As with
an autosomal dominant trait, the X-linked
dominant must be present in every generation of
the affected lineage. The genotype of an
individual with an X-linked dominant trait can
be inferred directly from inspection of the phenotype.
In a marriage between an
affected male (AY) (I-1) and an unaffected woman (aa),
the father's A will be passed to all daughters,
as will the mother's a, such that all daughters
are heterozygotes
(Aa). All sons will receive a Y
from their father and an a from their mother,
and will be unaffected (aY). An affected heterozygous
woman (Aa) (II-2)
who marries an unaffected man (aY) will pass the
A allele on to half
of her children, including sons (AY) and
daughters (Aa) in equal proportions.
Note that affected women
are almost always heterozygous (Aa). If an
affected woman were homozygous (AA), all of
her children (male and female) would be affected.
Homework:
(1) Expand the above pedigree, to show a cousin marriage
that could produce a homozygous (AA) woman.
(2) Write out the genotypes of every individual in the
tree.