A genetics problem...?

In bats, black, (B) is dominant to chesnut(b) and solid color (S) is dominant to spotted (s) .



What are the genotypes of the parents that would produce a cross with 3/8 black solid, 3/8 black spotted, 1/8 chestnut solid, and 1/8 chestnut spotted bats?



How would I figure this out?

I have no idea how to do this...

A genetics problem...?
Do the Punnet square with a pencil, so you can erase. Make a list of each genotype that can yield a certain phenotype. For example: black solid can be BBSS, BbSS, BBSs etc.



If you crossed 2 completely heterozygous parents (BbSs) with each other, your phenotypic ratios would yield a lot more black and solid phenotypes , so it's going to be a different answer. Also you have many more chestbut spotteds than that cross would yield. I would start with a cross which showed each one of the parents showing a different homozygous recessive genotype for one (but not both) of the traits and go from there.

The Website below shows Punnett Squares that might be helpful in getting you practice.

I'll see if I can do it myself when I get a break in a few mins - 8 PM EST.

OK. There are 16 squares in the dihybrid Punnett square cross. Since 1/8 are chestnut spotted, you must have two of the offspring with the genotype of bbss, so one parent must have ss and the other Ss. Bbss x BbSs doesn't work either, as there are too many black solids. Since you have both chestnuts and spotteds, you can't have either parent having BB or SS or all phenotypes would result in either blacks or solids accordingly.



If you try to finagle the square so you come out with 6 Black/Solid, 6 Black/smooth, 2 of each of the other categories, the only cross that seems to work is:



Bbss x BbSs
Reply:you have to make a square with each of the traits listed down each side and then cross them.
Reply:Diagram possible parent combinations and their progeny.

Start with the most basic which is a two heterozygotes.



ie. BbSs x BbSs (solid black)



=1 BBSS, 2 BbSS, etc.



Or figure out what kind of gametes were necessary from the description of the progeny.



ie solid black B?S? = B X ? and S x ?

spotted chestnut bbss = b x b and s x s
Reply:Both parents heterzygous for both traits