[Q8] Genome rearrangements: double-cut-and-join distance

Let b and c be the number of breakpoints and cycles, respectively, as pointed by the breakpoint graph G for two genomes. Consider the next cases for the DCJ operation:

1. Cutting two black lines belonging to different cycles and then joining
2. Improper joining
3. Proper joining with no successive lines
4. Proper joining with successive lines in a cycle with two or more black lines
5. Proper joining with successive lines in a 2-cycle

In which of the cases above, performing the DCJ operation would decrease c?

1. Just in I and V
2. Just in II, IV
3. Just in I, III and V
4. Just in II, III and IV
5. None of the above.

Original idea by: Juan Felipe Hernández Albarracín.

[Q7] Genome rearrangements: weighted median

For a given adjacency α to appear in the weighted median of n genomes, and assuming that all of them have the same weight, it is required for α to be present in:

1. All genomes
2. At least n - 1 genomes.
3. At least the half of the genomes.
4. At least one more genome than their conflicting adjacencies.
5. None of the above.

Original idea by: Juan Felipe Hernández Albarracín.

[Q6] Genome rearrangements: reversals

Given the fictitious genome G₁.

Which of the following adjacencies sets represents the genome with the minimum reversal distance from G₁?

1. {ND5_hND4_t, ND4_hND3_t, ND3_hCOIII_t, COIII_hCOI_h}
2. {ND5_hND3_t, ND3_hCOIII_t, COIII_hND4_h, ND4_tCOI_h}
3. {ND5_hND4_t, ND4_hCOIII_h, COIII_tND3_h, ND3_tCOI_h}
4. {ND5_hND4_h, ND4_tCOIII_h, COIII_tND3_h, ND3_tCOI_h}
5. None of the above.

Original idea by: Juan Felipe Hernández Albarracín.

[Q5] Phylogenetic trees: Parsimony

Given one rooted and one unrooted most parsimonious trees for the same group of species that consider only binary states, it is correct to say that:

1. The edge where an hypothetical root for the unrooted tree is placed would affect the number of changes of state of this most parsimonious tree.
2. Whether a tree is rooted or not, affects only in the possible positions of state changes for the most parsimonious tree.
3. The rooted and unrooted tree will never be able to have the same topology.¹
4. Nothing can be concluded from this, since the interpretation of a rooted and an unrooted tree is different.
5. None of the above.

¹ Assuming that two trees could be considered to have the same topology if the only difference in them is the fact that a root is present or not.

Original idea by: Juan Felipe Hernández Albarracín.

[Q4] DNA Evolution Models

According to the assumptions made for both Jukes-Cantor and Kimura 2 Parameter (K2P) models about the substitution rates (either α or β), and taking into account the restrictions for Markov models, it is correct to say about the range of possible values for them that:

1. Any of the rates α and β in K2P have a larger range than α in Jukes-Cantor.
2. The range of α in Jukes-Cantor is larger that any of the ranges for α and β in K2P.
3. Only depending on the value of α in K2P it is possible to conclude anything about the size of the ranges of β in K2P with respect to α in Jukes-Cantor.
4. Only depending on the value of β in K2P it is possible to conclude anything about the size of the ranges of α in K2P with respect to α in Jukes-Cantor.
5. None of the above

Original idea by: Juan Felipe Hernández Albarracín.

[Q3] Sequence Comparison

The main reason why it is acceptable a saving of space while increasing the execution time with the improvement for the basic sequence comparison algorithm is:

1. It is imperative to save space at any cost since the processing of normal sequences for real problems demands more RAM than any average computer can have nowadays.
2. The increment in execution time is merely theoretical, considering the processing capacity of current computers.
3. The increment in execution time is constant with this improvement.
4. Time complexity remains the same with this improvement.
5. None of the above

Original idea by: Juan Felipe Hernández Albarracín.

[Q2] Informatics Concepts

Let Q_s be the set containing all the subsequences of string S and T_s be the set of all the substrings of string S. It is correct to say that:

1. Q_s ⊆ T_s
2. T_s ⊆ Q_s
3. Q_s ⊂ T_s
4. T_s ⊂ Q_s
5. None of the above

Original idea by: Juan Felipe Hernández Albarracín.