3. Genome Annotation:
Gene Prediction
Gene Prediction: Computational Challenge
• Gene: A sequence of nucleotides coding
for protein
• Gene Prediction Problem: Determine the
beginning and end positions of genes in a
genome
Gene Prediction: Computational Challenge
aatgcatgcggctatgctaatgcatgcggctatgctaagctgggatccgatgacaa
tgcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgc
taagctgggatccgatgacaatgcatgcggctatgctaatgaatggtcttgggatt
taccttggaatgctaagctgggatccgatgacaatgcatgcggctatgctaatgaa
tggtcttgggatttaccttggaatatgctaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctgggat
ccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcct
gcggctatgctaatgaatggtcttgggatttaccttggaatgctaagctgggatcc
gatgacaatgcatgcggctatgctaatgaatggtcttgggatttaccttggaatat
gctaatgcatgcggctatgctaagctgggaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctcatgc
ggctatgctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaat
gcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgct
aagctgcggctatgctaatgcatgcggctatgctaagctcggctatgctaatgaat
ggtcttgggatttaccttggaatgctaagctgggatccgatgacaatgcatgcggc
tatgctaatgaatggtcttgggatttaccttggaatatgctaatgcatgcggctat
gctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaatgcatgc
ggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgctaagctg
cggctatgctaatgcatgcggctatgctaagctcatgcgg
Gene Prediction: Computational Challenge
aatgcatgcggctatgctaatgcatgcggctatgctaagctgggatccgatgacaa
tgcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgc
taagctgggatccgatgacaatgcatgcggctatgctaatgaatggtcttgggatt
taccttggaatgctaagctgggatccgatgacaatgcatgcggctatgctaatgaa
tggtcttgggatttaccttggaatatgctaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctgggat
ccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcct
gcggctatgctaatgaatggtcttgggatttaccttggaatgctaagctgggatcc
gatgacaatgcatgcggctatgctaatgaatggtcttgggatttaccttggaatat
gctaatgcatgcggctatgctaagctgggaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctcatgc
ggctatgctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaat
gcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgct
aagctgcggctatgctaatgcatgcggctatgctaagctcggctatgctaatgaat
ggtcttgggatttaccttggaatgctaagctgggatccgatgacaatgcatgcggc
tatgctaatgaatggtcttgggatttaccttggaatatgctaatgcatgcggctat
gctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaatgcatgc
ggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgctaagctg
cggctatgctaatgcatgcggctatgctaagctcatgcgg
Gene Prediction: Computational Challenge
aatgcatgcggctatgctaatgcatgcggctatgctaagctgggatccgatgacaa
tgcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgc
taagctgggatccgatgacaatgcatgcggctatgctaatgaatggtcttgggatt
taccttggaatgctaagctgggatccgatgacaatgcatgcggctatgctaatgaa
tggtcttgggatttaccttggaatatgctaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctgggat
ccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcct
gcggctatgctaatgaatggtcttgggatttaccttggaatgctaagctgggatcc
gatgacaatgcatgcggctatgctaatgaatggtcttgggatttaccttggaatat
gctaatgcatgcggctatgctaagctgggaatgcatgcggctatgctaagctggga
tccgatgacaatgcatgcggctatgctaatgcatgcggctatgcaagctgggatcc
gatgactatgctaagctgcggctatgctaatgcatgcggctatgctaagctcatgc
ggctatgctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaat
gcatgcggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgct
aagctgcggctatgctaatgcatgcggctatgctaagctcggctatgctaatgaat
ggtcttgggatttaccttggaatgctaagctgggatccgatgacaatgcatgcggc
tatgctaatgaatggtcttgggatttaccttggaatatgctaatgcatgcggctat
gctaagctgggaatgcatgcggctatgctaagctgggatccgatgacaatgcatgc
ggctatgctaatgcatgcggctatgcaagctgggatccgatgactatgctaagctg
cggctatgctaatgcatgcggctatgctaagctcatgcgg
Gene!
