Speciation Genomics · PB 495/595 Plant Evolutionary Biology
22 October 2026
By the end you should be able to:
Toy data: the 17-site A–E alignment, UPGMA tree, \(B=1000\) pseudoreplicates.
What it is
What it is not
The same recipe gives uncertainty on any estimate computed from the alignment — including node ages:
Toy example: root height \(=4.58\) substitutions, bootstrap 95% interval 3.67–5.67 — wide, as expected from 17 sites.
A branch length in substitutions per site is
\[ b = r \times t \quad(\text{rate} \times \text{time}). \]
The molecular clock hypothesis: \(r\) is roughly constant across lineages, so branch lengths are proportional to time. This hypothesis is based in the neutral theory of molecular evolution. Recall the neutral substitution rate is the mutation rate (\(\rho=\mu\)), which is independent of population size. If \(\mu\) is similar across lineages, the molecular clock should be constant.
Building off of neighbor-joining, we can use the pairwise distances to estimate node ages using least-squares.
Under a strict clock every lineage has the same rate \(r\), so all tips are the same distance from the root (the tree is ultrametric). Let \(m(i,j)\) be the most recent common ancestor (MRCA) of tips \(i\) and \(j\), and define its height \(h_m = r\,t_m\) (substitutions per site since the node). Then the distance between two tips is the path up to the MRCA and back down:
\[ d_{ij} \;=\; 2\,r\,t_{m(i,j)} \;=\; 2\,h_{m(i,j)}. \]
Choose node heights \(h\) to minimise the weighted sum of squared residuals
\[ S(h) \;=\; \sum_{i<j} w_{ij}\,\bigl(d_{ij} - 2h_{m(i,j)}\bigr)^2 . \]
With equal weights (\(w_{ij}=1\)) each node can be solved on its own: set \(\partial S/\partial h_m=0\),
\[ \hat h_m \;=\; \frac{1}{2}\,\overline{d}_m, \qquad \overline{d}_m=\text{mean of } d_{ij} \text{ over all pairs whose MRCA is } m . \]
Each node age is half the average distance across it. Ages must also satisfy the condition that a child is younger than its parent
Fix the topology \(((A,B),(C,(D,E)))\) and group the 10 distances by their MRCA:
| Node \(m\) | Pairs (distances) | \(\overline{d}_m\) | \(\hat h_m=\overline{d}_m/2\) |
|---|---|---|---|
| (D,E) | DE = 3 | 3 | 1.50 |
| (A,B) | AB = 5 | 5 | 2.50 |
| (C,(D,E)) | CD = 8, CE = 7 | 7.5 | 3.75 |
| root | AC = 9, AD = 9, AE = 8, BC = 10, BD = 10, BE = 9 | 9.17 | 4.58 |
Residual sum of squares \(=3.33\), almost all from the six root pairs (\(2.83\)) — the data are only approximately clock-like.
Real trees are rarely ultrametric. Instead of tip-to-tip distances, use the branch lengths \(b_i\) already estimated by ML (substitutions/site) on a rooted tree. Estimate node times \(t\) and a single rate \(r\) by
\[ \min_{r,\,t}\;\sum_{\text{branches } i} w_i\,\Bigl(b_i - r\,\bigl(t_{\text{child}(i)} - t_{\text{parent}(i)}\bigr)\Bigr)^2 \quad\text{subject to } t_{\text{parent}}<t_{\text{child}}. \]
Least squares gives heights \(\hat h_m\) in substitutions. Under a strict clock \(h_m=r\,t_m\), so any one node of known age fixes the scale:
\[ \hat r=\frac{\hat h_{\text{cal}}}{t_{\text{cal}}}, \qquad \hat t_m=\frac{\hat h_m}{\hat r}=\hat h_m\times\frac{t_{\text{cal}}}{\hat h_{\text{cal}}}. \]
Example tree, root calibrated to 100 My:
Calibrating (C,(D,E)) to 80 My instead gives a root of 97.8 My.
From sequence data we observe only the product \(b=rt\) — as the least-squares fit showed, the data alone give heights in substitutions, and the scale is set only by external time information. Doubling the rate and halving the time gives the identical likelihood:
\[ b = r\,t = (2r)\left(\tfrac{t}{2}\right). \]
External evidence fixes ages (or age distributions) at nodes/tips:
Calibrations are priors on time, layered onto the rate–time product. Their choice and placement is data driven but requires justification from organismal experts.
Crowns versus stems. Fig. 1 from Renner (2005)
Faster rates of evolution in herbaceous plants compared to woody plants. Fig. 1 from Smith and Donoghue (2008)
Effect of different calibrations on estimated node ages. Fig. 3 from Barba-Montoya et al. (2018)
Effect of different calibrations on estimated node ages. Fig. 3 from Wu et al. (2026)
\[ r_i \sim \text{Lognormal}(\mu_r, \sigma_r), \]
with \(\sigma_r\) measuring how clock-like the data are (Drummond et al. 2006).
Put it together in one hierarchical model, sampled by MCMC:
\[ \underbrace{P(\text{data}\mid \tau, t, r)}_{\text{substitution + clock}} \;\times\; \underbrace{P(\tau, t)}_{\text{tree prior}} \;\times\; \underbrace{P(\text{calibrations})}_{\text{time priors}}. \]
| Partition | Sites |
|---|---|
| plastid, codon positions 1+2 | 41,021 |
| plastid, codon position 3 | 20,360 |
| mitochondrial, 1+2 | 5,690 |
| mitochondrial, 3 | 2,878 |
| nuclear rDNA | 5,081 |
Same alignment, topology, tree prior and approximate likelihood. Only the calibrations and the clock change:
| Clock | Calibrations | |
|---|---|---|
| A reference | relaxed (independent log-normal) | all 52 |
| B | relaxed | angiosperm crown + deep backbone only (10) |
| C | strict | angiosperm crown + deep backbone only (10) |
Can you make any predictions about how ages and rates will change with fewer calibrations throughout the tree?
PB 495/595 · Plant Evolutionary Biology · Speciation Genomics