All Exams Test series for 1 year @ ₹349 only
Question

How many rooted and unrooted phylogenetic trees, respectively, are possible with four different sequences?

The correct answer is
15 and 3

Phylogenetic Tree Counting

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 '$n$'.

Unrooted Phylogenetic Trees Formula

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, $3!! = 3 \times 1$.

Rooted Phylogenetic Trees Formula

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 $(2n-3)!! = (2n-3) \times (2n-5)!!$.

Calculating Trees for Four Sequences

We need to calculate the number of rooted and unrooted trees for $n = 4$ different sequences.

  • Unrooted Trees Calculation: Applying the formula $(2n - 5)!!$ with $n=4$: $ (2 \times 4 - 5)!! = (8 - 5)!! = 3!! = 3 \times 1 = 3 $ Thus, there are 3 possible unrooted trees.
  • Rooted Trees Calculation: Applying the formula $(2n - 3)!!$ with $n=4$: $ (2 \times 4 - 3)!! = (8 - 3)!! = 5!! = 5 \times 3 \times 1 = 15 $ Thus, there are 15 possible rooted trees.

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.

Was this answer helpful?

Important Questions from Molecular Structure of Genes and Chromosomes

  1. All pseudogenes DO NOT code for a __________.
  2. C-value paradox refers to
  3. DNA sample collected from an unidentified bacterial species (Y) contains 13% of adenine. The G+C content (in percentage) of Y is ________
  4. 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)

  5. Which of the following methods is/are used for identifying histone modifications?
Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App