Central Dogma: DNA -> RNA -> Protein
DNA
CCTGAGCCAACTATTGATGAA
transcription
RNA
CCUGAGCCAACUAUUGAUGAA
translation
Protein
PEPTIDE
Prokaryotic and eukaryotic
organisms
Translating Nucleotides into Amino Acids
• Codon: 3 consecutive nucleotides
• 4 3 = 64 possible codons
• Genetic code is degenerative and redundant
– Includes start and stop codons
– An amino acid may be coded by more than
one codon
Codons
• In 1961 Sydney Brenner and Francis Crick
discovered frameshift mutations
• Systematically deleted nucleotides from DNA
– Single and double deletions dramatically
altered protein product
– Effects of triple deletions were minor
– Conclusion: every triplet of nucleotides, each
codon, codes for exactly one amino acid in a
protein
Genetic Code and Stop Codons
UAA, UAG and
UGA correspond to
3 Stop codons that
(together with Start
codon ATG)
delineate Open
Reading Frames
Six Frames in a DNA Sequence
CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC
CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC
CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC
CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC
GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG
GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG
GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG
GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG
• stop codons – TAA, TAG, TGA
• start codons - ATG
n
3
Open Reading Frames (ORFs)
• Detect potential coding regions by looking at ORFs
– A genome of length n is comprised of (n/3) codons
– Stop codons break genome into segments between
consecutive Stop codons
– The subsegments of these that start from the Start codon
(ATG) are ORFs
• ORFs in different frames may overlap
ATG
TGA
Genomic Sequence
Open reading frame
Long vs.Short ORFs
• Long open reading frames may be a gene
– At random, we should expect one stop codon
every (64/3) ~= 21 codons
– However, genes are usually much longer
than this
• A basic approach is to scan for ORFs whose
length exceeds certain threshold
– This is naïve because some genes (e.g. some
neural and immune system genes) are
relatively short
Testing ORFs: Codon Usage
• Create a 64-element hash table and count
the frequencies of codons in an ORF
• Amino acids typically have more than one
codon, but in nature certain codons are
more in use
• Uneven use of the codons may
characterize a real gene
• This compensate for pitfalls of the ORF
length test
Codon Usage in Human Genome
Codon Usage in Mouse Genome
AA codon
Ser TCG
Ser TCA
Ser TCT
Ser TCC
Ser AGT
Ser AGC
/1000
4.31
11.44
15.70
17.92
12.25
19.54
frac
0.05
0.14
0.19
0.22
0.15
0.24
Pro
Pro
Pro
Pro
6.33
17.10
18.31
18.42
0.11
0.28
0.30
0.31
CCG
CCA
CCT
CCC
AA codon
Leu CTG
Leu CTA
Leu CTT
Leu CTC
/1000
39.95
7.89
12.97
20.04
frac
0.40
0.08
0.13
0.20
Ala
Ala
Ala
Ala
GCG
GCA
GCT
GCC
6.72
15.80
20.12
26.51
0.10
0.23
0.29
0.38
Gln
Gln
CAG
CAA
34.18
11.51
0.75
0.25
Transcription in prokaryotes
Transcribed region
start codon
5’
stop codon
Coding region
3’
Untranslated regions
Promoter
Transcription start side
Transcription stop side
upstream
downstream
-k denotes kth base before transcription, +k denotes kth transcribed base
Microbial gene finding
• Microbial genome tends to be gene rich
(80%-90% of the sequence is coding)
• The most reliable method – homology
searches (e.g. using BLAST and/or
FASTA)
• Major problem – finding genes without
known homologue.
Open Reading Frame
Open Reading Frame (ORF) is a sequence of codons
which starts with start codon, ends with an end codon
and has no end codons in-between.
Searching for ORFs – consider all 6 possible
reading frames: 3 forward and 3 reverse
Is the ORF a coding sequence?
1. Must be long enough (roughly 300 bp or more)
2. Should have average amino-acid composition specific for a give
organism.
