Midterm Exam #1 MB 451Microbial Diversity

Honor pledge: “I have neither given nor received unauthorized aid on this test.”

Signed : ____________________________________________________________ Date : __Feb 5, 2007__________________

Name : ___KEY__________________________________________________________

1. What are the three primary evolutionary branches of life? (5 points)

Archaea, Bacteria, and Eukarya

Multiple choice (2 points each)

2. __D__ Which of the following is not an aspect of a tax-onomy?

A. Grouping

B. Naming

C. Identifying

D. Isolating

E. None of the above

3. __ E__ Which describes a phylogeny?

A. a genealogy

B. a tree

C. evolutionary pathways

D. natural relationships between organisms

E. All of the above

4. __ A__ Species evolve by ...

A. diversification

B. progression

C. linear advancement

D. magic

E. none of the above

5. _ D__ Which of these evolutionary representations is both objective and quantitative?

A. the 5-Kingdom tree

B. the Evolutionary Ladder

C. the Prokaryote/Eukaryote dichotomy

D. the 3-Domain molecular phylogenetic tree

E. None of the above

6. __ D__ Which if the following is not an algorithm for generating phylogenetic trees from molecular data?

A. Neighbor-joining

B. Parsimony

C. Maximum likelihood

D. Jukes & Cantor

E. all of the above

7. __ B__ A photoheterotroph gets its energy from X and its carbon from Y.

A. X= light ; Y= carbon dioxide

B. X= light ; Y= organic compounds

C. X= inorganic compounds ; Y= carbon dioxide

D. X and Y both = organic compounds

E. none of the above

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Short answer & fill in the blank (points as indicated)

8. Usually the best single molecular sequence for phylogenetic analysis is __ ssu-rRNA_______________________. (2 points)

9. A 2-dimensional matrix of molecular sequences is called a(n) ___sequence alignment______________________. (2 points)

10. Molecular sequences may be alignable using _______structural superimposition_____________________________ even if they are so different from one another that they cannot be reliably aligned based on the sequence alone. (2 points)

11. ___bootstrapping_____________________________ is a way to judge the reliability of the branches in a tree. (2 points)

12. Why is an understanding of phylogeny important? (4 points)

Both to make predictions about the properties of an organism, and to prevent inappropriate comparisons between unre-lated organisms.

13. What are the main steps of a molecular phylogenetic analysis? (4 points)

1. Decide on organisms and sequences

2. Obtain this data

3. Identify homologous residues

4. perform phylogenetic analysis

14. Why is raw sequence similarity an underestimate of evolutionary distance? (5 points)

Because of the occurrence of multiple substitutions at a single site that are only counted as a single difference, or in the case of reversion, as no difference at all.

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Tree interpretation : answer the following questions based on this ssu-rRNA-based phylogenetic tree. The scale bar (0.10) represents evolutionary distance. Sequences/organisms without names (i.e.pBB, Nak_9, SRI-240, and VC2.1bac27) are from uncultivated species. (2 points each)

D. Go! tz and others

.................................................................................................................................................

Fig. 2. Maximum-likelihood phylogenetic tree showing thepositions of strain EX-H1T and strain EX-H2T. The scale barrepresents the number of fixed mutations per nucleotideposition. The numbers at the branch nodes are bootstrap valuesbased on 100 bootstrap resamplings.

initially formed from pre-existing plasma membraneand were excised once elongated. Although internalmembranes are often used to align enzymes forcomplex metabolic reactions, the function of thesemembranous structures is not yet known. Similarmembranous structures have not previously beenreported in the Aquificales. Negative stains of cellsrevealed that thin, flexible, pili-like filaments extendedfrom the cell surface of many of the cells (Fig. 1,bottom).

Table 2. Electron donors and acceptors utilized by strain EX-H1T, strain EX-H2T andchemolithoautotrophic members of the order Aquificales.....................................................................................................................................................................................................................................

Strains (reference) : 1, Strain EX-H1T ; 2, Strain EX-H2T ; 3, Aquifex pyrophilus (Huber et al.,1992) ; 4, Thermocrinis ruber (Huber et al., 1998) ; 5, Calderobacterium hydrogenophilum (Kryukovet al., 1983) ; 6, Hydrogenobacter thermophilus (Kawasumi et al., 1984) ; 7, Hydrogenobacteracidophilus (Shima & Suzuki, 1993) ; 8, Hydrogenobacter halophilus (Nishihara et al., 1990) ;9, Hydrogenothermus marinus (Sto! hr et al., 2001). !", Not determined; !, growth;", no growth.

Electron donor Electron acceptor 1 2 3 4 5 6 7 8 9

O!

! ! ! ! ! ! ! ! !NO

"! ! ! " " " " " "

S# ! " " " " " " " "H

!S!O!!

"" " " " " " " " "

S!O!!

"O

!! ! ! ! " " " ! "

S!O!!

"NO

"" " ! " " " ! " "

S# O!

! ! ! ! " " ! ! "*Formate O

!" " !" ! !" !" " !" "

Formamide O!

