Page 195 - Applied Probability

9. Descent Graph Methods
180
genetic equilibrium, each founder gene is sampled independently; therefore,
m

=
Prior(G)
i=1
m Pr(a i )

= Pr(a ij ). (9.10)
i=1 j
By construction, the founder genes assigned to different components do not
impinge on one another. In other words, the set of founder genes consistent
with G and M is drawn from the Cartesian product of the sets S 1 ,... ,S m of
0
legal allele vectors for the components C 1 ,...,C m , respectively. Applying
the distributive rule to equation (9.10) consequently yields
m

Prior(G) = Pr(C i ), (9.11)
i=1
G à 0 G∩M
where

Pr(C i )= Pr(a ij ).
a i∈S i j
As mentioned earlier, an allele vector set S i contains either all allele vectors
or just two, one, or none. In the first case, Pr(C i )= Pr(a i )= 1,
a i ∈S i

and in the remaining three cases, Pr(C i )= Pr(a i ) contains only
a i ∈S i
two, one, or no terms. Hence, calculation of Prior(G) reduces

G à 0 G∩M
to easy component-by-component calculations.
Although likelihood calculation with non-codominant markers or incom-
pletely penetrant traits can be handled similarly, two complications intrude.
First, we need a systematic method of generating the set S i of allele vectors
for component C i . Second, we must include penetrance values in the like-
lihood calculation, assuming that each person’s phenotypes at the various
loci are independent conditional on his or her genotypes at the loci. Re-
garding the second complication, note that each component C i carries with
it a set Q i of phenotyped people through whom the founder genes pass.
Specifying an allele vector a i ∈ S i determines the genotype of each person

k ∈ Q i . In computing Pr(C i ), we must multiply the product Pr(a ij )by
j
the penetrance of each k ∈ Q i at the current locus.
The allele vectors a i ∈ S i can be generated efficiently by a backtracking
scheme [29]. This entails growing a compatible allele vector from partial
vectors that are compatible. The idea can be illustrated by reference to
component C 2 = {A, C, E} of Figure 9.2 We start with the assignment
(a A ) = (1), which is consistent with the phenotypes in the pedigree, grow
it to (a A ,a C )= (1, 1), which is inconsistent, discard all vectors begin-
ning with (a A ,a C )= (1, 1), move on to (a A ,a C )= (1, 2), which is consis-
tent, grow this to (a A ,a C ,a E )=(1, 2, 1), which is consistent, discard each