Download 52930 Protein informatics 09

52930 Protein informatics
Liisa Holm
Organization
• Lectures
– Wednesdays 6 September to 14 October
– Exam Friday 16 October
• Essay type question
• Numerical problems
• Textbook
– DW Mount: Bioinformatics. Sequence and genome
analysis. 2nd edition. Chapters 3-7,10-11
• Web site
– http://ekhidna.biocenter.helsinki.fi/teaching/winter20
09/proteiinianalyysi
Aims & scope
• Expose biology students to background of
methods
• Related practical course
– Practical course in protein informatics
(Proteiinianalyysin harjoitustyöt)
– Hands-on practice in using web servers that
implement methods
– Neither course required for the other
Topics
• Pairwise alignment
• Probability and statistical analysis of sequence
alignments
• Multiple sequence alignment
• Database searching
• Phylogenetic prediction
• Protein classification and structure prediction
• Genome annotation
Pairwise alignment
Why align sequences?
• Common ancestor
• Infer common evolutionary
origin from similarity
– Then can infer function and
structure
• Similarity can be due to
–
–
–
–
Gene duplication + speciation
Horizontal gene transfer
Gene fusion
Convergence (similarity
without homology)
Sequence A
x steps
Sequence B
y steps
Ancestral sequence
Similar sequences are likely
homologous
• Dissimilar sequences are less likely to be
homologous
4-letter word example
• This is not the usual substitution model
•
•
•
•
•
WORD
WORE
GORE
GONE
GENE
(d=0,
(d=1,
(d=2,
(d=3,
(d=4,
p=1/N^4)
p=4/N^3)
p=6/N^2)
p=4/N)
p=1)
Optimal alignment
• Assuming independence between scores for
each position, the optimal alignment can be
determined using dynamic programming
• Setup: scoring matrix, gap penalties
Dynamic programming
A
0
3
B
3
4
BEGIN
1
2
1
END
C
D
Maximal path sum BEGIN  END ?
(a) Enumerate every path brute force
(b) Use induction: only one optimal path up to any node in graph.
Example: all paths leading to B
A
3
0
3
3
8
4
BEGIN
7
1
1
2
3
B
1
END
C
D
Global alignment
• Needleman-Wunsch algorithm
• Maximal trace from beginning to end
• Global alignment score may be negative
Local alignment
• Aligned region truncated to segment giving
the largest positive contribution
Scoring alignments
• Substitution matrices
– Gap penalties
• Significance
– Aligning two sequences, would you expect the
same level of similarity by chance alone?
Conversion between odds score, log
odds and bit scores
• Odds score = ratio of likelihoods of two events or
outcomes. E.g. observed frequency of aligned A
and B in related sequences divided by the
frequency with which A and B align by chance
– f(A and B) / [ f(A) * f(B)]
• Odds scores are often converted to logarithms to
create log odds scores.
• Log odds scores are additive.
• Bit score = log odds score converted to a
logarithm to the base 2
Bit-scores
• The score needed to distinguish an MSP from chance
is approximately the number of bits needed to
specify where the MSP starts in each of the two
sequences being compared
– MSP = maximally scoring pair
– Ungapped alignment case
• Log2 N bits are needed to distinguish among N
possibilities
– Two proteins of 250 residues: 16 bits
– Database of 4M residues: 30 bits [160 M: 34 bits]
Dayhoff model
• Markov chain: mutations independent of
previous mutations
• Data: 71 groups of closely related sequnces
(>85 % similarity), yielding 1572 substitution
events
• Mutability of amino acid types (per 100
accepted point mutations)
PAM1 and PAM250 for Phe -> X
X
PAM1
PAM250 X
PAM1
PAM250
Ala
0.0002
0.04
Leu
0.0013
0.13
Arg
0.0001
0.01
Lys
0.0000
0.02
Asn
0.0001
0.02
Met
0.0001
0.02
Asp
0.0000
0.01
Phe
0.9946
0.32
Cys
0.0000
0.01
Pro
0.0001
0.02
Gln
0.0000
0.01
Ser
0.0003
0.03
Glu
0.0000
0.01
Thr
0.0001
0.03
Gly
0.0001
0.03
Trp
0.0001
0.01
His
0.0002
0.02
Tyr
0.0021
0.15
Ile
0.0007
0.05
Val
0.0001
0.05
These are mutation probabilities!
