The phylogeny below shows the relationship between six species based on whole genome data. The sequence of a part of a particular gene from each species is provided at the tips of the tree. Based on the principle of parsimony, which one of the following is the expected sequence for the ancestor X?
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Work out each ancestral position separately using the fewest base changes needed to explain the tips in each half of the tree, then combine the four positions.
Step 1: Note that positions 1 and 4 are already fixed.
Every one of the six tip sequences starts with $A$, and all end in $G$ except AACT, which ends in $T$ on its own separate branch. Since all four answer choices already agree on position 1 being $A$ and position 4 being $G$, these two positions do not distinguish the options, so the real work is in positions 2 and 3.
Step 2: Score position 2 by counting the minimum substitutions the tree needs.
At position 2 the tips read: ATGG has $T$, AAGG has $A$, AACG has $A$, AACT has $A$, ATGG has $T$, ATCG has $T$. Working up the tree, the ATGG/AAGG pair needs one change to explain their disagreement, and a further change links that pair's group with the AACT/ATGG/ATCG side. Setting ancestor $X$ to $A$ at this position keeps the total at 2 changes on the tree. Setting $X$ to $T$ instead forces one extra change near the root, giving 3 changes overall. So $A$ is the cheaper, more parsimonious choice at position 2.
Step 3: Score position 3 the same way.
At position 3 the tips read: ATGG has $G$, AAGG has $G$, AACG has $C$, AACT has $C$, ATGG has $G$, ATCG has $C$. Both halves of the tree independently settle on $C$ once the changes are traced through (one change between the ATGG-AAGG pair and AACG, one change between AACT and the ATGG-ATCG pair), and setting $X=C$ needs only these 2 changes. Setting $X=G$ instead adds an extra change at the root, for 3 changes overall. So $C$ is the cheaper choice at position 3.
Step 4: Add positions 1 and 4 back in.
Since position 1 is fixed at $A$ and position 4 is fixed at $G$, the full most-parsimonious ancestral sequence is $A$, $A$, $C$, $G$.
Step 5: Match against the options.
This sequence is AACG, option (B). The other three options each use the costlier base at position 2, position 3, or both, so each needs at least one extra evolutionary change compared to AACG.
\[ \boxed{\text{AACG}} \]
