A trihybrid plant is heterozygous for three genes, e.g. AaBbCc. Number of types of gametes $= 2^n$, where n is the number of heterozygous gene pairs: $2^3 = 8$ (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc). On self-fertilisation, each of the 8 types of male gametes can fuse with each of the 8 types of female gametes. Number of zygote combinations $= 8 \times 8 = 64$ (the squares of the Punnett square). (These give 27 different genotypes and 8 phenotypes in the ratio 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1.) So the plant forms 8 different gametes and 64 different zygotes.
ConceptGametes and zygotes in a trihybrid cross
Principles of Inheritance and Variation in past papers
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