How many type of gametes are found in `F_(1)` progeny of cross between AABBCC and aabbcc?
(a) 3
(b) 8
(c) 27
(d) 64

Options

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

Answer

To determine the number of types of gametes produced in the F1 progeny of a cross between AABBCC and aabbcc, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parent Genotypes**:
- The first parent is AABBCC, which has the following alleles:
- A (homozygous dominant)
- B (homozygous dominant)
- C (homozygous dominant)
- The second parent is aabbcc, which has the following alleles:
- a (homozygous recessive)
- b (homozygous recessive)
- c (homozygous recessive)

2. **Determine the Alleles Contributing to Gamete Formation**:
- In the F1 progeny, each parent contributes one allele for each gene. Since the first parent contributes only dominant alleles (A, B, C) and the second parent contributes only recessive alleles (a, b, c), the F1 progeny will have the genotype AaBbCc.

3. **Count the Number of Different Alleles (n)**:
- For the F1 progeny (AaBbCc), we have:
- 1 pair of alleles for gene A (A and a)
- 1 pair of alleles for gene B (B and b)
- 1 pair of alleles for gene C (C and c)
- Therefore, the total number of different genes (n) is 3.

4. **Use the Formula to Calculate the Number of Gametes**:
- The formula to calculate the number of different types of gametes is \(2^n\), where \(n\) is the number of different genes.
- Here, \(n = 3\), so we calculate:
\[ 2^3 = 8 \]

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

### Final Answer:
The number of types of gametes found in the F1 progeny is **8** (Option b).

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