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MBIO 3470- Final Exam Guide - Comprehensive Notes for the exam ( 28 pages long!)


Department
Microbiology
Course Code
MBIO 3470
Professor
Vladimir Yurkov
Study Guide
Final

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UofM
MBIO 3470
Final EXAM
STUDY GUIDE

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Microbial Systematics
January 26th 2018
Oligonucleotides Catalogues of 16S rRNA sequences
Dr. Woese (Univ. of Illinois, 1977) devised a method that more directly assesses
sequences similarities of rRNA of different organisms
o His work is summarized in aterial evolution: (Microbiol. Rev. 1987. V51,
p.221-271)
Instead of all the rRNAs, he studied only the rRNA extracted from the small subunit
(ssu) of ribosomes
From the prokaryotic ssu (30S ribosomal subunit) he isolated 16S Rrna (1500 bases) and
from the eukaryotic ssu (40S subunit) the 18S rRNA (2300 bases)
Organism ribosome ssu ssu-rna
Prokaryote 70S 30s 16S
Eukaryote 80S 40S 18S
**Make a table of this later
At the time, it was possible to sequence small oligonucleotides, but not the complete
molecule
o Isolate ssu-RNA from ribosomes and digest with RNAse T1. This enzyme cuts on
3’side of all guanine residues to generate oligonuleotides of various sizes
o The fragments were labeled with 32P and separated by electrophoresis and
purified. The sequence of each oligonucleotide with >5 bases is determined
o A list of these sequences is called the oligonucleotide catalogue
For 16S rRNA, the sequences of approximately 100 oligonucleotides
constitute a catalogue
o (table 1)
Oligonucleotide catalogues of 16s rRNAs of different organisms are compared
o E.g Bacillus sp.
Some oligonucleotides are present in species, others are common to 2 or more
o This is expected if the species evolved from common ancestral organism.
o If organism share a common ancestor, they would share common sequences and
also show distinct sequences that have accumulated during evolution
The greater the number of shared sequences, the greater is the sequence similarity
and the more closely related are the two organisms
Calculation of similarities. In a test group, the oligonucleotide catalogue of an organism
is compared with that of every other organism in the group in a pair wise manner
A binary association coefficient Sab is calculated for each pair
Sab = 2Nab/(Na+Nb)
Nab = number of bases in oligonucleotides that are identical in organisms a and b
Na(Nb) = number of bases in oligonucleotide catalogue of organism a (or b)
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Sab are analogous to similarity coefficients. In this case, each specific oligonucleotide is
analogous to a character in numerical taxonomy. Except, there are genotypic, not
phenotypic characters.
The value of Sab is directly related to the similarity in the sequences of ssu-RNA’s of
organism a and b and therefore the sequence similarity of the genes encoding the ssu-
RNA
If a and b are the same organisms, the Sab=1.0 the close the value is to 1.0, the more
similar are the sequences of the ssu-RNA gene
(table 2)
summary of results
the Sab values are generally quite low in this example because the organisms being
compared are quite different
o organisms 1,2 and 3 are the yeast, algae and animal cells, respectively, and
represent three completely different eukaryotic organism.
the rest are prokaryotic but widely different organisms
o 4 is gram negative chemoheterotroph, 5 is an anoxygenic phototroph, 6 is a
gram positive chemoheterotroph and 7 is a cyanobacterium, an oxygenic
phototroph
nevertheless, the organisms can be placed into 3 groups where Sab values for pairs
within a group are generally >0.2 while values for pairs formed by different groups are
<0.12
the eukaryotes fall into one group
o The Sab values for pars comprising organisms 1,2,3 (eukaryotes) are 0.29-0.36,
but range from 0.05 to 0.13 when compared with the remaining organism
(prokaryotes)
The prokaryotes also fall into 2 groups
o Sab values for pairs comprising organisms 4 to 8 range from 0.19 to 0.26 but
drop to 0.06-0.12 when compared to organisms 9-11 (also prokaryotes)
As stated previously, organisms 4-8 represent a wide range of bacteria
o Gram positive and gram negative, chemotrophs and phototrophs able to
quantify relatedness between them, which was not possible by rRNA-DNA
hybridization
Conversely, the Sab for pair from the last group (organisms 9-11) ranges from 0.25 to
0.32, but show low Sab values (0.06-0.13) with all other organisms in the test group
The first group of prokaryotes (organisms 4-8) belong to the eubacteria, while the last
group (organisms 9-11) are the archaebacterial
It was this study that led Woese to postulate the existence of a third kingdom of
organisms, the archaeebacteria
o According to this data, were a different from the eubacteria as eubacteria are
different from the eukaryotes
Division into 3 primary domains can be more clearly seen
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