!" !" !" ! !" !" !" !" "

*Sulfur is required for growth.

.................................................................................................................................................

Fig. 3. Effect of temperature on the specific growth rate (h!1)of strain EX-H1T (squares) and strain EX-H2T (circles) in thepresence of 3% (w/v) NaCl and at pH 6.

DNA base composition and 16S rRNA phylogeny

The G!C content of the DNA was 38#5 mol% forstrain EX-H1T and 37#4 mol% for strain EX-H2T.This is consistent with that of other members of theAquificales, whose G!C content varies from 35 mol%for Hydrogenobacter acidophilus to 40 mol% forAquifex pyrophilus and 47#5 mol% for T. ruber. Ad-ditionally, EX-H1T and EX-H2T form a phylo-genetically distinct lineage (Fig. 2) and are about 85%similar to A. pyrophilus (and 83% similar to Hydro-genobacter thermophilus), only 96% similar to eachother, and 92–94% similar to environmental sequencesthat fall within this lineage. They are 94#5% similar tothe recently described lineage Hydrogenothermus

1354 International Journal of Systematic and Evolutionary Microbiology 52

15. Is this a dendrogram or a phenogram? __________phenogram___________________________________________________

16. Which sequence(s) is/are presumably the outgroup? __Methanococcus jannashii___________________________________

17. Which sequence is most closely related to Bacillus subtilis? _____Escherichia coli__________________________________

18. What is the evolutionary distance between Persephonella marina and Hydrogenobacter thermophilus? _0.2__________

19. Circle (on the tree above) the last common ancestor of Bacillus subtilis and Thermus thermophilus.

20. Except for the outgroup, which two sequences in this tree are the most distantly related?

______Flexibacter flexilis_______________________ and ________pBB_________________________________________

21. Which branch(es) would you have the least confidence in?

The one between the 2 Persephonella species, and the one between E.coli and B.subtilis (the one between pBB and NAK_9 is also right up there).

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Problem-solving (points as indicated)

22. Align the following sequences: (8 points)

Sequence 1 G G G G U U C G C U C A - G G G G U U C G C U C A

Sequence 2 A G A G G U U G C U C U A G A G G U U - G C U C U

Sequence 3 C G A G G C U G C U C C G A G G C U - G C U C -

Sequence 4 U G A G G C U C G C U C A U G A G G C U C G C U C A

23. Generate a similarity matrix from the following alignment: (10 points)

Seq1 Seq2 Seq3 Seq4

Sequence 1 G A U C G A U C G A Seq1 XX XX XX XX

Sequence 2 G A U A G A U C G A Seq2 0.9 XX XX XX

Sequence 3 G A C C G A C C G A Seq3 _0.8_ _0.7_ XX XX

Sequence 4 G A G U G A C C G A Seq4 _0.7_ _0.7_ _0.8_ XX

24. Convert the following similarity matrix into a distance matrix using the Jukes & Cantor curve: (10 points)

Similarity matrix Distance matrix

Seq1 Seq2 Seq3 Seq4 Seq1 Seq2 Seq3 Seq4

Seq1 XX XX XX XX Seq1 XX XX XX XX

Seq2 0.90 XX XX XX Seq2 0.12 XX XX XX

Seq3 0.60 0.75 XX XX Seq3 0.55 0.30 XX XX

Seq4 0.50 0.45 0.35 XX Seq4 0.77 0.90 1.30 XX

10) (10 points) Convert the similarity matrix to a distance matrix using the Jukes & Cantor curve.

Similarity matrix Distance matrix

A B C D A B C D

A X X X X A X X X X

B 0.95 X X X B 0.05 X X X

C 0.75 0.80 X X C 0.30 0.25 X X

D 0.50 0.60 0.45 X D 0.75 0.55 0.90 X

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Sim

ilari

ty

Evolutionary distance

0.00.1

0.20.3

0.40.5

0.60.7

0.80.9

1.01.1

1.21.3

1.41.5

1.61.7

1.81.9

2.0

JUKES & CANTOR CURVE

11) (10 points) Redraw the dendrogram below, to the same approximate scale, as a phylogram.

A

B

C

D

EB

C

B

D

E

A

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25. Convert the following phenogram into a dendrogram. (10 points)

A

B

C

D

A

B

C

D

26. Generate a tree using the neighbor-joining method from the following distance matrix. SHOW YOUR WORK! You need only solve the structure of the tree; you do not need to solve the branch lengths. (10 points)

A B C D E reduced matrix

A X X X X X A/B C D E

B 0.3 X X X X A/B X X X X

C 0.6 0.5 X X X C 0.55 X X X

D 1.7 1.6 1.7 X X D 1.65 1.7 X X

E 1.0 0.9 1.0 1.7 X E 0.95 1.0 1.7 X

A

B

CD

EThe nearest neighbors areA and B, so these are joined:

A

B

CD

E Average distances in the matrix to A/B; the nearest neighbors are then between A/B and C, so these are joined:

A

B

CD

EThe nodes are completely resolved - you’re done!

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