Solutions For An Introduction to Genetic Analysis Chapter 17 Problem 71P

a) The species B. campestris was crossed with B. napus and the hybrid has 29 chromosomes consisting of 10 bivalents and 9 univalents. According to the table, the species B. napus should have contributed 19 chromosomes to the hybrid. Therefore, B. campestris should have contributed 10 chromosomes. Hence, the 2n number of B. campestris would be 20.

When B. nigra was crossed with B. napus, 8 chromosomes should have been contributed by B. nigra. Hence, the 2n number of B. nigra is 16.

Since the species of B. juncea has 18 bivalents, we minus this number from 27.When B. juncea is crossed with B. oleracea, 9 chromosomes should have been contributed by B. oleracea. Hence, the 2n number of B. oleracea is 18.

b) The haploid and diploid number for each species can be tabulated as seen:

The presence of a bivalent in a hybrid indicates that the chromosomes are identical. When more number of bivalents is formed in a hybrid, it indicates that the two parent species are evolutionary close to each other.

In the data given, the last three crosses result in no bivalents which indicate that the parent species are not closely related.

Three crosses result in bivalents of 8, 9 and 10. These suggest a closer relationship among the parents. These three crosses are:

If we see in each cross, the number of bivalents is equal to the number of haploid chromosomes of one species. This suggests that the species with the larger haploid number is a hybrid composed of the second species with some other species.

In each species, the haploid number of the unknown species is the number of univalents. Hence, we can deduce the following relationships:

B. juncea is an amphidiploid formed by the cross of B. nigra and B. campestris.

B. napus is an amphidiploid formed by the cross of B. campestris and B. oleracea.

B. carinata is an amphidiploid formed by the cross of B. nigra and B. oleracea.

These conclusions also match the last three crosses which have no bivalents.