Mass transfer in bioprocesses (kLa)

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.51)** **1. Governing Concept & Definition**: - **Topic**: Bioinformatics & Phylogenetics - For $n$ taxa/sequences: - Number of unrooted bifurcating trees: $N_U = (2n - 5)!!$ - Number of rooted bifurcating trees: $N_R = (2n - 3)!!$ - Each unrooted tree has $(2n - 3)$ branches on which a root can be placed. **2. Detailed Step-by-Step Calculation**: - **Given**: - Number of taxa $n = 4$ - Number of unrooted trees given as $N_U = 3$ - **Formula for Rooted Trees**: $N_R = (2n - 3)!! = (2(4) - 3)!! = 5!! = 5 \times 3 \times 1 = 15$ - Alternatively, multiplying unrooted trees by number of rootable branches: $\text{Branches per unrooted tree} = 2n - 3 = 2(4) - 3 = 5$ $N_R = N_U \times 5 = 3 \times 5 = 15$ **3. Final Answer**: - **Calculated Value**: **15** - **Official GATE Accepted Range**: **15 to 15**