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Table 5 Conditional probability distribution of Yj values for pair of sibs

From: Quantitative trait locus analysis of hybrid pedigrees: variance-components model, inbreeding parameter, and power

Genotypes of sib-pair

Y j

Conditional probability Pr (YjQTLj)

  

πQTLj = 0

πQTLj = 1/2

πQTLj = 1

AA-AA

ξj2

p(AAf)p(AAm) + τ

1/2 (p(AAf)p(Am) + p(Af)p(AAm)) + τ

p(Af) p(Am) + τ

BB-BB

ξj2

P(BBf)p(BBm) +τ

1/2 (p(BBf)p(Bm) + p(Bf)p(BBm))+τ

p(Bf)p(Bm)+ τ

AB-AB

ξj2

p(AAf)p(BBm) + p(BBf)p(AAm) + 1/2p(ABf) p(ABm) -2τ

p(Bf)p(Am) + p(Af)p(Bm) - 1/4 [p(ABf)+ p(ABm)] - 2 τ

p(Af)p(Bm) + p(Bf)p(Am)-2τ

AA-AB

(a-dj)2

1/2 (p(ABf)p(AAm) + p(AAf)p(ABm))

1/4(p(Af)p(ABm) + p(ABf)p(Am))

0

AB-AA

(-a + d + ξj)2

1/2 (p(ABf)p(AAm) + p(AAf)p(ABm))

1/4(p(Af)p(ABm)+ p(ABf)p(Am))

0

AB-BB

(a + dj)2

1/2 (p(ABf)p(BBm) + p(BBf)p(ABm))

1/4(p(Bf)p(ABm) + p(ABf)p(Bm))

0

BB-AB

(-a-dj)2

1/2 (p(ABf)p(BBm) + p(BBf)p(ABm))

1/4(p(Bf)p(ABm) + p(ABf)p(Bm))

0

AA-BB

(2aj)2

1/4 p(ABf) p(ABm)

0

0

BB-AA

(-2a + ξj)2

1/4 p(ABf) p(ABm)

0

0

Total

 

1

1

1