Question

A cell with three pairs of chromosomes has the genotype AaBbCc, such that each pair of alleles is on a different pair of chromosomes. If this cell were to undergo meiotic division, how many genetically different types of gametes could be produced?

Answer 1

8 genetically different types of gametes.

Step-by-step explanation:

This is typical heterozygotic trihybrid individual i.e. an individual with three pairs of contrasting characters possessing different alleles on each gene. This is similar to Mendel's observation of F1 hybrid.

According to Mendel's law of independent assortment, the alleles get sorted into gametes independently of one another. Hence, this hybrid (AaBbCc) with three unlinked genes produces 8 genetically different gametes after undergoing meosis viz:

ABC, ABc, AbC, Abc, aBC, aBc, abC, and abc.

The number of gametes formed by this trihybrid heterozygote (AaBbCc) is determined by the formula 2^n, where n represents the number of characters.

Since this is a trihybrid cell involving three unlinked genes, 2^3 = 8

i.e. 8 gametes will be produced as stated earlier.