STA6178_2005(4)

# STA6178_2005(4) - Computational Issues on Statistical...

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Computational Issues on Statistical Genetics Develop Methods Data Collection Analyze Data Write Reports/Papers Research Questions Review the Literature Test the power and robustness by computer simulation Database construction (Excel, Access) Translate data to analyzable form Preliminary results (figures, tables) Program languages Efficient, feasible Graphics Excel graphics Programmable graphics

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Program Languages Fortran, C, C++ Matrix language: MATLAB , S-Plus, R, SAS IML Symbolic Calculation: Mathematika ,Maple,Matlab Interface Programming: dotnet, C#, Visual Basic SAS , SPSS, BMDP Database: Access , Excel, SQL, SAS, Oracle MACRO Excel, Access, PowerPoint, Word Editor: WinEdt SAS Macro
Two Point Analysis in F2 Fully Informative Markers (codominant) BB Bb bb AA Obs n 22 n 21 n 20 Freq ¼(1-r) 2 ½r(1-r) ¼r 2 Recom. 0 1 2 Aa Obs n 12 n 11 n 10 Freq ½r(1-r) ½(1-r) 2 +½r 2 ½r(1-r) Recom. 1 2r 2 /[(1-r) 2 +r 2 ] 1 aa Obs n 02 n 01 n 00 Freq ¼r 2 ½r(1-r) ¼(1-r) 2 Recom. 2 1 0

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EM algorithm to estimate the recombination fraction r: 1. Given r(0), For t=0,1, 2,… 2. Do While abs[r(t+1)-r(t)]>1.e-8 E-step: Calculate φ (t) = r(t) 2 /[(1-r(t)) 2 +r(t) 2 ] (expected the number of recombination events for the double heterozygote AaBb) M-step: r(t+1)= 1/(2n)[2(n 20 +n 02 )+ (n 21 +n 12 +n 10 +n 01 )+2 φ (t)n 11 ]
Two Point Analysis in F2 Fully Informative Markers (codominant) AA Aa aa BB Bb bb n Start Input: Result: Reset r0 φ (t) = r(t) 2 /[(1-r(t)) 2 +r(t) 2 ] r(t+1)= 1/(2n)[2(n20+n02)+(n21+n12+n10+n01)+2 φ (t)n11]

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Two Point Analysis in F2 Fully Informative Markers (codominant) function r=rEstF2(n22,n21,n20,n12,n11,n10,n02,n01,n00) n=n22+n21+n20+n12+n11+n10+n02+n01+n00; r=0.2; r1=-1; while (abs(r1-r)>1.e-8) r1=r; %E-step phi=r^2/((1-r)^2+r^2); %M step r=1/(2*n)*(2*(n20+n02)+(n21+n12+n10+n01)+2*phi*n11); end Matlab program to estimate recombinant r
Log-likelihood ratio test statistic Two alternative hypotheses H0: r = 0.5 vs. H1: r 0.5 Likelihood value under H1 L 1 (r|n ij ) = n!/(n 22 !...n 00 !) × [ ¼(1-r) 2 ] n22+n00 [ ¼r 2 ] n20+n02 [ ½r(1-r) ] n21+n12+n10+n01 [ ½(1-r) 2 +½r 2 ] n11 Likelihood value under H0 L 0 (r=0.5|n ij ) = n!/(n 22 !...n 00 !) × [ ¼(1-0.5) 2 ] n22+n00 [ ¼0.5 2 ] n20+n02 [ ½0.5(1-0.5) ] n21+n12+n10+n01 [ ½(1-0.5) 2 +½0.5 2 ] n11 LOD = log 10 [L 1 (r|n ij )/L 0 (r=0.5|n ij )] = {(n 22 +n 00 )2[log 10 (1-r)-log 10 (1-0.5)+…} = 6.08 > critical LOD=3

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Two Point Analysis in F2 Fully Informative Markers (codominant)
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