Classical Statistics and Scoring of Alignments

Consider a probe of length l and a database of total length m.

How many subsequences of length n are there in l?

l-n+1

How many subsequences of length n are there in m?

m-n+1

SO THERE ARE EXACTLY

(l-m+1) (m-n+1) potential matches of length m between the probe and the database.

Probability: Coin Toss• There are only two outcomes

possible for a coin toss: heads and tails.

• For a fair coin, the probability of getting heads on any one toss is 0.5; p = 0.5

• The probability of getting tails on any one toss is 0.5; q = (1-p) = 0.5

• If you do lots of tosses, you expect heads and tails approximately half the time; the probability for one toss has meaning over many tosses, too.

• But if you do a finite number of tosses, you have a small chance of getting exactly half heads and and half tails; most of the time you’ll get almost half and half

• If you do many sets of tosses, you can find out empirically what your chances are of getting say 4 heads out of 10 tosses; P(k=4 heads out of N=10 tosses)

• Or you can figure it out mathematically by thinking about how many ways you could toss a coin 10 times and get 4 heads: combinations formula

Ways to get heads

ways to get dif #’s heads1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1

#trials123456

Binomial theorem• The combinations in the triangle come from the

binomial theorem: since we have only two outcomes, you can write the probability of all outcomes as (p + q)n, where n = the number of trials

(p+q)n = pn + n! p(n-1) q + n! p(n-2) q2…..+qn

i!j! i!j!• where i = the number of heads and j = the number

of tails• E.g., what’s the probability of 4 out of 6 heads?

n = 6, i = 4, j = 2: 6!/4!2!(0.5)4(0.5)2

• You can make a plot of all the possible probability scores against the number of heads you get

2 tosses

0123

0 1 2

Number of heads

Way

s of

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head

s

Series1

10 tosses

050

100150200250300

0 1 2 3 4 5 6 7 8 9 10

Number of heads

Way

s of

obt

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ng

head

sSeries1

• Notice that as n increases, the binomial distribution approaches a bell-shaped curve: the gaussian distribution

• The gaussian distribution has the familiar mean = x and standard deviation = s for common statistical analyses: t-test, etc.

Gaussian Distribution

What about statistics for alignments?• What we want is a way to find out the

significance of the score we assign to alignments – either the similarity search or the alignments that we will be talking about next week

• We need the probability of finding a particular score

Probability of finding a score – simplified example• Let’s consider sequences with an alphabet

of just 2 letters, A and B• Then p = probability of a match to A, and q

= probability of match to B (or no match to A); this is just like tossing a coin and p = q

• For random sequences, we expect a total score of ½ N (N = length of sequences)

In-class exercise I

• Assume identity scoring, a gapless alignment and a sequence alphabet of A and B. What is the probability of two random sequences of length 4 being identical?

• What is the probability of 3 out of 4 characters identical?

Probability of finding a score – almost realistic• So what if we let there be 4 characters (as

in the nucleic acid case)? We’ll keep identity scoring and no gaps

• Now p(match) = 0.25, and q(no match) = 0.75

• Since there’s still only 2 outcomes we can use the binomial theorem to calculate the distribution of scores just as before (this is like a “loaded” coin)

In-class exercise II

• Assume an alphabet of 4 letters A,G,C,T; a gapless alignment, and identity scoring. What is the probability of two random sequences of length 4 being identical?

• What is the probability of 3 out of 4 characters identical?

Distribution for 4-character, identity scoring• You can convince

yourself that the probability distributions for scores of alignments using a 4-character alphabet are not gaussian

• This means you can’t use t-tests, etc., to test significance of alignments

N = 2

00.10.20.30.40.50.6

0 1 2

number of identities

prob

abili

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f id

entit

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N = 5

00.10.20.30.40.5

0 1 2 3 4 5

number of identities

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Alignment statistics

• No theory for global alignments• Can use randomized sets• Statistics for ungapped local

alignments well-understood

FASTA statistics• FASTA generates a random set of proteins

sequences and reports the frequency of init1’s (modified hits) against the randomized set

• The tail of an actual search distribution usually contains more hits than are expected from the randomized set; some of those extra hits will represent homologous proteins

• Some of the other hits in the areas of higher probability of random hits will also represent homologous proteins

Karlin-Altschul statistics• BLAST initially generates ungapped alignments• Problem: How many hits of score S are generated using a probe

of length m vs. a database of total sequence length n? • Answer: for high scores the number of hits is determined by the

extreme value distribution:• E=Kmn exp(-S)

•

• So as the probe length and the data base size go up, the expected number of hits go up linearly. As the cutoff score is raised, the number decreases exponentially

and are parameters which depend on factors such as the scoring matrix used

Statistics of alignment• Consider a probe of length l and a database

of total length n. How many subsequences of length m are there in l?

• l-m+1• How many subsequences of length m are

there in n?• n-m+1• SO THERE ARE EXACTLY (l-m+1) (n-m+1)

potential matches of length m between the probe and the database.

• The number of matches expected is just the product of the number of potential matches and the probability of a pair matching:– Expectation value = (l-m+1) (n-m+1) pm

• where p is the probability of a match at each position; if each outcome (letter in the alphabet) is of equal probability, p is the inverse of the number of outcomes (letters).

• Using gapless alignments and identity scoring, the score of a particular comparison between a probe and a database is just the number of matches in the region considered.

• Suppose we extend any hits in both directions until we reach a mismatch; the score obtained will be just the number of consecutive hits m.

• We seek the number of random hits expected of score at least S=M. The expectation value is just

• (l-M+1) (n-M+1) pM

• for DNA (four bases) and large l and n, this is ~ n l 4-S

Statistics of alignment• Now suppose we modify our extension

rules so that to get score S we need to get only i out of m matches, with j misses allowed. The probability of i hits out of m is just

m! pi qj i!j!

• We can find the expectation value by multiplying by the number of potential m base comparisons. for DNA (four bases) and large l and n, this is

~ n l m! pi qj

i!j!

• The expectation value for the number of hits of score S in a database of length n with a probe of length l is

~ n l m! pi qj i!j!

• if it takes i hits out of i+j elements to reach S.• Using identity scoring, S=i and for q>>p and small j • this is close to ~ n l m! pi i!j!

• The expression ~ n l m! pi i!j! or ~ n l m! (1/p)-S

i!j!

• is a reasonable approximation for DNA sequences only for j=0 or 1, but for amino acid sequences q is large enough to tolerate several mismatches.

• More generally,

~ n l m! (1/p)-S i!j!

• is equivalent to K n l (1/p )-S , where K contains both the binomial coefficients and the terms in q.

• For non-identity scoring, we have K n l exp (-S)• which is the general form of the extreme value

distribution; is a scale factor for the score system.

Algorithms

• An algorithm is an exact set of instructions for carrying out a task using a limited menu of commands

• Algorithms are most often associated with computer programs, but they can also be used by humans to solve problems

• Small number of operations defined in the programming language compared to, say, words or phrases in English

Algorithms in minimal C• Instructions:• main () means program is beginning• Statements have operators:

• =, +, -, *, ++ (increment), % (modulus), () (separating terms)

• Statements are not algebra!»x = x + 1

• Statements end in ;• Statements are about variables

• Variables start with characters (i, x, b5)• Variables need to be declared so the program can

use them• float x (float means any real number); int I (int

means an integer); char b (char means character) ;• Arrays are a list of data that can be associated with

a variable; arrays run from 0 to n-1 where n is the number in brackets

• float x[]; int a[50]; char b[500]; float x [5][4];

• Variables need to be initialized (given a starting value) if they are going to be incremented

• Loops and conditions– For (conditions) { statements; } means do statement operations

while conditions are met– If (conditions) { statements; } else {statements; } means do first

statements while conditions are met, and when conditions are not met do second statements – this is a one-time thing

– While (conditions) { statements; } means keep doing statements while conditions are met, and as soon as they are not met, stop – this is continuous while conditions are met

– {continue; } go to next iteration of current loop skipping intervening statements (curly brackets)

– { break; } means get out of the current loop

• Loops have curly brackets around them– Loops can be nested– Loops are operations that continue until something

changes• Arithmetic: sqrt (x), exp (x), log (x), etc.• Logical operators: == (equals), != (not equal), >, <,

<=, >=, && (boolean AND), | | (boolean OR)• Real programs need an input and an output function;

input reads data into the program, and output records the results of the operations in the program. You will not need to worry about the input and output parts of the algorithms, we will provide them to you.

• Simple algorithm to find the average of N terms x(n)Main() { starts programFloat s, av, x[ ]; declares variablesint i, N; declares variables“Input” (x, N); gets datas = 0; initializes s (value 0)for (i = 0; i<N; i++) { loop and statements = s + x[i];}av = s/N arithmetic statement“Output” (av) average reported} end program

• How a simple loop works:for (i = 0; i<N; i++); {s = s + x[i];}

• Initially, we had set S=0, so the first time through the loop i=0 and S is set equal to X(0). Then we increment i by one, so I = 1, and s = x(0) + x(1). This process continues until i = N, at which point s =x(0) + x(1) + … x(N-1); in other words, s is the sum of all the elements of the array x, which is the input.

In-class exercise• Use the following data and the algorithm for the

average of N terms to find the average: 3,5,8,9Main() {Float s, av, x[ ];int i, N;“Input” (x, N);s = 0;for (i = 0; i<N; i++) {s = s + x[i];}av = s/N “Output” (av)}

Nested Loops• Often we need to cover a space defined by

multiple variables. Suppose we have a matrix with elements X(i,j):

• X(1,1) X(1,2) X(1,3) …… X (1,n)

• X(2,1) X(2,2) X(2,3) …….X (2,n) • .• .• X(m,1)

• To write the elements of X(i,j), we can use two nested loops, one indexed to i and the other to j:

• for (i=0, i<m , i++) {• for (j=0, j<n, j++) {• “output” x(i,j)• } • }

Outside loop

Inside loop

Another algorithm: simple sequence similarity search• Probe sequence w of length J;

elements w(j), j = 1,2, … J• Library sequence m of length I;

elements m(i), i = 1,2, … I• Problem: find all exact matches of for

the whole length of w in m

Search algorithm• Main () {• int i, j, I, J; char w[ ], m[ ];• “Input” (w, m, I, J);• i = 0, j = 0;• for (i=0; i<I; i ++){

for (j=0; j<J; j ++){if (w [ j ] != m [ i + j ]) {break;}if (j==J) {output i}

}}

}

Outer loop

Inner loop

Search algorithm• How it works: outside loop in i

controls the start of the sequence comparison in m: i is the # of the sequence element in m compared to the first element in w.

• inside loop in j increments the index of both w and m, allowing the comparison of successive elements for each starting point fixed by i.

Sliding Box Model• i controls the region of m selected for

comparison• W1 W2...

M1 M2 --------------------------------------------MI

M1 M2 -----------------------MI

Above: box represents the position of w on m for i=1, so that W1 gets compared to M1, W2 to M2, etc. as j increases

W1 W2...

Below: box represents the position of w on m for i=2, the second time through the outer loop. As j increases W1 is compared to M2, W2 to M3, etc.

In-class exercise

• W: AB• M : QLABCD• Use the simple sequence similarity

search algorithm to find matches between w and m