Number of gamete types formed by AaBbCc individuals/heterozygous for three gene loci is

Options

A. 2
B. 4
C. 8
D. 16

Answer

To determine the number of gamete types formed by an individual with the genotype AaBbCc (heterozygous for three gene loci), we can follow these steps:

### Step 1: Understand the Genotype
The genotype AaBbCc indicates that the individual is heterozygous for three different gene loci (A, B, and C). Each gene can have two alleles: one dominant (uppercase) and one recessive (lowercase).

### Step 2: Determine the Number of Alleles for Each Gene
For each gene locus:
- Gene A has 2 alleles: A (dominant) and a (recessive)
- Gene B has 2 alleles: B (dominant) and b (recessive)
- Gene C has 2 alleles: C (dominant) and c (recessive)

### Step 3: Calculate the Total Number of Gamete Types
The total number of different gametes that can be formed is calculated using the formula:
\[ \text{Number of gamete types} = 2^n \]
where \( n \) is the number of heterozygous gene loci.

In this case, \( n = 3 \) (for genes A, B, and C):
\[ \text{Number of gamete types} = 2^3 = 8 \]

### Step 4: List the Possible Gametes
The possible gametes formed by the genotype AaBbCc are:
1. ABC
2. ABc
3. AbC
4. Abc
5. aBC
6. aBc
7. abC
8. abc

### Conclusion
Thus, the number of gamete types formed by an individual with the genotype AaBbCc is **8**.

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