Introduction to

1. Introduction to Bioinformatics

2. Introduction to Bioinformatics. LECTURE 2: GENE FINDING * Chapter 2: All the sequence's men

3. 2.1 Human genome sweepstake • In 2003 Lee Rowen (Institute Systems Biology, Seattle) wins GeneSweep, the betting pool for the number of human genes • Her price: $1200 and a signed copy of Watson’s The Double Helix • Her guess: 25.947 genes Introduction to BioinformaticsLECTURE 2: GENE FINDING

4. 2.1 Human genome sweepstake • Total bets in the GeneSweep: • 2000: $1 • 2001: $5 • 2002: $20 • 2003: $1200 Introduction to BioinformaticsLECTURE 2: GENE FINDING

5. 2.1 Human genome sweepstake • The human genome counts 3.3 billion bp … • … but how to estimate the number of human genes ? Introduction to BioinformaticsLECTURE 2: GENE FINDING

6. 2.1 Human genome sweepstake • 1990: estimate human genes ~300,000 • 1995: estimate human genes ~100,000 • 2000: estimate human genes ~30,000 • 2004: estimate human genes ~25,000 • 2008: estimate human genes ~22,000 • 2009: known human genes :18,308 Introduction to BioinformaticsLECTURE 2: GENE FINDING

7. 2.1 Human genome sweepstake • The pie shows that we’re now down to just 18,308 genes. That’s over 8,000 genes fewer than six years ago. • Many sequences that once looked like full-fledged genes, capable of generating a protein, now don’t make the grade. Some genes turned out to be pseudogenes – vestiges of genes that once worked but have been since wrecked by mutations. • In other cases, DNA segments that appeared to be parts of separate genes have turned out to be part of the same gene. Introduction to BioinformaticsLECTURE 2: GENE FINDING

8. 2.1 Human genome sweepstake • In this lecture we will try to estimate the number of genes in a given DNA string • First however some biology … Introduction to BioinformaticsLECTURE 2: GENE FINDING

9. The human genome is stored on 23 chromosome pairs. 22 of these are autosomal chromosome pairs, while the remaining pair is sex-determining. The haploid human genome occupies a total of just over 3 billion DNA base pairs. The Human Genome Project produced a reference sequence of the euchromatic human genome, which is used worldwide in biomedical sciences. The haploid human genome contains an estimated 22,000 protein-coding genes, far fewer than had been expected before its sequencing. In fact, only about 1.5% of the genome codes for proteins, while the rest consists of RNA genes, regulatory sequences, introns and (controversially) "junk" DNA. Introduction to BioinformaticsLECTURE 2: GENE FINDING

10. 2.2 Genes and Proteins Introduction to BioinformaticsLECTURE 2: GENE FINDING • CENTRAL IDEA: • Genes code for proteins • There are fixed codes for START and STOP • We can use those to look for DNA words: [ START | n ×<triplet> | STOP ] • Such DNA words are s

11. 2.2 Genes and Proteins Introduction to BioinformaticsLECTURE 2: GENE FINDING • DRAW-BACKS: • Only candidate-genes are found • Most of the DNA is non-coding “junk DNA” (???... ) • Where to start reading … and in what direction? • Looooooooooooong computation times

12. 2.2 Genes and Proteins Introduction to BioinformaticsLECTURE 2: GENE FINDING DNA Deoxyribonucleic acid (DNA) is a nucleic acid that contains the genetic instructions specifying the biological development of all cellular forms of life (and most viruses). DNA is a long polymer of nucleotides and encodes the sequence of the amino acid residues in proteins using the genetic code, a triplet code of nucleotides.

14. DNA under electron microscope

15. 3D model of a section of the DNA molecule

18. Genetic code The genetic code is a set of rules that maps DNA sequences to proteins in the living cell, and is employed in the process of protein synthesis. Nearly all living things use the same genetic code, called the standard genetic code, although a few organisms use minor variations of the standard code. Fundamental code in DNA: {x(i)|i=1..N,x(i) in {C,A,T,G}} Human: N = 3.3 billion

19. Genetic code

20. Replication of DNA

21. Genetic code: TRANSCRIPTION DNA → RNA Transcription is the process through which a DNA sequence is enzymatically copied by an RNA polymerase to produce a complementary RNA. Or, in other words, the transfer of genetic information from DNA into RNA. In the case of protein-encoding DNA, transcription is the beginning of the process that ultimately leads to the translation of the genetic code (via the mRNA intermediate) into a functional peptide or protein. Transcription has some proofreading mechanisms, but they are fewer and less effective than the controls for DNA; therefore, transcription has a lower copying fidelity than DNA replication. Like DNA replication, transcription proceeds in the 5' → 3' direction (ie the old polymer is read in the 3' → 5' direction and the new, complementary fragments are generated in the 5' → 3' direction). in RNA Thymine (T) → Uracil (U)

22. Directionality: 5' to 3' direction Directionality, in molecular biology, refers to the end-to-end chemical orientation of a single strand of nucleic acid. The chemical convention of naming carbon atoms in the nucleotide sugar-ring numerically gives rise to a 5' end and a 3' end (usually pronounced "five prime end" and "three prime end"). The relative positions of structures along a strand of nucleic acid, including genes, transcription factors, and polymerases are usually noted as being either upstream (towards the 5' end) or downstream (towards the 3' end). The importance of having this naming convention lies in the fact that nucleic acids can only be synthesized in vivo in a 5' to 3' direction, as the polymerase used to assemble new strands must attach a new nucleotide to the 3' hydroxyl (-OH) group via a phosphodiester bond. By convention, single strands of DNA and RNA sequences are written in 5' to 3' direction.

