In maize, coloured endosprem (C) is dominant over colourless (c) , and full endosperm (R) is dominant over shrunken (r). When a dihybrid of `F_(1)` generation was test crossed, it produced four phenotypes in the following percentage :
Coloured full - 48 % Coloured shrunken - 5 %
Colourless full - 7 % Colourless shrunken - 40 %
From this data, what will be the distance between two non-allelic genes ?

Options

A. 48 units
B. 5 units
C. 7 units
D. 12 units

Explanation

To solve the problem of determining the distance between two non-allelic genes in maize based on the given phenotypic percentages from a test cross, we can follow these steps:

### Step 1: Identify the Phenotypes
From the test cross results, we have the following phenotypes:
- Coloured full (C R) - 48%
- Coloured shrunken (C r) - 5%
- Colourless full (c R) - 7%
- Colourless shrunken (c r) - 40%

### Step 2: Classify the Phenotypes
We can classify the phenotypes into parental types and recombinant types:
- **Parental types**:
- Coloured full (48%)
- Colourless shrunken (40%)

- **Recombinant types**:
- Coloured shrunken (5%)
- Colourless full (7%)

### Step 3: Calculate the Total Percentage of Recombinants
To find the total percentage of recombinant types, we add the percentages of the two recombinant phenotypes:
- Total recombinant percentage = Percentage of coloured shrunken + Percentage of colourless full
- Total recombinant percentage = 5% + 7% = 12%

### Step 4: Determine the Map Distance
The map distance between two genes is calculated based on the percentage of recombinant offspring. Each 1% of recombinants corresponds to 1 map unit (centiMorgan).
- Therefore, the map distance = Total recombinant percentage = 12%

### Conclusion
The distance between the two non-allelic genes in maize is **12 map units**.