cse182 l12
play

CSE182-L12 Gene Finding Quiz Who are these people, and what is - PowerPoint PPT Presentation

CSE182-L12 Gene Finding Quiz Who are these people, and what is the occasion? De novo Gene prediction: Summary Various signals distinguish coding regions from non-coding HMMs are a reasonable model for Gene structures, and provide


  1. CSE182-L12 Gene Finding

  2. Quiz • Who are these people, and what is the occasion?

  3. De novo Gene prediction: Summary • Various signals distinguish coding regions from non-coding • HMMs are a reasonable model for Gene structures, and provide a uniform method for combining various signals. • Further improvement may come from improved signal detection

  4. How many genes do we have? Nature Science

  5. Alternative splicing

  6. Comparative methods • Gene prediction is harder with alternative splicing. • One approach might be to use comparative methods to detect genes • Given a similar mRNA/protein (from another species, perhaps?), can you find the best parse of a genomic sequence that matches that target sequence • Yes, with a variant on alignment algorithms that penalize separately for introns, versus other gaps.

  7. Comparative gene finding tools • Procrustes/Sim4: mRNA vs. genomic • Genewise: proteins versus genomic • CEM: genomic versus genomic • Twinscan: Combines comparative and de novo approach.

  8. Course • Sequence Comparison (BLAST & other tools) • Protein Motifs: – Profiles/Regular Expression/HMMs • Protein Sequence Identification via Mass Spec. • Discovering protein coding genes – Gene finding HMMs – DNA signals (splice signals)

  9. Genome Assembly

  10. DNA Sequencing • DNA is double- stranded • The strands are separated, and a polymerase is used to copy the second strand. • Special bases terminate this process early.

  11. • A break at T is shown here. • Measuring the lengths using electrophoresis allows us to get the position of each T • The same can be done with every nucleotide. Color coding can help separate different nucleotides

  12. • Automated detectors ‘read’ the terminating bases. • The signal decays after 1000 bases.

  13. Sequencing Genomes: Clone by Clone • Clones are constructed to span the entire length of the genome. • These clones are ordered and oriented correctly (Mapping) • Each clone is sequenced individually

  14. Shotgun Sequencing • Shotgun sequencing of clones was considered viable • However, researchers in 1999 proposed shotgunning the entire genome.

  15. Library • Create vectors of the sequence and introduce them into bacteria. As bacteria multiply you will have many copies of the same clone.

  16. Sequencing

  17. Questions • Algorithmic: How do you put the genome back together from the pieces? Will be discussed in the next lecture. • Statistical? How many pieces do you need to sequence, etc.? – The answer to the statistical questions had already been given in the context of mapping, by Lander and Waterman.

  18. Lander Waterman Statistics G = Genome Length L = Clone Length N = Number of Clones T = Required Overlap c = Coverage = LN/G a = N/G q = T/L s = 1- q L G

  19. LW statistics: questions • As the coverage c increases, more and more areas of the genome are likely to be covered. Ideally, you want to see 1 island. • Q1: What is the expected number of islands? • Ans: N exp(-c s ) • The number increases at first, and gradually decreases.

  20. Analysis: Expected Number Islands • Computing Expected # islands. • Let X i =1 if an island ends at position i, X i =0 otherwise. • Number of islands = ∑ i X i • Expected # islands = E(∑ i X i ) = ∑ i E(X i )

  21. Prob. of an island ending at i L i T • E(X i ) = Prob (Island ends at pos. i) • = Prob(clone began at position i-L+1 AND no clone began in the next L-T positions) L - T = a e - c s ( ) E ( X i ) = a 1 - a G a e - c s = Ne - c s  Expected # islands = E ( X i ) = i

  22. LW statistics • Pr[Island contains exactly j clones]? • Consider an island that has already begun. With probability e -c s , it will never be continued. Therfore • Pr[Island contains exactly j clones]= (1 - e - c s ) j - 1 e - c s • Expected # j-clone islands = Ne - c s (1 - e - c s ) j - 1 e - c s

  23. Expected # of clones in an island e c s Why?

  24. Expected length of an island L e c s - 1 È ˘ Ê ˆ ˜ + (1 - s ) Í ˙ Á c Ë ¯ Î ˚

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend