Expert: Dana Krempels, Ph.D. - 12/2/2009

Question
An albino, nonhemophilic man, marries a normally pigmented, nonhemophilic woman whose father was hemophilic and whose mother was albino. What kind of children can they have and in what proportion? Hemophilia is a sex-linked trait.
(This makes no sense to me at all please explain thanks.)


Answer
Dear JT,

I will explain, but not do the square for you, since that's your homework.  :)

We're considering two traits here: albinism and hemophilia (the latter is X-linked).

If we assign the albinism locus the alleles A (normal) and a (albino), then the man is aa, and the woman--judging from her phenotype--could be either AA or Aa.  (But HINT:  If her mother was albino, what does this tell you about her genotype with respect to the A locus?)

If we assign the hemophilia locus the allele H (normal) and h (hemophilia), then the man MUST have only the H allele, since he has only one copy (he's male, and so has only one X chromosome).  The non-hemophiliac woman receives her X chromosome from her father, so that should tell you what's carried on at least ONE of her X chromosomes.  And if she doesn't express hemophilia, then you know the other one, right?

Given all this information, you should be able to determine the genotypes of the man and woman with respect to these two loci.  The possibilities:

AAHH
AAHh
AAhh
AaHH
AaHh
Aahh
aaHH
aaHh
aahh

...and from the information given, you should be able to determine which of these possible genotypes belong to the man and to the woman.  And that will allow you to set up the Punnett square, remembering that each gamete will cary only ONE allele of the A locus and ONE allele of the H locus.

Hope that helps!

Dana