An individual homozygous for genes cd is crossed with type and `F_1` Crossed back with the double recessive . The appearance of the offspring is as follows
`{:(++rarr,903),(cdrarr,897),(+drarr,98),(c+rarr,102):}`
The distance between the genes c and d is

Options

A. 20 map units
B. 9.8 map units
C. 10.2 map units
D. 10 map units

Answer

To solve the problem of determining the distance between genes c and d based on the given offspring phenotypes, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Parental and Recombinant Types**:
- The parental types are the phenotypes that resemble the original parents. In this case, they are:
- CD (homozygous parent) = 903
- ++ (wild type) = 897
- The recombinant types are the phenotypes that differ from the parental types due to crossing over:
- C+ d = 102
- c D+ = 98

2. **Calculate the Total Number of Offspring**:
- Add all the offspring counts together:
\[ \text{Total Offspring} = 903 + 897 + 102 + 98 = 2000 \]

3. **Calculate the Total Number of Recombinants**:
- Add the counts of the recombinant types:
\[ \text{Total Recombinants} = 102 + 98 = 200 \]

4. **Calculate the Distance Between Genes**:
- The distance between the genes is calculated using the formula:
\[ \text{Distance (in map units)} = \left(\frac{\text{Number of Recombinants}}{\text{Total Offspring}}\right) \times 100 \]
- Plugging in the values:
\[ \text{Distance} = \left(\frac{200}{2000}\right) \times 100 = 10 \text{ map units (MU)} \]

5. **Conclusion**:
- The distance between genes c and d is 10 map units.