Individuals homozygous for 'xy' genes were crossed with wild type '++' . The `F_1` dihybrid thus produced was test crossed . It produced progeny in the following in the following ratio `'+ + ' 900, ' +y ' 115, 'xy' 880, 'x+' 105. Find out the recombination frequency .

Options

A. 47 map units
B. 88 map units
C. 11 map units
D. 5.75 map units

Answer

To solve the problem, we need to calculate the recombination frequency based on the given progeny ratios from the test cross.

### Step-by-Step Solution:

1. **Identify Parental and Recombinant Types**:
- The parental types are the ones that resemble the original homozygous individuals. In this case, they are:
- '++' (900 individuals)
- 'xy' (880 individuals)
- The recombinant types are those that differ from the parental types:
- '+y' (115 individuals)
- 'x+' (105 individuals)

2. **Count the Total Number of Progeny**:
- Total progeny = 900 (++) + 880 (xy) + 115 (+y) + 105 (x+)
- Total progeny = 2000

3. **Calculate the Total Number of Recombinants**:
- Total recombinants = '+y' + 'x+'
- Total recombinants = 115 + 105 = 220

4. **Calculate the Recombination Frequency**:
- The formula for recombination frequency (RF) is:
\[ \text{Recombination Frequency (RF)} = \left( \frac{\text{Total number of recombinants}}{\text{Total number of progeny}} \right) \times 100 \]
- Substituting the values:
\[ RF = \left( \frac{220}{2000} \right) \times 100 = 11 \]

5. **Conclusion**:
- The recombination frequency is 11%. This can also be expressed as 11 centimorgans (cM) or map units.

### Final Answer:
The recombination frequency is **11%**.

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