MIDTERM #1 - MB409 : MICROBIAL DIVERSITY : FEB 12, 2001
1. What are the three primary evolutionary branches of life? (5 points)
2. What properties must a sequence have in order to be useful for molecular phylogenetic analysis? (5 points)
- Clock-like behavior
- Phylogenetic range
- No horizontal transfer
- Large existing database
3. Align these sequences: (10 points)
Sequence A : GGAUCGAGAGUUCC
Sequence B : UGGAACGAAAGAUCC
Sequence C : GGAUCGAUGAGAUCUSeq A : - G G A U C G A - G A G U U C C Seq B : U G G A A C G A - A A G A U C C Seq C : - G G A U C G A U G A G A U C U
4. Given the following alignment and the structure of RNA X, draw the structure of RNA Y and RNA Z. (10 points)
RNA X : G C A U A G C U U C G G C U G C A G A RNA Y : G G - - A A A U U C - U U U C C A - - RNA Z : A U A U A U A U A U A U A U A U A U A
5. Below are two RNase P RNAs. The first is from Bacillus subtilis, the reference sequence for the signature table below. The second is a partial sequence from a unknown organism. Use the signature table below to determine which genus of organism this unknown might belong to. Show your work! (10 points)
Position 64 78 100 245 317 318 339 340 Bacillus GX CX Gok AX UX Aok GX Aok 3/8 Staphylococcus Aok Uok GX Cok Cok Aok Cok Aok 7/8 Streptococcus Aok Uok UX AX UX Aok AX Aok 4/8 Lactobacillus Aok Uok Aok AX GX GX UX UX 3/8 Unknown A U A C C A C A
The best match, 7/8, suggests that the unknown is a member of the genus Staphylococcus.
6. Create a similarity matrix from this alignment: (10 points)
Sequence 1 : G A U C U U C G G A Sequence 2 : G A A C U C C G U A Sequence 3 : G A A C U U C G G A Sequence 4 : G G G G G G G G G G Sequence 5 : G A U U U U C G G U Seq1 Seq2 Seq3 Seq4 Seq5 Seq1 - - - - - Seq2 0.7 - - - - Seq3 0.8 0.8 - - - Seq4 0.3 0.2 0.3 - - Seq5 0.8 0.5 0.7 0.3 -
7. Explain why similarity between two sequences is an underestimate their evolutionary distance. (10 points)
Just counting differences between two sequences underestimates the number of changes that occured between them, because more than one evolutionary change at a single position (e.g. A -> G -> U) counts as only one difference between two sequences, and in the case of reversion counts as no change at all (e.g. A -> G -> A).
8. Convert this distance matrix into a tree. Be sure to include a scale bar. (10 points)
A B C D A - - - - B 0.1 - - - C 0.2 0.2 - - D 0.3 0.3 0.3 -
9. Answer the following questions about this tree. (10 points)
Which two sequences are the most closely related ? D & E
Which sequence is most closely related to C? A
Of sequences A - E, which is the most primitive? C
Circle the last common ancestor of sequences A, B, and C.
Which sequence is probably the outgroup? F
10. Describe the endosymbiont theory for the origin of the mitochondrion. How was this theory proven? (10 points)
The endosymbiont theory for the origin of mitochondria is that it was an endosymbiotic aerobic bacterium that has established a permanent relationship with it's eukaryotic host. This was proven by molecular phylogenetic analysis - the mitochondrion has it's own DNA, including small subunit rRNA genes, and trees generated with these sequences have the mitochondria forming a specific branch in the alpha-purple Bacteria.
11. In an episode of the X-files, Agent Scully sequences some alien DNA and finds 'missing bands' in the sequences that she interprets to correspond to bases that are unique to aliens (not found in earthling DNA). Why is this not technically feasible? (5 points)
DNA sequencing reactions contain all 4 "earthling" dNTPs (dATP, dGTP, dCTP, and dTTP) for DNA synthesis, and 4 chain-terminating ddNTPs (ddATP, ddGTP, ddCTP, ddTTP). If alien DNA contained additional bases, and even if DNA polymerase could use these bases for DNA synthesis, there are no dNTPs or ddNTPs to match these extra bases and so the sequencing reactions would most likely just fail to work at all, stopping at the first occurence of one of these extra bases, because there would be no dNTP to basepair to the alien base. If the alien bases could basepair to the standard 'earthling' bases, the sequencing reactions might work, but there would be no gaps - the alien bases would appear in the sequencing reactions to be replaced by standard earthling bases.
12. In the 3-Kingdom tree, most of the deep (early branching) species of eukaryotes are parasites - Giardia, microsporidia, trypanosomes, etc. Can you think of a reason why this might be? (5 points)
It seems likely that there are lots of non-parasitic early-branching eukaryotes, but they have yet to be discovered or characterized. Because they are likely all anaerobes (having diverged from the other eukaryotes before the origin of mitochondria), they would be difficult to grow in captivity. The reason the only such organisms to be characterized are parasites of humans or commercially-important animals is easy to explain - these are the only such species we, as humans, have had enough reason to go out and look hard for.