1. - Type of Study Composite Interval Mapping Program - Genetic Design

2. - Data and Options Map Function Parameters Here for Simulation Study Only QTL Searching StepcM Cumulative Marker Distance (cM) Composite Interval Mapping Program

3. - Data Put Markers and Trait Data into box below OR Composite Interval Mapping Program

4. - Analyze Data Composite Interval Mapping Program For CIM, Controlled Background Markers by Within cM Or Markers

5. - Profile Composite Interval Mapping Program

6. - Permutation Test #Tests Cut Off Point at Level Is Based on Tests. Composite Interval Mapping Program

7. Backcross Population – Two Point FreqQqqq Mm1/2(1-r)/2r/2 mm1/2r/2(1-r)/2

8. Backcross Population – Three Point FreqQqqq MmNn(1-r)/2(1-r 1 )(1-r 2 )r 1 *r 2 Mmnnr/2(1-r 1 )r 2 r 1 (1-r 2 ) mmNnr/2r 1 (1-r 2 )(1-r 1 )r 2 mmnn(1-r)/2r1*r2(1-r)/2 M Q N

9. F2 Population – Two Point FreqQQQqqq MM1/4(1-r) 2 /4(1-r)r/2r 2 /4 Mm1/2(1-r)r/2½-(1-r)r(1-r)r/2 mm1/4r 2 /4(1-r)r/2(1-r) 2 /4

10. F2 Population – Three Point FreqQQQqqq MMNN(1-r) 2 /4 1/4(1-a) 2 (1-b) 2 1/2a(1-a)b(1-b)1/4a 2 b 2 Nn(1-r)r/2 1/2(1-a) 2 b(1-b)1/2a(2b 2 - 2b+1)(1-a) 1/2a 2 b(1-b) nnr 2 /4 1/4(1-a) 2 b 2 1/2a(1-a)b(1-b)1/4a 2 (1-b) 2 MmNN(1-r)r/2 1/2a(1-a)(1-b) 2 1/2b(1- 2a+2a 2 )(1-b) 1/2a(1-a)b 2 Nn ½-(1-r)r a(1-a)b(1-b)1/2(2b 2 - 2b+1)(1-2a+2a 2 ) a(1-a)b(1-b) Nn(1-r)r/2 1/2a(1-a)b 2 1/2b(1- 2a+2a 2 )(1-b) 1/2a(1-a)(1-b) 2 mmNNr 2 /4 1/4a 2 (1-b) 2 1/2a(1-a)b(1-b)1/4(1-a) 2 b 2 Nn(1-r)r/2 1/2a 2 b(1-b)1/2a(2b 2 - 2b+1)(1-a) 1/2(1-a) 2 b(1-b) nn(1-r) 2 /4 1/4a 2 b 2 1/2a(1-a)b(1-b)1/4(1-a) 2 (1-b) 2 M a Q b N r=a+b-2ab

11. Composite model for interval mapping and regression analysis y i =  + a* z i +  k m-2 b k x ik + e i Expected means: Qq:  + a* +  k b k x ik = a* + X i B qq:  +  k b k x ik = X i B X i = (1, x i1, x i2, …, x i(m-2) ) 1x(m-1) B = ( , b 1, b 2, …, b m-2 ) T z i : QTL genotype x ik : marker genotype M 1 x 1 M 1 m 1 1  +b 1 m 1 m 1 0 

12. z i – conditional probability of Qq given markers of individual i and QTL position x ik – coding for ‘effect’ of k-th marker of i Backcross: x ik =1 if k-th marker of i is Mm =0 or –1 if k-th marker of i is mm  k m-2 – summation over all markers except two markers of current interval We want estimate a* and test if abs(a*) is big enough to claim that there is a QTL at the given location in an interval. The estimate B is not very important

13. Likelihood based CIM L(y,M|  ) =  i=1 n [  1|i f 1 (y i ) +  0|i f 0 (y i )] log L(y,M|  ) =  i=1 n log[  1|i f 1 (y i ) +  0|i f 0 (y i )] f 1 (y i ) = 1/[(2  ) ½  ]exp[-½(y-  1 ) 2 ],  1 = a*+X i B f 0 (y i ) = 1/[(2  ) ½  ]exp[-½(y-  0 ) 2 ],  0 = X i B Define  1|i =  1|i f 1 (y i )/[  1|i f 1 (y i ) +  0|i f 0 (y i )](1)  0|i =  0|i f 1 (y i )/[  1|i f 1 (y i ) +  0|i f 0 (y i )](2)

14. a* =  i=1 n  1|i (y i -a*-X i B)/  i=1 n  1|i (3) =  1 (Y-XB)´/c B = (X´X) -1 X´(Y-  1 a*) (4)  2 = 1/n (Y-XB)´(Y-XB) – a* 2 c (5)  = (  i=1 n2  1|i +  i=1 n3  0|i )/(n 2 +n 3 )(6) Y = {y i } nx1,  = {  1|i } nx1, c =  i=1 n  1|i

15. Hypothesis test H0: a*=0 vs H1: a*  0 L0 =  i=1 n f(y i )  B = (X´X) -1 X´Y,  2 = 1/n(Y-XB)´(Y-XB) L1=  i=1 n [  1|i f 1 (y i ) +  0|i f 0 (y i )] LR = -2(lnL0 – lnL1) LOD = -(logL0 – logL1)

16. Likelihood based CIM for BC and F2 L(y,M|  ) =  i=1 n  k=1 g [  k|i f k (y i )] log L(y,M|  ) =  i=1 n log[  k=1 g  k|i f k (y i )] g=2 for BC, 3 for F2 f k (y i ) = 1/[(2  ) ½  ]exp[-½(y-  k ) 2 ],  k =  g k +X i B k=1,…,g

17. Define  k|i =  k|i f 1 (y i )/[  k=1 g  k|i f 1 (y i )] (1) B =  k=1 g  k (X´X) -1 X´(Y-  g k ) (2)  g k =  i=1 n  k|i (y i -X i B)/  i=1 n  k|i (3)  2 = 1/n  i=1 n  k=1 g  k|i (y i -X i B -  g k ) 2 (4) Y = {y i } nx1,  k = {  k|i } nx1