The genetic distance between genes A and B is 10 cm. An organism with Ab combination of the alleles is crossed with the organism with aB combination of alleles. What will be the percentage of the gametes with AB alleles combinatory by an F1 individual ?

Options

A. 1
B. 5
C. 10
D. 50

Explanation

To solve the problem, we need to determine the percentage of gametes with the AB allele combination produced by an F1 individual when genes A and B are 10 centimorgans apart.

### Step-by-Step Solution:

1. **Understanding Genetic Distance**:
The genetic distance of 10 cm (centimorgans) between genes A and B indicates that there is a 10% chance of recombination occurring between these two genes during meiosis.

2. **Parental and Recombinant Gametes**:
- The parental gametes from the parents (Ab and aB) are AB and ab.
- The recombinant gametes that can be formed due to crossing over are Ab and aB.

3. **Gamete Combinations**:
- From the cross of Ab (parent 1) and aB (parent 2), the possible gametes produced by the F1 individual are:
- Parental types: AB and ab
- Recombinant types: Ab and aB

4. **Calculating Gamete Frequencies**:
- Since the genetic distance is 10 cm, the percentage of recombinant gametes (Ab and aB) is 10%.
- Therefore, the percentage of parental gametes (AB and ab) is 100% - 10% = 90%.

5. **Percentage of AB Gametes**:
- Since there are two parental types (AB and ab), the percentage of AB gametes specifically is half of the parental gametes.
- Thus, the percentage of AB gametes = 90% / 2 = 45%.

### Final Answer:
The percentage of gametes with the AB allele combination produced by an F1 individual is **45%**.