STA6178_2005(3) - Genetic design Testing Mendelian...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Genetic design Testing Mendelian segregation Consider marker A with two alleles A and a Backcross F 2 Aa aa AA Aa aa Observation n 1 n n 2 n 1 n Expected frequency ½ ½ ¼ ½ ¼ Expected number n/2 n/2 n/4 n/2 n/4 The x 2 test statistic is calculated by x 2 = Σ (obs – exp) 2 /exp = (n 1-n/2) 2 /(n/2) + (n-n/2) 2 /(n/2) =(n 1-n )2/n ~x 2 df=1 , for BC, (n 2-n/4) 2 /(n/4)+(n 1-n/2) 2 /(n/2)+(n-n/4) 2 /(n/4)~x 2 df=2 , for F 2 Examples Backcross F2 Aa aa AA Aa aa Observation 44 59 43 86 42 Expected frequency ½ ½ ¼ ½ ¼ Expected number 62.5 62.5 42.75 85.5 42.75 The x 2 test statistic is calculated by x 2 = Σ (obs – exp) 2 /exp = (44-59) 2 /103 = 2.184 < x 2 df=1 = 3.841, for BC, (43-42.75) 2 /42.75+(86-85.5) 2 /85.5+(42-42.75) 2 /42.75=0.018 < x 2 df=2 =5.991, for F2 The marker under study does not deviate from Mendelian segregation in both the BC and F2. Linkage analysis Backcross Parents AABB x aabb AB ab F1 AaBb x aabb AB Ab aB ab ab BC AaBb Aabb aaBb aabb Obs n 11 n 10 n 01 n 00 → n = Σ n ij Freq ½(1-r) ½r ½r ½(1-r) r is the recombination fraction between two markers A and B. The maximum likelihood estimate (MLE) of r is r^ = (n 10 +n 01 )/n. r has interval [0,0.5]: r=0 complete linkage, r=0.5, no linkage Proof of r^ = (n 10 +n 01 )/n The likelihood function of r given the observations: L(r|n ij ) = n!/(n 11 !n 10 !n 01 !n 00 !) × [ ½(1-r) ] n11 [ ½r ] n10 [ ½r ] n01 [ ½(1-r) ] n00 = n!/(n 11 !n 10 !n 01 !n 00 !) × [ ½(1-r) ] n11+n00 [ ½r ] n10+n01 log L(r|n ij ) = C+(n 11 +n 00 )log[ ½(1-r) ] +(n 10 +n 01 )log[½r] = C + (n 11 +n 00 )log(1-r) + (n 10 +n 01 )log r + nlog(½) Let the score ∂ logL(r|n ij )/ ∂ r = (n 11 +n 00 )[-1/(1-r)] +(n 10 +n 01 )(1/r) = 0, we have (n 11 +n 00 )[1/(1-r)]=(n 10 +n 01 )(1/r) → r^ = (n 10 +n 01 )/n Testing for linkage BC AaBb aabb Aabb aaBb Obs n 11 n 00 n 10 n 01 → n= Σ n ij Freq ½(1-r) ½(1-r) ½r ½r Gamete type n NR = n 11 +n 00 n R = n 10 +n 01 Freq with no linkage ½ ½ Exp ½n ½n x 2 = Σ (obs – exp) 2 /exp = (n NR- n R ) 2 /n ~ x 2 df=1 Example AaBb aabb Aabb aaBb 49 47 3 4 n NR = 49+47=96 n R = 3 + 4 = 7 n=96+7=103 x 2 = Σ (obs – exp) 2 /exp = (96-7)2/103 = 76.903 > x 2 df=1 = 3.841 These two markers are statistically linked. r^ = 7/103 = 0.068 Linkage analysis in the F2 BB Bb bb AA Obs n 22 n 21 n 20 Freq ¼(1-r) 2 ½r(1-r) ¼r 2 Aa Obs n 12 n 11 n 10 Freq ½r(1-r) ½(1-r) 2 +½r 2 ½r(1-r) aa Obs n 02 n 01 n 00 Freq ¼r 2 ½r(1-r) ¼(1-r) 2 Likelihood function L(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 Let the score = 0 so as to obtain the MLE of r, but this will be difficult because AaBb contains a mix of two genotype formation types (in the dominator we will have ½(1-r) 2 +½r 2 ). I will propose a shortcut EM algorithm for obtain the MLE of r BB Bb bb AA Obs n 22 n 21 n 20 Freq ¼(1-r) 2 ½r(1-r) ¼r 2 Recombinant 1 2 Aa Obs n 12 n 11 n 10 Freq ½r(1-r) ½(1-r) 2 +½r...
View Full Document

{[ snackBarMessage ]}

Page1 / 35

STA6178_2005(3) - Genetic design Testing Mendelian...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online