Midterm Exam #1
MB 451
Microbial Diversity
Honor pledge: “I have neither given nor received unauthorized aid on this test.”
Name : ___KEY___________________________________________________________
Date : _______________________________
Signed : _________________________________________________________________
PIN * : _______________________________
* Please give me a “PIN” (a short string of numbers, letters, and symbols, that are not personally identifiable) that I can use to
anonymously post your scores. Try to choose a PIN unlikely to be chosen by another student (e.g. not “666” or “123456”) or rejected by me (I’ll leave this to your imagination). If you do not wish your scores to be posted, leave this blank.
1. What are the 3 primary evolutionary branches of life? (5 points)
Archaea, Bacteria & Eukarya
__A__ 2. Many purple non-sulfur Bacteria are photoheterotrophic, meaning that they ... (2 points)
___E__ 5. The ssu-rRNA is a good molecule for generating phylogenetic trees because ... (2 points)
A. get energy from light and carbon from organics
A. It is composed of many independently-evolving parts
B. get energy from light and carbon from CO2
B. it is present in and has the same function in all cells
C. get energy from organics and carbon from light
C. it contains both rapidly and slowly evolving regions
D. get energy from organics and carbon from CO2
D. horizontal transfer (almost) never occurs
E. none of the above
E. all of the above
__C__ 3. Which if the following is not a part of a biological
taxonomy? (2 points)
__B__ 6. RNAs that are too different to align on the basis of
sequence might be alignable based on ... (2 points)
A. A naming method
A. three-dimensional structure determination
B. A grouping system
B. secondary structure superimposition
C. A phylogenetic tree
C. evolutionary history
D. An identification scheme
D. BLAST or CLUSTAL
E. All of the above are included in a taxonomy
E. there would be no way to align these sequences
___D__ 4. The term prokaryote means ... (2 points)
A. bacterial
B. pre-eukaryotic
C. lacking a nucleus or nucleoid
D. not a eukaryote
E. all of the above
__E__ 7. Which of the following is not a method for generating
phylogenetic trees? (2 points)
A. maximum likelihood
B. neighbor-joining
C. parsimony
D. Fitch
E. all of these methods are used to generate trees
Page 1 of 8
8. Charles Darwin described how species evolve by ___diversification_________________, rather than by progression. (2 points)
9. Molecular phylogenetic trees are subjective/objective (circle one) and qualitative/quantitative (circle one). (2 points)
10. Describe in one or two sentences what an outgroup is, and what the purpose of including one in a tree is. (2 points)
An outgroup is a sequence/organism that is known not to be related to the phylogenetic group being treed. The purpose of
including an outgroup in a tree is to place the location of the root in the rest of the tree.
11. What is clock-like behavior of a molecular sequence? What are some of the properties of a sequence required for it to have
clock-like behavior? (6 points)
Clock-like behavior in a gene means the the extent of sequence divergence between sequences from different sources (species) is approximately proportional to how long since they diverged from one another.
Some of the properties required for clock-like behavior are:
- Functional constancy
- a long enough sequence to be statistically meaningful
- the gene product must be composed of many independently-evolving domains
Page 2 of 8
12. What is a substitution model? (4 points)
A substitution model is an attempt to describe mathematically how sequences change, and is used to estimate evolutionary
distance in a distance matrix from distance matrices or alignments.
13. How was the root of the “universal” phylogenetic tree determined? Include a drawing to show how this works. (4 points)
By generating trees from pairs of genes that duplicated before the last common ancestor and are still similar enough to align,
e.g. methionyl-tRNA init and elong, ATPase alpha and beta subunits, or EF-Tu and EF-G. Each half of the tree serves as the
outgroup for the other half - e.g. the EF-Tu part of the tree as outgroup for the EF-G sequences.
Page 3 of 8
14. What is the purpose of bootstrapping. How is it done? (6 points)
Bootstrapping is a method for evaluating the reliability of the branches of a tree.
In a bootstrap analysis, an alignment of length X is randomly sampled X times (one column at a time). Some columns of the
original alignment will be used more than once, some not at all. This randomly sampled alignment is then used to generate a
tree. A large number of these random-sample trees are generated, and the percentage of these trees that contain each
branch are tabulated. These values are usually shown on the tree, and values above 50% are considered generally reliable.
Page 4 of 8
15. Describe in detail one method used to determine phylogenetic relationships, or distinguish or identify organism, other than ssurRNA molecular phylogenetic analysis. (10 points)
We talked about several methods, and any of these would be appropriate:
•DNA:DNA hybridization
•DNA base composition
•serology
•lipid profiling
•rRNA spacer sequence analysis
•other gene phylogenetic analysis
•ribotyping
•RFLP
•phenotype
e.g for DNA:DNA hybridization:
DNA:DNA hybridization is a method that was commonly used in the past. 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%.
