Kinship Coefficients. Biostatistics 666

Transcription

1 Knshp Coeffcents Bostatstcs 666

2 Today s Lecture Genetc analyses requre relatonshps to be specfed Msspecfed relatonshps lead to tests of napproprate sze Inflated Type I error Decreased power Knshp Coeffcents Usng data to verfy genetc relatonshps

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4 Knshp Coeffcents Summarze genetc smlarty between pars of ndvduals. Can be used to study relatonshp between genetc smlarty and phenotypc smlarty across ndvduals.

5 Knshp Coeffcents Defnton Gven two ndvduals One wth genes (g, g j ) The other wth genes (g k, g l ) The knshp coeffcent s: ¼P(g g k ) + ¼P(g g l ) + ¼P(g j g k ) + ¼P(g j g l ) where represents dentty by descent (IBD) Probablty that alleles sampled at random from each ndvdual are IBD

6 Some knshp coeffcents Sblngs (ϕ=1/4) Half-Sbs (ϕ=1/8) MZ Twns (ϕ=1/2) Unrelated (ϕ=0) Parent-Offsprng (ϕ=1/4)

7 What about other relatves? For any two related ndvduals and j use a recursve algorthm allows calculaton of knshp coeffcent Algorthm requres an order for ndvduals n the pedgree where ancestors precede descendants That s where for any >j, s not ancestor of j Such an order always exsts (e.g. the brth order!)

8 Computng Knshp Coeffcents The recursve defnton s then (for j): = + + = = j j j j father mother j father j mother j ) (1 ) ( s a founder, are founders and 0 ) ( ) ( 2 1 ) ( ) ( ϕ ϕ ϕ ϕ

9 An example pedgree Can you fnd Sutable orderng for recursve calculaton? Calculate knshp coeffcent between shaded ndvduals?

10 Inbreedng Coeffcents The knshp coeffcent s related to the nbreedng coeffcent If ϕ > 0.5, ndvdual s nbred The nbreedng coeffent s f = ϕ mother()father() = 2(ϕ 0.5) In most human populatons, f s small on the order of Modfes probablty of heterozygous genotypes to 2(1-f)p(1-p) Modfes probablty of homozygous genotypes to (1-f)p 2 + fp

11 Verfyng Relatonshps: Strategy I - Allele Sharng Methods For each par, summarze allele sharng across all markers Specfcally, average number of dentcal alleles at each marker par Number of alleles shared between two genotypes s the dentfy-by-state Compare observed values for each par to expected values Expected values derved by assessng all pars wth same putatve relatonshp

12 IBS Sharng Scores S k IBS score (0,1,2) for marker k S = k S k s 2 = k ( S k S ) 2 n markers n markers 1

13 Could construct a Z-score Comparng observed IBS score to expected values wthn class of relatves S E ( S R) Z = Var( S R)

14 Example ~800 marker genome scan Calculated IBS for each set of putatve relatonshps Unrelated pars Sblng pars Parent-offsprng pars

15 Putatve Unrelated Pars Mean = 0.87 St. Dev. = 0.07

16 Parent-Offsprng Pars Mean = 1.27 St. Dev. = 0.05

17 Putatve Sblng Pars Mean = 1.32 St. Dev. = 0.09

18 Problem Indvduals Are Outlers Crcled pars are lkely msclassfed

19 Problems wth IBS Scores Ineffcent Ignores nformaton on allele frequences Ignores correlatons between neghborng markers work well f large amounts of data avalable Cannot dstngush some types of relatves

20 Verfyng Relatonshps: Strategy I - Lkelhood Based Methods When evaluatng sharng, take allele frequency nto account Place greater mportance n sharng of rare alleles Recognze that sharng of common alleles can occur by chance Choce of parameters to maxmze and constrants on underlyng probabltes

21 P (X m IBD) IBD Sb CoSb (a,b) (c,d) 4papbpcpd 0 0 (a,a) (b,c) 2pa 2 pbpc 0 0 (a,a) (b,b) pa 2 pb (a,b) (a,c) 4pa 2 pbpc papbpc 0 (a,a) (a,b) 2pa 3 pb pa 2 pb 0 (a,b) (a,b) 4pa 2 pb 2 (papb 2 +pa 2 pb) 2papb (a,a) (a,a) pa 4 pa 3 pa 2

22 Example I Consder genotypes for one marker Let G = (1/1, 1/1) Assume p 1 =.5 Calculate P(G R) for each relatonshp MZ twn, Full Sbs, Half-Sbs, Unrelated How do results change wth p 1?

23 Lkelhood MM 2 LL = PP GG mm, GG jjmm IIIIII = kk PP(IIIIII = kk rrrrrrrrrrrrrrrrrrrrrrr) mm=1 kk=0 Lkelhood above assumes markers are ndependent Wth smaller amounts of data, mportant to model recombnaton Wth large amounts of data, ths works well Maxmze probablty of IBD=0, IBD=1, IBD=2 Or, often, just P(IBD=1) = 2Φ j and P(IBD=0) = 1-2 Φ j

24 Smulatons (M=50, 10 cm apart) Inferred R True R Full Sbs Half Sbs Unrelated Full Sbs Half Sbs Unrelated <

25 Smulatons (M=400, 10 cm apart) Inferred R True R Full Sbs Half Sbs Unrelated Full Sbs <.001 <.001 Half Sbs < <.001 Unrelated <.001 <

26 Weaknesses wth lkelhood approach One weakness s that the approach s senstve to genotypng error Consder some genome scan data 380 mcrosatellte markers Observed Sharng Identcal for 379/380 genotype pars L(G R=MZ Twns) = 0 L(G R=Any other) > 0 How to resolve?

27 Soluton: Allow for Genotypng Errors If lkelhood gnores errors, even a few errors can lead to msclassfcaton Need to update lkelhood to allow errors Introduce a dstncton between true genotypes G and observed genotypes X An error rate parameter, say ε, models the dfference between the two P( X ) = P( X = I G G (1 ε )² P( G, ε ) P( G I ) + I ) [ 1 (1 ε )²] P( G ) P( G ) 1 2

28 Weaknesses wth lkelhood approach Another weakness s that the approach s senstve to allele frequency estmates How can we make sure that we have chosen the rght allele frequences? Manchakul et al (2010) proposed focusng on marker pars that have confguraton (a/a, b/b) or (a/b, a/b) The rato of these two confguratons does not depend on allele frequences! However, t wll depend on the rato of P(IBD=1) to P(IBD=0)

29 Recommended Readng Boehnke and Cox (1997) Am J Hum Genet 61: Optonal Broman and Weber (1998), AJHG 63: McPeek and Sun (2000), AJHG 66: Epsten et al. (2000), AJHG 67: