More Genetics Help Please!?

Question

In humans, short fingers(brachydactyly) [B] and woolly hair [W] are dominant over the recessive alleles for normal fingers [b] & normal hair [w]. A man with brachydactyly & normal hair marries a woman with normal fingers & woolly hair. Their 1st child has normal fingers & normal hair.
a- What is the genotype of the man?
b- What is the genotype of the woman?
c- What is the genotype of the child?
d- What is the probability of any child of the above mentioned mating having normal fingers?
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I think I have to make a Dihybrid cross, I'm not sure what I'm using in it exactly. From the info given, I think the man is Bbww, the woman is bbWw & the child is bbww. What is my dihybrid supposed to be made of though?What goes on top- on the left?
a- Dad= heterozygous?
b- Mom= heterozygous?
c- Child=homozygous?
d-?? idk b/c idk how my punnett square should be set up

Please Help explain this to me, I am confused about what combinations should be in my punnett square. Thanks! = )

Answer 1

The kid must be bbww, because he has two recessive traits.

The father must be ww, or his hair would be wooly. He must have at least one B allele, because his fingers are short, but he must also have at least one b allele, because he passed it to his son. Therefore the father is Bbww.

The mother must be bb because she has normal fingers. She must have at least one W allele to account for her hair, and one w allele to pass to her son, so she's bbWw. So far you've gotten all that right.

Now for the Punnet square. It asked for the probability of normal fingers, so just forget the hair. On the side you have the mother who is bb, and on the top the father who is Bb. The mother always passes on a b allele, and the father has a 50-50 chance of passing on a b or a B. Therefore the chance of a child of theirs having normal fingers is 1/2.