Page 5 of 8
16. Align the following sequences: (5 points)
Sequence A A C A G A C U U U G U G U A G A C A G A C U U U G U G U A G
Sequence B G C G A C U U C G G U A C G C ‐ G A C U U C G ‐ G U A C
Sequence C G C A G A U U U C G U G U A U G C A G A U U U C G U G U A U
Sequence D A C G A U U U C G G U A A C ‐ G A U U U C G ‐ G U A ‐
17. Create a similarity matrix from the following alignment, using a 3-parameter substitution model in which transversions count as
a full “difference”, a transitions count as a half a difference, and any base mis-matched with a gap counts as 2 differences. In
other words, a match is +1, a transversion is -1, a transition is -0.5, and any base:gap comparison is -2. (6 points)
A B C D
Sequence A A A G C U U G A U C A X X X X Sequence B C A G C U U G A U C B 0.9 X X X Sequence C U A G C U ‐ G A G C C 0.6 0.65 X X Sequence D C A G C U G G A U C D 0.8 0.9 0.65 X 18. Convert the following similarity matrix to a distance matrix using the Jukes & Cantor curve: (6 points)
10) (10 points) Convert the similarity matrix to a distance matrix using the Jukes & Cantor curve.
Similarity
Similarity
matrix matrix
Distance
matrix
Distance
matrix
A B C D A B C D
A
B
C
D
A
X
X
X
X
B
0.95
X
X
X
A
B
C
D
A
X
X
X
X
B
0.05
X
X
X
A X X X X A _X_ _X_ _X_ _X_
B 0.9 X X X B 0.12 _X_ _X_ _X_
C 0.8 0.3 X X C 0.23 1.65 _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
D 0.6 0.5 0.4 X D 0.55 0.77 1.05 _X_
JUKES & CANTOR CURVE
1.0
0.9
Similarity
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
Evolutionary distance
11) (10 points) Redraw the dendrogram below, to the same approximate scale, as a phylogram.
Page 6 of 8
C
B
D
D
19. Use the distance matrix to create a tree using the neighbor-joining method. SHOW YOUR WORK! Intermediate stages will be
graded. Be sure to label the lengths of all the branches. (15 points)
This was supposed to be 0.9! A
Distance matrix
A B C D E
A _X_ _X_ _X_ _X_ _X_ E B
B 0.8 _X_ _X_ _X_ _X_
C 0.5 0.7 _X_ _X_ _X_
D 0.5 0.7 0.6 _X_ _X_ D C
E 0.7 0.9 0.8 0.4 _X_
A B C D/E A A
A _X_ _X_ _X_ _X_
B 0.8 _X_ _X_ _X_ E B E C
C 0.5 0.7 _X_ _X_
D/E 0.6 0.8 0.7 _X_
D C D B
OK, so there’s a mistake in this table ‐ B/C was supposed to be 0.9 rather than 0.7, a mis‐
take created when I changed the problem at the last minute. No tree can exactly satisfy these distances. A real computer program would adjust branch lengths to minimize the total error over the tree, but of course it wasn’t my intent do give you this. This error only affects this problem of branch lengths, it does not affect solving the tree structure. As a result, the 7 points that went with these branch lengths (1 point for each length) will be given to all students, and only the process of solving the tree structure will be graded, for 8 points.
Page 7 of 8
20. Convert the tree from a dendrogram into a phenogram. Be sure to maintain the same branch lengths! (5 points)
E
B
C
C
E
ROOT
B
A
D
A
D
Use the following tree to answer questions 21 - 25: (2 points each)
Evolutionary distance
0.1
32
100
50
96
50
100
24
53
100 40
100
90
98
60
95
100
Rhodospirillum rubrum
Campylobacter hominis
Comamonas testosteroni
Neisseria gonorrhoeae
Vibrio cholerae
Yersinia pestis
100
Escherichia coli
100
Salmonella enterica
Desulfovibrio africanus
Chondromyces lanuginosus
Chlamydia trachomatis
Anabaena sp. PCC 7108
Streptomyces bikiniensis
Micrococcus luteus
Mycobacterium tuberculosis
Corynebacterium diphtheriae
Clostridium putrefaciens
Bacillus subtilis
Lactobacillus vaginalis
Methanocaldococcus jannaschii
21. What is the closest relative of Bacillus subtilis? ___L. vaginalis_________________________________________________________
22. What are the two closest relatives on this tree? _______E.coli______________ and _________ S.enterica_____________________
23. Circle the last common ancestor of Neisseria gonorrhoeae and Escherichia coli
24. Which sequence(s) is/are presumably the outgroup? _____M. jannaschii_________________________________________________
25. What is the approximate evolutionary distance between Streptomyces bikiniensis and Bacillus subtilis? _____0.2______________
Page 8 of 8