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RNA Secondary Structure

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RNA Secondary Structure

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RNA Secondary Structure

aagacuucggaucuggcgacaccc

uacacuucggaugacaccaaagug

aggucuucggcacgggcaccauuc

ccaacuucggauuuugcuaccaua

aagccuucggagcgggcguaacuc

Given x = x1....xN, and a SCFG G,

Find the most likely parse of x

(the most likely alignment of G to x)

Dynamic programming variable:

(i, j, V):likelihood of the most likely parse of xi…xj,

rooted at nonterminal V

Then,

(1, N, S): likelihood of the most likely parse of x by the grammar

V

X

Y

j

i

Initialization:

For i = 1 to N, any nonterminal V,

(i, i, V) = log P(V xi)

Iteration:

For i = 1 to N – 1

For j = i+1 to N

For any nonterminal V,

(i, j, V) = maxXmaxYmaxik<j (i,k,X) + (k+1,j,Y) + log P(VXY)

Termination:

log P(x | , *) = (1, N, S)

Where * is the optimal parse tree (if traced back appropriately from above)

S a S | c S | g S | u S |

S a | S c | S g | S u

a S u | c S g | g S u | u S g | g S c | u S a

SS

- Adjust the probability parameters to reflect bond strength etc
- No distinction between non-paired bases, bulges, loops
- Can modify to model these events
- L: loop nonterminal
- H: hairpin nonterminal
- B: bulge nonterminal
- etc

Initialization:

(i, i-1) = log P()

Iteration:

For i = 1 to N

For j = i to N

(i+1, j–1) + log P(xi S xj)

(i, j–1) + log P(S xi)

(i, j) = max

(i+1, j) + log P(xi S)

maxi < k < j (i, k) + (k+1, j) + log P(S S)

Recall HMMs:

Forward:fl(i) = P(x1…xi, i = l)

Backward:bk(i) = P(xi+1…xN | i = k)

Then,

P(x) = k fk(N) ak0 = l a0l el(x1) bl(1)

Analogue in SCFGs:

Inside:a(i, j, V) = P(xi…xj is generated by nonterminal V)

Outside: b(i, j, V) = P(x, excluding xi…xj is generated by S and the excluded part is rooted at V)

To compute

a(i, j, V) = P(xi…xj, produced by V)

a(i, j, v) = X Y k a(i, k, X) a(k+1, j, Y) P(V XY)

V

X

Y

j

i

k

k+1

Initialization:

For i = 1 to N, V a nonterminal,

a(i, i, V) = P(V xi)

Iteration:

For i = 1 to N-1

For j = i+1 to N

For V a nonterminal

a(i, j, V) = X Y k a(i, k, X) a(k+1, j, X) P(V XY)

Termination:

P(x | ) = a(1, N, S)

b(i, j, V) = Prob(x1…xi-1, xj+1…xN, where the “gap” is rooted at V)

Given that V is the right-hand-side nonterminal of a production,

b(i, j, V) = X Y k<i a(k, i – 1, X) b(k, j, Y) P(Y XV)

Y

V

X

j

i

k

Initialization:

b(1, N, S) = 1

For any other V, b(1, N, V) = 0

Iteration:

For i = 1 to N-1

For j = N down to i

For V a nonterminal

b(i, j, V) = X Y k<i a(k, i – 1, X) b(k, j, Y) P(Y XV) +

X Y k<i a(j+1, k, X) b(i, k, Y) P(Y VX)

Termination:

It is true for any i, that:

P(x | ) = X b(i, i, X) P(X xi)

We can now estimate

c(V) = expected number of times V is used in the parse of x1….xN

1

c(V) = –––––––– 1iNijN a(i, j, V) b(i, j, v)

P(x | )

1

c(VXY) = –––––––– 1iNi<jN ik<j b(i,j,V) a(i,k,X) a(k+1,j,Y) P(VXY)

P(x | )

