If an individual of genotype aabbcc were intercrosed how many different phenotypes can appear in their offsprings
Answers
You see, in most cases, for the continuation of the diploid state, you gotta have two parents.
To answer the first question, be clear that we are considering a set of 3 genes, each gene having 2 alleles. Since all the 3 are heterozygous, a gamete can receive either
1. A or a,
2. B or b, and
3. C or c.
There are 3 cases, and each case has 2 possibilities.
So, the number of types of gametes is 2X2X2 = 8. (You could just list them out too, actually)
For the second question, there are 3 possibilities for the combination of alleles of each gene - AA, Aa, aa in case of the first gene.
Considering all the three, the total number would be 3X3X3 = 27.
Therefore, the number of genotypically distinct progeny would be 27, when you consider three genes.
The general formula for any number of genes(n) would be 3 raised to n.
Since you didn't specify, the number of phenotypically distinct progeny would be 2 raised to n( = 8 in this example), provided this is a case of complete dominance.
Hope this helped!