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Scoring matrices
Identity
PAM
BLOSUM
Scoring Matrices Types
• Identity matrix – exact matches receive one score and
non-exat matches a different score (say 1 and 0, or 6 and
–1 for local alignment.).
• Mutation data matrix – a scoring matrix compiled based
on observation of protein point mutation (PAM,
BLOSUM).
• Physical properties matrix – amino acids with with
similar properties (e.G. hydrophobicity ) receive high
score.
• Genetic code matrix – amino acids are scored based on
similarities in the coding triple (codons).
Substitution Matrix
 Amino acids substitute easily for another due to similar
physicochemical properties

Isoleucine for Valine (both small, hydrophobic)
Serine for Threonine (both polar)

Such changes – “conservative”

 Thus, need a way to increase sensitivity of the alignment
algorithm

Solution – substitution matrix
 Therefore, we need a range of values that depend on the
nature of sequences being compared
 Identical amino acids > Conservative substitutions >
Nonconservative substitutions
Choice of scoring matrix is
dictated by the alignment goals
• Two proteins are homologous if (and only if) they are
evolutionarily related (have a common ancestor)
• Homologous proteins are likely to have related functions
(and have the same fold)
• Scoring matrices must in some way model our
understanding of protein evolution.
• Based on the result of the search we have to be able to
decide if the discovered sequence similarity could happen by
chance or is a signature of likely homology.
BLOSUM
• Block – a short contiguous interval of multiple aligned
sequences.
• BLOCKS – data base of 3 000 blocks of highly conserved
sequences representing hundreds of protein groups.
•
Http://www.Blocks.fhcrc.Org/.
• BLOCKS  substitutions frequency  log odds score.
• Within each block cluster sequences within certain similarity
threshold (80% similarity yields BLOSUM80) and have such
cluster be represented by one sequence or average the
contribution.
• BLOSUM62 – most similar to PAM250 (believed to be
better).
BLOSUM METHOD
Data base

Deriving a frequency
tables from a data base
of blocks
1 .. .. w
1 A .. .. ..
.. .. .. .. ..
S .. .. .. ..
Data Base of blocks

Computing a logarithm of
odds matrix
1.2
7.5 6.3
1.9 5.5 3.1
6.5 2.0 8.1 4.3
3.7 5.8 2.9 7.7 3.2
Methods
Deriving a frequency table from a data base of
blocks.
1 2 … w
Seq 1 A
Seq 2 A
Seq 3 A
Seq 4 A
Seq 5 A
Seq 6 A
Seq 7 S
Seq 8 A
Seq 9 A
Seq 10 A
Seq s A
Frequency table consisting of all possible
amino acid pairs in a column
• 9A + 1S there are 8+7+…+1=36 AA pairs
• 9 AS or SA pairs
• no SS pairs
For a block : width of w and a depth of S, it
contribute WS(S-1)/2 [1.10.(10-1)]/2=45
METHODS
 The result of this counting is a frequency
table listing the number of time each of the
20+19+…+1=210 different amino acid pairs
occurs among the blocks.
 The table is used to calculate a matrix
representing odds ratio between these
observed frequency and those calculated by
chance.
METHODS
Observed probability qij :
20
i
qij  fij/  fij
i 1
j 1

fAA= 36, fAS = 9
qAA= 36/45 = 0.8
qAS = 9/45 = 0.2
Methods
Expected probability eij :
pi  qii   qij/ 2
j i

pA= [36 + (9/2)]/45 = 0.9
pS = [00 + (9/2) /45 = 0.1
• for i=j  eij = pi.pj ;
eAA = pA.pA = 0.9 x 0.9 = 0.81
• for ij  eij = pi.pj + pi.pj ;= 2 pi.pj
eAS = pA.pS + pA.pS = 2 pA.pS = 2 (0.9 x 0.1) = 0.18
Methods
The odds ratio
 An odds ratio matrix is calculated where each entry
is qij/eij
 The logarithm of odds ratio (Lod) in bit unit
 Sij = log2qij/eij
 A Lod is then calculated as score
 If the observed frequency is :



