In a Mendelian dihybrid cross, the probability of getting seeds with genotype Rryy, RrYy, rrYy and RrYY in `F_(2)` generation is respectively

Options

A. `(2)/(16):(4)/(16):(1)/(8):(1)/(8)`
B. `(2)/(16):(2)/(16):(2)/(16):(2)/(16)`
C. `(4)/(16):(4)/(16):(2)/(16):(2)/(16)`
D. `(1)/(8):(1)/(4):(2)/(8):(1)/(16)`

Answer

To solve the problem of determining the probability of getting seeds with specific genotypes in a Mendelian dihybrid cross, we can follow these steps:

### Step 1: Understand the Dihybrid Cross
A dihybrid cross involves two traits, each represented by two alleles. For this problem, we will consider two traits:
- Trait 1: R (dominant) and r (recessive)
- Trait 2: Y (dominant) and y (recessive)

### Step 2: Determine the Parental Genotypes
Assuming the parental genotypes are RrYy (heterozygous for both traits) crossed with RrYy, we can set up a Punnett square to visualize the offspring.

### Step 3: Create a Punnett Square
The gametes produced by each parent will be:
- Parent 1: RY, Ry, rY, ry
- Parent 2: RY, Ry, rY, ry

The Punnett square will show all possible combinations of these gametes.

### Step 4: Count the Offspring Genotypes
After filling out the Punnett square, we can count the occurrences of each genotype:
- RRYy: 1
- RrYY: 2
- RrYy: 4
- Rryy: 2
- rrYY: 1
- rrYy: 2
- Rryy: 2
- rryy: 1

### Step 5: Calculate the Ratios
Now, we need to find the probability of each specific genotype mentioned in the question:
1. Rryy
2. RrYy
3. rrYy
4. RrYY

From our counts:
- Rryy: 2 occurrences
- RrYy: 4 occurrences
- rrYy: 2 occurrences
- RrYY: 2 occurrences

The total number of offspring is 16.

### Step 6: Calculate the Probabilities
Now we can calculate the probabilities for each genotype:
- Probability of Rryy = 2/16 = 1/8
- Probability of RrYy = 4/16 = 1/4
- Probability of rrYy = 2/16 = 1/8
- Probability of RrYY = 2/16 = 1/8

### Final Answer
The probabilities of getting seeds with genotypes Rryy, RrYy, rrYy, and RrYY in the F2 generation are:
- Rryy: 1/8
- RrYy: 1/4
- rrYy: 1/8
- RrYY: 1/8