Please help with a genetics problem 10 points best answer for help

In cocker spaniels, black color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s). If the genes are unlinked, and two BbSs individuals are mated with each other, what fraction of their offspring will be black and spotted?

A. 1/16
B. 9/16
C. 1/9
D. 3/16
E. 3/4

To answer this question we need to form a punnett square. Since they are unlinked genes there is an equal chance of every possible allele. So form a grid with the following gametes on the rows and columns: BS, Bs, bS, and bs. The columns represent the gamete from the mother and the rows represent the gamete from the father. Look at the genomes of all possible children and look for those that are black and spotted (Bbss, and BBss). You can be heterozygous for color because black is dominant but you need to be homozygous for spots since it is recessive. Here is the punnet square (my apologies on the spacing issues with the table. but there are four columns and four rows, the first line is the headings to each of the four columns):

BS Bs bS bs

BS BBSS BBSs BbSS BbSs

Bs BBSs BBss BbSs Bbss

bS BbSS BbSs bbSS bbSs

bs BbSs Bbss bbSs bbss


answer= 3/16=D

If the genes were linked then you would eliminate two of the rows and columns from the above table. Say if BS were linked and bs were linked in the mother (the first two on the chromasome from the mother's one parent, the second form the other parent) then you would not have bS or Bs as possible options. The father would also have two of his options crossed out. So you would only have 4 possible outcomes instead of 16.