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Introduction to
Bioinformatics
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Introduction to Bioinformatics.
LECTURE 6: Natural selection at the
molecular basis
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Chapter 6: Fighting HIV
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.1 Acquired Immune Deficiency
Syndrome (AIDS)
* First noticed in 1979 as peculiar disease in US
* Only 1981 recognized as transmissible disease: AIDS
* Infectious agent: HIV (Human immunodeficiency Virus)
* Still not curable, more than 20 M victims, expensive
medication (eg AZT) to keep the virus in check
* How does HIV manage to evade our attempts to destroy it? 3
HIV virus
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
HIV is a retrovirus
A retrovirus is an enveloped virus possessing a RNA genome, and
replicate via a DNA intermediate.
Retroviruses rely on the enzyme reverse-transcriptase to perform the
reverse transcription of its genome from RNA into DNA, which can then
be integrated into the host's genome with an integrase enzyme.
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Scanning electron micrograph of HIV-1 budding from lymphocyte.
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
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THE WORLD
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Mark Newman (http://www-personal.umich.edu/~mejn/)
PEOPLE LIVING WITH HIV/AIDS
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Mark Newman (http://www-personal.umich.edu/~mejn/)
Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.2 Evolution and natural selection
1859: Charles Darwin: on the origin of species by means of
natural selection.
At the molecular level: natural selection :
* removes deleterious mutations: purifying or negative selection
* Promotes spread of advantageous mutation: positive selection
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.3 HIV and the human immune system
* HIV has a 9.5 Kb RNA genome - no DNA!!!
* HIV is a retro-virus: RNA  DNA  virus
* HIV recognizes helper T-cells of the human immune system
* Infected T-cells have viral proteins sticking out that can be
recognized by the immune system
* Short reproduction span: 1.5 days to reproduce
* RNA  High error rate
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Introduction to Bioinformatics
6.3: HIV and the human immune system
Fast reproduction + High error rate =
FAST EVOLUTION
Evolutionary arms race between human immune system
and HIV
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.4 Quantifying natural selection
on DNA sequences
* Mutations arise in the germ-line of one single individual and
eventually become fixed in the population
* We observe fixed mutations as differences between
individuals
* Most fixed mutations are neutral: genetic drift
* Some 80-90% of the non-neutral mutations are detrimental
to the organismal function.
* A very small fraction of mutations is advantageous – but this
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is the engine for evolution.
Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
* How to measure whether mutations are neutral, deleterious,
or advantageous?
* Experimentally very difficult: short-lived simple organisms,
and large populations (typical a virus)
* Alternative: count number of mutations that can change the
protein and those that don’t
* Synonymous and non-synonymous mutations.
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
Remember the translation from nucleotides to aminoacids
(read from centre
outwards)
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
* Synonymous mutation: the new codon translates for the
same amino-acid, example: GTT (Val) → GTA (Val).
* Non-synonymous mutations do not
* Mutations in the first position are sometimes synonymous
(5%)
* Mutations in the second position are never synonymous
* Mutations in the third position are mostly synonymous
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
* Almost all synonymous mutations are neutral.
* A priori, there are many more non-synonymous mutations
possible than synonymous.
* In most genes 70% of the mutations are non-synonymous
* KA: #non-synonymous substitutions per non-synonymous site
* KS: #synonymous substitutions per synonymous site
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
Motoo Kimura (1977):
Comparison of the non-synonymous to the synonymous
substitutions in a gene tells us about the strength and form
of the natural selection, i.e.: the ratio KA / KS.
Reasoning:
* Advantageous mutations are very rare
* Deleterious mutations will ‘not’ spread through a population
* Therefore, most mutations are neutral
Strong negative selection → Few non-synonymous substitutions
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
* f0 = fraction of non-synonymous mutations that are neutral.
* v = mutation rate
* # non-synonymous mutations after time t :
* # synonymous mutations after time t :
KA = v f0 t
KS = v t
* KA / KS = f0
* Strong negative selection: f0 is small thus KA / KS < 1
* If KA / KS is > 1 this is evidence for advantageous
non-synonymous mutations
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Introduction to Bioinformatics
6.4 QUANTIFYING NATURAL SELECTION ON DNA SEQUENCES
* Define: α = fraction of non-synonymous mutations that are
advantageous
* Then after time t : KA = v(f0 + α)t
* and: KA / KS = f0 + α
* Thus KA / KS is gauge for the natural selection on genes
* negative selection dominates: KA / KS < 1
* positive selection dominates: KA / KS > 1
* But averaged over the gene!
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.5 Estimating KA/KS
How to determine KA/KS?
Simplest way: just count and compare the number of
synonymous and non-synonymous sites and ditto
differences between two aligned strings
Correct for multiple substitutions (e.g. Jukes-Cantor)
Thus obtain a normalized ratio
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.5 Estimating KA/KS
Based upon this idea the algorithm of Masatoshi Nei and
Takashi Gojobori (1986):
Assume that rate of transitions and transversions is the
same
There is no bias towards codon usage (i.e. no information
on the ensuing protein)
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Introduction to Bioinformatics
6.5 ESTIMATING KA/KS
Nei-Gojobori algorithm
* Consider two aligned homologous sequences
without gaps s1 and s2
* Sc = #synonymous sites between s1 and s2
* Ac = #non-synonymous sites between s1 and s2
* Sd = #synonymous differences between s1 and s2
* Ad = #non-synonymous differences between s1 and s2
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Introduction to Bioinformatics
6.5 ESTIMATING KA/KS
Nei-Gojobori algorithm
* As the two sequences s1 and s2 are aligned there should be
a correspondence between their codons.
