Page 106 - Applied Probability

5. Genetic Identity Coefficients
90
1
=[1 − 2
2
1
s
Φ[{G j ,...}{} ...{}]
+
2
s
1

+ s ]Φ[{G j ,G k ,...}{} ...{}]
Φ[{G k ,...}{} ...{}].
2
1
s
In this rule the genes G ,...,G are replaced, respectively, by single
i i
genes from both j and k, by a single gene from j only, or by a single
1 s
gene from k only. The three corresponding coefficients 1 − 2( ) ,
2
1 s
1 s
( ) , and ( ) are determined by binomial sampling with s trials and
2 2
1
success probability .
2
s
1
Recurrence Rule 3 Assume that the genes G ,...,G ,G s+1 ,...,G s+t
i i i i
are sampled from i. If the first s genes occur in one block and the
remaining t genes occur in another block, then
s
1
Φ[{G ,... ,G ,...}{G s+1 ,... ,G s+t ,...}{} ...{}]
i
i
i
i
1
s+t
= Φ[{G j ,...}{G k ,...}{} ...{}]
2
1
s+t
+ Φ[{G k ,...}{G j ,...}{} ...{}].
2
This rule follows because neither the maternal gene nor the paternal
gene of i can be present in both blocks. Again binomial sampling
1 s+t
determines the coefficients ( ) .
2
Example 5.6.1 Sample Calculations for an Inbred Pedigree
Consider the inbred siblings 5 and 6 in Figure 5.1. Let us compute the
2
2
1
1
l
kinship coefficient 1 Ψ 8 =Φ({G ,G }{G }{G }), where G denotes the
4 5 6 5 6 k
lth sampled gene of person k. Recurrence rule 3 and symmetry imply
1 1 1 1 1 2 1 1 1 1 2
Ψ 8 = Φ({G ,G }{G }{G })+ Φ({G ,G }{G }{G })
4
6
3
6
6
3
6
4
4 4 4
1 1 1 1 2
= Φ({G ,G }{G }{G })
3
4
6
6
2
1 1 2 1 2 1 1 2 1 2
= Φ({G ,G }{G }{G })+ Φ({G ,G }{G }{G }).
4
4
3
3
3
3
4
4
8 8
Recurrence rules 2 and 3, boundary conditions 2 and 3, and symmetry
2
1
1
2
permit us to express A =Φ({G ,G }{G }{G })as
4
3
4
3
1 1 1 1 2 1 1 1 2
A = Φ({G ,G }{G }{G })+ Φ({G }{G }{G })
2
4
4
1
1
4
4
2 4
1
2
1
1
+ Φ({G }{G }{G })
4
4
2
4
1
2
1
1
=0 + Φ({G }{G }{G })
4
4
1
2