In a complimentary gene interaction calculate the number of phenotype and genotype produced in a cross AaBb `xx` aaBB

Options

A. 1 phenotype, 2 genotypes
B. 2 phenotypes, 4 genotypes
C. 4 phenotypes, 4 genotypes
D. 2 phenotypes, 2 genotypes

Answer

To solve the problem of determining the number of phenotypes and genotypes produced in the cross AaBb x aaBB, we will follow these steps:

### Step 1: Identify the Parent Genotypes
The two parents in this cross are:
- Parent 1: AaBb
- Parent 2: aaBB

### Step 2: Determine the Gametes Produced by Each Parent
Next, we will determine the gametes that each parent can produce:
- Parent 1 (AaBb) can produce the following gametes:
- AB
- Ab
- aB
- ab

- Parent 2 (aaBB) can produce the following gametes:
- aB
- aB (since it has two dominant alleles for B, it can only produce aB)

### Step 3: Create a Punnett Square
Now, we will set up a Punnett square to find the possible genotypes of the offspring.

| | aB | aB |
|----------|--------|--------|
| **AB** | AaBB | AaBB |
| **Ab** | Aabb | Aabb |
| **aB** | aaBB | aaBB |
| **ab** | aabb | aabb |

### Step 4: List the Offspring Genotypes
From the Punnett square, we can list the genotypes of the offspring:
1. AaBB
2. AaBB
3. Aabb
4. Aabb
5. aaBB
6. aabb

### Step 5: Count the Unique Genotypes
Now we will count the unique genotypes:
- AaBB (2 occurrences)
- Aabb (2 occurrences)
- aaBB (2 occurrences)
- aabb (1 occurrence)

This gives us a total of 4 unique genotypes.

### Step 6: Determine the Phenotypes
Next, we will determine the phenotypes based on the dominant and recessive traits:
- AaBB and aaBB will show one phenotype (dominant for A and B).
- Aabb and aabb will show another phenotype (recessive for A and dominant for B).

Thus, we have 2 unique phenotypes.

### Conclusion
In this cross (AaBb x aaBB), the number of unique phenotypes is 2, and the number of unique genotypes is 4.

### Final Answer
- **Number of Phenotypes:** 2
- **Number of Genotypes:** 4

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