Page 141 - Applied Probability

7. Computation of Mendelian Likelihoods
TABLE 7.2. Phenotypes for a Paternity-Testing Problem
Person
1/1
Mother
AB
B
1/2
Child
B
Putative Father ABO Phenotype ADA Phenotype 125
1/2
Weinberg and linkage equilibrium. The paternity index for the jth locus is
j
j
Pr(X | Ped 1 )/ Pr(X | Ped 2 ). Over all loci it is
j
Pr(X | Ped 1 )
j
j . (7.6)
Pr(X | Ped 2 )
j
Let α be the prior probability that the putative father is the actual father
based on the nongenetic evidence; let β be the posterior probability that
the putative father is the actual father based on both the nongenetic and
the genetic evidence. Then a convenient form of Bayes’ theorem is
j
β α j Pr(X | Ped 1 )
= .
j
1 − β (1 − α) Pr(X | Ped 2 )
j
The exclusion probability for the jth locus can be found by carrying out
the genotype elimination algorithm on Ped 2 for this locus. Let S j be the
set of non-excluded genotypes for the random male. Then the exclusion

probability for locus j is 1 − Pr(G j ). The exclusion probability
G j ∈S j

over all loci typed is 1 − [ Pr(G j )].
j G j ∈S j
TABLE 7.3. Statistics for the Paternity-Testing Example
Locus Paternity Index Exclusion Probability
ABO 1.39 .078
ADA 7.58 .872
Both Loci 10.5 .882
As a simple numerical example, consider the phenotype data in Table 7.2.
At the ABO locus, suppose the three alleles A, B, and O have population
frequencies of .28, .06, and .66, respectively. At the ADA locus, suppose
the two codominant alleles 1 and 2 have population frequencies of .934
and .066, respectively. It is evident in this case that the only excluded
genotype for the father at the ABO locus is A/A; at the ADA locus, the only
excluded genotype is 1/1. Table 7.3 lists the computed paternity indices and
exclusion probabilities. Although the ADA locus is less polymorphic than
the ABO locus, in this situation it yields the more decisive statistics. In
practice, a larger number of individually more polymorphic loci would be
used.