A male rabbit of genotype AABBDDEE is crossed with a female rabbit of genotype aabbddee to produce `F_(1)` hybrid offspring. How many genetically different gametes can be produced by this `F_(1)` hybrid

Options

A. 4
B. 8
C. 16
D. 32

Answer

To solve the problem of how many genetically different gametes can be produced by the F1 hybrid offspring resulting from the cross of a male rabbit with genotype AABBDDEE and a female rabbit with genotype aabbddee, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes of Parents**:
- Male Rabbit: AABBDDEE (homozygous dominant for all traits)
- Female Rabbit: aabbddee (homozygous recessive for all traits)

2. **Determine the Genotype of F1 Hybrid Offspring**:
- When these two rabbits are crossed, the F1 offspring will inherit one allele from each parent for each trait.
- The F1 genotype will be: AaBbDdEe (heterozygous for all traits).

3. **Count the Number of Traits**:
- There are four traits being considered: A/a, B/b, D/d, and E/e.

4. **Calculate the Number of Gametes**:
- The number of different gametes that can be produced by an organism is given by the formula \(2^n\), where \(n\) is the number of heterozygous gene pairs.
- In this case, since the F1 hybrid is heterozygous for all four traits (A/a, B/b, D/d, E/e), we have \(n = 4\).

5. **Apply the Formula**:
- Using the formula \(2^n\):
\[ 2^4 = 16 \]
- Therefore, the F1 hybrid can produce 16 genetically different gametes.

### Final Answer:
The F1 hybrid offspring can produce **16 genetically different gametes**.

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