Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with the following circumstance:

linkage with approximately 33 map units between the two gene loci

Explanation:

If two genes are not linked, number of recombinants and parental offspring will be equal. Here it is clearly visible that recombinants are less than parental offspring hence the genes are linked. Given, the offspring are in following numbers:

AaBb = 106 = Parental

aabb = 94 = Parental

Aabb = 48 = Recombinant

aaBb = 52 = Recombinant

Recombination frequency = (Number of recombinants / Total progeny) * 100 = (100/300) * 100 = 33.33 %

1% recombination frequency = 1 map unit of distance between the two gene loci. So here the distance between the two gene loci is approximately 33 map units.

Hence, these results are consistent with linkage with approximately 33 map units between the two gene loci