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Statistics methods in bioinformatics
Yiannis Kourmpetis
Eric Boer
29 Jan 2007
Biometris
quantitative methods in life and earth sciences
Statistics methods in bioinformatics
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The analysis of One DNA Sequence
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5.2 – Modeling DNA (Eric)
5.3 – Modeling signals in DNA (Eric)
5.4 – Long Repeats (Eric)
5.6 – The analysis of patterns
• 5.7 – Overlaps counted (Yiannis)
• 5.8 – Overlaps not counted (Yiannis)
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5.9 – Motifs (Yiannis)
5.1 – Shotgun sequencing (Eric if time left)
Biometris
quantitative methods in life and earth sciences
Modeling DNA
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Hypothesis testing (Paragraph 3.4)
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Null hypothesis: p = 0.25
Alternative hypothesis: p > 0.25
Y = number of matches
If null hypothesis is correct, Y ~ Binomial(26, 0.25)
Biometris
quantitative methods in life and earth sciences
Modeling DNA
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Hypothesis testing (Paragraph 3.4)
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Calculating P-value:
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11 Matches, Y = 11; P-value = P(Y ≥ 11) = 0.04
Compare 0.04 with Type I error of testing:
• Type I error = 0.05; rejecting the null hypothesis
• Type I error = 0.01; not rejecting the null hypothesis
Biometris
quantitative methods in life and earth sciences
Modeling DNA
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DNA: long sequence of nucleotides
Test for independence (Table 5.4)
Biometris
quantitative methods in life and earth sciences
Modeling DNA

H0:
no association in nucleotide present
and preceding site
there is an association in nucleotide
present and preceding site
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Ha:
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See paragraph 3.5.5; Chi-square test
Biometris
quantitative methods in life and earth sciences
Modeling DNA
Chi-square test
Biometris
quantitative methods in life and earth sciences
Modeling signals in DNA
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Signal: short sequence of DNA having a specific
purpose
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Signals also occur in nonfunctional DNA
Assumption all signals have the same length n
Biometris
quantitative methods in life and earth sciences
Modeling signals in DNA

Signal probabilities; Weight matrices: independence
Signal length = 5
Prob(s | M)
For example:
s = aacct
Prob(s | M) = 0.33 ×…
Biometris
quantitative methods in life and earth sciences
Modeling signals in DNA
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Weight matrices: test of independence
Similar as Table 5.4
Position 1 in
the signal
Perform test for
every combination
Biometris
quantitative methods in life and earth sciences
Position 2 in
the signal
Modeling signals in DNA
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Markov Dependencies:
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If the nucleotides at the sites in a signal are not
indepent  It can be modelled by first or higher order
Markov dependencies (chapter 4).
By higher order dependencies, important to know
the most informative dependencies in our model:
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Maximal Dependence decomposition
Biometris
quantitative methods in life and earth sciences
Modeling signals in DNA
Pos. 1
Maximal Dependence decomposition
Biometris
quantitative methods in life and earth sciences
Pos. 2
Modeling signals in DNA
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Maximal Dependence decomposition
Row with highest sum, indication position with
highest influence.
Biometris
quantitative methods in life and earth sciences
Modeling signals in DNA

Position 4 greatest influence:
• The model M then consists of the distribution:
– {pa, pg, pc, pt}
– {Ma, Mg, Mc, Mt}
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If nucleotide x occurs in position 4, Matrix Mx is used
to assign probability pk, k = 1,2,3, 5.
Prob(s | M) = px × p1 × p2 × p3 × p5
Biometris
quantitative methods in life and earth sciences
Long Repeats
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Repeats of a nucleotide: for example cgtataaaaaagg
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Test of randomness
Suppose:
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N is length DNA sequence
P(a) = p
Occurrence a is called: “succes”
Occurrence other nucleotide at any site: “failure”
Apply geometric distribution
Biometris
quantitative methods in life and earth sciences
Long Repeats
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Apply geometric distribution:
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Sequence of success (for example nucleotide a)
Prob(Y = y) = (1-p) py,
y = 0,1,2,…..
y = 0,
y = 1,
y = 2,
for example
for example
for example
(a)c
(a)ag
(a)aag
Biometris
quantitative methods in life and earth sciences
Prob = 1-p
Prob = (1-p) p
Prob = (1-p) p2
Long Repeats
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Suppose that we have Y1,….,Yn independent random
variables:
In general: Ymax is the maximum of n discrete
independent variables, given a common cumulative
distribution function FY(y)
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Prob(Ymax ≥ y) =1- (FY(y-1))n
For geometric distribution:
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Prob(Ymax ≥ y) =1- (1-py)n
Biometris
quantitative methods in life and earth sciences
Long Repeats
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For geometric distribution:
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Prob(Ymax ≥ y) =1- (1-py)n
(5.13)
P-value of any observed value ymax can be calculated by
5.13, but which number should be n (number of
sequences of successes of length 0 or more)
Any sequences of success most be precede by a failure
Under
H0
(randomness),
(1-p)×N
failures,
so
approximately n = (1-p)N
Biometris
quantitative methods in life and earth sciences
Long Repeats
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So, p-value:
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Suppose: N = 100.000, p = 0.25 and the observed ymax of
the longest repeated sequence (ymax) is 10.
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P-value ~ =1- (1-pymax)(1-p)N
P-value is equal to 0.0690.  H0 not rejected
Suppose ymax = 12; p-value = 0.0045  H0 rejected
Approximation possible:
Biometris
quantitative methods in life and earth sciences
Long repeats
Parameters for normal distribution
Euler’s constant ≈ 0.5772
Long repeats
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Now, looking at runs of any nucleotide.
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p = 0.25 (probability that any nucleotide arises at any site)
Mean and variance longest repeated sequence:
Mean is 1 higher
as previous slide
Variance is the
equal
Long repeats
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Suppose the probability (p*) of one nucleotide is bigger than
others:
When n is large intuition tells us that the longest sequence will
be of the nucleotide with the largest probability
r =2, for example pa = 0.3, pc = 0.3, pg = 0.2 pt = 0.2
Shot sequencing
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Contigs
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N = 17 fragments, 7 contigs, overlapping fragments
N fragments of length L and full length of DNA is G (G is chosen
large compared to L)
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Coverage a = N L / G
Fragments are random placed on G, left-hand ends uniform
distribution over [0,G]
Shot sequencing
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Binomial and poisson distribution:
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Probability that left-hand side is in interval (x, x+h) is equal
to h/G.
Number of fragments in this interval ~ Bin(N, h/G)
When N is large and h is small poisson distribution can be
used as approximation (N > 20 and p < 0.05) with mean
(intensity) equal to Nh/G
The number Y of fragments whose left-hand end is located
in a interval of length L left to randomly chosen point is
Poisson distributed with parameter NL/G = a
Probability that at least one fragment arises in interval:
• 1- Prob(Y=0) = 1 – e-a
Shot sequencing
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Mean proportion of the genome covered by contigs:
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This is equal 1 – e-a
Shot sequencing
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Mean number of contigs:
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Each contig has a unique rightmost fragment
N fragments
Probability that no other fragment has its left-hand end point on
fragment in question
Mean number of contigs = N e-a
N small, small number of contigs
N large, large number of long contigs
Maximum at a=1, 1X coverage
Shot sequencing
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Mean contig size: