Surely, what has caught the attention of this problem have been a couple of things:
- That there are three different phenotypes: mink can be black, platinum and sapphire.
- That there are three crosses of the same phenotypes with different results. In addition, in the third crossing a new phenotype appears different from that of the parents.
Due to the problems that we have seen so far, we could think that it is a case of intermediate inheritance, but the first crossing makes that we have to rule out. Therefore, it seems that the three different phenotypes are due to three different alleles for the same gene and we are facing a multiple allelism problem.
Now we will have to find out which allele dominates over the others. From the first crossing we deduce that the black color dominates over the sapphire (N>z), since from the crossing between one black (NN) and another sapphire (zz) they all come out black (Nz).
Cross two is the same phenotypically as the first, but the results are not the same. From the crossing of a black mink and another sapphire, half come out black and the other half sapphire. Therefore, as in crossing 1, it seems clear that black dominates over sapphire, that this case would be homozygous recessive (zz), and black, heterozygous (Nz).
In the third crossing, of a black mink and another sapphire, you get half black minks and the other half platinum. The sapphire phenotype is replaced by the platinum phenotype, which may make us think that the platinum allele dominates over sapphire (P>z), but black (N) dominates over them. ( N>P>z). According to this, from the crossing of a black mink (NP) with a sapphire (zz), a descendant of black (Nz) and platinum (Pz) color arises .
In the fourth crossing, from a sapphire mink (zz) and another sapphire (zz) all the sapphire-colored offspring (zz) emerge.
In the fifth crossing, of a platinum-colored mink (P) and a sapphire-colored one (zz), an offspring formed by half platinum-colored (Pz) and the other half sapphire (zz) is obtained, so the parent must be Pz.