Log odds form of PAM250
• Unit is 10 * logarithm to the base 10 of ratio
• S(A,B) = ½ * (10 * log10(p(A->B)/f(A)) + 10 *
log10 (p(B->A)/f(B))
• Range -8 … +17
• Local alignment scores are maximal, when
PAM distance corresponds to the similarity of
the target sequences
BLOSUM matrices
• The BLOSUM matrix assigns a probability
score for each residue pair in an
alignment based on:
–the frequency with which that pairing is
known to occur within conserved blocks of
related proteins.
• BLOSUM matrices are constructed from
observations which lead to observed
probabilities
BLOSUM substitution matrices
BLOSUM matrices are used in
‘log-odds’ form based on
actually observed substitutions.
This is because:
Ease of use: ‘Scores’ can be just
added (the raw probabilities
would have to be multiplied)
Ease of interpretation:
S=0 : substitution is just as likely
to occur as random
S<0 : substitution is more likely to
occur randomly than observed
S>0 : substitution is less likely to
occur randomly than observed
Unit is half-bits (odds ratio to
logarithm base 2, multiplied by
2)
Information content
• Using a standard measure for overall amino acid frequencies gives the
information content of a random protein sequence as 4.19 bits/residue.
• Thus, for an average size protein domain (150 residues), the message length
is ~630 bits and the probability that 2 random sequences would specify
the same message is 2-630 (10-190).
> Database searching for protein similarities is doable, even for fairly short
sequences
• BUT, for a transcription binding site of 8-10 bp, the odds of 2 random
sequences arriving at the same message is 10-5.
> Database searching for regulatory elements does not work well as
databases get larger
Relative entropy H of target and
background distributions
• Scale score matrix s to bits
qij
• H = S qij sij = S qij log ----------pi pj
q = target frequencies of amino acids
p = background frequencies
H measures the average information available per
position to distinguish the alignment from chance
qij
• Score = S fij sij ~ S fij ln ----------pi pj
Optimal scoring matrix: target distribution q =
frequencies in alignment f
Affine gap penalties
• Gap opening penalty (g)
• Gap extension penalty (r)
• W(x) = g + rx
• X is the length of the gap
• Well working gap penalties:
• BLSOUM62 (-11,-1)
Statistical Significance
• A good way to determine if an alignment score has
statistical meaning is to compare it with the score
generated from the alignment of two random
sequences
• A model of ‘random’ sequences is needed. The
simplest model chooses the amino acid residues in a
sequence independently, with background
probabilities (Karlin & Altschul (1990) Proc. Natl.
Acad. Sci. USA, 87 (1990) 2264-2268)
Alignment score
• Optimal alignment scores follow extreme
value distribution
– Exact theory for ungapped local alignments
• There is at least one positive score sij
• Average score is negative
– Results hold empirically for gapped alignments
Probability and statistics
The need for statistics
• Statistics is very important for bioinformatics.
– It is very easy to have a computer analyze the data
and give you back a result.
– Problem is to decide whether the answer the
computer gives you is any good at all.
• Questions:
– How statistically significant is the answer?
– What is the probability that this answer could have
been obtained by random? What does this depend
on?
Basics
N

n X
Sample
Population
S
Basics
N

Descriptive statistics
n X
Sample
Population
Probability
Substitution matrices
Score of amino acid a
with amino acid b
Pab is the observed frequency that
residues a and b are correlated
because of homology
S ( a , b)   log
1
Lambda is a scaling
factor equal to 0.347,
set so that the scores
can be rounded off to
sensible integers
pa b
fa fb
fafb is the expected frequency of seeing residues a and b
paired together, which is just the product of the frequency of
residue a multiplied by the frequency of residue b
Source: Where did the BLOSUM62 alignment score matrix come from?
22 Aug 2004
Eddy S., Nat. Biotech.
Substitution matrices
Pab is the observed frequency that
residues a and b are correlated
because of homology
pa b
fa fb
fafb is the expected frequency of seeing
residues a and b paired together, which is just
the product of the frequency of residue a
multiplied by the frequency of residue b
e
S
Lambda is a scaling
factor equal to
0.347, set so that the
scores can be
rounded off to
sensible integers
i) S=0 : O/E ratio=1
ii) Compare S=5 and S=10.
Ratio is based on
exponential function
iii) S=-10: O/E ratio =
0.031 ≈ 1/32.
iv) Ratio of scores S1, S2
in terms of probabilities
of observed/random =
i) S=0 : O/E ratio=1
ii) Compare S=5 and S=10.
