Pedigree analysis of an X-linked recessive

    Marriage between a male with an X-linked recessive trait (a- or aY) and an unaffected woman (AA) produces children with one of two genotypes. All of the sons will are A- (AY), with the Y chromosome from the father and an A allele from the mother. All of the daughters are Aa (ordinarily shown as a circle & dot), with the a allele from the father and an A allele from the mother. They are therefore heterozygous carriers for the trait: they do not show the trait, but can pass it along to their sons.

    When an Aa carrier woman marries an unaffected man (A- or AY), four genotypes are produced, in equal proportions. Half of the sons will show the trait (aY) and half will not (AY), half the daughters will be carriers like their mother (Aa) and half will not (AA).

    Note the pattern of "criss-cross inheritance," where an affected male has an unaffected daughter, who in turn has an affected son. The trait "skips a generation."

    Homework: How might a daughter be born with an aa genotype [disregarding brother / sister, or father / daughter matings]?

Homework: Write out the genotypes of every individual in the tree.

Figure ©2002 by Griffiths et al;. all text material ©2012 by Steven M. Carr