Genes a, b and c assort independently and are recessive to their respective alleles A, B and C. Two triple heterozygous AaBbCc individual are crossed. What is the probability that a given offspring will be genotypically homozygous for all three dominant alleles :-

Options

A. `2//64`
B. `27//64`
C. `1//64`
D. `6//64`

Answer

To solve the problem of determining the probability that a given offspring from a cross between two triple heterozygous individuals (AaBbCc) will be genotypically homozygous for all three dominant alleles (A, B, and C), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parental Genotypes**:
The parents are both AaBbCc, meaning they are heterozygous for all three traits.

2. **Determine the Possible Gametes**:
Each parent can produce gametes that carry one allele from each gene. The possible gametes from each parent are:
- ABC
- ABc
- AbC
- Abc
- aBC
- aBc
- abC
- abc

Since there are 3 genes and each can be either dominant or recessive, the total number of different gametes is \(2^3 = 8\).

3. **Set Up a Punnett Square**:
To find the offspring's genotypes, we can set up a Punnett square with the gametes from one parent on the top and the gametes from the other parent on the side. This will create a 8x8 square.

4. **Calculate the Probability for Each Gene**:
For each gene (A, B, and C), we need to find the probability of getting homozygous dominant (AA, BB, CC):
- For gene A (A from both parents):
- Possible combinations: AA (1 way), Aa (2 ways), aa (1 way)
- Probability of AA = 1/4
- For gene B (B from both parents):
- Probability of BB = 1/4
- For gene C (C from both parents):
- Probability of CC = 1/4

5. **Combine the Probabilities**:
Since the genes assort independently, we multiply the probabilities of getting homozygous dominant for each gene:
\[ P(AA) \times P(BB) \times P(CC) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64} \]

6. **Conclusion**:
The probability that a given offspring will be genotypically homozygous for all three dominant alleles (A, B, and C) is \( \frac{1}{64} \).

### Final Answer:
The probability that a given offspring will be genotypically homozygous for all three dominant alleles is \( \frac{1}{64} \).