In a cross between two individuals with the genotype AaBbccDdEeFf, the probability that an offspring will be heterozygous at all these genes:-

Options

A. 0
B. `1//16`
C. `1//32`
D. `1//64`

Answer

To solve the problem of determining the probability that an offspring will be heterozygous at all genes in a cross between two individuals with the genotype AaBbccDdEeFf, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**: The genotypes of both parents are AaBbccDdEeFf. This means:
- A: heterozygous (A or a)
- B: heterozygous (B or b)
- C: homozygous recessive (c or c)
- D: heterozygous (D or d)
- E: heterozygous (E or e)
- F: heterozygous (F or f)

2. **Analyze Each Gene**: We need to determine the probability of being heterozygous for each gene:
- For genes A, B, D, E, and F, the offspring can be heterozygous if they inherit one dominant and one recessive allele.
- For gene C, since both parents are homozygous recessive (cc), the offspring will always inherit a recessive allele from both parents, resulting in a homozygous recessive genotype (cc).

3. **Calculate Probability for Each Gene**:
- **Gene A**: Probability of being heterozygous (Aa) = 1/2
- **Gene B**: Probability of being heterozygous (Bb) = 1/2
- **Gene C**: Probability of being heterozygous (Cc) = 0 (since both are cc)
- **Gene D**: Probability of being heterozygous (Dd) = 1/2
- **Gene E**: Probability of being heterozygous (Ee) = 1/2
- **Gene F**: Probability of being heterozygous (Ff) = 1/2

4. **Combine Probabilities**: To find the total probability of being heterozygous at all genes, we multiply the probabilities:
- Total Probability = (1/2) * (1/2) * (0) * (1/2) * (1/2) * (1/2)

5. **Final Calculation**: Since one of the probabilities is 0 (for gene C), the total probability of being heterozygous at all genes is:
- Total Probability = 0

6. **Conclusion**: Therefore, the probability that an offspring will be heterozygous at all these genes is 0.