How many types of gametes are formed by a plant having AaBbCc genotype, in which A & B genes shows complete linkage.

Options

A. 2
B. 4
C. 6
D. 8

Answer

To determine how many types of gametes are formed by a plant with the genotype AaBbCc, where genes A and B show complete linkage, we can follow these steps:

### Step 1: Identify the genotype and linked genes
The given genotype is AaBbCc. In this case, genes A and B are linked, which means they will be inherited together. The genotype can be broken down as follows:
- A (dominant allele) and a (recessive allele)
- B (dominant allele) and b (recessive allele)
- C (dominant allele) and c (recessive allele)

### Step 2: Determine the number of heterozygous alleles
In the genotype AaBbCc:
- The A gene has one heterozygous allele (A/a).
- The B gene has one heterozygous allele (B/b).
- The C gene has one heterozygous allele (C/c).

However, since A and B are completely linked, we will treat them as a single unit. Thus, we only consider the heterozygosity of C.

### Step 3: Calculate the number of gametes using the formula
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 gene pairs.

In this case:
- Since A and B are linked, we treat them as one unit (AB or ab).
- Therefore, we have:
- 1 unit from A and B (AB or ab)
- 1 heterozygous pair from C (C/c)

Thus, we have a total of 2 heterozygous units (1 for the linked genes and 1 for C).

### Step 4: Substitute into the formula
Now we can substitute into the formula:
\[ n = 2 \]
\[ \text{Number of gametes} = 2^2 = 4 \]

### Step 5: Conclusion
Therefore, the number of types of gametes formed by the plant with the genotype AaBbCc, where genes A and B show complete linkage, is **4**.

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