Then, we can re-estimate the parameters with EM, by:

c(VXY)

Pnew(VXY) = ––––––––––––

c(V)

c(V a) i: xi = a b(i, i, V) P(V a)

Pnew(V a) = –––––––––– = ––––––––––––––––––––––––––––––––

c(V) 1iNi<jN a(i, j, V) b(i, j, V)

GOALHMM algorithmSCFG algorithm

Optimal parseViterbiCYK

EstimationForwardInside

BackwardOutside

LearningEM: Fw/BckEM: Ins/Outs

Memory ComplexityO(N K)O(N2 K)

Time ComplexityO(N K2)O(N3 K3)

Where K: # of states in the HMM

# of nonterminals in the SCFG

position i

length l

5’

5’

position i

3’

3’

position j

position j

Models energy of a fold in terms of specific features:

- Pairs of base pairs (stacked pairs)
- Bulges
- Loops (size, composition)
- Interactions between stem and loop

position j’

positions i

5’

position j

3’

1

4

3

5

2

5

2

3

1

4

Phylogeny Tree Reconstruction

Trees can be inferred by several criteria:

- Morphology of the organisms
- Can lead to mistakes!

- Sequence comparison
Example:

Orc:ACAGTGACGCCCCAAACGT

Elf:ACAGTGACGCTACAAACGT

Dwarf:CCTGTGACGTAACAAACGA

Hobbit:CCTGTGACGTAGCAAACGA

Human:CCTGTGACGTAGCAAACGA

During infinitesimal time t, there is not enough time for two substitutions to happen on the same nucleotide

So we can estimate P(x | y, t), for x, y {A, C, G, T}

Then let

P(A|A, t) …… P(A|T, t)

S(t) = ……

……

P(T|A, t) ……P(T|T, t)

x

x

t

y

A

C

Reasonable assumption: multiplicative

(implying a stationary Markov process)

S(t+t’) = S(t)S(t’)

That is, P(x | y, t+t’) = z P(x | z, t) P(z | y, t’)

Jukes-Cantor: constant rate of evolution

1 - 3

For short time , S() = I+R = 1 - 3

1 - 3

1 - 3

T

G

Jukes-Cantor:

For longer times,

r(t)s(t) s(t) s(t)

S(t) = s(t)r(t) s(t) s(t)

s(t)s(t) r(t) s(t)

s(t)s(t) s(t) r(t)

Where we can derive:

r(t) = ¼ (1 + 3 e-4t)

s(t) = ¼ (1 – e-4t)

S(t+) = S(t)S() = S(t)(I + R)

Therefore,

(S(t+) – S(t))/ = S(t) R

At the limit of 0,

S’(t) = S(t) R

Equivalently,

r’ = -3r + 3s

s’ = -s + r

Those diff. equations lead to:

r(t) = ¼ (1 + 3 e-4t)

s(t) = ¼ (1 – e-4t)

Kimura:

Transitions: A/G, C/T

Transversions: A/T, A/C, G/T, C/G

Transitions (rate ) are much more likely than transversions (rate )

r(t)s(t)u(t)s(t)

S(t) = s(t)r(t)s(t)u(t)

u(t)s(t)r(t)s(t)

s(t)u(t)s(t)r(t)

Wheres(t) = ¼ (1 – e-4t)

u(t) = ¼ (1 + e-4t – e-2(+)t)

r(t) = 1 – 2s(t) – u(t)

Basic principles:

- Degree of sequence difference is proportional to length of independent sequence evolution
- Only use positions where alignment is pretty certain – avoid areas with (too many) gaps

Given sequences xi, xj,

Define

dij = distance between the two sequences

One possible definition:

dij = fraction f of sites u where xi[u] xj[u]

Better model (Jukes-Cantor):

f = 3 s(t) = ¾ (1 – e-4t)

¾ e-4t = ¾ – f log (e-4t) = log (1 – 4/3 f)

