If an individual having the genotype AaBbCc is selfed, how many squares will be required to determined the genotype of the population by Punnett's checker board method?

Options

A. `8 xx 8`
B. `2 xx 2`
C. `4 xx 4`
D. `1 xx 8`

Answer

To determine the number of squares required in a Punnett square for the genotype AaBbCc, we can follow these steps:

### Step 1: Identify the Genotype
The genotype given is AaBbCc. This indicates that there are three genes involved, each with two alleles (one dominant and one recessive).

### Step 2: Count the Heterozygous Conditions
In the genotype AaBbCc:
- A is heterozygous (Aa)
- B is heterozygous (Bb)
- C is heterozygous (Cc)

Thus, we have three heterozygous conditions (n = 3).

### Step 3: Use the Formula to Calculate Genotypes
To find the number of different genotypes that can be produced, we use the formula:
\[ 2^n \]
Where \( n \) is the number of heterozygous gene pairs.

Substituting \( n = 3 \):
\[ 2^3 = 8 \]
This means that 8 different gametes can be formed from each parent.

### Step 4: Determine the Size of the Punnett Square
Since each parent can produce 8 different gametes, we need a Punnett square that can accommodate all combinations of these gametes. Therefore, the size of the Punnett square will be:
\[ 8 \times 8 \]
This results in a total of 64 squares in the Punnett square.

### Conclusion
To determine the genotype of the population from the selfing of an individual with the genotype AaBbCc, a Punnett square of size 8 x 8 will be required.

**Final Answer: 8 x 8 (64 squares)**

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