Answer:
The genes are linked and 10 mu apart.
Explanation:
A female AaBb rabbit is test crossed with a male rabbit (aabb). The male can only produce<em> ab</em> gametes (all the progeny will have ab on one of the homologous chromosomes).
If the genes assorted independently, the female would produce 4 types of gametes with the same frequency: 1/4 AB, 1/4 Ab, 1/4 aB and 1/4 ab.
However, the observed AB and ab gametes were much more frequent than Ab and aB, which means that the genes are linked and alleles on the same chromosome do not assort independently during meiosis.
Recombination is a rare event, so the most abundant gametes are the parentals. That is how we know that the mother had the AB/ab genotype. The recombinant gametes therefore are Ab and aB.
Distance (mu) = # Recombinants × 100/ Total progeny
Distance = (54 + 46) × 100/ 1000
Distance = 100 × 100/1000
Distance = 10 mu
