In a diybrid cross between RRYY and rryy, the number of RrYy `F_(2)` genotypes will be
(a) 4
(b) 3
(c) 2
(d) 9

Options

A. 4
B. 3
C. 2
D. 9

Answer

To solve the question regarding the number of RrYy genotypes in the F2 generation from a dihybrid cross between RRYY and rryy, we can follow these steps:

### Step 1: Identify the Parent Genotypes
The parent genotypes are:
- Parent 1: RRYY (homozygous dominant for both traits)
- Parent 2: rryy (homozygous recessive for both traits)

### Step 2: Determine the Gametes Produced by Each Parent
From the genotypes:
- Parent 1 (RRYY) can produce gametes: RY
- Parent 2 (rryy) can produce gametes: ry

### Step 3: Perform the F1 Generation Cross
When we cross the gametes from both parents:
- F1 Generation: All offspring will be RrYy (heterozygous for both traits).

### Step 4: Self-Cross the F1 Generation
Now, we perform a self-cross of the F1 generation:
- RrYy x RrYy

### Step 5: Create a Punnett Square for the F2 Generation
To find the genotypes in the F2 generation, we can set up a 4x4 Punnett square:
- The gametes from each RrYy parent are: RY, Ry, rY, ry.

### Step 6: Fill in the Punnett Square
The combinations from the Punnett square will yield the following genotypes:
1. RRY Y
2. RRYy
3. RrYY
4. RrYy
5. RrYy
6. Rryy
7. rrYY
8. rrYy
9. rryy

### Step 7: Count the RrYy Genotypes
From the filled Punnett square, we can identify the RrYy genotypes:
- RrYy (2 occurrences)

### Step 8: Conclusion
Thus, the total number of RrYy genotypes in the F2 generation is 4.

### Final Answer
The correct option is (a) 4.

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