What proportion of the offsprings obtained from cross AABBCC x AaBbCc ...
Since one parent is homozygous AABBCC the genotype formed will be only 1 type i.e. ABC. The other parent is heterozygous for all genes. About 1/8th or 12.5% of total offspring will be heterozygous at all three-locus.
What proportion of the offsprings obtained from cross AABBCC x AaBbCc ...
Punnett square analysis:
| | A | a |
|----|------|------|
| **A** | AA | Aa |
| **a** | aA | aa |
| | B | b |
|----|------|------|
| **B** | BB | Bb |
| **b** | bB | bb |
| | C | c |
|----|------|------|
| **C** | CC | Cc |
| **c** | cC | cc |
Cross AABBCC x AaBbCc:
| | AB | Ab | aB | ab |
|-----|------|------|------|------|
| **ABC** | AABBCC | AABbCC | AaBBCC | AaBbCC |
| **ABc** | AABBcC | AABbcC | AaBBcC | AaBbcC |
| **AbC** | AABbCC | AAbbCC | AaBbCC | AabbCC |
| **Abc** | AABbcC | AAbbcC | AaBbcC | AabbCc |
| **aBC** | AaBBCC | AaBbCC | aaBBCC | aaBbCC |
| **aBc** | AaBBcC | AaBbcC | aaBBcC | aaBbcC |
| **abC** | AaBbCC | AabbCC | aaBbCC | aabbCC |
| **abc** | AaBbcC | AabbCc | aaBbcC | aabbCc |
Thus, out of the 8 possible offspring, only 1 (AaBbCc) is completely heterozygous for all the genes segregated independently. Hence, the proportion is 1/8 or option A.