How many different types of gametes can be formed by `F_(1)` progent, resulting from the following cross : AA BB CC x aa bb cc

Options

A. 3
B. 8
C. 27
D. 64

Answer

To determine how many different types of gametes can be formed by the F1 progeny resulting from the cross AA BB CC x aa bb cc, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parental Genotypes**:
- The parental genotypes are AA BB CC (homozygous dominant) and aa bb cc (homozygous recessive).

2. **Determine the F1 Progeny Genotype**:
- When these two parental genotypes are crossed, the F1 progeny will be heterozygous for each trait. Therefore, the genotype of the F1 progeny will be:
- A/a, B/b, C/c
- This can be represented as: Aa Bb Cc.

3. **Count the Number of Traits**:
- There are three traits involved in this cross: A, B, and C.

4. **Calculate the Number of Gametes**:
- The number of different types of gametes that can be formed is given by the formula \(2^n\), where \(n\) is the number of heterozygous gene pairs.
- In this case, since there are 3 heterozygous pairs (A/a, B/b, C/c), we have:
\[ n = 3 \]
- Therefore, the number of different types of gametes is:
\[ 2^3 = 8 \]

5. **Conclusion**:
- The F1 progeny can produce 8 different types of gametes.

### Final Answer:
The number of different types of gametes that can be formed by the F1 progeny is **8**.