Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 2 / b / Solution

Only heterozygous parents contribute an alternative-allele transmission contrast to the transmission disequilibrium test. Both parents are homozygous in rows 1, 13, 14 and 15, so rows 1, 13, 14 and 15 are uninformative. Their counts total 15 families. Every other row has at least one heterozygous parent. When both parents are heterozygous and the child is heterozygous, there is one transmission of each allele, regardless of the unresolved parent-specific phase; these transmissions contribute to both discordant cells.