3. Should have codon use specific for the given organism.
Recognition of gene related signals
• Promoter: bacteria TATAAT – Pribnow box at about –10;
• Terminator: Rho-independent (intrinsic) transcription
termination – G-C reach inverted repeat.
• Other signal recognition –e.g binding motifs
Codon frequency
Input sequence
frequency in coding region
frequency in non-coding region
Compare
Coding region or non-coding region
Example
codon
position
1
A
C
T
G
28% 33% 18% 21%
2
32% 16% 21% 32%
3
33% 15% 14% 38%
Assume: bases making
codon are independent
P(x|in coding)
P(x|random)
=
P(Ai at ith position)
P
freqency 31% 18% 19% 31%
i P(Ai in the sequence)
in
Score of AAAGAT:
genome
.28*.32*.33*.21*.26*.14
.31*.31*.31*.31*.31*.19
Using codon frequency to find
correct reading frame
Consider sequence x1 x2 x3 x4 x5 x6 x7 x8 x9….
where xi is a nucleotide
let p1 = p x1 x2 x3 p x3 x4 x5….
p2 = p x2 x3 x4 p x5 x6 x7….
p3 = p x3 x4 x5 p x6 x7 x8….
then probability that ith reading frame is the coding frame is:
Algorithm:
pi
• slide a window along the sequence and
Pi = p + p + p
1
2
3
compute Pi
•Plot the results
First order Markov chain: formal
def.
M = (Q, p, P)
Q – finite set of states, say |Q|= n
a – n x n transition probability matrix
a(i,j)= P[q t+1=j|g t=i]
p – n-vector, starting probability vector p(i) = Pr[q 0=i]
For any row of a the sum of entries = 1
Sp (i) = 1
Comment: Frequently extra start and end states are
added and vector p is not needed.
Hidden Markov Model
Hidden Markov Model is a Markov model in
which one does not observe a sequence
of states but results of a function
prescribed on states – in our case this is
emission of a symbol (amino acid or a
nucleotide).
States are hidden to the observers.
Introducing emission
probabilities
• Assume that at each state a Markov
process emits (with some distribution) a
symbol from alphabet S.
• Rather than observing a sequence of
states we observe a sequence of
emitted symbols.
Example:
S ={A,C,T,G}.
Generate a sequence
where A,C,T,G have
frequency p(A) =.33,
p(G)=.2, p(C)=.2, p(T)
= .27 respectively
A
T
C
G
one state
.33
.27
.2
.2
1.0
emission probabilities
HMM – formal definition
HMM is a Markov process that at each time step generates a
symbol from some alphabet, S, according to emission
probability that depends on state.
M = (Q, S, p, a,e)
Q – finite set of states, say n states ={q0,q1,…}
a – n x n transition probability matrix: a(i,j) = Pr[q t+1=j|g t=i]
p – n-vector, start probability vector: p (i) = Pr[q 0=i]
S = {s1, …,sk}-alphabet
e(i,j) = P[ot=sj|qt = i]; ot –tth element of generated sequences
= probability of generating oj in state qi (S=o0,…oT the output
sequence)
Algorithmic problems related to HMM
• Given HMM M and a sequence S compute
Pr(S|M) – probability of generating S by M.
• Given HMM M and a sequence S compute
most likely sequence of states generating
S.
• What is the most likely state at position i.
• Given a representative sample (training set)
of a sequence property construct HMM that
best models this property.
Probability of generating a sequence
S by HMM
Recall: probability of a sequence of states
S=q0,…,qn in Markov chain :
P(S|M) =a(q0,q1) a(q1 , q2) a(q2 , q3) … a(qn-1 , qn)
Now: Our sequence S = o0,…,oT is a word over S
and there can be many paths leading to the
generation of this sequence. Need to find all these
paths and sum the probabilities.
Formal definition of P(S|M)
M = (Q, S, a,e) ;
(p is dropped: q0 is the starting state)
S = o0…on
P[S|M] = S pQn+1P[p|M]P[S|p,M]
(sum over all possible paths p=q0,…qn of product: probability of the path
times the probability of generating given sequence assuming given
path.)
