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Two Point Analysis in F2 Fully Informative Markers (codominant) BBBbbb AAObsn 22 n 21 n 20 Freq ¼ (1-r) 2 ½ r(1-r) ¼ r 2 Recom.012 AaObs n 12 n 11 n 10 Freq ½ r(1-r) ½ (1-r) 2 + ½ r 2 ½ r(1-r) Recom.12r 2 /[(1-r) 2 +r 2 ]1 aaObs n 02 n 01 n 00 Freq ¼ r 2 ½ r(1-r) ¼ (1-r) 2 Recom.210

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  (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  (t)n 11 ]

Two Point Analysis in F2 Fully Informative Markers (codominant) AA Aa aa BBBbbb n Input:Result: r0  (t) = r(t) 2 /[(1-r(t)) 2 +r(t) 2 ] r(t+1)= 1/(2n)[2(n20+n02)+(n21+n12+n10+n01)+2  (t)n11]

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 [ ¼ r 2 ] n20+n02 [ ½ r(1-r)] n21+n12+n10+n01 [ ½ (1-r) 2 + ½ r 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 [ ¼ ] n20+n02 [ ½ 0.5(1-0.5)] n21+n12+n10+n01 [ ½ (1- 0.5) 2 + ½ ] 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

Two Point Analysis in F2 Fully Informative Markers (codominant) function LOD=calcLOD_F2(r,n22,n21,n20,n12,n11,n10,n02,n01,n00) %log likelihood under H1 LOD=(n22+n00)*log10((1-r)^2/4)... +(n20+n02)*log10(r^2/4)... +(n21+n12+n10+n01)*log10(r*(1-r)/2)... +n11*log10((1-r)^2/2+r^2/2); %log likelihood under H0 r=0.5; LOD0=(n22+n00)*log10((1-r)^2/4)... +(n20+n02)*log10(r^2/4)... +(n21+n12+n10+n01)*log10(r*(1-r)/2)... +n11*log10((1-r)^2/2+r^2/2); LOD=LOD-LOD0; Matlab program to calculate log likelihood test score (LOD)

Two Point Analysis in F2 Partial Informative Markers (codominant X dominant) BBBbbb AAObsn 22 n 21 n 20 Freq ¼ (1-r) 2 ½ r(1-r) ¼ r 2 Recom.012 AaObs n 12 n 11 n 10 Freq ½ r(1-r) ½ (1-r) 2 + ½ r 2 ½ r(1-r) Recom.12r 2 /[(1-r) 2 +r 2 ]1 aaObs n 02 n 01 n 00 Freq ¼ r 2 ½ r(1-r) ¼ (1-r) 2 Recom.210

Two Point Analysis in F2 Partial Informative Markers (codominant X dominant) B_bb AAObs n 2_ =n 22 +n 21 n 20 Freq ¼ (1-r) 2 + ½ r(1-r) ¼ r 2 Recom.C 1 = ½ r(1-r)/[ ¼ (1-r) 2 + ½ r(1-r)]2 AaObs n 1_ =n 12 +n 11 n 10 Freq ½ r(1-r)+ ½ (1-r) 2 + ½ r 2 ½ r(1-r) Recom.C 2 =[ ½ r(1-r) +r 2 ]/ [ ½ r(1-r)+ ½ (1-r) 2 + ½ r 2 ] 1 aaObs n 0_ =n 02 +n 01 n 00 Freq ¼ r 2 + ½ r(1-r) ¼ (1-r) 2 Recom.C 3 =[2* ¼ r 2 + ½ r(1-r)]/[ ¼ r 2 + ½ r(1-r)]0 Estimate of r=(c1* n 2_ +c2* n 1_ +c3* n 0_ +2* n 20 + n 00 )/(2n)

Two Point Analysis in F2 Partial Informative Markers (codominant X dominant) E-Step C 1 = ½ r(1-r)/[ ¼ (1-r) 2 + ½ r(1-r)] C 2 =[ ½ r(1-r) +r 2 ]/ [ ½ r(1-r)+ ½ (1-r) 2 + ½ r 2 ] C 3 =[2* ¼ r 2 + ½ r(1-r)]/[ ¼ r 2 + ½ r(1-r)] M-Step r=(c1* n 2_ +c2* n 1_ +c3* n 0_ +2* n 20 + n 00 )/(2n)

Two Point Analysis in F2 Partial Informative Markers (codominant X dominant) AA Aa aa B_bb n Input:Result: r0

