Individuals homozyous for cd genes were crossed with wid type (++). The `F _(1)` dihybrid thus produced was test crossed It produced progeny in the following ratio-
`{:(++, 900),(cd,880) ,(+d ,115), (+c,105):}`
What is distance between c and d

Options

A. `5.75` unit
B. 11 unit
C. 27 units
D. 88 units

Answer

To determine the distance between the genes c and d based on the given progeny ratios, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Parental and Recombinant Types**:
- From the progeny ratios provided, we can identify the parental types and the recombinant types.
- Parental types:
- Wild type (++): 900
- Homozygous for cd (cd): 880
- Recombinant types:
- (+d): 115
- (+c): 105

2. **Calculate the Total Number of Progeny**:
- Total progeny = Number of wild type + Number of homozygous cd + Number of (+d) + Number of (+c)
- Total = 900 + 880 + 115 + 105 = 2000

3. **Calculate the Total Number of Recombinants**:
- Total recombinants = (+d) + (+c)
- Total recombinants = 115 + 105 = 220

4. **Calculate the Recombination Frequency**:
- Recombination frequency (RF) = (Number of recombinants / Total progeny) × 100
- RF = (220 / 2000) × 100 = 11%

5. **Determine the Distance Between Genes**:
- The distance between genes c and d is equivalent to the recombination frequency, which is expressed in centiMorgans (cM).
- Therefore, the distance between c and d is 11 cM.

### Final Answer:
The distance between genes c and d is **11 cM**.

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