What proportion of the offsprings obtained from cross AABBCC `xx` AaBbCc will be completely heterozygous for all the genes segregated independently ?

Options

A. `1//8`
B. `1//4`
C. `1//2`
D. `1//16`

Answer

To solve the problem of determining the proportion of offspring from the cross AABBCC x AaBbCc that are completely heterozygous for all genes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**:
- The first parent is AABBCC (homozygous for all three genes).
- The second parent is AaBbCc (heterozygous for all three genes).

2. **Determine the Gametes**:
- The first parent (AABBCC) can only produce one type of gamete: ABC.
- The second parent (AaBbCc) can produce three types of gametes: ABC, Abc, aBC, abC, etc. However, for our purpose, we will focus on the combinations that yield heterozygous offspring.

3. **Cross the Gametes**:
- When we cross the gametes from both parents, we can only get offspring that are heterozygous for the genes contributed by the second parent.
- The offspring will inherit the alleles from the first parent (A, B, C) and the second parent (a, b, c).

4. **Determine Heterozygous Combinations**:
- For the A gene: The offspring will be Aa (heterozygous).
- For the B gene: The offspring will be Bb (heterozygous).
- For the C gene: The offspring will be Cc (heterozygous).

5. **Calculate the Probability of Heterozygous Offspring**:
- The probability of getting Aa from AABBCC x AaBbCc is 100% (1).
- The probability of getting Bb is 1/2 (since the second parent can contribute either B or b).
- The probability of getting Cc is also 1/2 (for the same reason as above).

6. **Multiply the Probabilities**:
- To find the total probability of obtaining an offspring that is completely heterozygous (AaBbCc), we multiply the probabilities:
\[ P(Aa) \times P(Bb) \times P(Cc) = 1 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

7. **Conclusion**:
- Therefore, the proportion of the offspring that will be completely heterozygous for all the genes is **1/4**.