GATE BT 2021 • Q13
MCQ • 1 Mark
Section 2: General Biotechnology
Biomolecules, enzyme kinetics
Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
Number of unrooted trees in a phylogeny of five sequences is
Options:
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(A) 3
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(B) 115
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(C) | 105
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(D) 945
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2021 Q.13)**
**1. Governing Concept**:
- **Topic**: Bioinformatics & Phylogenetic Tree Topologies
- The number of unrooted bifurcating phylogenetic trees $N_U$ for $n$ distinct taxa/sequences is given by the formula:
$N_U = (2n - 5)!! = \prod_{i=3}^n (2i - 5) = 1 \times 3 \times 5 \times \cdots \times (2n - 5)$
Alternatively:
$N_U = \frac{(2n - 4)!}{2^{n - 2} (n - 2)!}$
- For $n = 5$ sequences:
$N_U = (2 \times 5 - 5)!! = 5!! = 1 \times 3 \times 5 = 15$
- Note: Option (B) in the OCR text represents 15 (OCR misread "115" for "15"). The official GATE key is **B** ($15$).
**2. Option Analysis**:
- **Option (A) [3] is INCORRECT**: Number of unrooted trees for $n = 4$ ($3!! = 3$).
- **Option (B) [15] is CORRECT**: $N_U = (2 \times 5 - 5)!! = 5 \times 3 \times 1 = 15$.
- **Option (C) [105] is INCORRECT**: Number of rooted trees for $n = 5$ ($(2n - 3)!! = 7!! = 105$).
- **Option (D) [945] is INCORRECT**: Number of rooted trees for $n = 6$.
**3. Final Answer**:
- **Correct Choice**: **B**
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