Ratio is based on
exponential function
iii) S=-10: O/E ratio =
0.031 ≈ 1/32.
32.1
5.7
iv) Ratio of scores S1, S2
in terms of probabilities
of observed/random =
i) S=0 : O/E ratio=1
ii) Compare S=5 and S=10.
Ratio is based on
exponential function
iii) S=-10: O/E ratio =
0.031 ≈ 1/32.
32.1
5.7
iv) Ratio of scores S1, S2
in terms of probabilities
of observed/random =
i) S=0 : O/E ratio=1
ii) Compare S=5 and S=10.
Ratio is based on
exponential function
iii) S=-10: O/E ratio =
0.031 ≈ 1/32.
32.1
iv) Ratio of scores S1, S2
in terms of probabilities
of observed/random =
5.7
e
S1
/e
S2
e
 ( S1  S2 )
Example: BLAST
• Motivations
–Exact algorithms are exhaustive but
computationally expensive.
–Exact algorithms are impractical for comparing
a query sequence to millions of other
sequences in a database (database scanning),
–and so, database scanning requires heuristic
alignment algorithm (at the cost of optimality).
Interpret BLAST results - Description
ID (GI #, refseq #, DB-specific Gene/sequence Bit score – higher, better.
ID #) Click to access the
Definition
Click to access the
record in GenBank
pairwise alignment
Links
Expect value – lower, better. It tells
the possibility that this is a random hit
Problems with BLAST
• Why do results change?
• How can you compare results from different
BLAST tools which may report different types
of values?
• How are results (eg evalue) affected by query
• There are _many_ values reported in the
output – what do they mean?
Example: Importance of Blast statistics
But, first a review.
Review
• What is a distribution?
– A plot showing the frequency of a given variable or
observation.
Review
• What is a distribution?
– A plot showing the frequency of a given variable or
observation.
Features of a Normal Distribution
Symmetric Distribution
Has an average or mean
value at the centre
Has a characteristic width
called the standard
deviation (S.D. = σ)
Most common type of
distribution known
 = mean
Standard Deviations (Z-score)
 ± 1.0 S.D. 0.683
>  + 1.0 S.D.
0.158
 ± 2.0 S.D. 0.954
>  + 2.0 S.D.
0.023
 ± 3.0 S.D. 0.9972
>  + 3.0 S.D.
0.0014
 ± 4.0 S.D. 0.99994
>  + 4.0 S.D.
0.00003
 ± 5.0 S.D. 0.999998
>  + 5.0 S.D.
0.000001
Mean, Median & Mode
Mode
Median
Mean
Mean, Median, Mode
• In a Normal Distribution the mean, mode and
median are all equal
• In skewed distributions they are unequal
• Mean - average value, affected by extreme
values in the distribution
• Median - the “middlemost” value, usually half
way between the mode and the mean
• Mode - most common value
Different Distributions
Unimodal
Bimodal
Other Distributions
•
Binomial Distribution
•
•
Poisson Distribution
•
•
the discrete probability distribution of the number of successes in a sequence
of n independent yes/no experiments, each of which yields success with
probability p.
expresses the probability of a number of events occurring in a fixed period of time if these
events occur with a known average rate and independently of the time since the last event.
Extreme Value Distribution
– Gumbel distribution
– used to model the distribution of the maximum (or the minimum)
of a number of samples of various distributions.
Binomial Distribution
P(x) = (p + q)n
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Poisson Distribution
P( x ) 
 =0.1
 =1
Proportion of samples
P(x)
 =2
 =3
 = 10
x
x e  
x!
Gumbel distribution
Review
• What is a distribution?
– A plot showing the frequency of a given variable or observation.
• What is a null hypothesis?
– A statistician’s way of characterizing “chance.”
– Generally, a mathematical model of randomness with respect to
a particular set of observations.
– The purpose of most statistical tests is to determine whether the
observed data can be explained by the null hypothesis.
Review
• What is a distribution?
– A plot showing the frequency of a given variable or observation.
• What is a null hypothesis?
– A statistician’s way of characterizing “chance.”
– Generally, a mathematical model of randomness with respect to
a particular set of observations.
– The purpose of most statistical tests is to determine whether the
observed data can be explained by the null hypothesis.
Review
• Examples of null hypotheses:
– Sequence comparison using shuffled sequences.
