Introduction to Bioinformatics 1 Introduction to Bioinformatics LECTURE

Introduction to Bioinformatics 1

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

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 1 Human genome sweepstake In 2003 Lee Rowen (Institute Systems Biology, Seattle) wins Gene. Sweep, 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 3

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 1 Human genome sweepstake Total bets in the Gene. Sweep: 2000: $1 2001: $5 2002: $20 2003: $1200 4

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 1 Human genome sweepstake The human genome counts 3. 3 billion bp … … but how to estimate the number of human genes ? 5

Introduction to Bioinformatics LECTURE 2: GENE FINDING 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 6

Introduction to Bioinformatics LECTURE 2: GENE FINDING 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. 7

Introduction to Bioinformatics LECTURE 2: GENE FINDING 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 … 8

Introduction to Bioinformatics LECTURE 2: GENE FINDING 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. 9

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 2 Genes and Proteins 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 10

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

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 2 Genes and Proteins 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. 12

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DNA under electron microscope 14

3 D model of a section of the DNA molecule 15

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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 18

Genetic code 19

Replication of DNA 20

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 m. RNA 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) 21

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 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. 22

Genetic code: TRANSCRIPTION DNA → RNA 23

Genetic code: TRANSLATION RNA → protein 24

Genetic code: exons/introns 25

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 Leu L Arg R Lys K Asn N Met M Asp D Phe F Cys C Pro P. . . Start Stop GCU, GCC, GCA, GCG UUA, UUG, CUU, CUC, CUA, CUG CGU, CGC, CGA, CGG, AGA, AGG AAA, AAG AAU, AAC AUG GAU, GAC UUU, UUC UGU, UGC CCU, CCC, CCA, CCG AUG, GUG UAG, UGA, UAA 26

Protein Structure: primary structure 27

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

Protein Structure: super-secondary Structure 29

Protein Structure = protein function: 30

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

Introduction to Bioinformatics LECTURE 2: GENE FINDING intron - exon 32

Introduction to Bioinformatics LECTURE 2: GENE FINDING 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 33

Change points in Labda-phage 34

Introduction to Bioinformatics LECTURE 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. 35

Introduction to Bioinformatics LECTURE 2: Section 2. 3 Gene annotation: gene finding READING FRAMES The DNA is translated per codon = nucleotide-triplet. The sequence: …ACGTACGTACGT… 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-… 36

Introduction to Bioinformatics LECTURE 2: Section 2. 3 Gene annotation: gene finding OPEN READING FRAMES: ORF An open reading frame or 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 startcode sequence (initiation codon) and the stop -code sequence (termination codon). 37

Introduction to Bioinformatics LECTURE 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 38

Introduction to Bioinformatics LECTURE 2: Section 2. 3 Gene annotation: gene finding OPEN READING FRAMES: ORF 39

Introduction to Bioinformatics LECTURE 2: Section 2. 3 Gene annotation: gene finding Introns and Exons 40

Introduction to Bioinformatics LECTURE 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. 41

Introduction to Bioinformatics LECTURE 2: GENE FINDING 2. 4 Detecting spurious signals • a pattern in DNA can arise from pure chance • hypothesis testing with null-hypothesis H 0 • test statistics • p-value = probability-value • significance level • type I-error (FP = False Positive) of H 0 • type II-error (FN = False Negative) of H 0 42

Hypothesis testing with H 0 43

Introduction to Bioinformatics LECTURE 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 = Methionine • 3 stop-codons TAA, TAG, and TGA • a priori probability non stop-codon = 61/64 44

Introduction to Bioinformatics LECTURE 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 45

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals 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 46

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals 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 47

Introduction to Bioinformatics LECTURE 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 48

Introduction to Bioinformatics LECTURE 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 • H 0 = randomized DNA sequence 49

Introduction to Bioinformatics LECTURE 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. 50

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals ORF length in Mycoplasma genitalium • H 0 = randomized DNA sequence • Keep ORFs in original DNA that are longer than (most) ORFs in the randomized DNA • max(ORF-length) in random seq. = 402 bp • in original DNA 326 ORFs longer than 402 bp • Good estimate: M. genitalium has 470 genes 51

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals Here follows the ORF-length distribution in the original and the randomized DNA * Note the long tail of the real DNA! * 52

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals 53

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals Example 2: ORF length in Haemophilus influenzae • threshold = max(ORF-length) in random seq. = 537 bp • in original DNA 1182 ORFs longer than 537 bp • this is about the real number of genes: 1428 54

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals Problems with multiple testing • the α-significance (e. g. 5%) represents the false positive rate of one single test • If we conduct – say – 100 tests this means that 5 false positives are expected • therefore, if 5 significant genes were found out of 100 tests with α = 0. 05 this does not mean anything biologically! 55

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals Problems with multiple testing • The ORFs found in this way are only candidate genes! • it is not clear at this point whether these ORFS are actually translated to proteins! • Using sequence alignment (chapter 3) the case for a candidate gene can be tightened 56

Introduction to Bioinformatics LECTURE 2: Section 2. 4 Spurious signals 57

END of LECTURE 2 58

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