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.