You are supposed to say hich allele is dominant fo the 2 characteristics of your mice hair (smooth/curly and Black/white). and then draw the punnet squares for the first generation (F1): [white smooth]x[black curly] and then the same thing for a [black smooth]x[black smooth] (the F2) based on the genotype you have determined on the preceding step.

You might considered find a middle school life science textbook. They explain this type of problem in very simple terms.

If the dominant and recessive traits are not given, I don’t think you can tell which traits are dominant and which are recessive with this particular cross. Seems like some information is missing. If both parents in this cross are homozygous then you would know that black(B) and smooth(S)are dominant because all of the offspring were black and smooth.

However, I do not see that given in the problem.

If I do assume that they are homozygous the parents would have the genotypes BBss and bbSS because the offspring are all black with smooth coats. There is still much to understand to solve this problem.
1. parents must have two alleles for each trait such as (BB) (SS) (traditional caps are used for the dominant trait lower case for recessive trait)
2. The alleles separate during meiosis (part of sexual reproduction)
3. The alleles are reunited in offspring, but one allele must come from the father and one from the mother for each pair. In this case one allele for coat color and one allele for fur type from each parent. This means each offspring will end up with two alleles for coat color and two for fur type.

To set up the problem, again assuming the parents are homozygous
1. determine genotype of each parent, remember must be in pairs
Black curly parent White smooth parent
BBss bbSS
2. determine the possible gametes of each parent, one allele from each set
Black curly parent White smooth parent
Bs bS
3. multiply the gametes to find the offspring use a Punnent square with the gametes of one parent on the horizontal and the other parent on the vertical axis of the square. Or use simple algebra. Tradition the dominant trait appears first in a pair.
(Bs)(bS) = BbSs

For the second generation you repeat the three steps given above.
The gametes for both male and female would be BS, Bs, bS, bs
(BS+Bs+bS+bs)(BS+Bs+bS+bs) = BBSS + BBSs + BbSS +BbSs + BBSs+BBss+BbSs+Bbss+BbSS+BbSs+bbSS+bbSs +BbSs+Bbss+bbSs+bbss
identify phenotype (traits)
BBSS Black Smooth
BBSs Black Smooth
BbSS Black Smooth
BbSs Black Smooth
BBSs Black Smooth
BBss Black curly
BbSs Black Smooth
Bbss Black curly
BbSS Black Smooth
BbSs Black Smooth
bbSS white Smooth
bbSs White Smooth
BbSs Black Smooth
Bbss Black Curly
bbSs White smooth
bbss White curly
Add up all of the like phenotypes 9 Black Smooth, 3 Black curly, 3 white smooth, one white curly. The same steps apply to the Punnet square. Many people find the square an easier way to multiply. My guess it is because it is algebra in disguise. 🙂