as the expected, then Sij = 0
if less than expected Sij < 0
if more than expected Sij > 0
METHODS
Clustering segment within blocks
 Sequences are clustered within blocks, and each
cluster is weighted. This is done by specifying a
clustering percentage in which sequence segments
that are identical for at least that percentage of
amino acids are grouped together.
 The lod matrix derived from a database of blocks
in which sequences that are identical at  80% of
aligned residues are clustered is referred to as
BLOSUM 80, and so forth.
The Dayhoff Matrix (PAM)
 Developed by Margaret Dayhoff, 1978.
 Counted likelihood of all possible substitutions in
closely related proteins.
 Derived mutability matrix Mi,j:
 Probability that Ai mutates to Aj in one
evolutionary unit, PAM.
 Multiplying M by itself extrapolate to higher
evolutionary orders (Mk).
PAM units
 Log-odds approach: Scores proportional to the log of the
ratio of target frequencies to background frequencies
 PAM – Point Accepted Mutation /Percent Accepted
Mutation
 Two sequences S and T are defined to be one PAM unit
diverged if a series of accepted point mutation (and no
insertion/deletion) can convert S to T with an average of one
mutation per 100 res.
 Point accepted mutation – mutation of one residue accepted
by evolution.
PAM units
 Problem 1: given two sequences you cannot tell
their PAM distance in the strict sense of the above
definition since one residue could mutate more than
once
 BUT: If you take sequences that are closely related
then problem above is unlikely to occur.
 Problem 2 : A change could happen by
deletion/insertion
PAM Matrices - Summary
• There is a sequence of PAM matrices
• PAMn attempts to provide proper scoring for sequences that
diverged n PAM units.
• PAMn matrix is obtained from PAM1 assuming Markov model
of protein evolution where transition probabilities in 1 PAM
step are given by PAM1.
•
PAMn = PAM1 n
• PAM1 is constructed based on highly similar sequences
(believed to be apart at most few PAM units) so that Problems1
& 2 are unlikely to occur.)
Computation representation
 Define:




fp(a) = probabilities of occurrence for each
amino acid a.
f(a,b) = the number of times the mutation a↔b
( f(a,b) = f(b,a) )
f(a) = ∑f(a,b) ( b≠a )
m(a) = mutability of amino acid a = f(a) / fp(a)
b
Computation
representation ,cnd



M(a,b) = the probability of amino acid a changing to
amino acid b
M(a,b) = Pr(a↔b)
= Pr(a↔b | a changed)Pr(a changed)
= f(a,b)* m(a) / f(a)
(the conditional probability above is estimated as the
ratio between the a↔b mutations and the total number of
mutations involving a )
M(a,a) = 1- m(a) unchange probability
(the diagonal elements)
Relatedness odds Matrix
 M(a,b) gives the probability that amino acid
a will change to b in a related sequence in a
interval
 f(b) is the chance of a random occurrence of
amino acid b
 Score(a,b) = 10log[M(a,b)/f(b)]
(symmetric matrix)
PAM

Let us assume to AA (or nucleotides) i and j, with
frequency fi and fj.
P(random alignment of i and j)=fi fj.

P( i and j have a common ancestor x ) 
f
Pr(x  i)Pr(x  j)   f x M ix M jx   M ix f x M j x 
x
x
 f j  M ix M xj  f j M ij2
x
x
x
PAM
P( i and j have a common ancestor x )

P(randomalig nment )
Dij 
f j M ij2
fif j

M ij2
fi
Long Distance Evolution
 There is a different mutation probability matrix for
each evolutionary interval. These can be derived
from the one for 1 PAM by matrix multiplication.

e.g.
in 2 PAM units of evolution
a→c→b
(c can be anything including a or b)
 In general Mⁿ is the transition probability matrix for
a period of n units of evolution
Estimation of Evolutionary
Distance
 Different mutation probability matrix for
each evolutionary interval measured in PAMs.
 Calculate the percentage of amino acids that
will be observed to change on the average in
the interval
P = 100(1 – ∑f(i)M(i,i))
 A PAM250 matrix usually represents two
sequences which have about 20% identity
Nucleotide PAM scoring
matrices
Assuming equal probability for each mutation PAM1 would be:
A T
G
A .99 .0033 .0033
T .0033 .99 .0033
G .0033 .0033 .99
C .0033 .0033 .0033
C
.0033
.0033
.0033
.99
Some models would score higher transitions (purine into purine
pirimidine into pirimidine) that transversions:
A T
G
A .99 .0002 .0006
T .0002 .99 .0002
G .0006 .0002 .99
C .0002 .0006 .0002
C
.0002
.0006
.0002
.99
Discrimination of real local alignment
from “by chance” alignment
Method: Compute mutual information:
Sx Syp(x,y) log (p(x,y)/ p(x)p(y))
Recall that score s(x,y) = log (p(x,y)/ p(x)p(y))
Thus we simply compute:
Sx=1..20 Sy=1,..20 p(x,y) s(x,y)
Examples (in bits):
PAM160 = .7; PAM250 = .36
Higher mutual information  better discrimination
between true and by chance alignment.
Problems with PAM
 Defining PAM 1 in terms of amino acid mutation
rather than number of nucleotide changes.
 Some mutation may be rare and underrepresented in
PAM1 (which is based on closely related proteins
only).
 The mutation rate depends on the position of an
amino-acid in the structure.
 Require construction phylogenic tree which in turn
need scoring matrices for proper construction.
(remains a problem for many other methods)
Some more problems with
PAM Matrices
 Derived from global alignments of closely related
sequences.
 Matrices for greater evolutionary distances are
extrapolated from those for lesser ones.
 The number with the matrix (PAM40, PAM100)
refers to the evolutionary distance; greater
numbers are greater distances.
 Does not take into account different evolutionary
rates between conserved and non-conserved
regions.
BLOSUM matrices
BLOcks SUbstitution Matrix
 Amino acid substitution matrices
from protein blocks
S. HENIKOFF and J. HENIKOFF
Proc. Natl. Acad. Sci.USA
Vol.89, pp. 10915-10919, November 1992
Biochmistry
Comparison to PAM
 The BLOSUN series derived from alignments in
blocks is fundamentally different from the Dayhoff
PAM series, which is derived from the estimation of
mutation rates.
 Nevertheless, the BLOSUM series based on percent
clustering of aligned segments in blocks, can be
compared to the Dayhoff matrices based on percent
accepted mutation (PAM) using the measure of
average information per residue pair in bits units
called relative entropy.
Comparison between
BLOSUM 62 and PAM 160
 The BLOSUM 62 is less tolerant to substitutions
involving hydrophilic amino acids, while it is more
tolerant to substitutions involving hydrophobic
amino acids.
 For rare amino acids especially cysteine and
tryptophane, BLOSUM 62 is typically more
tolerant to mismatches than is PAM 160.
PAM vs BLOSUM
 Dayhoff estimated mutation rates from substitutions
observed in closely related proteins and
extrapolated those rates to models distant
relationships.
 In BLOSUM approach, frequencies were obtained
directly from relationships represented in the block,
regardless of evolutionary distance.
 The Dayhoff frequency table included 36 pairs in
which no accepted point mutations.
Differences Between the PAM
and BLOSUM Approach
 In contrast, the pairs counted with BLOSUM,
included no fewer than 2369 occurrences of any
particular substitution.
• The BLOSUM matrices depend only on the identity
and composition of groups protein in Prosite.
• Therefore, there is no expectation that these
substitution matrices will change significantly in
the future.
PAM Versus BLOSUM
 PAM is based on an evolutionary model.
 BLOSUM is based on protein families.
 PAM is based on global alignment.
 BLOSUM is based on local alignment.