Q.Enlist the seven traits of the pea plant selected/studied by Mendel.

The number of gamete types = 2^(number of heterozygous gene pairs). With three heterozygous pairs (Xx, Yy, Zz), that is 2³ = 8.

By the law of independent assortment, each heterozygous locus contributes 2 possible alleles to a gamete. For $n$ heterozygous gene pairs, the number of genetically different gametes is $2^n$.

Here the genotype XxYyZz is heterozygous at all three loci, so $n = 3$:

$2^n = 2^3 = 8$

The 8 gamete combinations are: XYZ, XYz, XyZ, Xyz, xYZ, xYz, xyZ, xyz.

A classic NCERT/CBSE Class 12 Biology genetics numerical from Principles of Inheritance and Variation.

Statement (c) wrongly gives the monohybrid F2 phenotypic ratio as 1:2:1; the true phenotypic ratio is 3:1 (1:2:1 is the genotypic ratio).

Checking each option against Mendelian genetics:

• (a) Dihybrid cross F2 phenotypic ratio = 9:3:3:1 — correct.

• (b) Incomplete dominance F2 phenotypic ratio = 1:2:1 (because heterozygotes show a distinct intermediate phenotype, so genotypic ratio = phenotypic ratio) — correct.

• (c) Standard Mendelian monohybrid cross (complete dominance) F2 phenotypic ratio is 3:1 (3 dominant-phenotype : 1 recessive-phenotype). The value 1:2:1 is the genotypic ratio (1 homozygous dominant : 2 heterozygous : 1 homozygous recessive), not the phenotypic ratio. So this statement is incorrect.

• (d) Monohybrid cross F2 genotypic ratio = 1:2:1 — correct.