Two Point Analysis in F2 Partial Informative Markers (co dominant X dominant) function r=rEstF2CoXdomin(n2_,n1_,n0_,n20,n10,n00) n=n2_+n1_+n0_+n20+n10+n00; r=0.2;r1=-1; while(abs(r1-r)>1.e-8) r1=r; %E-step c1= 1/2*r*(1-r)/[1/4*(1-r)^2+ 1/2*r*(1-r)]; c2=[1/2*r*(1-r)+r^2]/[1/2*r*(1-r)+1/2*(1-r)^2+1/2*r^2]; c3=[2*1/4*r^2+1/2*r*(1-r)]/[1/4*r^2+1/2*r*(1-r)]; %M-step r=(c1*n2_+c2* n1_ +c3* n0_+2* n20 + n00)/(2*n); end Matlab program to estimate recombinant r

Two Point Analysis in F2 Partial Informative Markers (co dominant X dominant) Matlab program to calculate log likelihood test score (LOD) function LOD=calcLOD_F2CoXdomin(r, n2_,n1_,n0_,n20,n10,n00) %log likelihood under H1 LOD=log([1/4*(1-r)^2+ 1/2*r*(1-r)])*n2_... +log([1/2*r*(1-r)+1/2*(1-r)^2+1/2*r^2])*n1_... +log([1/4*r^2+1/2*r*(1-r)])*n0_... +log(r^2/4)*n20+log(r*(1-r)/2)*n10+log((1-r)^2/4)*n00; %log likelihood under H0 r=0.5; LOD0=log([1/4*(1-r)^2+ 1/2*r*(1-r)])*n2_... +log([1/2*r*(1-r)+1/2*(1-r)^2+1/2*r^2])*n1_... +log([1/4*r^2+1/2*r*(1-r)])*n0_... +log(r^2/4)*n20+log(r*(1-r)/2)*n10+log((1-r)^2/4)*n00; LOD=LOD-LOD0; LOD=LOD/log(10);

Two Point Analysis in F2 Partial Informative Markers (dominant) BBBbbb AAObsn 22 n 21 n 20 Freq ¼ (1-r) 2 ½ r(1-r) ¼ r 2 Recom.012 AaObs n 12 n 11 n 10 Freq ½ r(1-r) ½ (1-r) 2 + ½ r 2 ½ r(1-r) Recom.12r 2 /[(1-r) 2 +r 2 ]1 aaObs n 02 n 01 n 00 Freq ¼ r 2 ½ r(1-r) ¼ (1-r) 2 Recom.210

Two Point Analysis in F2 Partial Informative Markers (dominant) B_bb A_Obs n 1 =n 22 +n 21 +n 12 + n 11 n 2 =n 20 +n 10 Freq ¼ (1-r) 2 +r(1-r) + ½ (1-r) 2 + ½ r 2 ¼ r 2 Recom.c1c2 aaObs n 3 =n 02 +n 01 n 4 = n 00 Freq ¼ r 2 + ½ r(1-r) ¼ (1-r) 2 Recom.C2= (2( ¼ r 2 )+ ½ r(1-r)) 0 /( ¼ r 2 + ½ r(1-r)) where C1=[r 2 +r(1-r)]/[ ¼(1-r) 2 +r(1-r) + ½(1-r) 2 +½r 2 ], expected number of recombinant gametes Estimate of r=(c1* n 1 +c2* n 2 +c2* n 3 )/(2n)

Two Point Analysis in F2 Fully Informative Markers (codominant) A_ aa B_bb n Input:Result: r0 C1=[r 2 +r(1-r)]/[ ¼(1-r) 2 +r(1-r) + ½(1-r) 2 +½r 2 ], C2= (2( ¼ r 2 )+ ½ r(1-r)) /( ¼ r 2 + ½ r(1-r)) Estimate of r=(c1* n 1 +c2* n 2 +c2* n 3 )/(2n)

Two Point Analysis in F2 Partial Informative Markers (dominant) function r=rEstF2Partial(n1,n2,n3,n4) n=n1+n2+n3+n4; r=0.2;r1=-1; while (abs(r1-r)>1.e-8) r1=r; %E-step c1=(r^2+r*(1-r))/((1-r)^2/4+r*(1-r)+(1-r)^2/2+r^2/2); c2=(r^2/2+r*(1-r)/2)/(r^2/4+r*(1-r)/2); %M-step r=1/(2*n)*(c1*n1+c2*n2+c2*n3); end Matlab program to estimate recombinant r

