Von Bing Yap
Statistics Department, UC Berkeley yapvb@stat.berkeley.edu

Human genome project furnishes sequences from a few individuals, but is relevant to all of us: we are very similar.
We are also slightly different: polymorphisms.
Species variations.

Mutation : (a) Inherent: DNA replication errors are not always corrected. (b) External: exposure to chemicals and radiation.
Selection : Deleterious mutations are removed quickly. Neutral and rarely, advantageous mutations, are tolerated and stick around …
Fixation : It takes time for a new variant to be established (having a stable frequency) in a population.

(Relatively) early days … to test hypotheses from population genetics. Do rodents evolve faster than primates (the generation time hypothesis) ? Which sites in a protein are under much selective pressure?
More recently (from 70’s) …to derive better substitution matrices like the PAM and BLOSUM series for sequence alignment and building phylogenetic trees.

Strictly speaking, only applicable to regions undergoing little selection.
Assumptions
1 . Site independence.
2 . Site homogeneity.
3 . Markovian: given current base, future substitutions independent of past.
4 . Temporal homogeneity: stationary Markov chain.

A stationary Markov chain is a family of transition probability matrices P(t), t

0, satisfying
P(t)P(s) = P(t+s), t,s

0.
P(t,a,b) = Pr(b at time s+t | a at time s).
Let Q be a rate matrix , i.e., it has positive offdiagonal entries and each row sum is 0.
Eg., Q(1,2) is the instantaneous rate of A going to C.
Q defines an MC by
P(t) = exp(Qt) = I + Q t + Q 2 t 2 /2! + …

Let Q =
-3
   

-3
  
 
-3
 
  
-3

Then P(t) = r s s s s r s s s s r s s s s r r = (1+3exp(-4
 t))/4, s = (1- exp(-4
 t))/4.

A probability distribution on {A,C,G,T},

, is a stationary distribution if
 a

(a) P(t,a,b) =

(b), b, t

0, or

P(t) =

, t

0, or

Q = 0 (global balance).
Facts: MC built from our Q has a unique stationary distribution. Let X(t) be the base at time t. If X(0)
~

, then X(t) ~

. Given any initial distribution, the distribution of X(t)
  as t
 
.
For the JC model,
 is the uniform distribution.

ancestor
T years
Q h
Q m
C
A
Typically, ancestor is unknown.

5 . Q h
= s h
Q and Q m
= s m
Q, for some positive s h and s m
, and some rate matrix Q.
6 . The ancestor is sampled from the stationary distribution
 of Q.
7 . Q is reversible :

(a) P(t,a,b) = P(t,b,a)

(b), b, t

0
(detailed balance).

s h
T
PAMs
ancestor ~

A
Q
Q s m
T
PAMs
C

Under the model, the joint probability is
 a

(a) P(s h
T,a,A) P(s m
T,a,C)
=
 a

(A) P(s h
T,A,a) P(s m
T,a,C)
=

(A) P(s h
T+ s m
T,A,C) = F(t,A,C).
The matrix F(t) is symmetric. It is equally valid to view A as the ancestor of C or vice versa. t = s h
T+ s m
T is the “ distance ” between A and C.
Note: Q and t are identifiable from F(t), but s h
, s m and T are not.

Joint probability of a pair of homologous sequences
Pr(a
1
…a n
,b
1
...b
n
)
=
 k
=
 a,b
F(t,a k
,b k
)
F(t,a,b) c(a,b) , where c(a,b) = # {k : a k
= a, b k
= b}.

By convention (Dayhoff 1978), Q is chosen to satisfy
–
 a

(a) Q(a,a) = 0.01, or the expected number of substitutions in 1 time unit is 0.01 per site.
This new time scale is called evolutionary time , measured in PAM ( P oint A ccepted
M utation).

The k th pair of sequences, separated by t k
PAMs, gives the count matrix c(k). Assuming that the pairs are independent and underwent the same MC defined by Q over pair-specific distances t k
’s.
Pr(all pairs) =
 k
 a,b
F(t k
,a,b) c(k,a,b) .
The pair-specific distances allow for different rates of substitution across pairs.
Parameters : Q, t
1
, t
2
,…,t
M
.
Maximum likelihood estimation by numerical methods.

This model is general enough to include almost all models in use as special cases.
For DNA base, there are JC, Kimura and Hasegawa-
Kishino-Yano models.
.
For amino acid, the PAM (Dayhoff 1978) and BLOSUM
(Henikoff 1993) substitution matrices are derived based on the model. Müller and Vingron (2000) presents an interesting method of estimation.
For codon, Z Yang and collaborators use constrained versions to estimate ratio of syn/nonsyn substitutions, among other things.

A
C
G
A
.

(1)

C
(2)
.

(1)

(2)


G
.
(3)
(3)



T
(4)
(4)
(4)

(1)

(2)

(3) .
T

: stationary distribution of Q;
 ,…, 
: positive constants.

It seems plausible that the substitution process in noncoding regions is strand-symmetric , i.e.,
Q(A,C) = Q(T,G), Q(C,C) = Q(G,G), etc.
From the point of view of sequence alignment, SS implies that the same substitution matrix can be used for aligning either strand.
Strand-symmetry (of the lack of) is well-studied in bacteria (Francino, Ochman).

A
C
G
T
A
.
C G T

(2)

(2)

(1)

(1) .

(2)

(1)

(1)

(2) .

(1)

(1)

(2)

(2) .

: stationary distribution; 2(

(1) +

(2)) = 1.
 ,…, 
: positive constants.

Staining chromosomes with the Giemsa dye produces characteristic banding patterns. G-bands:
G1, G2, G3 or G4 (Francke,1994). The unstained regions are H3- or H3+.
Data: 22,490 local alignments (4.4 Mbp) of human chromosome 22 with homologous mouse reads, produced by blastz, and approximate boundary coordinates of the bands on chr 22, obtained from the UCSC Human genome browser.

Only alignments in noncoding regions are used. Alignments that are closer than 10 bases are glued. Of the resultant alignments, those longer than 100 bases are used to fit band-specific substitution models.

R,SS
R
G3 G4, H3-, H3+ Band
A C G T A C G T
A .
18 62 15 .
21 67 15
C 21 .
16 71 19 .
16 63
G 71 16 .
21 63 16 .
19
T 15 62 18 .
15 67 21 .
A .
18 61 15 .
21 67 15
C 21 .
16 70 19 .
16 63
G 71 16 .
20 63 16 .
20
T 16 62 18 .
15 68 21 .
GC 46.7
51.7

Exactly the same formulation applies to modeling amino acid substitution.
Fast-evolving and slow-evolving proteins are known to have different substitution rates.
How can we summarize such observations?
Buried amino acids evolve differently from exposed residues. HMM with 2 hidden state?

Bergman and Kreitman (2001) showed that the substitution patterns between intergenic and intronic regions are similar. The combined estimated rate matrix is
A C G T
A .
21 42 20
C 31 .
32 62
G 62 32 .
31
T 20 42 21 .

Study other human chromosomes, and isochore-specific substitution patterns.
Multiple species. Holmes (2002) has a neat
EM algorithm for estimating substitution rates if the process is reversible. What to do if irreversible?
Even for 2 species, irreversible substitution models are not fully explored.

Terry Speed
Webb Miller
David Haussler
Terry Furey
Casey Bergman
Anne Yap