To determine the distance between linked genes, we need to analyze the results of a test cross. In this case, the test cross was conducted between F1 individuals with genotypes A/ and B. The offspring produced from this test cross were as follows:

- /ab = 9
- ab/ab = 9
- b/ab = 41
- a/ab = 41

To calculate the distance between the linked genes A and B, we need to compare the number of recombinant offspring (those with new combinations of alleles) to the total number of offspring.

The recombinant offspring in this case are those with the genotype /ab and ab/ab, which add up to 9 + 9 = 18.

The total number of offspring is the sum of all four genotypes, which is 9 + 9 + 41 + 41 = 100.

To calculate the distance between the linked genes, we use the formula:

Distance = (Recombinant offspring / Total offspring) x 100

Plugging in the values we obtained earlier:

Distance = (18 / 100) x 100 = 18 cM

Therefore, the distance between the linked genes A and B is 18 cM.

Now, we need to determine if the linkage is in cis or trans configuration. The cis configuration refers to the two linked genes being on the same chromosome, while the trans configuration refers to them being on different chromosomes.

To determine the configuration, we compare the numbers of offspring with the genotypes /ab and ab/ab. If the numbers are roughly equal, the genes are likely in the cis configuration. If one genotype is significantly more common than the other, the genes are likely in the trans configuration.

In this case, we have 9 offspring with the genotype /ab and 9 offspring with the genotype ab/ab. Since the numbers are equal, we can conclude that the genes are in the cis configuration.

Therefore, the correct answer is D) 18 cM (cis).