NOTE: point mutations only act on nucleotides and not on
codons but here we analyse whether a mutation results in
different aminoacids
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Introduction to Bioinformatics
6.5 ESTIMATING KA/KS
Nei-Gojobori algorithm
STEP 1: Count A and S sites
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Introduction to Bioinformatics
6.5 NEI-GOJOBORI ALGORITHM
STEP 1: Count A and S sites
Example:
Consider the alignment :
TTT
TTA
This is – say – the k-th codon of a sequence.
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Now define:
sc(ck) = #synonymous sites in this codon
ac(ck) = 1 - sc(ck) = #non-synonymous sites in this codon
fi :
fraction of changes in at i-th position of codon
that result in a synonymous change (i=1,2,3)
Then:
sc(ck) = ∑ fi and:
ac(ck) = 3 - sc(ck) = 3 - ∑ fi
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
In our example:
Codon: TTA codes for: Leucine
The 6 synonyms for Leucine (table 2.2 chapter 2, p. 27):
CTA CTG CTC CTT TTA TTG
f1 : 1 (ATA(-),GTA(-),CTA(+) from 3 changes, so: 1/3
f2 : 0 (TAA(-),TGA(-),TCA(-)) from 3 changes, so: 0/3
f3 : 1 (TTG(+),TTC(-),TTT(-)) from 3 changes, so: 1/3
So:
sc(ck) = ∑ fi = 2/3
ac(ck) = 3 - sc(ck) = 3 - ∑ fi = 7/3
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
For a DNA sequence of r codons:
Sc = ∑k=1:r sc(ck)
Ac = 3r - Sc
For multiple sequences: average these quantities
Note: do not include the STOP codon
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Nei-Gojobori algorithm
STEP 2: Count A and S differences
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Now define:
sd(ck) = #synonymous differences in this codon
ad(ck) = 1 - sd(ck) = #non-synonymous differences
Example:
sequence 1: GTT (Val)
sequence 2: GTA (Val)
there is only 1 difference and it is synonymous, so:
sd = 1 and ad = 0
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Multiple nucleotide differences between two
codons:
If there are n differences between two codons (n=0,1,2,3)
then there are n! pathways from the first to the second
codon
Example:
sequence 1: TTT (Phe)
sequence 2: GTA (Val)
the two possible pathways are :
pathway 1 : TTT (Phe) ↔ GTT (Val) ↔ GTA (Val)
pathway 2 : TTT (Phe) ↔ TTA (Leu) ↔ GTA (Val)
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Example (Continued):
the two possible pathways are :
pathway 1 : TTT (Phe) ↔ GTT (Val) ↔ GTA (Val)
pathway 2 : TTT (Phe) ↔ TTA (Leu) ↔ GTA (Val)
Pathway 1 has: 1 non-syn and 1 syn substitution
Pathway 2 has: 2 non-syn and 0 syn substitutions
Assume that both pathways occur with same probability
Therefore:
sd = 1 syn / 2 pathways = 0.5
ad = 3 non-syns / 2 pathways = 1.5
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
For a codon with n differences:
* Consider all n! pathways of n point-mutations
* Evaluate sd and ad as above:
* Average over all paths with equal weights
* The total number of syn and non-syn differences is:
Sd = ∑k=1:r sd(ck)
Ad = ∑k=1:r ad(ck)
Note: Sd + Ad is the total number of differences
between the two sequences
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
Nei-Gojobori algorithm
STEP 3: Compute KA and KS
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
* Approximate the proportion of synonymous (ds) and
non-synonymous differences by:
Sd
ds 
Sˆ and
c
Ad
da 
Aˆ
c
* Use the Jukes-Cantor correction to find the number
of substitutions:
K   34 ln 1  43 d 
For both ds and da to obtain KS and KA.
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Introduction to Bioinformatics
6.5: NEI-GOJOBORI ALGORITHM
SUMMARY of Nei-Gojobori algorithm:
see box on page 105 of the book
Remark: the algorithm is linear in the size of the
sequences
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.6 Case study: natural selection
and the HIV genome
* HIV is a fast evolving virus
* HIV is a different kind of virus and has RNA and no DNA
* An analysis of KA/KS over a gene is not so informative
as it averages over positive and negative selection
* Sliding window plot gives information on smaller scale of
evolution pressure.
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.6 Case study: natural selection
and the HIV genome
* STEP 1: ORF finding
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
HIV-I genome
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
6.6 Case study: natural selection
and the HIV genome
* STEP 1: ORF finding
* STEP 2: Nei-Gojobori to find high KA/KS ratios with
sliding window plot.
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
HIV epitopes: the ENV gene
An epitope is the part of a macromolecule that is recognized by the immune
system, specifically by antibodies.
ENV: Envelope and docking: strong selection pressure from human immune
system
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
HIV epitopes: the GAG polyprotein
1500 bp : viral core
Strong selection pressure from human immune system
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
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Introduction to Bioinformatics
LECTURE 6: NATURAL SELECTION AT THE MOLECULAR BASIS
Visualisation of the
fast evolution of
the HIV virus with a
phylogenetic tree
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END of LECTURE 6
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