HELP with GENETICS 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.From the info given, I think the man is Bbww, the woman is bbWw & the child is bbww. I have to SHOW it though,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- 50-50 chance

Please Help explain this to me, I Don't know what combinations should be in my punnett square, what letters go where? Thanks so Much! = )

Answer 1

Yes. You need to do a dihybrid cross.
The genotype of the parents is correct (Bbww x bbWw). The gamete combinations will look like this (on the left) Bw Bw bw bw and (on the top) bW bw bW bw. The offspring genotypic possibilities will include (running top left to right): 1st row: BbWw Bbww BbWw Bbww, 2nd row: BbWw Bbww BbWw Bbww, 3rd row: bbWw bbww bbWw bbww, 4th row: bbWw bbww bbWw bbww. Total genotype count 4BbWw 4Bbww 4bbWw 4bbww / phenotypes 4 brachydactyly woolly hair, 4 brachydactyly normal hair, 4 normal fingers woolly hair, 4 normal fingers normal hair.

The 1st child's genotype is bbww (homozygous recessive). Fifty percent of the offspring should have normal figers.

Hope this helps.

Answer 2

You've got all your answers right. You'd set up the cross like this:
Dad on top, mom on the left

Bw bw
bW
bw

Since those are the only alleles the parents could make

Answer 3

get a life.