By the cross of `F_(1)` linked dihybrid plant with a plant of genotype aabb, we get the 1300 plants out of 2000 of parental type. The distance between A and B gene is :-

Options

A. 35 map unit
B. 15 map unit
C. 65 map unit
D. 65 kello meter

Answer

To solve the problem of determining the distance between genes A and B based on the given data, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Total Number of Plants**:
We are given that the total number of plants produced from the cross is 2000.

**Hint**: Always start by identifying the total number of individuals in the experiment.

2. **Identify the Number of Parental Type Plants**:
According to the problem, out of the 2000 plants, 1300 are of the parental type.

**Hint**: Look for the number of plants that exhibit the parental phenotype, as this is crucial for calculating recombination frequency.

3. **Calculate the Number of Recombinant Plants**:
To find the number of recombinant plants, subtract the number of parental type plants from the total number of plants:
\[ \text{Number of Recombinant Plants} = \text{Total Plants} - \text{Parental Type Plants} = 2000 - 1300 = 700 \]

**Hint**: Remember that recombinant plants are those that do not show the parental phenotype.

4. **Use the Formula for Recombination Frequency**:
The formula to calculate the distance between the two genes (in map units) is:
\[ \text{Recombination Frequency} = \left( \frac{\text{Number of Recombinant Plants}}{\text{Total Number of Plants}} \right) \times 100 \]
Plugging in the values:
\[ \text{Recombination Frequency} = \left( \frac{700}{2000} \right) \times 100 \]

**Hint**: This formula helps to convert the proportion of recombinant plants into a percentage, which is expressed in map units.

5. **Calculate the Recombination Frequency**:
Performing the calculation:
\[ \text{Recombination Frequency} = 0.35 \times 100 = 35 \]

**Hint**: Make sure to multiply by 100 to convert the decimal into a percentage.

6. **Interpret the Result**:
The distance between gene A and gene B is 35 map units.

**Hint**: Map units are a measure of genetic distance based on recombination frequency.

### Final Answer
The distance between gene A and gene B is **35 map units**.