In a cross between genotype AB and ++,650 out of 1000 individuals were of parental type The distance between A and B

Options

A. (a) 35 map units
B. (b) 45 map units
C. (c) 15 map units
D. (d) 30 map units

Explanation

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

### Step-by-Step Solution:

1. **Identify the Total Number of Individuals**:
We are given that there are 1000 individuals in total.

2. **Identify the Number of Parental Type Individuals**:
Out of the 1000 individuals, 650 are of the parental type.

3. **Calculate the Number of Recombinant Individuals**:
To find the number of recombinant individuals, we subtract the number of parental type individuals from the total:
\[ \text{Number of Recombinants} = \text{Total Individuals} - \text{Parental Type Individuals} = 1000 - 650 = 350 \]

4. **Determine the Recombinant Frequency**:
The recombinant frequency (RF) is calculated by dividing the number of recombinant individuals by the total number of individuals, and then multiplying by 100 to express it as a percentage:
\[ \text{Recombinant Frequency (RF)} = \left( \frac{\text{Number of Recombinants}}{\text{Total Individuals}} \right) \times 100 = \left( \frac{350}{1000} \right) \times 100 = 35\% \]

5. **Convert Recombinant Frequency to Map Units**:
In genetics, the distance between two genes is often expressed in map units (centimorgans), where 1% recombinant frequency equals 1 map unit. Therefore, the distance between genes A and B is:
\[ \text{Distance between A and B} = \text{Recombinant Frequency} = 35 \text{ map units} \]

6. **Conclusion**:
The distance between genes A and B is 35 map units.

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

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