The number of possible phylogenetic trees varies depending on whether the trees are rooted or unrooted and the number of distinct sequences (taxa), represented by '$n$'.
For $n$ distinct sequences, the number of possible unrooted binary phylogenetic trees is given by the formula:
$ \text{Number of unrooted trees} = (2n - 5)!! $
This formula is valid for $n \ge 3$. The double factorial $k!!$ involves multiplying integers down to 1 with the same parity as $k$. For instance, $3!! = 3 \times 1$.
For $n$ distinct sequences, the number of possible rooted binary phylogenetic trees is calculated using:
$ \text{Number of rooted trees} = (2n - 3)!! $
This formula applies for $n \ge 2$. It's worth noting that $(2n-3)!! = (2n-3) \times (2n-5)!!$.
We need to calculate the number of rooted and unrooted trees for $n = 4$ different sequences.
The question asks for the number of rooted and unrooted trees, in that specific order (respectively). Therefore, the result is 15 rooted trees and 3 unrooted trees.
Consider the following multiple sequence alignment of four DNA sequences.
$\begin{matrix}A & C & T & A \\ A & C & T & G \\ A & G & T & C \\ A & G & C & T \end{matrix} $
Shannon's entropy of the above alignment is ________.
The transcription factor X binds a 10 base pair DNA stretch. In the DNA of an organism, X was found to bind at 20 distinct sites. An analysis of these 20 binding sites showed the following distribution:
| Base | Position in the binding site | |||||||||
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
| A | 11 | 0 | 0 | 0 | 16 | 2 | 4 | 0 | 4 | 3 |
| T | 3 | 0 | 19 | 0 | 1 | 3 | 4 | 20 | 2 | 4 |
| G | 4 | 20 | 0 | 0 | 2 | 4 | 6 | 0 | 12 | 2 |
| C | 2 | 0 | 1 | 20 | 1 | 11 | 6 | 0 | 2 | 11 |
What is the consensus sequence for the binding site of X?