Question

The phylogenetic tree depicts the relationship between 5 species of snakes (labelled A to E) and provides information about their habitat specialization. Given the principle of parsimony (least number of evolutionary changes required) and that ancestor Y was terrestrial, which one of the options given is correct?

• A. X was more likely to be aquatic than terrestrial.
• B. X was more likely to be terrestrial than aquatic
• C. X was equally likely to be aquatic or terrestrial
• D. X was neither aquatic nor terrestrial.

Hint:

In Wright’s island model, \(F_{st}\) increases as either population size \(N\) or migration rate \(m\) decreases; the key driver is the product \(Nm\).

Answer 1

The Correct Option is C

Step 1: Read tip states on the tree.
Leaves show: \(A=\) Aquatic, \(B=\) Terrestrial, \(C=\) Aquatic, \(D=\) Aquatic, \(E=\) Terrestrial. Node \(Y\) (ancestor of all) is given Terrestrial. Node \(X\) is the common ancestor of \(A,B,C,D\).

Step 2: Assume \(X=\) Terrestrial and count changes.
If \(X\) is Terrestrial, then to match leaves we need:
\(\bullet\) One change \(T\to A\) on branch to \(A\).
\(\bullet\) One change \(T\to A\) on the branch ancestral to \(C\) and \(D\) (shared), making both Aquatic.
\(\bullet\) \(B\) stays Terrestrial; \(Y\to X\) has no change; \(E\) stays Terrestrial.
\(\Rightarrow\) Total changes \(=2\).

Step 3: Assume \(X=\) Aquatic and count changes.
If \(X\) is Aquatic, then:
\(\bullet\) One change \(T\to A\) on the branch \(Y\to X\).
\(\bullet\) One change \(A\to T\) on the branch to \(B\) to recover Terrestrial.
\(\bullet\) \(A,C,D\) remain Aquatic without extra changes; \(E\) remains Terrestrial from \(Y\).
\(\Rightarrow\) Total changes \(=2\).

Step 4: Parsimony decision.
Both assignments of \(X\) (Aquatic or Terrestrial) require the same minimum number of changes (2).
\(\Rightarrow\) Under parsimony, \(X\) is \emph{equally likely} to be Aquatic or Terrestrial. Final Answer:\quad \(\boxed{\text{(C) X was equally likely to be aquatic or terrestrial.}}\)