Question

A female rabbit heterozygous at two genes (named A and B) is test-crossed with a male rabbit. Their progeny include 442 A/a;B/b, 458 a/a;b/b, 46 A/a;b/b, and 54 a/a;B/b. Explain these results. Identify which alleles are recessive, whether genes A and B are linked, and, if so, how far apart they are on the chromosome

Answer 1

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 ab 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