-4t = log(1 – 4/3 f)

dij = t = - ¼ -1 log(1 – 4/3 f)

UPGMA (unweighted pair group method using arithmetic averages)

Or the Average Linkage Method

Given two disjoint clusters Ci, Cj of sequences,

1

dij = ––––––––– {p Ci, q Cj}dpq

|Ci| |Cj|

Claim that if Ck = Ci Cj, then distance to another cluster Cl is:

dil |Ci| + djl |Cj|

dkl = ––––––––––––––

|Ci| + |Cj|

Proof

Ci,Cl dpq + Cj,Cl dpq

dkl = ––––––––––––––––

(|Ci| + |Cj|) |Cl|

|Ci|/(|Ci||Cl|) Ci,Cl dpq + |Cj|/(|Cj||Cl|) Cj,Cl dpq

= ––––––––––––––––––––––––––––––––––––

(|Ci| + |Cj|)

|Ci| dil + |Cj| djl

= –––––––––––––

(|Ci| + |Cj|)

1

4

Initialization:

Assign each xi into its own cluster Ci

Define one leaf per sequence, height 0

Iteration:

Find two clusters Ci, Cj s.t. dij is min

Let Ck = Ci Cj

Define node connecting Ci, Cj,

& place it at height dij/2

Delete Ci, Cj

Termination:

When two clusters i, j remain,

place root at height dij/2

3

5

2

5

2

3

1

4

4

3

2

1

v

w

x

y

z

Definition:

A distance function d(.,.) is ultrametric if for any three distances dij dik dij, it is true that

dij dik = dij

The Molecular Clock:

The evolutionary distance between species x and y is 2 the Earth time to reach the nearest common ancestor

That is, the molecular clock has constant rate in all species

The molecular clock results in ultrametric distances

years

1

4

2

3

5

Average Linkage is guaranteed to reconstruct correctly a binary tree with ultrametric distances

Proof: Exercise

5

1

4

2

3

Molecular clock: all species evolve at the same rate (Earth time)

However, certain species (e.g., mouse, rat) evolve much faster

Example where UPGMA messes up:

AL tree

Correct tree

3

2

1

3

4

4

2

1

d1,4

1

Given a tree, a distance measure is additive if the distance between any pair of leaves is the sum of lengths of edges connecting them

Given a tree T & additive distances dij, can uniquely reconstruct edge lengths:

- Find two neighboring leaves i, j, with common parent k
- Place parent node k at distance dkm = ½ (dim + djm – dij) from any node m

4

12

8

3

13

7

9

5

11

10

6

2

x

T

D

y

5

4

3

z

3

4

w

7

6

v

If we know T and D, but do not know the length of each leaf, we can reconstruct those lengths

x

T

D

y

z

w

v

D

x

T

y

z

a

w

D1

v

dax = ½ (dvx + dwx – dvw)

day = ½ (dvy + dwy – dvw)

daz = ½ (dvz + dwz – dvw)

D1

x

T

y

5

4

b

3

z

3

a

4

c

w

7

D2

6

d(a, c) = 3

d(b, c) = d(a, b) – d(a, c) = 3

d(c, z) = d(a, z) – d(a, c) = 7

d(b, x) = d(a, x) – d(a, b) = 5

d(b, y) = d(a, y) – d(a, b) = 4

d(a, w) = d(z, w) – d(a, z) = 4

d(a, v) = d(z, v) – d(a, z) = 6

Correct!!!

v

D3

- Guaranteed to produce the correct tree if distance is additive
- May produce a good tree even when distance is not additive
Step 1: Finding neighboring leaves

Define

Dij = dij – (ri + rj)

Where

1

ri = –––––k dik

|L| - 2

Claim: The above “magic trick” ensures that Dij is minimal iffi, j are neighbors

Proof: Very technical, please read Durbin et al.!

