Key for Midterm #1 - MB 409 - Feb 11, 2002


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

Bacteria, Archaea, and Eukarya

2. The structure and biochemistry of all living things are very much the same, even in the smallest details. What do we conclude from this? (10 points)

1. All organisms share a common ancestry. In other words, all known organisms can trace their past back to a single origin of life. This might not have been; other lineages, if they ever existed, seem to be extinct.
   
2. The last common ancestor of all known living things was a complex organism - most of biochemical evolution predates this organism! The last common ancestor had all of the biochemistry that is now universal, which means nearly everything! Biochemical evolution occured very early in the emergence of life. The diversity in extant life (i.e. known modern life) is in peripheral biochemistry.

3. What is a taxonomy? (5 points)

Taxonomies are classification schemes for species. There are three parts to taxonomy:

1. The organization of organisms into groups based on similarity
2. Naming organisms & groups of organisms
3. The identification of organisms when they're found

Taxonomies are artificial constructions, methods for structuring & organizing species. Any self-consistent taxonomy is valid, whether it reflects the natural history of the organisms or not.

4. What is a phylogeny? (5 points)

Phylogenies are evolutionary pathways, i.e. geneologies. Phylogenies represent the natural relationships between organisms.

5. Describe the properties of a macromolecular sequence that are important if it is to be used for molecular phylogenetic analysis. (10 points)

1. Clock-like behavior, i.e. sequence divergence in the gene between two organisms should be proportional to how long ago they diverged. Clock-like behavior depends mostly on functional constancy of the sequence - function change leads to large, directed sequence change. Clock-like behavior also depends on the sequence being long enough to provide statistically-significant information, and be comprized of many independently-evolving domains so that random changes in one domain don't affect other domains.
2. Phylogenetic range. The sequence must be present in all of the organisms to to analysed and must have retained it's structure & function in these organisms. Watch out for gene families, because each member of the family is probably specialized for a slightly different function. The gene must have enough variation in sequence to evaluate statistically but must be similar enough to that homologous residues can be identified. Non-functional sequences (e.g. introns) usually change too fast for analysis except of the very closest of relatives.
3. No horizontal transfer. This means that the gene must be aquired only by inheretance, not by transfer from another organism. An example of frequently horizontally- transferred genes are those encoding antibiotic resistance. You can still generate a tree with these sequences, but the do not reflect the geneology of the organism as a whole.
4. Must have a large existing data set to compare your sequences with!

6. In what situation is ssu-rRNA not suitable for molecular phylogenetic analysis? Describe another method that could be used to determine the relationships between organisms in this situation. (10 points)

Very close relatives cannot be distinguished using rRNA sequences.

e.g.DNA:DNA hybridization

The extent that the genomic DNAs of 2 species will hybridize is a general measure of how much sequence similarity there is between the genomes, and therefore how closely related they are. This method is widely used to define bacterial species - in general, two organisms are considered to be the same species if the DNA:DNA hybridization is 70% or greater, or different species of the same genus if they have measurable hybridization less than 70%.

7. Answer the questions below based on this tree (2 points each for 10 points):



a. Which of the organisms shown are the closest relatives? H. halobium and N. gregoryi
b. What is the approximate evolutionary distance between T. celer and P. furiosus? 0.10
c. Circle the last common ancestor of M. barkeri and T. celer.
d. Which organism is probably the outgroup? S. acidocaldarius
e. If this is an outgroup, which of the other organisms is the most primative? M. thermoautotrophicus


8. Align the following sequences (10 points):

Sequence A    A C A U G G U G U U C G C A C C A U G U
Sequence B    A C A U G G U G U U U C A C C A U G U
Sequence C    A C G G G U G G A A A C A C C C G U
Sequence D    C A U A G U G C U U G C A C U A U G

A

A

C

A

U

G

G

U

G

U

U

C

G

C

A

C

C

A

U

G

U

-

-

-

-

-

-

-

-

-

B

A

C

A

U

G

G

U

G

U

U

U

-

C

A

C

C

A

U

G

U

-

-

-

-

-

-

-

-

-

C

A

C

G

-

G

G

U

G

G

A

A

A

C

A

C

C

-

C

G

U

-

-

-

-

-

-

-

-

-

D

-

C

A

U

A

G

U

G

C

U

U

G

C

A

C

U

A

U

G

-

-

-

-

-

-

-

-

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-

9. Draw the phenogram below as a dendrogram. Be careful to maintain the distance scale! (5 points)


10. Align the following RNAs on the basis of their structure and sequence (10 points):

     RNA W      RNA X      RNA Y      RNA Z

                                       A C
     A A        U C          U        U   G
    G   A      U   G       U   U       G-C
     G-C        G-C         G-C        G-C
     G-C        G-C         G-C        U-A
     G-C        G-C         G-C        G-C
     G-C        G-C         G-C        G-C
     C-GUA     AC-GUA      GC-GUAC    AC-GUA

w

-

C

G

G

-

G

G

G

A

A

A

C

C

-

C

C

G

U

A

-

-

-

-

-

-

-

-

-

-

x

A

C

G

G

-

G

G

U

U

C

G

C

C

-

C

C

G

U

A

-

-

-

-

-

-

-

-

-

-

y

G

C

G

G

-

G

G

U

U

U

-

C

C

-

C

C

G

U

A

C

-

-

-

-

-

-

-

-

-

z

A

C

G

G

U

G

G

U

A

C

G

C

C

A

C

C

G

U

A

-

-

-

-

-

-

-

-

-

-

11. Create a similarity matrix from the following alignment (10 points):

   Sequence A     G G G G A A A C C C
   Sequence B     G C G G A G A C G C
   Sequence C     G A G G U G A C U C
   Sequence D     G U G G C A A C A C
   Sequence E     C A G G C G A C U G

	Seq A	Seq B	Seq C	Seq D	Seq E
Seq A	X	X	X	X	X
Seq B	0.7	X	X	X	X
Seq C	0.6	0.7	X	X	X
Seq D	0.7	0.6	0.6	X	X
Seq E	0.4	0.5	0.7	0.5	X

12. On the next page there is a portion of the secondary structure of the E. coli ssu-rRNA and an unknown ssu-rRNA. Below is a short signature table designed to discriminate between bacterial, archaeal, and eukaryotic ssu-rRNAs. The nucleotide numbering on the signature table is based on the E. coli ssu-rRNA (i.e. the E. coli RNA is the reference sequence). Perform a signature analysis, and identify which kind of organism the unknown is. Show your work - no credit will be awarded without it! (10 points)

• E. coli reference structure
• unknown structure

     Position    Bacteria    Archaea    Eukarya  Unknown
     537         A or GX     C+         C+         C
     551         UX          A or G+    UX         G
     585         GX          C+         UX         C
     675         AX          U+         U+         U
     716         AX          C+         C or U+    C
     756         CX          G+         AX         G
     912         CX          U+         U+         U
     923         AX          G+         AX         G
     930         CX          A+         GX         A
     931         CX          G+         G+         G
     952         UX          C+         C+         C
     962         CX          G+         UX         G
     966         GX          U+         U+         U
     973         GX          C+         GX         C
     975         AX          G+         G+         G
     1060        UX          C+         C+         C
     1086        UX          C+         C or U+    C
     1087        GX          C+         UX         C
     1109        CX          A+         A+         A
     1110        AX          G+         G+         G

              0/20=0%   20/20=100%  11/20=55%

This is organism is clearly an archaeon. (in fact, it is Methanococcus jannaschii)