Question

Two parents are AabbCc and aaBbcC.

What is the probability that their child will have the dominant phenotype for all three alleles shown above. Alleles are independently assorted.

Answer 1

The genotypes of the parents are AabbCc and aaBbcC

As the alleles (genes) are independently assorted, the inheritance of one gene for a character into the next generation is independent of the inheritance of another gene for another character, into the next generation. Thus, during the gamete formation and inheritance into the next generation, there is an equal chance for each allele of a gene to occur with any another allele of another gene.

As the genes assort independently during gamete formation, there is a possibility of the following gametes to be formed:

Genotype of Parents	AabbCc	aaBbcC
Gametes	AbC	aBc
Abc	aBC
abC	abc
abc	abC

To study the genotype of the offspring, let us consider a Punnett square analysis:

	AbC	Abc	abC	abc
aBc	AaBbCc	AaBbcc	aaBbCc	aaBbcc
aBC	AaBbCC	AaBbCc	aaBbCC	aaBbCc
abc	AabbCc	Aabbcc	aabbCc	aabbcc
abC	AabbCC	AabbCc	aabbCC	aabbCc

From the table above, it can be concluded that in the case of the three genotypes AaBbCc, AaBbCC and AaBbCc; the child will have a dominant phenotype for all the three genes.

Therefore the probabbility that the child will have a dominant phenotype for all the three alleles is 3/16.

(The table might get distorted. Attaching alongwith an image file of both the above tables.)