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 expected distance between the two gene loci based on the given offspring proportions, we can follow these steps:

### Step 1: Identify the Parental and Recombinant Offspring
From the given offspring proportions:
- Parental types: `++//ab` (200) and `ab//ab` (100)
- Recombinant types: `a+//ab` (50) and `+b//ab` (30)

### Step 2: Calculate the Total Number of Offspring
Total offspring = Parental offspring + Recombinant offspring
- Total = 200 (++//ab) + 100 (ab//ab) + 50 (a+//ab) + 30 (+b//ab)
- Total = 200 + 100 + 50 + 30 = 380

### Step 3: Calculate the Total Number of Recombinants
Total Recombinants = `a+//ab` + `+b//ab`
- Total Recombinants = 50 + 30 = 80

### Step 4: Calculate the Recombinant Frequency
Recombinant Frequency = (Total Recombinants / Total Offspring) × 100
- Recombinant Frequency = (80 / 380) × 100

### Step 5: Simplify the Calculation
- Recombinant Frequency = (80 / 380) × 100 = 21.05 (approximately 21)

### Step 6: Interpret the Result
The recombinant frequency represents the map distance between the two gene loci in map units (centiMorgans). Therefore, the expected distance between the two gene loci is approximately 21 units.

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

---