Log-likelihood ratio test statistic Partial Informative Markers (dominant) Two alternative hypotheses H0: r = 0.5 vs. H1: r  0.5 Likelihood value under H1 L 1 (r|n ij ) = n!/(n 1 !...n 4 !)  [3/4(1-r) 2 +r(1-r) + ½ r 2 ] n1 [ ¼ r 2 + ½ r(1-r)] n2+n3 [ ¼ (1-r) 2 ] n4 Likelihood value under H0 L 0 (r=0.5|n ij ) = n!/(n 1 !...n 4 !)  [3/4(1-.5) 2 +.5(1-.5) + ½.5 2 ] n1 [ ¼ ½.5(1-.5)] n2+n3 [ ¼ (1-.5) 2 ] n4 LOD = log 10 [L 1 (r|n ij )/L 0 (r=0.5|n ij )] = 3.17 > critical LOD=3

Two Point Analysis in F2 Partial Informative Markers (dominant) function LOD=calcLOD_F2Partial(r,n1,n2,n3,n4) %log likelihood under H1 LOD=(n1)*log10((1-r)^2*3/4+r^2/2+r*(1-r))... +(n2+n3)*log10(r^2/4+r*(1-r)/2)... +(n4)*log10((1-r)^2/4); %log likelihood under H0 r=0.5; LOD0=(n1)*log10((1-r)^2*3/4+r^2/2+r*(1-r))... +(n2+n3)*log10(r^2/4+r*(1-r)/2)... +(n4)*log10((1-r)^2/4); LOD=LOD-LOD0; Matlab program to calculate log likelihood test score (LOD)

Three Point Analysis in Backcross a rice data

RG472 RG K5 U10 RG532 W1 RG173 RZ276 Amy1B RG146 RG345 RG381 RZ19 RG690 RZ730 RZ801 RG810 RG RG437 RG544 RG171 RG157 RZ318 Pall RZ58 CDO686 Amy1A/C RG95 RG654 RG256 RZ213 RZ123 RG RG104 RG348 RZ329 RZ892 RG100 RG191 RZ678 RZ574 RZ284 RZ394 pRD10A RZ403 RG179 CDO337 RZ337A RZ448 RZ519 Pgi -1 CDO87 RG910 RG418A RG218 RZ262 RG190 RG908 RG91 RG449 RG788 RZ565 RZ675 RG163 RZ590 RG214 RG143 RG chrom1chrom2chrom3chrom4

Three Point Analysis in Backcross Summarized the data as A,B,C Obs.A & BB & C 111abcn abc abCn abC aBcn aBc aBCn aBC Abcn Abc AbCn AbC ABcn ABc ABCn ABC 00

Rice Data A,B,C Obs.A & BB & C 111abcn abc = abCn abC = aBcn aBc = aBCn aBC = Abcn Abc = AbCn AbC = ABcn ABc = ABCn ABC =3800 Marker RG472 denoted by A, RG246 by B, K5 by C

Multilocus likelihood – determination of a most likely gene order Consider three markers A, B, C, with no particular order assumed. A triply heterozygous F1 ABC/abc backcrossed to a pure parent abc/abc GenotypeABC or abc ABc or abC Abc or aBC AbC or aBc Obs. n 00 =69 n 01 =12 n 10 =16 n 11 =3 Frequency under Order A-B-C (1-r AB )(1- r BC ) (1-r AB ) r BC r AB (1- r BC ) r AB r BC Order A-C-B (1-r AC )(1- r BC ) r AC r BC r AC (1-r BC ) (1-r AC )r BC Order B-A-C (1-r AB )(1- r AC ) (1-r AB ) r AC r AB r AC r AB (1-r AC ) r AB = the recombination fraction between A and B= (n 10 + n 11 )/n=0.19 r BC = the recombination fraction between B and C= (n 01 + n 11 )/n=0.15 r AC = the recombination fraction between A and C= (n 01 + n 10 )/n=0.28

What order is the mostly likely? L ABC  (1-r AB ) n00+n01 (1-r BC ) n00+n10 (r AB ) n10+n11 (r BC ) n01+n11 L ACB  (1-r AC ) n00+n11 (1-r BC ) n00+n10 (r AC ) n01+n10 (r BC ) n01+n11 L BAC  (1-r AB ) n00+n01 (1-r AC ) n00+n11 (r AB ) n10+n11 (r AC ) n01+n10 Log(LABC) = Loo(LACB) = Log(LBAC) = According to the maximum likelihood principle, the linkage order that gives the maximum likelihood for a data set is the best linkage order supported by the data. the best linkage order A B C 20cM 15cM

GenotypeABC or abc ABc or abC Abc or aBC AbC or aBc Obs. n 00 =69 n 01 =12 n 10 =16 n 11 =3 DATA Result: r AB = =0.19 r BC = =0.15 r AC = =0.28 d AB =1/4*ln[(1+2 r AB )/(1-2 r AB )]=20 d BC =1/4*ln[(1+2 r BC )/(1-2 r BC )]=15 Log(LABC) = Loo(LACB) = Log(LBAC) = the best linkage order A B C 20cM 15cM