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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Understand how scored characters, change models and support measures contribute to a proposed tree.
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.
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
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.
| Description | Pattern X: 0,0,1,1 | Pattern Y: 0,1,0,1 |
|---|---|---|
| A | 0 | 0 |
| B | 0 | 1 |
| C | 1 | 0 |
| D | 1 | 1 |
| Description | Relationship |
|---|---|
| Internal node U ↔ Taxon A | Connection |
| Internal node U ↔ Taxon B | Connection |
| Internal node U ↔ Internal node V | Internal split |
| Internal node V ↔ Taxon C | Connection |
| Internal node V ↔ Taxon D | Connection |
| Description | One minimum internal-state assignment |
|---|---|
| Internal node U | 0 |
| Internal node V | 1 |
| Description | Outcome |
|---|---|
| Internal node U ↔ Internal node V | One state change |
| Description | Relationship |
|---|---|
| Internal node U ↔ Taxon A | Connection |
| Internal node U ↔ Taxon C | Connection |
| Internal node U ↔ Internal node V | Internal split |
| Internal node V ↔ Taxon B | Connection |
| Internal node V ↔ Taxon D | Connection |
| Description | One minimum internal-state assignment |
|---|---|
| Internal node U | 0 |
| Internal node V | 0 |
| Description | Outcome |
|---|---|
| Internal node U ↔ Taxon C | One state change |
| Internal node V ↔ Taxon D | One state change |
| Description | Relationship |
|---|---|
| Internal node U ↔ Taxon A | Connection |
| Internal node U ↔ Taxon D | Connection |
| Internal node U ↔ Internal node V | Internal split |
| Internal node V ↔ Taxon B | Connection |
| Internal node V ↔ Taxon C | Connection |
| Description | One minimum internal-state assignment |
|---|---|
| Internal node U | 0 |
| Internal node V | 0 |
| Description | Outcome |
|---|---|
| Internal node U ↔ Taxon D | One state change |
| Internal node V ↔ Taxon C | One state change |
| Description | Pattern X: 0,0,1,1 | Pattern Y: 0,1,0,1 | Total |
|---|---|---|---|
| A/B versus C/D split | 1 | 2 | 3 |
| A/C versus B/D split | 2 | 1 | 3 |
| A/D versus B/C split | 2 | 2 | 4 |
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.
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
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.
| Description | Probability |
|---|---|
| U: 0 | 0.5 |
| U: 1 | 0.5 |
| Description | Probability |
|---|---|
| Matching edge-end states | 0.9 |
| Differing edge-end states | 0.1 |
| Description | Relationship |
|---|---|
| Internal node U ↔ Taxon A | Connection |
| Internal node U ↔ Taxon B | Connection |
| Internal node U ↔ Internal node V | Internal split |
| Internal node V ↔ Taxon C | Connection |
| Internal node V ↔ Taxon D | Connection |
| Description | Observed tip state |
|---|---|
| Taxon A | 0 |
| Taxon B | 0 |
| Taxon C | 1 |
| Taxon D | 1 |
| Unobserved U/V states | Probability of U state | U ↔ A | U ↔ B | U ↔ V | V ↔ C | V ↔ D | Outcome |
|---|---|---|---|---|---|---|---|
| U/V: 0/0 | 0.5 | 0.9 | 0.9 | 0.9 | 0.1 | 0.1 | 0.003645 |
| U/V: 0/1 | 0.5 | 0.9 | 0.9 | 0.1 | 0.9 | 0.9 | 0.032805 |
| U/V: 1/0 | 0.5 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.000005 |
| U/V: 1/1 | 0.5 | 0.1 | 0.1 | 0.9 | 0.9 | 0.9 | 0.003645 |
| Description | Probability of observed pattern given fixed tree/model |
|---|---|
| A/B versus C/D split | 0.0401 |
| Description | Relationship |
|---|---|
| Internal node U ↔ Taxon A | Connection |
| Internal node U ↔ Taxon C | Connection |
| Internal node U ↔ Internal node V | Internal split |
| Internal node V ↔ Taxon B | Connection |
| Internal node V ↔ Taxon D | Connection |
| Description | Observed tip state |
|---|---|
| Taxon A | 0 |
| Taxon B | 0 |
| Taxon C | 1 |
| Taxon D | 1 |
| Unobserved U/V states | Probability of U state | U ↔ A | U ↔ C | U ↔ V | V ↔ B | V ↔ D | Outcome |
|---|---|---|---|---|---|---|---|
| U/V: 0/0 | 0.5 | 0.9 | 0.1 | 0.9 | 0.9 | 0.1 | 0.003645 |
| U/V: 0/1 | 0.5 | 0.9 | 0.1 | 0.1 | 0.1 | 0.9 | 0.000405 |
| U/V: 1/0 | 0.5 | 0.1 | 0.9 | 0.1 | 0.9 | 0.1 | 0.000405 |
| U/V: 1/1 | 0.5 | 0.1 | 0.9 | 0.9 | 0.1 | 0.9 | 0.003645 |
| Description | Probability of observed pattern given fixed tree/model |
|---|---|
| A/C versus B/D split | 0.0081 |
| Description | Outcome |
|---|---|
| 0.0401 / 0.0081 | 4.950617283950617 |
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.
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
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%.
| Description | Column 1 | Column 2 | Column 3 | Column 4 |
|---|---|---|---|---|
| A | 0 | 0 | 0 | 0 |
| B | 0 | 1 | 0 | 0 |
| C | 1 | 0 | 1 | 0 |
| D | 1 | 1 | 1 | 0 |
| Description | Column 1 | Column 1 | Column 3 | Column 4 |
|---|---|---|---|---|
| A | 0 | 0 | 0 | 0 |
| B | 0 | 0 | 0 | 0 |
| C | 1 | 1 | 1 | 0 |
| D | 1 | 1 | 1 | 0 |
| Position in resample | Original column |
|---|---|
| 1 | Column 1 |
| 2 | Column 1 |
| 3 | Column 3 |
| 4 | Column 4 |
| Assigned analysis output | Specified A/B versus C/D split |
|---|---|
| 1 | ✓ |
| 2 | ✓ |
| 3 | ✓ |
| 4 | × |
| 5 | ✓ |
| 6 | ✓ |
| 7 | ✓ |
| 8 | × |
| 9 | ✓ |
| 10 | ✓ |
| Description | Count |
|---|---|
| Recovered outputs | 8 |
| Total | 10 |
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
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.
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.
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.
- 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 support
- 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 bootstrap
- 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 · PDF