– A normal distribution of log ratios from a
microarray experiment.
– LOD scores from genetic linkage analysis when the
relevant loci are randomly sprinkled throughout
the genome.
Empirical score distribution
The picture shows a
distribution of scores
from a real database
search using BLAST.
This distribution
contains scores from
non-homologous and
homologous pairs.
High scores from homology.
Empirical null score distribution
This distribution is
similar to the
previous one, but
generated using a
randomized
sequence database.
Review
• What is a p-value?
– The probability of observing an effect as strong or
stronger than you observed, given the null
hypothesis. I.e., “How likely is this effect to occur
by chance?”
– Pr(x > S|null)
Review
What is the name of the
distribution created by
sequence similarity
scores, and what does it
look like?
Extreme value distribution,
or Gumbel distribution.
It looks similar to a normal
distribution, but it has a
larger tail on the right.
Review
What is the name of the
distribution created by
sequence similarity
scores, and what does it
look like?
Extreme value
distribution, or Gumbel
distribution.
It looks similar to a
normal distribution, but it
has a larger tail on the
right.
8000
7000
6000
5000
4000
3000
2000
1000
0
<20
30
40
50
60
70
80
90
100
110
>120
Statistics
• BLAST (and also local i.e. Smith-Waterman and BLAT scores)
between random, unrelated sequences follow the Gumbel Extreme
Value Distribution (EVD)
• Pr(s>S) = 1-exp(-Kmn e-S)
–
–
–
–
This is the probability of randomly encountering a score greater than S.
S alignment score
m,n query sequence lengths, and length of database resp.
K,  parameters depending on scoring scheme and sequence composition
• Bit score : S’ = S – log(K)
log(2)
BLAST output revisited
S’ S

K
From: Expasy BLAST
E
n
m
Review
EVD for random blast
Upper tail behaviour:
Pr( s > S ) ~ Kmn e-S
This is the EXPECT
value = Evalue
8000
7000
6000
5000
4000
3000
2000
1000
0
<20
30
40
50
60
70
80
90
100
110
>120
P-value in Sequence Matching
• P(s > S) = .01
– P-value of .01 occurs at score threshold S (392 below) where score s
from random comparison is greater than this threshold 1% of the
time
• Likewise for P=.001 and so on.
What Distribution Really Looks Like
• N Dependence
• True Positives
A most important caveat...
• For database searches, the ONLY criteria available to
judge the likelihood of a structural or evolutionary
relationship between 2 sequences is an estimate of
statistical significance
• For a medium-sized protein using default parameters
(Blosum62, E = 10), the cut-off for statistical
significance is P =10-7-10-5
• Statistical significance and biological significance are
NOT necessarily the same
P-value
P<=10-100
exact match
10-100 < P < 10-50
10-5 < P < 10-1
sequences very nearly
identical, e.g., alleles or
SNPs
closely related sequences,
homology certain
usually distant relatives
P > 0.1
match probably insignificant
10-50 < P < 10-10
Significance Depends
on Database Size
• The Significance of Similarity Scores Decreases with Database Growth
–
–
–
–
The score between any pair of sequence pair is constant
The number of database entries grows exponentially
The number of non-homologous entries >> homologous entries
Greater sensitivity is required to detect homologies
Greater s
• Score of 100 might rank as best in database of 1000 but only in top-100 in
database of 1000000
– expectation value
DB-1
DB-2
Summary
• Want to be able to compare scores in
sequences of different compositions or
different scoring schemes
• Score: S = sum(match) – sum(gap costs)
Summary
• Want to be able to compare scores
in sequences of different
compositions or different scoring
schemes
• Score: S = sum(match) – sum(gap
costs)
• Bit score
– S’ = S – log(K)
log(2)
Summary
Score and bit score
grow linearly with
the length of the
alignment
• Want to be able to compare scores
in sequences of different
compositions or different scoring
schemes
• Score: S = sum(match) – sum(gap
costs)
• Bit score
– S’ = S – log(K)
log(2)
Summary
Score and bit score
grow linearly with
the length of the
alignment
• Want to be able to compare scores in
sequences of different compositions
or different scoring schemes
• Score: S = sum(match) – sum(gap
costs)
• Bit score
– S’ = S – log(K)
log(2)
• E-value of bit score
– E = mn2-S’
Summary
Score and bit score
grow linearly with
the length of the
alignment
• Want to be able to compare scores in E-Value shrinks
sequences of different compositions really fast as bit
score grows
or different scoring schemes
• Score: S = sum(match) – sum(gap
costs)
• Bit score
– S’ = S – log(K)
log(2)
• E-value of bit score
– E = mn2-S’
Summary
Score and bit score
grow linearly with
the length of the
alignment
• Want to be able to compare scores in E-Value shrinks
sequences of different compositions really fast as bit
score grows
or different scoring schemes
• Score: S = sum(match) – sum(gap
E-Value grows
costs)
linearly with the
• Bit score
product of target
– S’ = S – log(K)
log(2)
• E-value of bit score
– E = mn2-S’
and query sizes.
