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Introduction to. Bioinformatics. Introduction to Bioinformatics. LECTURE 2: GENE FINDING * Chapter 2: All the sequence\'s men. 2.1 Human genome sweepstake In 2003 Lee Rowen (Institute Systems Biology, Seattle) wins GeneSweep, the betting pool for the number of human genes

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introduction to

Introduction to

Bioinformatics

introduction to bioinformatics

Introduction to Bioinformatics.

LECTURE 2: GENE FINDING

* Chapter 2: All the sequence\'s men

introduction to bioinformatics lecture 2 gene finding

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

introduction to bioinformatics lecture 2 gene finding1

2.1 Human genome sweepstake

  • Total bets in the GeneSweep:
  • 2000: $1
  • 2001: $5
  • 2002: $20
  • 2003: $1200

Introduction to BioinformaticsLECTURE 2: GENE FINDING

introduction to bioinformatics lecture 2 gene finding2

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

introduction to bioinformatics lecture 2 gene finding3

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

introduction to bioinformatics lecture 2 gene finding4

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

introduction to bioinformatics lecture 2 gene finding5

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

introduction to bioinformatics lecture 2 gene finding6

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

introduction to bioinformatics lecture 2 gene finding7

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
introduction to bioinformatics lecture 2 gene finding8

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
introduction to bioinformatics lecture 2 gene finding9

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.

slide18

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

slide21

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)

slide22

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.

slide26

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

slide27

Protein Structure:

primary structure

slide28

Protein

Structure:

secondary

Structure

a: Alpha-helix,

b: Beta-sheet

slide29

Protein Structure:

super-secondary Structure

introduction to bioinformatics lecture 2 gene finding12

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

introduction to bioinformatics lecture 2 section 2 3 gene annotation gene finding

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.
introduction to bioinformatics lecture 2 section 2 3 gene annotation gene finding1

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

introduction to bioinformatics lecture 2 section 2 3 gene annotation gene finding2

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).

introduction to bioinformatics lecture 2 section 2 3 gene annotation gene finding3

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

introduction to bioinformatics lecture 2 gene finding13

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.

introduction to bioinformatics lecture 2 gene finding14

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

slide43

Hypothesis

testing with H0

introduction to bioinformatics lecture 2 section 2 4 spurious signals

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
introduction to bioinformatics lecture 2 section 2 4 spurious signals1

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
introduction to bioinformatics lecture 2 section 2 4 spurious signals2

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

introduction to bioinformatics lecture 2 section 2 4 spurious signals3

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

introduction to bioinformatics lecture 2 section 2 4 spurious signals4

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
introduction to bioinformatics lecture 2 section 2 4 spurious signals5

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
introduction to bioinformatics lecture 2 section 2 4 spurious signals6

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.
introduction to bioinformatics lecture 2 section 2 4 spurious signals7

ORF length in Mycoplasma genitalium

  • H0 = 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

Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals

introduction to bioinformatics lecture 2 section 2 4 spurious signals8

Introduction to BioinformaticsLECTURE 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! *

introduction to bioinformatics lecture 2 section 2 4 spurious signals10

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

Introduction to BioinformaticsLECTURE 2: Section 2.4 Spurious signals

introduction to bioinformatics lecture 2 section 2 4 spurious signals11

Introduction to BioinformaticsLECTURE 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!
introduction to bioinformatics lecture 2 section 2 4 spurious signals12

Introduction to BioinformaticsLECTURE 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
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