Explanation:

Given:
- An individual with cd genes is crossed with wild type.
- On test crossing F, the progeny was c 105, d 115, cd 880, and 900.

To determine the distance between the cd genes, we need to analyze the results of the test cross progeny.

1. Determine the parental genotypes:
- The individual with cd genes is the parent with genotype cd.
- The wild type individual is the parent with genotype ++ (wild type for both genes).

2. Determine the genotypes of the test cross progeny:
- The progeny includes c 105, d 115, cd 880, and 900.
- The numbers indicate the number of individuals with each genotype.

3. Calculate the recombination frequency:
- The recombination frequency is the percentage of recombinant offspring.
- Recombinant offspring are those with a different combination of alleles than the parents.

- The total number of recombinant offspring is the sum of c d and cd (105 + 880 = 985).
- The total number of offspring is the sum of all genotypes (105 + 115 + 880 + 900 = 2000).

Recombination frequency = (Total number of recombinant offspring / Total number of offspring) * 100
= (985 / 2000) * 100
= 49.25%

4. Convert the recombination frequency to map units:
- 1% recombination frequency is equal to 1 map unit or centimorgan.
- Therefore, 49.25% recombination frequency is equal to 49.25 map units.

5. Determine the distance between the cd genes:
- The distance between two genes can be measured in terms of the frequency of recombination between them.
- Since the cd genes are on the same chromosome, the recombination frequency represents the distance between them.

Therefore, the distance between the cd genes is 49.25 map units or centimorgans.

Answer: The distance between the cd genes is 49.25 map units or centimorgans, which is approximately 49 map units.

Note: The options provided in the question may have been rounded for simplicity, so the closest option to the calculated distance is option C, 11 map units. However, the correct answer should be approximately 49 map units.