P[p|M] =a(q0,q1) a(q1 , q2) a(q2 , q3) … a(qn-1 , qn)
P[S|p,M] = e(q0,s1) e(q1,s2) e(q2 , s3) … e(qn , sn)
Computing LS(M) from the definition (enumerating all sequences p)
would take exponential time!
Example
Sample sequence:
S=ATCACAT
Assume: 0 start state
4 – end state
.4
3
.5
0
A .5
T .25
C .25
1
1.0
A
T
C
G
.25
.25
.25
.25
2
.5
A .3
T .3
C .3
G .1
A .3
T .3
C .1
G .3
.6
.6
4
A .5
T .5
.4
Two possible paths: p1=0,1,3,3,3,3,4 and p2=0,1,2,2,2,4
P(p1|M) = 1*.5*.4*.4*.4*.6=.0192;
P(S|p1) = .5*.25*.3*.3*.3*.3*.5 =.00050625; P(p1|M) * P(S|p1) = .00000972
P(p2|M) =.0192;
P(S|p2) =1*.5*.25*.1*.3*.1*.3*.5 =.00005625; P(p2|M) * P(S|p2) = .000000198
P(S|M) = .00000972 + 000000198 = 0.000009828
Observe how small are the numbers – we need to switch from probabilities to log.
Most likely path
• Most likely path for sequence S is a path
p in M that maximizes P(S|p).
• Why it may be interesting to know most
likely path:
– Frequently HMM are designed in such a
way that one can assign biological
relevance to a state: e.g. position in protein
sequence, beginning of coding region,
beginning/end of an intron.
Example: GpC island
Model for general
sequence
Mode for GpC island
Transitions between any pair of states
ATCAGTGCGCAAATAC
If this is most likely path we have estimated
position if GpC island
Simple HMM for prokaryotic genes
p
Position 1
1
Position 2
1
Position 3
1-q
1-p
q
Non-Coding Region
HMM-based Gene finding programs
for microbial genes
• GenMark- [Borodovsky, McInnich –
1993, Comp. Chem., 17, 123-133] 5th
order HMM
• GLIMMER[Salzberg, Delcher,
Kasif,White 1998, Nucleic Acids
Research, Vol 26, 2 55408] –
Interpolated Markov Model (IMM)
Eukaryotic gene finding
• On average, vertebrate gene is about 30KB
long
• Coding region takes about 1KB
• Exon sizes vary from double digit numbers to
kilobases
• An average 5’ UTR is about 750 bp
• An average 3’UTR is about 450 bp but both
can be much longer.
Exons and Introns
• In eukaryotes, the gene is a combination
of coding segments (exons) that are
interrupted by non-coding segments
(introns)
• This makes computational gene prediction
in eukaryotes even more difficult
• Prokaryotes don’t have introns - Genes in
prokaryotes are continuous
Central Dogma and Splicing
exon1
intron1
exon2
intron2
exon3
transcription
splicing
exon = coding
intron = non-coding
translation
Gene Structure
Splicing Signals
Exons are interspersed with introns and
typically flanked by GT and AG
Splice site detection
Donor site
5’
3’
Position
%
A
C
G
T
-8 … -2 -1
26
26
25
23
…
…
…
…
0
1
2
… 17
60 9 0 1 54 … 21
15 5 0 1 2 … 27
12 78 99 0 41 … 27
13 8 1 98 3 … 25
Consensus splice sites
Donor: 7.9 bits
Acceptor: 9.4 bits
Promoters
• Promoters are DNA segments upstream
of transcripts that initiate transcription
Promoter
5’
3’
• Promoter attracts RNA Polymerase to the
transcription start site
Splicing mechanism
(http://genes.mit.edu/chris/)
Splicing mechanism
• Adenine recognition site marks intron
• snRNPs bind around adenine recognition
site
• The spliceosome thus forms
• Spliceosome excises introns in the mRNA