Summary
Score and bit score
grow linearly with
the length of the
alignment
• Want to be able to compare scores in E-Value shrinks
sequences of different compositions really fast as bit
score grows
or different scoring schemes
• Score: S = sum(match) – sum(gap
E-Value grows
costs)
linearly with the
• Bit score
product of target
– S’ = S – log(K)
log(2)
• E-value of bit score
– E = mn2-S’
and query sizes.
Doubling target set size
and doubling query
length have the same
effect on e-value
Conclusion
• You should now be able to compare BLAST results from different
databases, converting values if they are reported differently (which
happens frequently)
• You should now know why BLAST results might change from one day to
the next, even on the same server
• You should understand also the dependence of query length on Evalue.
• Statistical rankings are reported for (almost) every database search
tool. When making comparisons between databases, between
sequences it is useful to know how the statistics are derived to know if
comparisons are meaningful.
Exercises
Exercise 1: Calculation of Log Odds and
Odds Scores by the BLOSUM Method
• In one column of an alignment of a set of related, similar
sequences, amino acid D changes to amino acid E at a frequency of
0.10, and the number of times this change is expected based on the
number of occurrences of D and E in the column is 0.05.
• What is the odds score of finding a D-to-E substitution in an
alignment?
• What is the log odds score for the D-to-E substitution in bits? (Note:
log to base 2 = natural log / 0.693.)
• What would be the entry in the BLOSUM amino acid scoring matrix
for this substitution? Compare your result to the actual entry in the
BLOSUM62 matrix (D to E in BLOSUM62 : +2).
• In the same column, D does not change at all at a frequency of 0.80,
and the expected frequency of D not changing is 0.10. Calculate the
corresponding log odds score and the BLOSUM62 entry for D not
changing (D to D in BLOSUM62: +6).
Exercise 2: USING THE DYNAMIC PROGRAMMING METHOD TO
CALCULATE THE LOCAL ALIGNMENT OF TWO SHORT SEQUENCES
BY HAND
• The BLASTP algorithm performs a local alignment between a query
sequence and a matching database sequence using the dynamic
programming algorithm with the BLOSUM62 scoring matrix, a gap opening
penalty of –11, and a gap extension penalty of –1 (i.e., a gap of length 1
has a penalty of –11, one of length 2, –12, etc.). Align the sequences
MDPW and MEDPW using the Smith–Waterman algorithm described in
the dynamic programming notes by following the global alignment
example given in the notes, but using the Smith–Waterman algorithm.
• Make a matrix for keeping track of best scores and a second matrix to
keep track of the moves that give the best scores. (Hint: The alignment of
M's, P's, and W's all give high scores, so the problem boils down to how to
align D with ED and is actually quite a trivial problem.)
• Use the BLOSUM62 matrix and BLASTP gap penalties of –11,–1. What is
the optimal alignment and score between these two sequences?
Exercise 3
• What is the odds score and log odds score of the following
alignment? Blosum (D,D)=+6, (D,E)=+2
DEDEDEDE
DDDDDDDD
• Using the approximation S ~ log2(nm) and assuming that the above
alignment was found by aligning two sequences of length 250, is
the alignment significant at the 0.05 level? (That is, could an
alignment of two random sequences of the same length achieve
such a score with a probability of 0.05?)
• If the gap penalty was very high, e.g., gap opening of 8 and gap
extension of 8, so that no gaps were produced, and the BLOSUM62
scoring matrix was used, calculate the significance of the alignment
using the equation P(S>x) ~ 1-exp(-Kmn exp(- λ x)). Use K=0.060 and
λ=0.270; note that this λ assumes that the alignment score is in
half-bits so that the alignment score must be in these units also.