Maximum Parsimony Scores with nucleotide data

Informative SNP positions of types 5, 6, & 7 favor the first, second, and third hypotheses of relationship, respectively, because each of these types requires a single nucleotide change in that hypothesis, and two changes in the two alternative hypotheses.

Then, in a hypothetical data set [not shown] with 10 Type 5 positions, 5 Type 6 positions, and 2 Type 7 positions, 
    Hypothesis #1 then requires  N = 10 + 2(  5 + 2) = 24 changes
    Hypothesis #2 would require N =   5 + 2(10 + 2) = 29 changes
    Hypothesis #3 would require N =   2 + 2(10 + 5) = 32 changes

Thus, the first hypothesis requires the fewest changes and is therefore the most parsimonious explanation of the data. Note that this hypothesis has the largest number of parsimony sites that favor it.

Uninformative SNP positions of types 1, 2, 3, & 4 uniformly require 0, 1, 3, & 2 nucleotide changes, respectively, in any of the hypotheses. They therefore do not contribute any information towards the relative ranking of the three hypotheses.

Homework: SHOW that positions of types 1, 2, 3, & 4 require 0, 1, 3, & 2 nucleotide changes, respectively, for any of the three hypotheses above. [Hint: Draw the three hypotheses, and find at least one assignment of changes to the branches or internode that satisfies the statement.