A double homozygous yellow round seeded plant of Pea is crossed with green wrinkled seeded plant. The offspring shall be of

Options

A. One type
B. Two types
C. Four types
D. Several types

Answer

To solve the problem of crossing a double homozygous yellow round seeded plant with a green wrinkled seeded plant, we can follow these steps:

### Step 1: Identify the Genotypes of the Parents
- The double homozygous yellow round seeded plant has the genotype **RRYY** (where R = round seed shape, Y = yellow seed color).
- The green wrinkled seeded plant has the genotype **rryy** (where r = wrinkled seed shape, y = green seed color).

### Step 2: Determine the Gametes Produced by Each Parent
- The yellow round seeded plant (RRYY) can only produce one type of gamete: **RY**.
- The green wrinkled seeded plant (rryy) can only produce one type of gamete: **ry**.

### Step 3: Perform the Cross
- When we cross the gametes from both parents, we combine them:
- **RY (from RRYY) × ry (from rryy)** results in all offspring having the genotype **RrYy**.

### Step 4: Analyze the Offspring Phenotype
- The genotype **RrYy** indicates that all offspring will express the dominant traits:
- R (round seed shape) is dominant over r (wrinkled seed shape).
- Y (yellow seed color) is dominant over y (green seed color).
- Therefore, all offspring will be round and yellow seeded.

### Step 5: Conclusion
- The offspring from this cross will all be round yellow seeded plants. Thus, the answer to the question is that the offspring shall be of one type: **round yellow seeded plants**.

### Final Answer
The offspring shall be all round yellow seeded plants.

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