 Solution

Count one transmitted/untransmitted allele pair per observable parent. A homozygous parent contributes to $a$ or $d$; a heterozygous parent contributes to $b$ or $c$. When two heterozygotes have a heterozygous child, the parent-specific origin is ambiguous, but the combined contribution is unambiguously one to $b$ and one to $c$.

The derived contributions, ordered as $(a,b,c,d)$, areIn family 7, the observed father's transmission cannot be determined from the heterozygous child without the mother's genotype. In family 10, the homozygous child reveals the father's allele-1 transmission, but not the missing mother's complete transmitted/untransmitted pair.

Keeping only unambiguous observed-parent transmissions and omitting family 7 gives the raw tableThis bookkeeping convention is not the same as a valid ordinary transmission disequilibrium test sampling rule. Incomplete-trio transmission-selection bias arises if only homozygous children make a missing-parent family usable. The usual complete-trio analysis excludes both families 7 and 10, givingNeither assigning family 7 a half transmission each way nor inventing its mother's genotype is justified by the supplied data. A method using all twelve families must model or validly condition on the missing information.