Brazhyphalangy in man causes shortening of the fingers in the
heterozygous condition and is lethal in the homozygous recessive
condition. If the fene is homozygous dominant, the hands are normal. A
man who has brachyphalangy hands and type A blood has a mother who has
normal fingers and type O blood. He is married to a brachyphalangyed
wife with type B blood who has a mother with type AB blood and a
father with type O blood. How many children would you expect to have
shortening of the fingers and type O blood?

HI again!

I'm calling the fingers trait F and f (because of the blood type, b
would just be confusing!).  This question is tricky because there are
three rather than two blood alleles - A, O, and B.  A and B are equal
to each other but dominant to O - so a type O person must be
homozygous for O.

Man: AO Ff
Woman: BO Ff

So, this is another 4x4 cross.  When you write out each of the
possibilities you'll see that 4 children will have type O blood; 1
will have normal fingers (FF), 2 will have short fingers (Ff), and one
will die (ff).  This is out of 16 possible combinations:

ABFF (1)
ABFf (2)
ABff (1)
AOFF (1)
AOFf (2)
AOff (1)
BOFF (1)
BOFf (2)
BOff (1)
OOFF (1)
OOFf (2)
OOff (1)

- Librariankt