Q.13 Number of unrooted trees in a phylogeny of five sequences is
(A) 3
(B) 15
(C) 105
(D) 945
(2n−5)!! = ∏i=3n(2i−5), where !! denotes the double factorial for odd numbers.
Formula Breakdown
For unrooted bifurcating phylogenetic trees with n labeled taxa (sequences), the count is given by the product 1×3×5×⋯×(2n−5), or equivalently (2n−5)! / 2n−3(n−3)! for n≥3. Substituting n=5 yields 1×3×5=15 unrooted trees.
Actually for unrooted with 5: start from 3 taxa (1), add branches: ×3 (for 4 taxa=3), ×5 (for 5 taxa=15).
This assumes fully resolved (bifurcating) trees with distinct leaf labels, common in molecular phylogenetics for sequence data. The exponential growth underscores computational challenges in tree search algorithms like heuristic methods in tools such as PHYLIP or RAxML.
Option Analysis
Each option represents tree counts at different taxa levels or types:
| Option | Value | Represents | Correct for Five Sequences? |
|---|---|---|---|
| (A) | 3 | Unrooted trees for 4 taxa: 1×3=3 |
No, too low |
| (B) | 15 | Unrooted trees for 5 taxa: 1×3×5=15 |
Yes |
| (C) | 105 | Unrooted for 6 taxa or rooted for 5 taxa: 1×3×5×7=105 |
No, confuses rooted/unrooted |
| (D) | 945 | Unrooted for 7 taxa or rooted for 6 taxa | No, for more taxa |
Why It Matters
In exams like CSIR NET, GATE Biotechnology, or your molecular biology coursework, this tests understanding of phylogenetic space complexity. For five sequences (e.g., gene orthologs), exhaustive search of 15 topologies is feasible, but beyond 10 taxa (over 2 million), stochastic optimization is essential.
Rooted trees are 2n−3 times more numerous due to internal branch rooting options.
