How many different types of gametes can be formed by `"F"_(1)` progeny resulting from the following cross?
AABBCC `xx` aabbcc

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 AABBCC x aabbcc, we can follow these steps:

### Step 1: Identify the Parental Genotypes
The parental genotypes are:
- Parent 1: AABBCC (homozygous dominant for all traits)
- Parent 2: aabbcc (homozygous recessive for all traits)

### Step 2: Determine the Gametes of Each Parent
Since Parent 1 is homozygous for all alleles (A, B, and C), it can only produce one type of gamete:
- Parent 1 Gamete: ABC

Since Parent 2 is also homozygous for all alleles (a, b, and c), it can only produce one type of gamete:
- Parent 2 Gamete: abc

### Step 3: Form the F1 Generation
When we cross these two parents, the F1 progeny will be heterozygous for all traits:
- F1 Genotype: AaBbCc

### Step 4: Determine the Number of Heterozygous Alleles
In the F1 generation (AaBbCc), we can see that there are three heterozygous pairs:
- A/a
- B/b
- C/c

### Step 5: Calculate the Number of Different Gametes
The formula to calculate the number of different types of gametes is:
\[ \text{Number of Gametes} = 2^n \]
where \( n \) is the number of heterozygous pairs.

In this case, \( n = 3 \) (since we have A/a, B/b, and C/c):
\[ \text{Number of Gametes} = 2^3 = 8 \]

### Conclusion
The F1 progeny can produce 8 different types of gametes.

### Summary of Gametes
To list the possible gametes:
1. ABC
2. ABc
3. AbC
4. Abc
5. aBC
6. aBc
7. abC
8. abc

Thus, the answer is **8**.

---