Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Notes on Selection and Test Neutrality - Applied Bioinformatics | BIT 150, Exams of Bioinformatics

Material Type: Exam; Class: Applied Bioinformatics; Subject: Biotechnology; University: University of California - Davis; Term: Winter 2006;

Typology: Exams

Pre 2010

Uploaded on 07/31/2009

koofers-user-a6i
koofers-user-a6i 🇺🇸

10 documents

1 / 16

Toggle sidebar

Related documents


Partial preview of the text

Download Notes on Selection and Test Neutrality - Applied Bioinformatics | BIT 150 and more Exams Bioinformatics in PDF only on Docsity! Lecture 18 - Selection and Tests of Neutrality • Gibson and Muse, chapter 5 • Nei and Kumar, chapter 12.6 p. 258- 264 • Hartl, chapter 3, p. 122-127 The Usefulness of Theta • Under evolution by genetic drift (i.e., neutral evolution), each estimator of theta is an unbiased estimator of the true value of theta: • Therefore, differences between estimates from each estimator can be used to infer non-neutral evolution (i.e., natural selection). • Keep in mind that there are other estimators of theta as well. Each estimator, however, has the same expectation (i.e., the true value of theta) ! E( ˆ " W ) = E( ˆ " # ) =" Nucleotide diversity in loblolly pine candidate genes for drought Total Synonymous Non-synonymous Silent (non-coding + synonymous)Gene n S θ π S θ π S θ π S θ π Tajima’s D lp3-1 32 18 8.77 12.70 1 9.31 2.34 1 2.29 0.58 17 17.40 12.47 -1.05088 lp3-3 32 3 1.59 0.97 0 0 0 2 2.10 1.42 1 1.08 0.53 -0.89756 dhn-1 32 13 5.08 4.15 4 8.88 7.15 3 1.83 1.72 10 11.30 8.81 -0.59854 dhn-2 31 14 6.64 7.76 6 15.31 16.58 4 3.03 2.87 10 13.17 16.56 0.56236 mtl-like 32 9 5.55 5.10 0 0 0 2 6.77 2.50 7 5.27 5.68 -0.24719 sod-chl 32 19 6.88 7.80 2 12.61 7.54 2 3.86 4.11 17 7.57 8.65 0.45770 ferritin 32 7 2.92 1.28 0 0 0 1 2.07 0.52 6 3.14 1.48 -1.63069 rd21A-2 31 26 7.02 7.68 7 13.28 9.66 5 2.84 3.53 21 11.40 11.51 0.33170 sams-2 32 6 2.75 3.60 2 6.07 9.07 0 0 0 6 5.40 7.06 0.86302 pal-1 31 6 3.81 2.62 1 4.13 2.06 1 1.35 0.35 5 6.00 4.64 -0.88514 ccoaomt-1 32 13 6.68 11.86 4 18.07 26.56 0 0 0 13 11.67 19.23 2.52548* cpk3 32 8 3.16 3.55 3 9.49 13.82 1 0.84 0.59 7 5.26 6.22 0.36974 pp2c 32 1 0.39 0.10 1 2.20 0.55 0 0 0 1 0.86 0.22 -1.14244 Aqua-MIP 32 5 2.06 1.74 2 7.03 10.50 0 0 0 5 3.03 2.55 -0.42191 erd3 32 6 1.70 0.43 0 0 0 2 1.04 2.26 4 2.48 0.62 -2.10198* ug-2_498 32 10 8.65 5.26 - - - - - - - - - -1.22364 AVERAGE 10 4.6 4.79 2.2 7.09 7.06 1.6 1.87 1.36 9 7 7.08 * P < 0.05; ** P < 0.01 Diversity values are multiplied by 103 Gonzalez-Martinez et al. Genetics 2006 Tajima’s D - An Example 1 0 5 1 6 5 1 7 1 1 7 3 2 1 2 2 1 3 2 6 1 2 7 9 3 3 1 3 7 0 3 7 4 3 9 8 4 1 6 5 1 5 C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C HAP1 C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C C A T A G A T T T A T A I C T G G T T C C C C C C T C T G G T T C C C C C C T C T G G T T C C C C C C T C T G G T T C C C C C C T C HAP2 T G G T T C C C C C C T C T G G T T C C C C C C T C T G G T T C C C C C C T C T G G T T C C C C C C T C C A T A G A T T T C T A I C C A T A G A T T T C T A I C HAP3 C A T A G A T T T C T A I C C A T A G A T T T C T A I C C A T A G A T T T C T A I C HAP4 C A T A G A T T T C T A C C A T A G A T T T C T A C HAP5 T G G T T G C C C C C T C T G G T T G C C C C C T C HAP6 T G G T T C C C C C C T T HAP7 C A T A G A T T T A T A C INDELS I TGAGGAAACAGGGGTG 3 0 8 3 9 4 5 6 3 5 6 9 5 7 8 7 6 5 8 5 6 A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C HAP1 A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G T G C A G G I T G C HAP2 A G G I T G C A G G I T G C A G G I T G C HAP3 A A T T C C HAP4 A G G C G C HAP5 A G G I T G T HAP6 G G G T G C INDELS I ACTT ccoaomt-1, excess of high frequency variants. Balancing selection? erd3, excess of rare variants. Genetic hitchhiking? Tajima’s D = 2.52 Tajima’s D = -2.10 Other Neutrality Tests Using Theta ! D* = ˆ " W # ˆ " $ var( ˆ " % # ˆ " $ ) ! H norm = ˆ " # $ ˆ " L var( ˆ " # $ ˆ " L ) Fu and Li’s D* Zeng et al.’s H Both of these tests require an outgroup for their estimation. Expectations from Genealogies Comparisons of within to between species polymorphisms, so-called polymorphism-divergence tests, can remove the dependency upon the genealogy for tests of selection. The Hudson, Kreitman, and Agaude (HKA) Test • The HKA test uses data from two different species. Under neutral evolution, the expected number of differences (D) between two homologous sequences (one from each species) is given by: • Similarly, the expected number of polymorphisms within each species are given by: • The rationale for the HKA test is that under neutral evolution, the diversity within each species depends upon theta, but that the divergence between species depends on theta AND time (T). ! E(D) = k"(T +1) var(D) = k"(T +1)+ k 2" 2 ! E(S) = ˆ " W a n var(S) = ˆ " W a n + b n ( ˆ " W ) 2 The HKA Test Statistic • The equations from the last slide can be used to construct a X2 statistic with 2L-2 degrees of freedom (L = number of loci, L must be ≥ 2): • If X2 is large enough, the null hypothesis of neutral evolution can be rejected. (If E(S) and E(D) are estimated from the same data as that being used in the test, X2 must be bigger than 3.841 for L = 2 using a critical p-value of 0.05 ). ! " 2 = [S #E(S)]2 var(S) + [D #E(D)]2 var(D)i $ j $ i $ The MK Table and Tests for Independence Fixed Polymorphic Non-synonymous NF NP Synonymous SF SP Ratio NF / SF = NP /SP Under neutrality, NF/ SF = NP/ SP Under positive directional selection, NF/ SF > NP/ SP A likelihood ratio test for independence in a contingency table is then used to test the null hypothesis of neutrality. If we assume that SF = SP under neutrality, the test statistic is given as: ! G = 2 [N F N P ]ln [N F N P ] avg[N F N P ] " # $ % & ' all loci ( dN/dS • A related concept to the MK test are tests based on the dN/dS ratio. • These tests assume that under neutrality, dN/dS = 1.0 • Statistical tests are then constructed to test whether or not an observed dN/dS deviates from 1
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved