1. Bioinformatics: Buzzword or Discipline (???)

2. Outline of the course
Analysis of one DNA sequence: Shotgun sequencing, Markov-Chain modeling, patterns and repeats.
Analysis of multiple DNA or protein sequences: Dynamic programming alignments, substitution matrices.
BLAST: Algorithm for sequence retrieval and comparison.
Refresher on Markov Chains: Capsule theory, Markov-Chain Monte Carlo algorithms.
Hidden Markov Models: Viterbi Algorithm and its applications.
Evolutionary Models: Models of nucleotide mutation and substitution, recombination and genetic drift, with applications to genome evolution and gene mapping.
Molecular phylogenetics (tree making): distance matrix, maximum likelihood and parsimony.
Special topics: Gene and protein networks, analysis of DNA-microarray data, …

3. 30,000
Genes make up
only 3%
of the genome
BCM-
HGSC

4. Genome Sizes Human 3.0 x 109 base pairs Mouse 3.0 x 109
Drosophila x 108
Worm x 108
Dictyostelium x 107
Yeast x 107
Bacteria x 106

5. Shotgun Sequencing
High Accuracy Sequence:
< 1 error/ 10,000 bases

6. The Human Genome: 3 Billion Base Pairs Whole Genome Shotgun Strategy
3 billion bases
Libraries of clones
3kb, 10kb, 50kb base pairs
DNA sequence reads
500 bases each
AGGCTCACTG
BCM-
HGSC

7. Statistical issues in shotgun strategy
Model for the random fragments: Binomial/Poisson process
Coverage of sequence by random fragments
Mean number of contigs
Mean size of contigs
Coverage by anchored contigs

8. Binomial/Poisson Process
N fragments, of length L each, randomly scattered in the interval of length G.
Coverage a = NL/G
Contig: Union of overlapping fragments. We want to have them cover as much of G as possible.
Pr[#frags with left end in (x, x-h) = k] “is” binomial(N,h/G) or approximately Poisson(Nh/G) (when?).

9. Mean number of contigs
E[#contigs] = N  Pr[a frag is rightmost in a contig]
= N  Pr[frag does not include the left end of any other frag]
= N  exp(- NL/G) = (aG/L)  exp(- a)
L = 800
G = 100,000

10. Mean contig size
E[S] = E[#frags-1] E[inter-epoch distance] + L

11. Mean contig size
E(S) a

12. Number of anchored contigs
#anchors = M
#frags = N
a = NL/G
b = ML/G
E[#anchored contigs]
=Nb [exp(-a)-exp(-b)]/(b-a)

13. Conclusions
Expected number of contigs first increases, then decreases with coverage.
Expected size of contig increases with coverage.
Expected number of anchored contigs first increases then decreases with anchor density .
Attention: Computations do not involve boundary effects.