1 / 15

Taylor’s Cycle & Genetic Code

Taylor’s Cycle & Genetic Code. vif,. p23. rev,. p19. 5' LTR. gag. 3' LTR. tat,. p14. pol. env. vpr vpu. p15 p16. nef. , p27. Case study: Selection on HIV-1 nef Gene.

jadzia
Download Presentation

Taylor’s Cycle & Genetic Code

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Taylor’s Cycle& Genetic Code

  2. vif, p23 rev, p19 5' LTR gag 3' LTR tat, p14 pol env vpr vpu p15 p16 nef , p27 Case study: Selection on HIV-1 nef Gene

  3. dN e dS along sequences by the Nei-Gojobori method (Nei & Kumar, 2000) are the total of synonymous (S) and non-synonynous (N) substitutions over the number of potentially synonimous (s) and non-synonimos (n) sites, respectively. Ex.: Phenylalanine (TTT), substitutions at the 1st and 2nd positions are non-synonimous 1 in 3 substitutions are synonimous (TTC): (s) is 0+0+1/3 and (n): 3-1/3=8/3. The proportion of synonymous (ps) and non-synonymous (pn) difference are ps = sd /S e pn = sn /N, respectively. To estimate (dS) and (dN) we can use Jukes-Cantor model:dS=-(3/4ln[1-(4/3)ps]dN=-(3/4ln[1-(4/3)pn] Pairwise determination of dN & dS dN/dS or >1 “Positive selection” dN/dS or  < 1 “Negative selection” dN/dS or  = 1 “Neutrality”

  4. Using the pairwise determination of dN & dS: Looking at comparisons

  5. Maximum likelihood determination of selected codon sites M0 assumes a single value of dN/dS for all codon sites. M1 (“Neutral”) 2 classes of codons: “conserved” (dN/dS= 0) and neutral (dN/dS= 1), distributed in diferent proportions (p0 e p1). M2 (“Selection”) adds a new codon class (p2) which can assume values greater than 1 for dN/dS ( 2). M3 (“Discrete”) assumes n categorias of códons for a given number of classes (K), the values of dN/ dS () per codon-class are gamma-distributed (K categories). For K=8, (Yang et al., 2000) we get  0 até  7 and (p0 até p7). The proportion p8, of sites with higher dN/dS indicates intense positive selection. M7 assumes 10 codon classes, with dN/dS being beta-distributed, with values lower than 1 (explicit negative selection model). M8 adds an eleventh class to model M7 with dN/dS greater than 1. LRT rationale : (i) M0 vs. M2 with 2 d.f.(ii) M0 vs. M3 with 4 d.f.(iii) M1 vs. M2 with 2 d.f.(iv) M1 vs. M3 with 4 d.f.(v) M2 vs. M3 with 3 d.f. (vi) M7 vs. M8 with 2 d.f.

  6. M2 better than M0 e M1 indicates positive selection (>1).M3 better than M1 and M2 points to which codons are positively selected. M0 vs. M2 e M1 vs. M2, since M2 gives conservative estimates of  (Yang, 2000), only then we can trust codons under selection found with M3 and M8. Pros:Yang method allows fitting adequate substitution models using different amino acids transition matrices and rates of heterogeneity when estimating  for each codon, Uses genealogical information (Felsenstein, 1973; Yang, 1994). Cons:May inflate dN, when (n < 20) and when phylogenetic signal is low (Suzuki & Nei, 2001). Considerations about the maximum likelihood method

  7. Using the maximum likelihood method

  8. Using the maximum likelihood method: data

  9. HIV-1 nef gene: Positively seleted sites

  10. HIV-1 Subtype B: Positively seleted sites

  11. Gene substitution and trees: A cladistic view

  12. Gene substitution in the HIV-1 nef gene

  13. Positive selection in the HIV-1 pol gene: immune pressure? 211 214 122, 123 135

More Related