Pedigree analysis of an X-linked recessive trait
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."
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.
Homework: Write out the genotypes of every individual in the tree.