Total10

Original illustrative matrix and one resample. The ten output flags are independently assigned teaching results, not outputs computed from this matrix or the preceding two-character parsimony example. Conventional character bootstrap assumes suitable character independence; no posterior truth probability is calculated.

Worked example

Keep a step count and a support fraction separate

For the binary character 0,0,1,1, enumerate internal states on each quartet. The minimum is one step for A/B versus C/D and two for each alternative. Add a second character 0,1,0,1 with equal weight: it requires two steps on the first split and one on A/C versus B/D. The combined totals are 3 and 3, a tie; A/D versus B/C totals 4.

In a separate supplied resampling ledger, the chosen split appears in 8 of 10 analysis outputs. Its recovery fraction is 8/10 = 0.8, or 80%. These ten outputs are illustrative assigned results, not inferred from the two-character parsimony case. A drawing that relabelled 80% as “probability the history is true” would add an unsupported interpretation.

The resampling panel shows four original columns and one replicate draw with indices 1, 1, 3, 4. Column 1 is used twice, column 2 is absent, and all four taxa remain. Replicates can differ while retaining the same number of sampled columns; replacing taxa instead would be a different procedure.

Check your understanding

  1. Why does adding the 0,1,0,1 character create a parsimony tie between the first two splits?

    Show reasoning

    The first pattern costs 1 versus 2; the second costs 2 versus 1. Equal-weight sums are 3 versus 3. Neither wins from these two characters under the stated rule, and the tie should not be hidden by choosing a preferred drawing.

  2. What does the likelihood ratio 0.0401/0.0081 establish?

    Show reasoning

    The observed single-site pattern is about 4.95 times as probable under the first fixed tree/model combination as under the second. It does not make the first tree 4.95 times as likely to be true without further assumptions and inferential treatment.

  3. If a split occurs in 8 of 10 bootstrap outputs, what is scored and what uncertainty remains?

    Show reasoning

    Its recovery frequency is 80% under that resampling and analysis. The data, independence assumptions, alignment, model and sampling can still be wrong or incomplete; the fraction is not automatically a probability of biological truth.

What to carry forward

A phylogenetic result depends on comparable evidence, an explicit inference rule and a support measure whose meaning is preserved.

References

Resources supporting this lesson’s explanations and examples.

  1. The Tree Room — Reading trees and comparing evidence (Live teaching text; inspected October 2026)Understanding Evolution team · UC Museum of Paleontology, University of California, BerkeleyInstitutional scientific educationThe Tree Room — Reading trees and comparing evidence (Live teaching text; inspected October 2026)Common ancestry and relationshipsTip order, rotation and progress misconceptionsComparing charactersParsimony and its limitsResolution and branch supportevolution.berkeley.edu
  2. Foundational likelihood and bootstrap methods (Journal of Molecular Evolution 17:368–376, 1981; Evolution 39:783–791, 1985)Joseph FelsensteinResearch abstractEvolutionary trees from DNA sequences: a maximum likelihood approachConfidence limits on phylogenies: an approach using the bootstrappubmed.ncbi.nlm.nih.gov
  3. Gene tree discordance, phylogenetic inference and the multispecies coalescent (Trends in Ecology & Evolution 24:332–340, 2009)James H. Degnan and Noah A. RosenbergAuthor research review — selected author-hosted text · PDFweb.stanford.eduRead source