23. Genetic code: TRANSCRIPTION DNA → RNA

24. Genetic code: TRANSLATION RNA → protein

25. Genetic code: exons/introns

26. Genetic code: TRANSLATION DNA-triplet → RNA-triplet = codon → amino acid RNA codon table There are 20 standard amino acids used in proteins, here are some of the RNA-codons that code for each amino acid. Ala A GCU, GCC, GCA, GCG Leu L UUA, UUG, CUU, CUC, CUA, CUG Arg R CGU, CGC, CGA, CGG, AGA, AGG Lys K AAA, AAG Asn N AAU, AAC Met M AUG Asp D GAU, GAC Phe F UUU, UUC Cys C UGU, UGC Pro P CCU, CCC, CCA, CCG ... Start AUG, GUG Stop UAG, UGA, UAA

27. Protein Structure: primary structure

28. Protein Structure: secondary Structure a: Alpha-helix, b: Beta-sheet

29. Protein Structure: super-secondary Structure

30. Protein Structure = protein function:

31. Standard Genetic Code Introduction to BioinformaticsLECTURE 2: GENE FINDING note: RNA ‘U’ ~ DNA ‘A’

32. Introduction to BioinformaticsLECTURE 2: GENE FINDING intron - exon

33. 2.3 Gene annotation: gene finding • Statistical analysis (eg GC-content) can identify different regions on a DNA strand • ab initio methods (=statistical analysis) • Markov sequence model Introduction to BioinformaticsLECTURE 2: GENE FINDING

34. Change points in Labda-phage

35. Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding • Ab initio methods suffice for finding genes on Prokaryotic DNA • For more complex Eukaryotic DNA we need sequence alignment methods and Markov sequence models.

36. READING FRAMES The DNA is translated per codon = nucleotide-triplet. The sequence: …ACGTACGTACGTACGTACGT… Can thus be read as: …-ACG-TAC-GTA-CGT-ACG-TAC-GT… or: …A-CGT-ACG-TAC-GTA-CGT-ACG-T… or: …AC-GTA-CGT-ACG-TAC-GTA-CGT-… Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding

37. Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding OPEN READING FRAMES: ORF An open reading frameor ORF is a portion of an organism's genome which contains a sequence of bases that could potentially encode a protein In a gene, ORFs are located between the start-code sequence (initiation codon) and the stop-code sequence (termination codon).

38. Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding OPEN READING FRAMES: ORF As we saw, we can distinguish 3 possible ORFs on one strand (5’ to 3’). On the complementary strand (5’ to 3’) we can also look for 3 possibiloties – but these can be reconstructed from the first strand. So, we can distinguish 6 possible ORFs

39. OPEN READING FRAMES: ORF Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding

40. Introns and Exons Introduction to BioinformaticsLECTURE 2: Section 2.3 Gene annotation: gene finding

41. Introduction to BioinformaticsLECTURE 2: GENE FINDING Algorithm 2.1: ORF-finder Given a DNA sequence s and a positive integer k, for each possible reading frame decompose the sequence into triplets, and find all stretches of triplets starting with a START-codon and ending with a STOP-codon. Repeat also for the reverse compliment of s. The Output consists of all ORFS longer than or equal to the prefixed threshold k.

42. 2.4 Detecting spurious signals • a pattern in DNA can arise from pure chance • hypothesis testing with null-hypothesis H0 • test statistics • p-value = probability-value • significance level  • type I-error (FP = False Positive) of H0 • type II-error (FN = False Negative) of H0 Introduction to BioinformaticsLECTURE 2: GENE FINDING

43. Hypothesis testing with H0

44. Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals • Computing a p-value for ORFs • Translation table : triplet → aminoacid (AA) • 64 possible triplets (for 20 AAs) • 1 start-codon ATG = M = Met = Methionine • 3 stop-codons TAA, TAG, and TGA • a priori probability non stop-codon = 61/64

45. Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals • Computing a p-value for ORFs • a priori probability non stop-codon = 61/64 • P(k non-stopcodons) = (61/64)k • 95%-significance : p = 0.05 • (61/64)k ≈ 0.05  k≈ 62  +/- 64 codons • 99%-significance : p = 0.01 • (61/64)k ≈ 0.01  k≈ 100  +/- 102 codons

46. Non-uniform codon distribution • Pstop = P(TAA) + P(TAG) + P(TGA) • P(k non-stop codons) = (1 - Pstop)k • For a significance-level α we need k* codons with: (1 - Pstop)k*= α Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals

47. Randomization tests • Generate a string with the same statistical properties of the original data • Per nucleotide? per triplet? per … ? • p-value: find the rank of observed test statistic in null distribution: if its percentile is less than α then it is significant Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals

48. Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals • Randomization tests • Another method is bootstrapping • No permutation but sampling with replacement • Again: per nucleotide? per triplet? per … ? • p-value: find the rank of observed test statistic in null distribution: if its percentile is less than α then it is significant

49. Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals • Example: ORF length in Mycoplasma genitalium • original DNA sequence: 11,922 ORFs • single-nucleotide permutation test = multinomial distribution • permute, search ORFs, record their length • randomized DNA sequence: 17,367 ORFs • H0 = randomized DNA sequence

50. Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals • ORF length in Mycoplasma genitalium • This approach does not identify short genes • Smaller threshold for ORF-length: the upper 5% of randomized DNA • In original DNA 1520 ORFs in this upper 5% • Many FALSE POSITIVES but still better than the original 11,922 in the DNA of M.gen.