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.
The contour length of a B-DNA molecule that encodes a bacterial protein of 33 kDa is _________ nm.
Consider the average molecular weight of an amino acid as 110 Da and helix rise per base pair for B-DNA as 0.34 nm.
(Round off to the nearest integer)