1

3

0.1

0.1

0.1

0.4

0.4

4

2

Initialization:

Define T to be the set of leaf nodes, one per sequence

Let L = T

Iteration:

Pick i, j s.t. Dij is minimal

Define a new node k, and set dkm = ½ (dim + djm – dij) for all m L

Add k to T, with edges of lengths dik = ½ (dij + ri – rj)

Remove i, j from L;

Add k to L

Termination:

When L consists of two nodes, i, j, and the edge between them of length dij

- One of the most popular methods:
- GIVEN multiple alignment
- FIND tree & history of substitutions explaining alignment
Idea:

Find the tree that explains the observed sequences with a minimal number of substitutions

Two computational subproblems:

- Find the parsimony cost of a given tree (easy)
- Search through all tree topologies (hard)

{A}

Final cost C = 1

{A}

{A, B}

Cost

C+=1

A

A

B

A

{B}

{A}

{A}

{A}

Given a tree, and an alignment column u

Label internal nodes to minimize the number of required substitutions

Initialization:

Set cost C = 0; k = 2N – 1

Iteration:

If k is a leaf, set Rk = { xk[u] }

If k is not a leaf,

Let i, j be the daughter nodes;

Set Rk = Ri Rj if intersection is nonempty

Set Rk = Ri Rj, and C += 1, if intersection is empty

Termination:

Minimal cost of tree for column u, = C

{B}

{A,B}

{A}

{B}

{A}

{A,B}

{A}

A

A

A

A

B

B

A

B

{A}

{A}

{A}

{A}

{B}

{B}

{A}

{B}

A more refined measure of evolution along a tree than parsimony

P(x1, x2, xroot | t1, t2) = P(xroot) P(x1 | t1, xroot) P(x2 | t2, xroot)

If we use Jukes-Cantor, for example, and x1 = xroot = A, x2 = C, t1 = t2 = 1,

= pA¼(1 + 3e-4α) ¼(1 – e-4α) = (¼)3(1 + 3e-4α)(1 – e-4α)

xroot

t1

t2

x1

x2

xroot

- If we know all internal labels xu,
P(x1, x2, …, xN, xN+1, …, x2N-1 | T, t) = P(xroot)jrootP(xj | xparent(j), tj, parent(j))

- Usually we don’t know the internal labels, therefore
P(x1, x2, …, xN | T, t) = xN+1 xN+2 … x2N-1P(x1, x2, …, x2N-1 | T, t)

xu

x2

xN

x1

Let P(Lk | a) denote the prob.

of all the leaves below node k,

given that the residue at k is a

To calculate P(x1, x2, …, xN | T, t)

Initialization:

Set k = 2N – 1

Iteration: Compute P(Lk | a) for all a

If k is a leaf node:

Set P(Lk | a) = 1(a = xk)

If k is not a leaf node:

1. Compute P(Li | b), P(Lj | b) for all b, for daughter nodes i, j

2. Set P(Lk | a) = b,c P(b | a, ti) P(Li | b) P(c | a, tj) P(Lj | c)

Termination:

Likelihood at this column = P(x1, x2, …, xN | T, t) = aP(L2N-1 | a)P(a)

Given M (ungapped) alignment columns of N sequences,

- Define likelihood of a tree:
L(T, t) = P(Data | T, t) = m=1…M P(x1m, …, xnm, T, t)

Maximum Likelihood Reconstruction:

- Given data X = (xij), find a topology T and length vector t that maximize likelihood L(T, t)

HUNDREDS of programs available!

http://evolution.genetics.washington.edu/phylip/software.html#methods

Some recommended programs:

- Discrete—Parsimony-based
- Rec-1-DCM3
http://www.cs.utexas.edu/users/tandy/mp.html

Tandy Warnow and colleagues

- Rec-1-DCM3
- Probabilistic
- SEMPHY
http://www.cs.huji.ac.il/labs/compbio/semphy/

Nir Friedman and colleagues

- SEMPHY