Female AaBb is crossed with male Aabb. The gametes shall be:

Options

A. Female AB and ab, male AA and bb
B. Female Aa and Bb, male AA and bb
C. Female AB, Ab, aB and ab, male Ab
D. Female AA,bb, AB and ab, male Ab.

Answer

To solve the problem of determining the gametes produced by the female AaBb and male Aabb, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Genotypes**:
- Female: AaBb
- Male: Aabb

2. **Determine Heterozygous Loci**:
- For the female (AaBb):
- A is heterozygous (A/a)
- B is heterozygous (B/b)
- Total heterozygous loci (n) = 2
- For the male (Aabb):
- A is heterozygous (A/a)
- b is homozygous (b/b)
- Total heterozygous loci (n) = 1

3. **Calculate Number of Gametes**:
- The formula for the number of gametes is \(2^n\), where n is the number of heterozygous loci.
- For the female: \(2^2 = 4\) types of gametes.
- For the male: \(2^1 = 2\) types of gametes.

4. **List the Gametes**:
- Female (AaBb) can produce:
- AB (from A and B)
- Ab (from A and b)
- aB (from a and B)
- ab (from a and b)
- Male (Aabb) can produce:
- Ab (from A and b)
- ab (from a and b)

5. **Final List of Gametes**:
- Female produces: AB, Ab, aB, ab
- Male produces: Ab, ab

### Conclusion:
The gametes produced by the female AaBb are AB, Ab, aB, and ab. The gametes produced by the male Aabb are Ab and ab.