A heterozygous individual `++ab` is crossed to its revessive parent and has produced the offsprings in the following proportion.
`++//ab-200`
`a+//ab-50`
`+b//ab-30`
`ab//ab-100`
What is the expected distance between the two gene loci ?

Options

A. 47 cm
B. 21 cm
C. `2.1` cm
D. 80 cm

Answer

To solve the problem of determining the distance between the two gene loci based on the offspring proportions from a cross between a heterozygous individual (`++ab`) and a recessive parent (`abab`), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**:
The heterozygous parent has the genotype `++ab`, meaning it carries two dominant alleles (let's denote them as A and B) and two recessive alleles (a and b). The recessive parent has the genotype `abab`.

2. **Determine Possible Gametes**:
The heterozygous parent can produce four types of gametes:
- `AB` (both dominant)
- `Ab` (dominant A and recessive b)
- `aB` (recessive a and dominant B)
- `ab` (both recessive)

3. **Cross with Recessive Parent**:
When these gametes are crossed with the recessive parent's gametes (which can only be `ab`), the following offspring genotypes can be produced:
- `AB//ab` (from `AB` gamete)
- `Ab//ab` (from `Ab` gamete)
- `aB//ab` (from `aB` gamete)
- `ab//ab` (from `ab` gamete)

4. **Count Offspring**:
The offspring proportions given are:
- `++//ab` (200)
- `a+//ab` (50)
- `+b//ab` (30)
- `ab//ab` (100)

5. **Calculate Total Offspring**:
Total offspring = 200 + 50 + 30 + 100 = 380.

6. **Identify Parental and Recombinant Types**:
- Parental types: `++//ab` (200) and `ab//ab` (100) = 300
- Recombinant types: `a+//ab` (50) and `+b//ab` (30) = 80

7. **Calculate Recombination Frequency**:
Recombination frequency (RF) = (Number of recombinant offspring / Total offspring) × 100
\[ RF = \frac{80}{380} \times 100 \approx 21.05\% \]

8. **Determine Distance Between Gene Loci**:
The distance between the two gene loci can be measured in centimorgans (cM), where 1% recombination frequency corresponds to 1 cM. Therefore, a recombination frequency of approximately 21% corresponds to a distance of about 21 cM.

### Final Answer:
The expected distance between the two gene loci is **21 cM**.